INSTRUCTOR SOLUTIONS MANUAL [PDF]

INSTRUCTOR SOLUTIONS MANUAL

INSTRUCTOR’S SOLUTIONS MANUAL to accompany

ADAMS / ESSEX

CALCULUS: A COMPLETE COURSE; CALCULUS: SINGLE VARIABLE; and CALCULUS: SEVERAL VARIABLES Eighth Edition

Prepared by Robert A. Adams University of British Columbia Christopher Essex University of Windsor Ontario

Toronto Copyright © 2014 Pearson Canada Inc., Toronto, Ontario. Pearson Canada. All rights reserved. This work is protected by Canadian copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course, Eighth Edition, by Adams/Essex to post this material online only if the use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and provided the reproduced material bears this copyright notice.

FOREWORD These solutions are provided for the benefit of instructors using the textbooks:

Calculus: A Complete Course (8th Edition), Single-Variable Calculus (8th Edition), and Calculus of Several Variables (8th Edition) by R. A. Adams and Chris Essex, published by Pearson Education Canada. For the most part, the solutions are detailed, especially in exercises on core material and techniques. Occasionally some details are omitted — for example, in exercises on applications of integration, the evaluation of the integrals encountered is not always given with the same degree of detail as the evaluation of integrals found in those exercises dealing specifically with techniques of integration. Instructors may wish to make these solutions available to their students. However, students should use such solutions with caution. It is always more beneficial for them to attempt exercises and problems on their own, before they look at solutions done by others. If they examine solutions as “study material” prior to attempting the exercises, they can lose much of the benefit that follows from diligent attempts to develop their own analytical powers. When they have tried unsuccessfully to solve a problem, then looking at a solution can give them a “hint” for a second attempt. Separate Student Solutions Manuals for the books are available for students. They contain the solutions to the even-numbered exercises only. Apr, 2012. R. A. Adams [email protected]

Chris Essex [email protected]

Copyright © 2014 Pearson Canada Inc.

CONTENTS

Solutions for Chapter 18 extended

1 23 40 82 109 179 215 270 318 353 393 421 450 494 541 583 614 641 648 669

Solutions for Appendices

699

Solutions for Chapter P Solutions for Chapter 1 Solutions for Chapter 2 Solutions for Chapter 3 Solutions for Chapter 4 Solutions for Chapter 5 Solutions for Chapter 6 Solutions for Chapter 7 Solutions for Chapter 8 Solutions for Chapter 9 Solutions for Chapter 10 Solutions for Chapter 11 Solutions for Chapter 12 Solutions for Chapter 13 Solutions for Chapter 14 Solutions for Chapter 15 Solutions for Chapter 16 Solutions for Chapter 17 Solutions for Chapter 18

NOTE: Chapter 18 extended is only needed by users of

Calculus of Several Variables (Eighth Edition)

Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

CHAPTER P.

SECTION P.1 (PAGE 10)

PRELIMINARIES

19. Given: 1/(2 − x) < 3.

Section P.1 Real Numbers and the Real Line (page 10) 1.

2 = 0.22222222 · · · = 0.2 9

2.

1 = 0.09090909 · · · = 0.09 11

20. Given: (x + 1)/x ≥ 2.

3. If x = 0.121212 · · ·, then 100x = 12.121212 · · · = 12 + x. Thus 99x = 12 and x = 12/99 = 4/33.

4. If x = 3.277777 · · ·, then 10x − 32 = 0.77777 · · · and 100x − 320 = 7 + (10x − 32), or 90x = 295. Thus x = 295/90 = 59/18.

5.

CASE I. If x < 2, then 1 < 3(2 − x) = 6 − 3x, so 3x < 5 and x < 5/3. This case has solutions x < 5/3. CASE II. If x > 2, then 1 > 3(2 − x) = 6 − 3x, so 3x > 5 and x > 5/3. This case has solutions x > 2. Solution: (−∞, 5/3) ∪ (2, ∞).

CASE I. If x > 0, then x + 1 ≥ 2x, so x ≤ 1. CASE II. If x < 0, then x + 1 ≤ 2x, so x ≥ 1. (not possible) Solution: (0, 1].

21. Given: x 2 − 2x ≤ 0. Then x(x − 2) ≤ 0. This is only possible if x ≥ 0 and x ≤ 2. Solution: [0, 2].

22. Given 6x 2 − 5x ≤ −1, then (2x − 1)(3x − 1) ≤ 0, so

1/7 = 0.142857142857 · · · = 0.142857

either x ≤ 1/2 and x ≥ 1/3, or x ≤ 1/3 and x ≥ 1/2. The latter combination is not possible. The solution set is [1/3, 1/2].

3/7 = 0.428571428571 · · · = 0.428571

23. Given x 3 > 4x, we have x(x 2 − 4) > 0. This is possible

2/7 = 0.285714285714 · · · = 0.285714 4/7 = 0.571428571428 · · · = 0.571428 note the same cyclic order of the repeating digits 5/7 = 0.714285714285 · · · = 0.714285 6/7 = 0.857142857142 · · · = 0.857142

6. Two different decimal expansions can represent the same number. For instance, both 0.999999 · · · = 0.9 and 1.000000 · · · = 1.0 represent the number 1.

7.

x ≥ 0 and x ≤ 5 define the interval [0, 5].

8.

x < 2 and x ≥ −3 define the interval [−3, 2).

9.

x > −5 or x < −6 defines the union (−∞, −6) ∪ (−5, ∞).

10.

x ≤ −1 defines the interval (−∞, −1].

11.

x > −2 defines the interval (−2, ∞).

12.

x < 4 or x ≥ 2 defines the interval (−∞, ∞), that is, the whole real line.

14. If 3x + 5 ≤ 8, then 3x ≤ 8 − 5 − 3 and x ≤ 1. Solution: (−∞, 1]

15. If 5x − 3 ≤ 7 − 3x, then 8x ≤ 10 and x ≤ 5/4. Solution: (−∞, 5/4]

6−x 3x − 4 ≥ , then 6 − x ≥ 6x − 8. Thus 14 ≥ 7x 4 2 and x ≤ 2. Solution: (−∞, 2]

16. If

17. If 3(2 − x) < 2(3 + x), then 0 < 5x and x > 0. Solution: (0, ∞)

18. If

24. Given x 2 −x ≤ 2, then x 2 −x −2 ≤ 0 so (x −2)(x +1) ≤ 0. This is possible if x ≤ 2 and x ≥ −1 or if x ≥ 2 and x ≤ −1. The latter situation is not possible. The solution set is [−1, 2].

x 4 ≥1+ . 2 x CASE I. If x > 0, then x 2 ≥ 2x + 8, so that x 2 − 2x − 8 ≥ 0, or (x − 4)(x + 2) ≥ 0. This is possible for x > 0 only if x ≥ 4. CASE II. If x < 0, then we must have (x − 4)(x + 2) ≤ 0, which is possible for x < 0 only if x ≥ −2. Solution: [−2, 0) ∪ [4, ∞).

25. Given:

3 2 < . x −1 x +1 CASE I. If x > 1 then (x − 1)(x + 1) > 0, so that 3(x +1) < 2(x −1). Thus x < −5. There are no solutions in this case. CASE II. If −1 < x < 1, then (x − 1)(x + 1) < 0, so 3(x + 1) > 2(x − 1). Thus x > −5. In this case all numbers in (−1, 1) are solutions. CASE III. If x < −1, then (x − 1)(x + 1) > 0, so that 3(x + 1) < 2(x − 1). Thus x < −5. All numbers x < −5 are solutions. Solutions: (−∞, −5) ∪ (−1, 1).

26. Given:

13. If −2x > 4, then x < −2. Solution: (−∞, −2)

x2

if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The possibilities are, therefore, −2 < x < 0 or 2 < x < ∞. Solution: (−2, 0) ∪ (2, ∞).

< 9, then |x| < 3 and −3 < x < 3. Solution: (−3, 3)

27. If |x| = 3 then x = ±3. 28. If |x − 3| = 7, then x − 3 = ±7, so x = −4 or x = 10. 29. If |2t + 5| = 4, then 2t + 5 = ±4, so t = −9/2 or t = −1/2.

30. If|1 − t| = 1, then 1 − t = ±1, so t = 0 or t = 2.

1 Copyright © 2014 Pearson Canada Inc.

SECTION P.1 (PAGE 10)

ADAMS and ESSEX: CALCULUS 8

31. If |8 − 3s| = 9, then 8 − 3s = ±9, so 3s = −1 or 17, and 32. 33.

s = −1/3 or s = 17/3. s s If − 1 = 1, then − 1 = ±1, so s = 0 or s = 4. 2 2 If |x| < 2, then x is in (−2, 2).

2. From A(−1, 2) to B(4, −10), 1xp = 4 − (−1) = 5 and

3. From A(3, 2) to B(−1, −2), 1x and p = −1 − 3 = −4 √ 1y = −2 − 2 = −4. | AB| =

35. If |s − 1| ≤ 2, then 1 − 2 ≤ s ≤ 1 + 2, so s is in [−1, 3]. 36. If |t + 2| < 1, then −2 − 1 < t < −2 + 1, so t is in (−3, −1).

4. From A(0.5, 3) to B(2, 3), 1x = 2 − 0.5 = 1.5 and

5. Starting point: (−2, 3). Increments 1x = 4, 1y = −7. New position is (−2 + 4, 3 + (−7)), that is, (2, −4).

6. Arrival point: (−2, −2). Increments 1x = −5, 1y = 1. Starting point was (−2 − (−5), −2 − 1), that is, (3, −3).

37. If |3x − 7| < 2, then 7 − 2 < 3x < 7 + 2, so x is in

7.

38. If |2x + 5| < 1, then −5 − 1 < 2x < −5 + 1, so x is in

8.

(5/3, 3).

40.

(−4)2 + (−4)2 = 4 2.

1y = 3 − 3 = 0. | AB| = 1.5.

34. If |x| ≤ 2, then x is in [−2, 2].

39.

52 + (−12)2 = 13.

1y = −10 − 2 = −12. | AB| =

(−3, −2). x x If − 1 ≤ 1, then 1 − 1 ≤ ≤ 1 + 1, so x is in [0, 4]. 2 2 x 1 If 2 − < , then x/2 lies between 2 − (1/2) and 2 2 2 + (1/2). Thus x is in (3, 5).

41. The inequality |x + 1| > |x − 3| says that the distance

from x to −1 is greater than the distance from x to 3, so x must be to the right of the point half-way between −1 and 3. Thus x > 1.

42. |x − 3| < 2|x| ⇔ x 2 − 6x + 9 = (x − 3)2 < 4x 2

⇔ 3x 2 + 6x − 9 > 0 ⇔ 3(x + 3)(x − 1) > 0. This inequality holds if x < −3 or x > 1.

43. |a| = a if and only if a ≥ 0. It is false if a < 0.

9. 10. 11. 12.

x 2 + y 2 = 1 represents a circle of radius 1 centred at the origin. √ x 2 + y 2 = 2 represents a circle of radius 2 centred at the origin. x 2 + y 2 ≤ 1 represents points inside and on the circle of radius 1 centred at the origin. x 2 + y 2 = 0 represents the origin.

y ≥ x 2 represents all points lying on or above the parabola y = x 2 .

y < x 2 represents all points lying below the parabola y = x 2.

13. The vertical line through (−2, 5/3) is x = −2; the hori14.

zontal line through that point is y = 5/3. √ √ The vertical line through ( 2, −1.3) is x = 2; the horizontal line through that point is y = −1.3.

44. The equation |x − 1| = 1 − x holds if |x − 1| = −(x − 1),

15. Line through (−1, 1) with slope m = 1 is

45. The triangle inequality |x + y| ≤ |x| + |y| implies that

16. Line through (−2, 2) with slope m = 1/2 is

that is, if x − 1 < 0, or, equivalently, if x < 1.

|x| ≥ |x + y| − |y|.

y = 2 + (1/2)(x + 2), or x − 2y = −6.

17. Line through (0, b) with slope m = 2 is y = b + 2x.

Apply this inequality with x = a − b and y = b to get

18. Line through (a, 0) with slope m = −2 is y = 0 − 2(x − a), or y = 2a − 2x.

19. At x = 2, the height of the line 2x + 3y = 6 is

|a − b| ≥ |a| − |b|. Similarly, |a − b| = |b − a| ≥ |b| − |a|. Since |a| − |b| is equal to either |a| − |b| or |b| − |a|, depending on the sizes of a and b, we have

y = (6 − 4)/3 = 2/3. Thus (2, 1) lies above the line.

20. At x = 3, the height of the line x − 4y = 7 is

y = (3 − 7)/4 = −1. Thus (3, −1) lies on the line.

21. The line through (0, 0) and (2, 3) has slope m = (3 − 0)/(2 − 0) = 3/2 and equation y = (3/2)x or 3x − 2y = 0.

|a − b| ≥ |a| − |b| .

22. The line through (−2, 1) and (2, −2) has slope

Section P.2 Cartesian Coordinates in the Plane (page 16) 1. From A(0, 3) to B(4, 0), 1xp= 4 − 0 = 4 and 1y = 0 − 3 = −3. | AB| =

y = 1 + 1(x + 1), or y = x + 2.

42 + (−3)2 = 5.

m = (−2 − 1)/(2 + 2) = −3/4 and equation y = 1 − (3/4)(x + 2) or 3x + 4y = −2.

23. The line through (4, 1) and (−2, 3) has slope m = (3 − 1)/(−2 − 4) = −1/3 and equation 1 y = 1 − (x − 4) or x + 3y = 7. 3

2 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.2 (PAGE 16)

y

24. The line through (−2, 0) and (0, 2) has slope 25.

m = (2 − 0)/(0 + 2) = 1 and equation y = 2 + x. √ If m = −2 and √ b = 2, then the line has equation y = −2x + 2.

1.5x − 2y = −3

26. If m = −1/2 and b = −3, then the line has equation

x

y = −(1/2)x − 3, or x + 2y = −6.

27. 3x + 4y = 12 has x-intercept a = 12/3 = 4 and yintercept b = 12/4 = 3. Its slope is −b/a = −3/4. y

Fig. P.2.30

31. line through (2, 1) parallel to y = x + 2 is y = x − 1; line perpendicular to y = x + 2 is y = −x + 3.

32. line through (−2, 2) parallel to 2x + y = 4 is

3x + 4y = 12

2x + y = −2; line perpendicular to 2x + y = 4 is x − 2y = −6.

33. We have x

3x + 4y = −6 2x − 3y = 13

Fig. P.2.27

28.

x + 2y = −4 has x-intercept a = −4 and y-intercept b = −4/2 = −2. Its slope is −b/a = 2/(−4) = −1/2. y

x

34. We have 2x + y = 8 5x − 7y = 1

H⇒

14x + 7y = 56 5x − 7y = 1.

a straight line that is neither horizontal nor vertical, and does not pass through the origin. Putting y = 0 we get x/a = 1, so the x-intercept of this line is x = a; putting x = 0 gives y/b = 1, so the y-intercept is y = b.

√

√ √ √ 2x − 3y = 2 has x-intercept a = 2/ 2 = 2 √ and y-intercept √ b √= −2/ 3. Its slope is −b/a = 2/ 6 = 2/3. y

2x −

Subtracting these equations gives 17y = −51, so y = −3 and x = (13−9)/2 = 2. The intersection point is (2, −3).

35. If a 6= 0 and b 6= 0, then (x/a) + (y/b) = 1 represents

Fig. P.2.28

√

6x + 8y = −12 6x − 9y = 39.

Adding these equations gives 19x = 57, so x = 3 and y = 8 − 2x = 2. The intersection point is (3, 2).

x + 2y = −4

29.

H⇒

√

36. The line (x/2) − (y/3) = 1 has x-intercept a = 2, and y-intercept b = −3.

y 2

x

x y − =1 2 3

x 3y = 2

−3

Fig. P.2.36

Fig. P.2.29

30. 1.5x − 2y = −3 has x-intercept a = −3/1.5 = −2 and yintercept b = −3/(−2) = 3/2. Its slope is −b/a = 3/4.

37. The line through (2, 1) and (3, −1) has slope

m = (−1 − 1)/(3 − 2) = −2 and equation y = 1 − 2(x − 2) = 5 − 2x. Its y-intercept is 5.

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SECTION P.2 (PAGE 16)

ADAMS and ESSEX: CALCULUS 8

38. The line through (−2, 5) and (k, 1) has x-intercept 3, so

43.

also passes through (3, 0). Its slope m satisfies 1−0 0−5 =m= = −1. k −3 3+2 Thus k − 3 = −1, and so k = 2.

39. C = Ax + B. If C = 5, 000 when x = 10, 000 and C = 6, 000 when x = 15, 000, then

10, 000 A + B = 5, 000 15, 000 A + B = 6, 000

placement of M from P1 equals the displacement of P2 from M:

40. −40◦ and −40◦ is the same temperature on both the

xm − x1 = x2 − xm ,

45. If Q = (xq , yq ) is the point on P1 P2 that is two thirds of the way from P1 to P2 , then the displacement of Q from P1 equals twice the displacement of P2 from Q: xq − x1 = 2(x2 − xq ),

30 C=F

-30

46. Let the coordinates of P be (x, 0) and those of Q be

C=

(x + X)/2 = 2,

5 (F − 32) 9

47.

Fig. P.2.40

42.

(X, −2X). If the midpoint of P Q is (2, 1), then

10 20 30 40 50 60 70 80 F

-40 (−40, −40) -50

41.

yq − y1 = 2(y2 − yq ).

Thus xq = (x1 + 2x2 )/3 and yq = (y1 + 2y2 )/3.

10

-20

ym − y1 = y2 − ym .

Thus xm = (x1 + x2 )/2 and ym = (y1 + y2 )/2.

Fahrenheit and Celsius scales. C 40

-50 -40 -30 -20 -10 -10

B = (1, 3), C = (−3, 2) p √ | AB| = (1 − 2)2 + (3 + 1)2 = 17 p √ √ √ | AC| = (−3 − 2)2 + (2 + 1)2 = 34 = 2 17 p √ |BC| = (−3 − 1)2 + (2 − 3)2 = 17. √ Since | AB| = |BC| and | AC| = 2| AB|, triangle ABC is an isosceles right-angled triangle with right angle at B. Thus ABC D is a square if D is displaced from C by the same amount A is from B, that is, by increments 1x = 2 − 1 = 1 and 1y = −1 − 3 = −4. Thus D = (−3 + 1, 2 + (−4)) = (−2, −2).

44. If M = (xm , ym ) is the midpoint of P1 P2 , then the dis-

Subtracting these equations gives 5, 000 A = 1, 000, so A = 1/5. From the first equation, 2, 000 + B = 5, 000, so B = 3, 000. The cost of printing 100,000 pamphlets is $100, 000/5 + 3, 000 = $23, 000.

20

A = (2, −1),

A = (2, 1),

B = (6, 4), C = (5, −3) p √ | AB| = (6 − 2)2 + (4 − 1)2 = 25 = 5 p √ | AC| = (5 − 2)2 + (−3 − 1)2 = 25 = 5 p √ √ |BC| = (6 − 5)2 + (4 + 3)2 = 50 = 5 2. Since | AB| = | AC|, triangle ABC is isosceles. √ A = (0, 0), B = (1, 3), C = (2, 0) q √ √ | AB| = (1 − 0)2 + ( 3 − 0)2 = 4 = 2 p √ | AC| = (2 − 0)2 + (0 − 0)2 = 4 = 2 q √ √ |BC| = (2 − 1)2 + (0 − 3)2 = 4 = 2. Since | AB| = | AC| = |BC|, triangle ABC is equilateral.

48.

(0 − 2X)/2 = 1.

The second equation implies that X = −1, and the second then implies that x = 5. Thus P is (5, 0). p (x − 2)2 + y 2 = 4 says that the distance of (x, y) from (2, 0) is 4, so the equation represents a circle of radius 4 centred at (2, 0). p p (x − 2)2 + y 2 = x 2 + (y − 2)2 says that (x, y) is equidistant from (2, 0) and (0, 2). Thus (x, y) must lie on the line that is the right bisector of the line from (2, 0) to (0, 2). A simpler equation for this line is x = y.

49. The line 2x + ky = 3 has slope m = −2/k. This line

is perpendicular to 4x + y = 1, which has slope −4, provided m = 1/4, that is, provided k = −8. The line is parallel to 4x + y = 1 if m = −4, that is, if k = 1/2.

50. For any value of k, the coordinates of the point of intersection of x + 2y = 3 and 2x − 3y = −1 will also satisfy the equation

4 Copyright © 2014 Pearson Canada Inc.

(x + 2y − 3) + k(2x − 3y + 1) = 0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.3 (PAGE 22)

because they cause both expressions in parentheses to be 0. The equation above is linear in x and y, and so represents a straight line for any choice of k. This line will pass through (1, 2) provided 1 + 4 − 3 + k(2 − 6 + 1) = 0, that is, if k = 2/3. Therefore, the line through the point of intersection of the two given lines and through the point (1, 2) has equation 2 x + 2y − 3 + (2x − 3y + 1) = 0, 3

13. Together, x 2 + y 2 > 1 and x 2 + y 2 < 4 represent annulus (washer-shaped region) consisting of all points that are outside the circle of radius 1 centred at the origin and inside the circle of radius 2 centred at the origin.

14. Together, x 2 + y 2 ≤ 4 and (x + 2)2 + y 2 ≤ 4 represent the region consisting of all points that are inside or on both the circle of radius 2 centred at the origin and the circle of radius 2 centred at (−2, 0).

15. Together, x 2 + y 2 < 2x and x 2 + y 2 < 2y (or, equivalently, (x − 1)2 + y 2 < 1 and x 2 + (y − 1)2 < 1) represent the region consisting of all points that are inside both the circle of radius 1 centred at (1, 0) and the circle of radius 1 centred at (0, 1).

or, on simplification, x = 1.

Section P.3 Graphs of Quadratic Equations (page 22) 1.

x 2 + y 2 = 16

2.

x 2 + (y − 2)2 = 4, or x 2 + y 2 − 4y = 0

3. (x + 2)2 + y 2 = 9, or x 2 + y 2 + 4y = 5 4. (x − 3)2 + (y + 4)2 = 25, or x 2 + y 2 − 6x + 8y = 0. 5.

2

x 2 − 2x + 1 + y 2 = 4

18. The exterior of the circle with centre (2, −3) and ra-

x 2 + y 2 + 4y = 0

19.

x 2 + y 2 < 2,

x ≥1

20.

x 2 + y 2 > 4,

(x − 1)2 + (y − 3)2 < 10

dius 4 is given by (x − 2)2 + (y + 3)2 > 16, or x 2 + y 2 − 4x + 6y > 3.

2

x + y + 4y + 4 = 4

x 2 + (y + 2)2 = 4 centre: (0, −2); radius 2.

21. The parabola with focus (0, 4) and directrix y = −4 has equation x 2 = 16y.

x 2 + y 2 − 2x + 4y = 4

22. The parabola with focus (0, −1/2) and directrix y = 1/2

x 2 − 2x + 1 + y 2 + 4y + 4 = 9

has equation x 2 = −2y.

(x − 1)2 + (y + 2)2 = 9 centre: (1, −2); radius 3.

8.

x 2 + y 2 − 2x − y + 1 = 0

x 2 − 2x + 1 + y 2 − y + 14 =
2 (x − 1)2 + y − 21 = 14 centre: (1, 1/2); radius 1/2.

9. 10.

6 is given by (x + 1)2 + (y − 2)2 < 6, or + y 2 + 2x − 4y < 1.

x2

x + y − 2x = 3

2

7.

x 2 + y 2 − 4x + 2y > 4 can be rewritten (x −2)2 +(y +1)2 > 9. This equation, taken together with x + y > 1, represents all points that lie both outside the circle of radius 3 centred at (2, −1) and above the line x + y = 1.

17. √ The interior of the circle with centre (−1, 2) and radius

2

(x − 1)2 + y 2 = 4 centre: (1, 0); radius 2.

6.

16.

23. The parabola with focus (2, 0) and directrix x = −2 has equation y 2 = 8x.

24. The parabola with focus (−1, 0) and directrix x = 1 has

1 4

equation y 2 = −4x.

25.

y = x 2 /2 has focus (0, 1/2) and directrix y = −1/2. y

x 2 + y 2 > 1 represents all points lying outside the circle of radius 1 centred at the origin. x2

y=x 2 /2

(0,1/2)

y2

+ < 4 represents the open disk consisting of all points lying inside the circle of radius 2 centred at the origin.

x y=−1/2

11. (x + 1)2 + y 2 ≤ 4 represents the closed disk consisting of all points lying inside or on the circle of radius 2 centred at the point (−1, 0).

12.

x 2 + (y − 2)2 ≤ 4 represents the closed disk consisting of all points lying inside or on the circle of radius 2 centred at the point (0, 2).

Fig. P.3.25

26.

y = −x 2 has focus (0, −1/4) and directrix y = 1/4.

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SECTION P.3 (PAGE 22)

ADAMS and ESSEX: CALCULUS 8

y

y

Version (c) y=1/4

y = x2

x

(3, 3) Version (b)

(0,−1/4) y=−x 2

Fig. P.3.26

x 4 Version (d)

27.

Version (a) −3

x = −y 2/4 has focus (−1, 0) and directrix x = 1. y

(4, −2)

Fig. P.3.29 a) has equation y = x 2 − 3.

x=1

b) has equation y = (x − 4)2 or y = x 2 − 8x + 16.

(−1,0) x

c) has equation y = (x − 3)2 + 3 or y = x 2 − 6x + 12.

d) has equation y = (x − 4)2 − 2, or y = x 2 − 8x + 14.

x=−y 2 /4

30. Fig. P.3.27

28.

x = y 2 /16 has focus (4, 0) and directrix x = −4. y

31. (4,0) x

32. 33.

x=−4 x=y 2 /16

34. Fig. P.3.28

b) If y = mx is shifted vertically by amount y1 , the equation y = mx + y1 results. If (a, b) satisfies this equation, then b = ma + y1 , and so y1 = b − ma. Thus the shifted equation is y = mx + b − ma = m(x − a) + b, the same equation obtained in part (a). p y = (x/3) + 1 √ 4y = x + 1 p y = (3x/2) + 1 √ (y/2) = 4x + 1

35.

y = 1 − x 2 shifted down 1, left 1 gives y = −(x + 1)2 .

36.

x 2 + y 2 = 5 shifted up 2, left 4 gives (x + 4)2 + (y − 2)2 = 5.

37.

y = (x − 1)2 − 1 shifted down 1, right 1 gives y = (x − 2)2 − 2. √ √ y = x shifted down 2, left 4 gives y = x + 4 − 2.

38.

29.

a) If y = mx is shifted to the right by amount x1 , the equation y = m(x − x1 ) results. If (a, b) satisfies this equation, then b = m(a − x1 ), and so x1 = a −(b/m). Thus the shifted equation is y = m(x − a + (b/m)) = m(x − a) + b.

6 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

39.

SECTION P.3 (PAGE 22)

y

y = x 2 + 3, y = 3x + 1. Subtracting these equations gives x 2 − 3x + 2 = 0, or (x − 1)(x − 2) = 0. Thus x = 1 or x = 2. The corresponding values of y are 4 and 7. The intersection points are (1, 4) and (2, 7).

9x 2 +16y 2 =144

x

40.

41.

y = x 2 − 6, y = 4x − x 2 . Subtracting these equations gives 2x 2 − 4x − 6 = 0, or 2(x − 3)(x + 1) = 0. Thus x = 3 or x = −1. The corresponding values of y are 3 and −5. The intersection points are (3, 3) and (−1, −5). x 2 + y 2 = 25, 3x + 4y = 0. The second equation says that y = −3x/4. Substituting this into the first equation gives 25x 2 /16 = 25, so x = ±4. If x = 4, then the second equation gives y = −3; if x = −4, then y = 3. The intersection points are (4, −3) and (−4, 3). Note that having found values for x, we substituted them into the linear equation rather than the quadratic equation to find the corresponding values of y. Had we substituted into the quadratic equation we would have got more solutions (four points in all), but two of them would have failed to satisfy 3x + 4y = 12. When solving systems of nonlinear equations you should always verify that the solutions you find do satisfy the given equations.

Fig. P.3.44

45.

(x − 3)2 (y + 2)2 + = 1 is an ellipse with centre at 9 4 (3, −2), major axis between (0, −2) and (6, −2) and minor axis between (3, −4) and (3, 0). y

x

(3,−2)

(x−3)2 (y+2)2 9 + 4 =1

42. 2x 2 + 2y 2 = 5, x y = 1. The second equation says that

y = 1/x. Substituting this into the first equation gives 2x 2 + (2/x 2 ) = 5, or 2x 4 − 5x 2 + 2 = 0. This equation factors to √ (2x 2 − 1)(x 2 −√2) = 0, so its solutions are x = ±1/ 2 and x = ± 2. The corresponding values of y are given√by √ y = 1/x. √ Therefore, √ the √ intersection √ points are (1/ 2, 2), (−1/ 2, − 2), ( 2, 1/ 2), and √ √ (− 2, −1/ 2).

43. (x 2 /4) + y 2 = 1 is an ellipse with major axis between (−2, 0) and (2, 0) and minor axis between (0, −1) and (0, 1).

Fig. P.3.45

(y + 1)2 = 4 is an ellipse with centre at 4 (1, −1), major axis between (1, −5) and (1, 3) and minor axis between (−1, −1) and (3, −1).

46. (x − 1)2 +

y

(x−1)2 +

(y+1)2 4 =4

y x2 2 4 +y =1

x (1,−1) x

Fig. P.3.46 Fig. P.3.43

44. 9x 2 + 16y 2 = 144 is an ellipse with major axis between (−4, 0) and (4, 0) and minor axis between (0, −3) and (0, 3).

47. (x 2 /4) − y 2 = 1 is a hyperbola with centre at the origin and passing through (±2, 0). Its asymptotes are y = ±x/2.

7 Copyright © 2014 Pearson Canada Inc.

SECTION P.3 (PAGE 22)

ADAMS and ESSEX: CALCULUS 8

y

y x2 2 4 −y =1 y=−x/2

x (x − 1)(y + 2) = 1 y = −2

x y=x/2

x =1

Fig. P.3.50

Fig. P.3.47

51. 48.

x 2 − y 2 = −1 is a rectangular hyperbola with centre at the origin and passing through (0, ±1). Its asymptotes are y = ±x. y x 2 −y 2 =−1

a) Replacing x with −x replaces a graph with its reflection across the y-axis. b) Replacing y with −y replaces a graph with its reflection across the x-axis.

52. Replacing x with −x and y with −y reflects the graph in both axes. This is equivalent to rotating the graph 180◦ about the origin.

53. |x| + |y| = 1.

y=x

In In In In

x y=−x

the the the the

first quadrant the equation is x + y = 1. second quadrant the equation is −x + y = 1. third quadrant the equation is −x − y = 1. fourth quadrant the equation is x − y = 1. y

Fig. P.3.48 1 |x| + |y| = 1

49.

1

x y = −4 is a rectangular hyperbola with centre at the origin and passing through (2, −2) and (−2, 2). Its asymptotes are the coordinate axes.

x

−1

y

−1

Fig. P.3.53 x

Section P.4 (page 32)

xy=−4

Fig. P.3.49

1. 2. 3.

50. (x − 1)(y + 2) = 1 is a rectangular hyperbola with centre at (1, −2) and passing through (2, −1) and (0, −3). Its asymptotes are x = 1 and y = −2.

4.

Functions and Their Graphs

f (x) = 1 + x 2 ; domain R, range [1, ∞)

√ f (x) = 1 − x; domain [0, ∞), range (−∞, 1] √ G(x) = 8 − 2x; domain (−∞, 4], range [0, ∞) F(x) = 1/(x − 1); domain (−∞, 1) ∪ (1, ∞), range (−∞, 0) ∪ (0, ∞)

8 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.4 (PAGE 32)

t ; domain (−∞, 2), range R. (The equa2−t tion y = h(t) can be squared and rewritten as t 2 + y 2 t − 2y 2 = 0, a quadratic equation in t having real solutions for every real value of y. Thus the range of h contains all real numbers.)

y

5. h(t) = √

y

graph (a)

graph (b)

x

y

6.

x

y

graph (c)

1 ; domain (2, 3) ∪ (3, ∞), range √ 1− x −2 (−∞, 0) ∪ (0, ∞). The equation y = g(x) can be solved for x = 2 − (1 − (1/y))2 so has a real solution provided y 6= 0.

graph (d)

g(x) =

x

x

Fig. P.4.8 a) is the graph of x(1−x)2 , which is positive for x > 0.

7.

y

b) is the graph of x 2 − x 3 = x 2 (1 − x), which is positive if x < 1.

y graph (i)

c) is the graph of x − x 4 , which is positive if 0 < x < 1 and behaves like x near 0.

graph (ii)

d) is the graph of x 3 − x 4 , which is positive if 0 < x < 1 and behaves like x 3 near 0. x

x

9.

y

y graph (iii)

x

graph (iv)

x

x

0 ±0.5 ±1 ±1.5 ±2

f (x) = x 4 0 0.0625 1 5.0625 16

Fig. P.4.7

y y = x4

Graph (ii) is the graph of a function because vertical lines can meet the graph only once. Graphs (i), (iii), and (iv) do not have this property, so are not graphs of functions.

x

8.

Fig. P.4.9

9 Copyright © 2014 Pearson Canada Inc.

SECTION P.4 (PAGE 32)

ADAMS and ESSEX: CALCULUS 8

24.

10. x 0 ±0.5 ±1 ±1.5 ±2

f (x) =

y

x 2/3

0 0.62996 1 1.3104 1.5874

y=1−x 2

x

y y=

x 2/3

25. y

y=(x−1)2

x Fig. P.4.10

11.

f (x) = x 2 + 1 is even: f (−x) = f (x)

12. 13.

f (x) = x 3 + x is odd: f (−x) = − f (x) x f (x) = 2 is odd: f (−x) = − f (x) x −1

14.

f (x) =

15.

f (x) =

16.

f (x) =

17.

f (x) = x 2 − 6x is even about x = 3: f (3 − x) = f (3 + x)

18.

f (x) = x 3 − 2 is odd about (0, −2): f (−x) + 2 = −( f (x) + 2)

19.

f (x) = |x 3 | = |x|3 is even: f (−x) = f (x)

20.

f (x) = |x + 1| is even about x = −1: f (−1 − x) = f (−1 + x) √ f (x) = 2x has no symmetry. p f (x) = (x − 1)2 is even about x = 1: f (1 − x) = f (1 + x)

21. 22. 23.

x

26. y

1 is even: f (−x) = f (x) x2 − 1

1 is odd about (2, 0): f (2 − x) = − f (2 + x) x −2

1 is odd about (−4, 0): x +4 f (−4 − x) = − f (−4 + x)

y=(x−1)2 +1 x

27. y

y=1−x 3

x

28.

y

y y=−x 2

x

x y=(x+2)3

10 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.4 (PAGE 32)

29.

34. y

y y=1+|x−2|

√ y= x+1

x x

30.

35. y

y

x=−2 √ y= x+1 x

x

2 y= x+2

36.

31. y

y x

x=2

x y=−|x| 1 y= 2−x

32.

37. y

y x y= x+1

y=|x|−1

y=1 x x x=−1

33.

38. y

y x=1 y=|x−2| x y=−1 x y= 1−x x

11 Copyright © 2014 Pearson Canada Inc.

SECTION P.4 (PAGE 32)

ADAMS and ESSEX: CALCULUS 8

45.

39.

y

y

y (1,3) y= f (x)+2 2

(2,2)

(1,1) x

2

Fig. P.4.39(a)

(3,1)

y= f (4−x)

y= f (x)

2

x

4

x

Fig. P.4.39(b)

46.

40. y

(1,3) y= f (x)+2 2

y

y

(2,2)

(1,1)

(−1,1)

y=1− f (1−x)

1 x

Fig. P.4.40(a)

x

y= f (x)−1 x (2,−1)

−1

Fig. P.4.40(b)

41. y

47. Range is approximately [−0.18, 0.68]. y 0.8 0.6

(−1,1) y= f (x+2)

0.4 0.2

x

−2

42.

-5

y

(2,1) 1

-4

-3

-2

-1 -0.2

y = 0.68 x + 2 y= 2 x + 2x + 3

1y = 2−0.18 3

4

x

4

x

-0.4 -0.6 -0.8 -1.0 Fig. P.4.47

y= f (x−1) 3

x

43. y

48. Range is approximately (−∞, 0.17]. y -5

2 y=− f (x) (1,−1)

-4

x

-3

-2

-1

1

2

y=

x −1 x2 + x

-3

y

-4 -5

(−1,1) y= f (−x) −2

-6 x

12 Copyright © 2014 Pearson Canada Inc.

3

-1 -2

44.

y = 0.17

-7 Fig. P.4.48

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.5 (PAGE 38)

49.

Apparent symmetry about (2, 1), and about the lines y = x − 1 and y = 3 − x.

y

1 , x −2 so the graph is that of 1/x shifted right 2 units and up one.

5

These can be confirmed by noting that f (x) = 1 +

4 3

52.

2

y=

1 -5

-4

-3

-2

-1

1

2

3

4

2x 2 + 3x + 4x + 5

y

x2

5 4

x

3

-1

y = x 4 − 6x 3 + 9x 2 − 1 Fig. P.4.49

2 1

Apparent symmetry about x = 1.5. This can be confirmed by calculating f (3 − x), which turns out to be equal to f (x).

-7

-6

-5

-4

-3

-2

-1

1

2

x

-1 -2

50.

Fig. P.4.52

y y= 2

3 − 2x + x 2 2 − 2x + x 2

Apparent symmetry about (−2, 2). This can be confirmed by calculating shifting the graph right by 2 (replace x with x − 2) and then down 2 (subtract 2). The result is −5x/(1 + x 2 ), which is odd.

53.

1

-5

-4

-3

-2

-1

1

2

3

4

Section P.5 Combining Functions to Make New Functions (page 38)

x

-1 Fig. P.4.50

1.

Apparent symmetry about x = 1. This can be confirmed by calculating f (2 − x), which turns out to be equal to f (x).

51. y 4 y = x −1 x −1 y= x −2

3 2

2.

1 -3

-2

-1

1

2

3

-1

4

5

6

x

y = −x + 3

-2 Fig. P.4.51

If f is both even and odd the f (x) = f (−x) = − f (x), so f (x) = 0 identically.

√ f (x) = x, g(x) = x − 1. D( f ) = R, D(g) = [1, ∞). D( f + g) = D( f − g) = D( f g) = D(g/ f ) = [1, ∞), D( f /g) = (1, ∞). √ ( f + g)(x) = x + x − 1 √ ( f − g)(x) = x − x − 1 √ ( f g)(x) = x x − 1 √ ( f /g)(x) = x/ x − 1 √ (g/ f )(x) = ( 1 − x)/x √ √ f (x) = 1 − x, g(x) = 1 + x. D( f ) = (−∞, 1], D(g) = [−1, ∞). D( f + g) = D( f − g) = D( f g) = [−1, 1], D( f /g) = (−1, f ) = [−1, 1). √1], D(g/√ ( f + g)(x) = 1 − x + 1 + x √ √ ( f − g)(x) = 1 − x − 1 + x p ( f g)(x) = 1 − x 2 p ( f /g)(x) = (1 − x)/(1 + x) p (g/ f )(x) = (1 + x)/(1 − x) 13

Copyright © 2014 Pearson Canada Inc.

SECTION P.5 (PAGE 38)

3.

ADAMS and ESSEX: CALCULUS 8

7.

y

y=x−

x2

f (g(x)) = f (x 2 − 3) = x 2 + 2

x

g ◦ f (x) = g( f (x)) = g(x + 5) = (x + 5)2 − 3 f ◦ f (−5) = f (0) = 5, g(g(2)) = g(1) = −2 f ( f (x)) = f (x + 5) = x + 10

y=x

g ◦ g(x) = g(g(x)) = (x 2 − 3)2 − 3

8. y=

−x 2

4.

y y = −x 1

9. -2

x 1 y = x3 − x

-1 -1 y = x3

5.

-2 y

y = x + |x|

10.

y = |x|

y = x = |x| x

y=x

6.

y 4

y = |x|

2

y = |x − 2|

1 -1

1 -1

2

3

4

f (x) = 2/x, g(x) = x/(1 − x). f ◦ f (x) = 2/(2/x) = x; D( f ◦ f ) = {x : x 6= 0} f ◦ g(x) = 2/(x/(1 − x)) = 2(1 − x)/x; D( f ◦ g) = {x : x 6= 0, 1} g ◦ f (x) = (2/x)/(1 − (2/x)) = 2/(x − 2); D(g ◦ f ) = {x : x 6= 0, 2} g ◦ g(x) = (x/(1 − x))/(1 − (x/(1 − x))) = x/(1 − 2x); D(g ◦ g) = {x : x 6= 1/2, 1}

√ f (x) = 1/(1 − x), g(x) = x − 1. f ◦ f (x) = 1/(1 − (1/(1 − x))) = (x − 1)/x; D( f ◦ f ) = {x : x 6= 0, 1} √ f ◦ g(x) = 1/(1 − x − 1); D( f ◦ g) = {x : x ≥ 1, x 6= 2} p p g ◦ f (x) = (1/(1 − x)) − 1 = x/(1 − x); D(g ◦ f ) = [0, 1) q √ g ◦ g(x) = x − 1 − 1; D(g ◦ g) = [2, ∞)

f (x) = (x + 1)/(x − 1) = 1 + 2/(x − 1), g(x) = sgn (x). f ◦ f (x) = 1 + 2/(1 + (2/(x − 1) − 1)) = x; D( f ◦ f ) = {x : x 6= 1} sgn x + 1 f ◦ g(x) = = 0; D( f ◦ g) = (−∞, 0) sgn x − 1   n x +1 1 if x < −1 or x > 1 g ◦ f (x) = sgn = ; −1 if −1 < x < 1 x −1 D(g ◦ f ) = {x : x 6= −1, 1} g ◦ g(x) = sgn (sgn (x)) = sgn (x); D(g ◦ g) = {x : x 6= 0}

11. 12. 13. 14. 15. 16.

y = |x| + |x − 2|

3

-2

f (x) = x + 5, g(x) = x 2 − 3. f ◦ g(0) = f (−3) = 2, g( f (0)) = g(5) = 22

5

17. x

f (x)

g(x)

f ◦ g(x)

x2 x −4 √ x 2x 3 + 3 (x + 1)/x 1/(x + 1)2

x +1 x +4 x2 x 1/3 1/(x − 1) x −1

(x + 1)2 x |x| 2x + 3 x 1/x 2

√ y = x. √ y = 2 + √ x: previous graph is raised 2 units. y = 2 + 3√ + x: previous graph is shiftend left 3 units. y = 1/(2 + 3 + x): previous graph turned upside down and shrunk vertically.

14 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION P.5 (PAGE 38)

y

21. y

y =2+ y =2+

√

x +3

y = 1/(2 +

√

√

(1/2,1) y= f (2x)

x

y=

x + 3)

x

1

√

x

22. y

x y= f (x/3)

Fig. P.5.17

18. y=

√ 1 − 2x

23.

y

y

y = 2x y= √

6 x

3

(−2,2)

y = 2x − 1

1

y=1+ f (−x/2)

1 − 2x

x

x y= √

1 1 − 2x

24. y

−1

y=2 f ((x−1)/2)

y = 1 − 2x Fig. P.5.18

1

x

5

19. y

25.

y

(1,2)

y = f (x) (1, 1)

y=2 f (x)

82

x

2

20.

26. y

x

-2 y y = g(x) (1, 1)

2 x y=−(1/2) f (x)

x

15 Copyright © 2014 Pearson Canada Inc.

SECTION P.5 (PAGE 38)

27.

ADAMS and ESSEX: CALCULUS 8

35. a) Let E(x) = 21 [ f (x) + f (−x)].

F(x) = Ax + B (a) F ◦ F(x) = F(x) ⇒ A(Ax + B) + B = Ax + B ⇒ A[(A − 1)x + B] = 0 Thus, either A = 0 or A = 1 and B = 0. (b) F ◦ F(x) = x ⇒ A(Ax + B) + B = x ⇒ (A2 − 1)x + (A + 1)B = 0 Thus, either A = −1 or A = 1 and B = 0

Then E(−x) = 21 [ f (−x) + f (x)] = E(x). Hence, E(x) is even. Let O(x) = 21 [ f (x) − f (−x)]. Then O(−x) = 21 [ f (−x) − f (x)] = −O(x) and O(x) is odd. E(x) + O(x)

= 21 [ f (x) + f (−x)] + 21 [ f (x) − f (−x)]

28. ⌊x⌋ = 0 for 0 ≤ x < 1; ⌈x⌉ = 0 for −1 ≤ x < 0.

= f (x).

29. ⌊x⌋ = ⌈x⌉ for all integers x.

Hence, f (x) is the sum of an even function and an odd function.

30. ⌈−x⌉ = −⌊x⌋ is true for all real x; if x = n + y where n 31.

is an integer and 0 ≤ y < 1, then −x = −n − y, so that ⌈−x⌉ = −n and ⌊x⌋ = n. y

b) If f (x) = E 1 (x) + O1 (x) where E 1 is even and O1 is odd, then

y = x − ⌊x⌋

E 1 (x) + O1 (x) = f (x) = E(x) + O(x). Thus E 1 (x) − E(x) = O(x) − O1 (x). The left side of this equation is an even function and the right side is an odd function. Hence both sides are both even and odd, and are therefore identically 0 by Exercise 36. Hence E 1 = E and O1 = O. This shows that f can be written in only one way as the sum of an even function and an odd function.

x

32.

f (x) is called the integer part of x because | f (x)| is the largest integer that does not exceed x; i.e. |x| = | f (x)| + y, where 0 ≤ y < 1. y

Section P.6 Polynomials and Rational Functions (page 45) x y = f (x)

1.

x 2 − 7x + 10 = (x + 5)(x + 2) The roots are −5 and −2.

2.

x 2 − 3x − 10 = (x − 5)(x + 2) The roots are 5 and −2.

3. If x 2 + 2x + 2 = 0, then x =

and f ◦ f are all even. f g, f /g, g/ f , and g ◦ g are odd, and f + g is neither even nor odd. Here are two typical verifications:

34.

f even ⇔ f (−x) = f (x) f odd ⇔ f (−x) = − f (x) f even and odd ⇒ f (x) = − f (x) ⇒ 2 f (x) = 0 ⇒ f (x) = 0

2

4−8

= −1 ± i .

4. Rather than use the quadratic formula this time, let us complete the square. x 2 − 6x + 13 = x 2 − 6x + 9 + 4

= (x − 3)2 + 22 = (x − 3 − 2i )(x − 3 + 2i ).

f ◦ g(−x) = f (g(−x)) = f (−g(x)) = f (g(x)) = f ◦ g(x) ( f g)(−x) = f (−x)g(−x) = f (x)[−g(x)] = − f (x)g(x) = −( f g)(x). The others are similar.

√

The roots are −1 + i and −1 − i . x 2 + 2x + 2 = (x + 1 − i )(x + 1 + i ).

Fig. P.5.32

33. If f is even and g is odd, then: f 2 , g 2 , f ◦ g, g ◦ f ,

−2 ±

The roots are 3 + 2i and 3 − 2i .

5. 16x 4 − 8x 2 + 1 = (4x 2 − 1)2 = (2x − 1)2 (2x + 1)2 . There are two double roots: 1/2 and −1/2.

6.

x 4 + 6x 3 + 9x 2 = x 2 (x 2 + 6x + 9) = x 2 (x + 3)2 . There are two double roots, 0 and −3.

16 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

7.

SECTION P.6 (PAGE 45)

x 3 + 1 = (x + 1)(x 2 − x + 1). One root is −1. The other two are the solutions of x 2 − x + 1 = 0, namely x=

1±

√ √ 1 1−4 3 = ± i. 2 2 2

15. The denominator is x 3 + x 2 = x 2 (x + 1) which is zero only if x = 0 or x = −1. Thus the rational function is defined for all real numbers except 0 and −1.

16. The denominator is x 2 + x − 1, which is a quadratic

polynomial whose roots can be found with the quadratic √ formula. They are x = (−1 ± 1 + 4)/2. Hence the given rational √ function is defined√for all real numbers except (−1 − 5)/2 and (−1 + 5)/2.

We have √ ! √ ! 1 3 1 3 x + 1 = (x + 1) x − − i x− + i . 2 2 2 2 3

8. 9. 10.

17.

x 4 − 1 = (x 2 − 1)(x 2 + 1) = (x − 1)(x + 1)(x − i )(x + i ). The roots are 1, −1, i , and −i . x 6 − 3x 4 + 3x 2 − 1 = (x 2 − 1)3 = (x − 1)3 (x + 1)3 . The roots are 1 and −1, each with multiplicity 3. 5

4

4

x − x − 16x + 16 = (x − 1)(x − 16)

18.

= (x − 1)(x 2 − 4)(x 4 + 4) = (x − 1)(x − 2)(x + 2)(x − 2i )(x + 2i ).

11.

12.

The roots are 1, 2, −2, 2i , and −2i .

19.

= (x + 2)(x − i )(x + i )(x 2 − 2x + 4) Three of the five roots are −2, i and −i . The remain2 ing two are √ solutions of x − 2x + 4 = 0, namely √ 2 ± 4 − 16 = 1 ± 3 i . We have x= 2 √ √ x 5 +x 3 +8x 2 +8 = (x+2)(x−i )(x+i )(x−a+ 3 i )(x−a− 3 i ).

20.

x 5 + x 3 + 8x 2 + 8 = (x 2 + 1)(x 3 + 8)

x 9 − 4x 7 − x 6 + 4x 4 = x 4 (x 5 − x 2 − 4x 3 + 4) = x 4 (x 3 − 1)(x 2 − 4)

= x 4 (x − 1)(x − 2)(x + 2)(x 2 + x + 1).

Seven of the nine roots are: 0 (with multiplicity 4), 1, 2, and −2. The other two roots are solutions of x 2 + x + 1 = 0, namely x=

−1 ±

√ √ 1 1−4 3 =− ± i. 2 2 2

The required factorization of x 9 − 4x 7 − x 6 + 4x 4 is √ ! √ ! 1 3 1 3 i x− − i . x (x−1)(x−2)(x+2) x − + 2 2 2 2 4

13. The denominator is x 2 + 2x + 2 = (x + 1)2 + 1 which is never 0. Thus the rational function is defined for all real numbers.

14. The denominator is x 3 − x = x(x − 1)(x + 1) which

is zero if x = 0, 1, or −1. Thus the rational function is defined for all real numbers except 0, 1, and −1.

x 3 − 2x + 2x − 1 x3 − 1 = 2 x −2 x2 − 2 2 x(x − 2) + 2x − 1 = x2 − 2 2x − 1 =x+ 2 . x −2 x2

x2 x 2 + 5x + 3 − 5x − 3 = + 5x + 3 x 2 + 5x + 3 −5x − 3 =1+ 2 . x + 5x + 3

x 3 + 2x 2 + 3x − 2x 2 − 3x x3 = x 2 + 2x + 3 x 2 + 2x + 3 x(x 2 + 2x + 3) − 2x 2 − 3x = x 2 + 2x + 3 2(x 2 + 2x + 3) − 4x − 6 + 3x =x− x 2 + 2x + 3 x +6 = x −2+ 2 . x + 2x + 3

x(x 3 + x 2 + 1) − x 3 − x + x 2 x4 + x2 = x3 + x2 + 1 x3 + x2 + 1 −(x 3 + x 2 + 1) + x 2 + 1 − x + x 2 =x+ x3 + x2 + 1 2 2x − x + 1 = x −1+ 3 . x + x2 + 1

4 2 21. As in Example √ 6, we want√a = 4, so a = 2

and a = 2, b = ± 2a = ±2. Thus P(x) = (x 2 − 2x + 2)(x 2 + 2x + 2).

22. Following the method of Example 6, we calculate (x 2 −bx+a 2 )(x 2 +bx+a 2 ) = x 4 +a 4 +(2a 2 −b2 )x 2 = x 2 +x 2 +1 √ provided a = 1√and b2 = 1 + √ 2a 2 = 3, so b = 3. Thus P(x) = (x 2 − 3x + 1)(x 2 + 3x + 1).

23. Let P(x) = an x n + an−1 x n−1 + · · · + a1 x + a0 , where n ≥ 1. By the Factor Theorem, x − 1 is a factor of P(x) if and only if P(1) = 0, that is, if and only if an + an−1 + · · · + a1 + a0 = 0.

24. Let P(x) = an x n + an−1 x n−1 + · · · + a1 x + a0 , where

n ≥ 1. By the Factor Theorem, x + 1 is a factor of P(x) if and only if P(−1) = 0, that is, if and only if a0 − a1 + a2 − a3 + · · · + (−1)n an = 0. This condition says that the sum of the coefficients of even powers is equal to the sum of coefficients of odd powers.

17 Copyright © 2014 Pearson Canada Inc.

SECTION P.6 (PAGE 45)

ADAMS and ESSEX: CALCULUS 8

25. Let P(x) = an x n + an−1 x n−1 + · · · + a1 x + a0 , where the coefficients ak , 0 ≤ k ≤ n are all real numbers, so that a¯k = ak . Using the facts about conjugates of sums and products mentioned in the statement of the problem, we see that if z = x + i y, where x and y are real, then

4.

P(z) = an z n + an−1 z n−1 + · · · + a1 z + a0 = an z¯ n + an−1 z¯ n−1 + · · · + a1 z¯ + a0 = P(¯z ).

5.

If z is a root of P, then P(¯z ) = P(z) = 0¯ = 0, and z¯ is also a root of P.

26. By the previous exercise, z¯ = u − i v is also a root of

P. Therefore P(x) has two linear factors x − u − i v and x − u + i v. The product of these factors is the real quadratic factor (x − u)2 − i 2 v 2 = x 2 − 2ux + u 2 + v 2 , which must also be a factor of P(x).

sin

P(x) = Q m (x), (x 2 − 2ux + u 2 + v 2 )m

π 4 3 π π π π = sin cos + cos sin 4 3 √ 4 3 √ 1 1 1 3 1+ 3 = √ +√ = √ 22 2 2 2 2 = sin

1. cos 2. tan 3.



3π 4



 π π 1 = cos π − = − cos = − √ 4 4 2

−3π 3π = − tan = −1 4 4

+

11π π = sin 12 12π π = sin − 3 4 π π π π = sin cos − cos sin 3 4 3 4 √ !     1 1 3 1 = √ √ − 2 2 2 2 √ 3−1 √ = 2 2

7.

  cos(π + x) = cos 2π − (π − x)   = cos −(π − x)

= cos(π − x) = − cos x

8. sin(2π − x) = − sin x 9.

sin



3π −x 2



where Q m no longer has z (or z¯ ) as a root. Thus z and z¯ must have the same multiplicity as roots of P.

Section P.7 The Trigonometric Functions (page 57)

π

sin

P(x) Q 1 (x) = 2 = Q 2 (x), 2 2 2 − 2ux + u + v ) x − 2ux + u 2 + v 2

where Q 2 is a polynomial with real coefficients. We can continue in this way until we get



6.

P(x) P(x) = = Q 1 (x), x 2 − 2ux + u 2 + v 2 (x − u − i v)(x − u + i v)

(x 2

7π 12

  5π 2π π cos = cos − 12 3 4 2π π 2π π = cos cos + sin sin 3 4 3 ! 4 √      3 1 1 1 =− √ + √ 2 2 2 2 √ 3−1 = √ 2 2

27. By the previous exercise

where Q 1 , being a quotient of two polynomials with real coefficients, must also have real coefficients. If z = u +i v is a root of P having multiplicity m > 1, then it must also be a root of Q 1 (of multiplicity m − 1), and so, therefore, z¯ must be a root of Q 1 , as must be the real quadratic x 2 − 2ux + u 2 + v 2 . Thus



10.

11.

  π  = sin π − x − 2  π = sin x − π 2  = − sin −x 2 = − cos x

  3π 3π 3π + x = cos cos x − sin sin x cos 2 2 2 = (−1)(− sin x) = sin x sin x cos x + cos x sin x sin2 x + cos2 x = cos x sin x 1 = cos x sin x

tan x + cot x =

√  2π π π 3 sin = sin π − = sin = 3 3 3 2 18 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

12.

13.

SECTION P.7 (PAGE 57)

 sin x cos x  − tan x − cot x x sin x =  cos cos x  sin x tan x + cot x + cos x sin x  2  sin x − cos2 x cos x sin x =  2  sin x + cos2 x cos x sin x = sin2 x − cos2 x

y

π/2 π

20. sin

2

= cos x − sin x = cos(2x)

14. (1 − cos x)(1 + cos x) = 1 −

15.

16.

17.

cos2

2π x

Fig. P.7.19

cos4 x − sin4 x = (cos2 x − sin2 x)(cos2 x + sin2 x) 2

y = cos(2x)

1

x =

sin2

x has period 4π . 2 y 1

x implies

sin x 1 − cos x = . Now sin x 1 + cos x   x 1 − cos 2 1 − cos x  x 2 = sin x sin 2  2  x  1 − 1 − 2 sin2 2 = x x 2 sin cos 2 2 x sin 2 = tan x = x 2 cos 2   2 x 2 sin x  1 − cos x  2x  = tan2 = 1 + cos x 2 2 cos2 2

π

2π x

−1

Fig. P.7.20

21. sin π x has period 2. y

y = sin(π x)

1

1

2

4 x

3

−1

Fig. P.7.21

cos x − sin x (cos x − sin x)2 = cos x + sin x (cos x + sin x)(cos x − sin x) cos2 x − 2 sin x cos x + sin2 x = cos2 x − sin2 x 1 − sin(2x) = cos(2x) = sec(2x) − tan(2x)

22. cos

πx has period 4. 2 y 1 1

5 3

x

−1

sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x

Fig. P.7.22

= 2 sin x cos2 x + sin x(1 − 2 sin2 x)

= 2 sin x(1 − sin2 x) + sin x − 2 sin3 x

23.

y

= 3 sin x − 4 sin3 x

18.

2

cos 3x = cos(2x + x) = cos 2x cos x − sin 2x sin x

 π y = 2 cos x − 3

1

= (2 cos2 x − 1) cos x − 2 sin2 x cos x

= 2 cos3 x − cos x − 2(1 − cos2 x) cos x = 4 cos3 x − 3 cos x

−π

π x -1 -2

19. cos 2x has period π .

-3

19 Copyright © 2014 Pearson Canada Inc.

SECTION P.7 (PAGE 57)

24.

ADAMS and ESSEX: CALCULUS 8

y

y = 1 + sin

2

29.

π  4

1

√ − 3

x

π

−π

1 3π sin x = − , π < x < 2√ 2 3 cos x = − 2 1 tan x = √ 3

−1

-1

25.

π 3 , x 3 on (−∞, −1) and (0, 1). The graphs of x 1/3 and x 3 intersect at (−1, −1), (0, 0), and (1, 1). If the graph of h(x) lies between those of x 1/3 and x 3 , then we can determine limx→a h(x) for a = −1, a = 0, and a = 1 by the squeeze theorem. In fact

x→−1

x

-0.1

79.

1 is defined for all x 6= 0; its domain is x (−∞, 0) ∪ (0, ∞). Since | sin t| ≤ 1 for all t, we have | f (x)| ≤ |x| and −|x| ≤ f (x) ≤ |x| for all x 6= 0. Since limx→0 = (−|x|) = 0 = limx→0 |x|, we have limx→0 f (x) = 0 by the squeeze theorem. | f (x)| ≤ g(x) ⇒ −g(x) ≤ f (x) ≤ g(x) Since lim g(x) = 0, therefore 0 ≤ lim f (x) ≤ 0. x→a

x→a

x→a

If lim g(x) = 3, then either −3 ≤ lim f (x) ≤ 3 or

√ √ Since √ 5 − 2x 2 ≤ f (x) ≤ 5√− x 2 for −1 √ ≤ x ≤ 1, and 2 = lim 2 limx→0 5 − 2x√ 5, we have x→0 5 − x = limx→0 f (x) = 5 by the squeeze theorem.

Since 2 − x 2 ≤ g(x) ≤ 2 cos x for all x, and since limx→0 (2 − x 2 ) = limx→0 2 cos x = 2, we have limx→0 g(x) = 2 by the squeeze theorem.

lim h(x) = 1.

x→1

Hence, lim f (x) = 0.

x→a

f (x) does not exist.

Section 1.3 Limits at Infinity and Infinite Limits (page 78)

1. 75.

lim h(x) = 0,

x→0

f (x) = s sin

x→a limx→a

74.

x

Fig. 1.2.76

√ x− x lim √ = −1 x→0+ sin x

y = −x

1

2.

lim

x 1 1 = lim = 2x − 3 x→∞ 2 − (3/x) 2

lim

x 1/x 0 = lim = =0 1 x 2 − 4 x→∞ 1 − (4/x 2 )

x→∞

x→∞

28 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

3.

4.

SECTION 1.3

3x 3 − 5x 2 + 7 x→∞ 8 + 2x − 5x 3 7 5 3− + 3 3 x x = lim =− 2 x→∞ 8 5 + 2 −5 x3 x

12.

lim

lim

x→−∞

13. 14.

x2 − 2 x − x2

15.

2 1− 2 1 x = lim = = −1 x→−∞ 1 −1 −1 x

5.

6.

7.

8.

9.

10. 11.

16.

1 3 + 3 x2 + 3 x =0 = lim x lim 2 x→−∞ x→−∞ x 3 + 2 1+ 3 x sin x 1+ 2 1 x 2 + sin x x = lim lim cos x = 1 = 1 x→∞ x→∞ x 2 + cos x 1+ 2 x sin x We have used the fact that limx→∞ 2 = 0 (and simix larly for cosine) because the numerator is bounded while the denominator grows large. √ 3x + 2 x lim x→∞ 1−x 2 3+ √ x = −3 = lim x→∞ 1 −1 x 2x − 1 lim √ x→∞ 3x 2 + x + 1   1 x 2− x (but |x| = x as x → ∞) r = lim x→∞ 1 1 |x| 3 + + 2 x x 1 2− 2 x = lim r = √ x→∞ 1 1 3 3+ + 2 x x lim √

lim

x→−∞

lim

x→3

18. 19. 20. 21. 22. 23. 24. 25.

26.

2x − 1

3x 2 + x + 1 1 2− 2 x = lim r = −√ , x→−∞ 1 1 3 − 3+ + 2 x x √ because x → −∞ implies that x < 0 and so x 2 = −x. x→−∞

17.

27.

2x − 5 2x − 5 2 = lim =− |3x + 2| x→−∞ −(3x + 2) 3

1 does not exist. 3−x

28.

lim

x→3

(PAGE 78)

1 =∞ (3 − x)2

lim

x→3−

lim

x→3+

1 =∞ 3−x

1 = −∞ 3−x

2x + 5 0 = =0 −25 5x + 2 +2 2 2x + 5 lim does not exist. x→−2/5 5x + 2 lim

x→−5/2

lim

x→−(2/5)−

2x + 5 = −∞ 5x + 2

2x + 5 =∞ 5x + 2 x lim = −∞ x→2+ (2 − x)3 x lim √ =∞ x→1− 1 − x 2 1 =∞ lim x→1+ |x − 1| lim

x→−2/5+

lim

x→1−

1 =∞ |x − 1|

x −3 x −3 = lim = −∞ x 2 − 4x + 4 x→2 (x − 2)2 √ x2 − x −1 = lim √ = −∞ lim x→1+ x→1+ x − x 2 x2 − x lim

x→2

x + x3 + x5 x→∞ 1 + x 2 + x 3 1 + 1 + x2 2 x = lim =∞ 1 x→∞ 1 + + 1 x3 x 3 x+ 2 x3 + 3 x lim = lim =∞ 2 x→∞ x 2 + 2 x→∞ 1+ 2 x √ √
x x + 1 1 − 2x + 3 lim 2 x→∞ 7r − 6x + 4x ! ! r 1 1 3 2 x 1+ √ − 2+ x x x   = lim x→∞ 7 6 x2 − +4 x x2 √ 1(− 2) 1√ = =− 2 4 4  2  x x2 −2x 2 lim − = lim 2 = −2 x→∞ x + 1 x→∞ x −1 x −1 lim

29 Copyright © 2014 Pearson Canada Inc.

SECTION 1.3

29.

30.

lim

x→−∞

p

(PAGE 78)

x 2 + 2x −

p

x 2 − 2x

ADAMS and ESSEX: CALCULUS 8



36.

(x 2 + 2x) − (x 2 − 2x) = lim √ √ x→−∞ x 2 + 2x + x 2 − 2x 4x ! = lim r r x→−∞ 2 2 (−x) 1+ + 1− x x 4 =− = −2 1+1 p  p lim x 2 + 2x − x 2 − 2x

37.

lim f (x) = ∞

x→1

y 3

y = f (x)

2 1

x→∞

x 2 + 2x − x 2 + 2x √ x 2 + 2x + x 2 − 2x 4x = lim r r x→∞ 2 2 x 1+ +x 1− x x 4 4 = lim r r = =2 x→∞ 2 2 2 1+ + 1− x x = lim √

1

x→∞

31.

lim √

x→∞

√

x 2 − 2x + x √ √ 2 ( x − 2x + x)( x 2 − 2x − x) √ x 2 − 2x + x = lim 2 x→∞ x − 2x − x 2 √ x( 1 − (2/x) + 1) 2 = lim = = −1 x→∞ −2x −2

33.

lim √

x→−∞

1 x 2 + 2x − x

= lim

x→−∞

1 =0 √ |x|( 1 + (2/x) + 1

By Exercise 35, y = −1 is a horizontal asymptote (at the 1 . Since right) of y = √ 2 x − 2x − x 1

1 lim √ = lim √ = 0, 2 x→−∞ x→−∞ |x|( 1 − (2/x) + 1 x − 2x − x y = 0√is also a horizontal asymptote (at the left). Now x 2 − 2x − x = 0 if and only if x 2 − 2x = x 2 , that is, if and only if x = 0. The given function is undefined at x = 0, and where x 2 − 2x < 0, that is, on the interval [0, 2]. Its only vertical asymptote is at x = 0, where 1 limx→0− √ = ∞. 2 x − 2x − x

34.

35.

2x − 5 2 2x − 5 2 = and lim = − , x→∞ |3x + 2| x→−∞ |3x + 2| 3 3 y = ±(2/3) are horizontal asymptotes of y = (2x − 5)/|3x + 2|. The only vertical asymptote is x = −2/3, which makes the denominator zero.

39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52.

Since lim

lim f (x) = 1

x→0+

5

limx→2+ f (x) = 1

x→∞

32.

4

Fig. 1.3.37

38.

= lim

3

-1

1 x 2 − 2x − x

2

53.

lim f (x) = 2

x→2−

lim f (x) = −∞

x→3−

lim f (x) = ∞

x→3+

lim f (x) = 2

x→4+

lim f (x) = 0

x→4−

lim f (x) = −1

x→5−

lim f (x) = 0

x→5+

lim f (x) = 1

x→∞

horizontal: y = 1; vertical: x = 1, x = 3. lim ⌊x⌋ = 3

x→3+

lim ⌊x⌋ = 2

x→3−

lim ⌊x⌋ does not exist

x→3

lim ⌊x⌋ = 2

x→2.5

lim ⌊2 − x⌋ = lim ⌊x⌋ = 1

x→0+

x→2−

lim ⌊x⌋ = −4

x→−3−

lim C(t) = C(t0 ) except at integers t0

t→t0

lim C(t) = C(t0 ) everywhere

t→t0 −

lim C(t) = C(t0 ) if t0 6= an integer

t→t0 +

lim C(t) = C(t0 ) + 1.5 if t0 is an integer

t→t0 +

30 Copyright © 2014 Pearson Canada Inc.

6

x

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 1.4

y 6.00

3.

g has no absolute maximum value on [−2, 2]. It takes on every positive real value less than 2, but does not take the value 2. It has absolute minimum value 0 on that interval, assuming this value at the three points x = −2, x = −1, and x = 1.

4.

Function f is discontinuous at x = 1, 2, 3, 4, and 5. f is left continuous at x = 4 and right continuous at x = 2 and x = 5. y

4.50

y = C(t)

3.00

1.50

1

2

3

4

(PAGE 87)

x

3

y = f (x)

Fig. 1.3.53 2

54.

lim f (x) = L

x→0+

(a) If f is even, then f (−x) = f (x). Hence, lim f (x) = L.

1

x→0−

(b) If f is odd, then f (−x) = − f (x). Therefore, lim f (x) = −L.

1

x→0−

55.

lim f (x) = A,

x→0+

2

6

5

x

-1

lim f (x) = B

Fig. 1.4.4

x→0+

b) lim f (x 3 − x) = A (because x 3 − x > 0 if

5.

f cannot be redefined at x = 1 to become continuous there because limx→1 f (x) (= ∞) does not exist. (∞ is not a real number.)

c) lim f (x 2 − x 4 ) = A

6.

sgn x is not defined at x = 0, so cannot be either continuous or discontinuous there. (Functions can be continuous or discontinuous only at points in their domains!)  x if x < 0 f (x) = is continuous everywhere on the x 2 if x ≥ 0 real line, even at x = 0 where its left and right limits are both 0, which is f (0).  x if x < −1 f (x) = is continuous everywhere on the x 2 if x ≥ −1 real line except at x = −1 where it is right continuous, but not left continuous.

x→0−

−1 < x < 0) x→0−

d) lim f (x 2 − x 4 ) = A (since x 2 − x 4 > 0 for x→0+

0 < |x| < 1)

7.

Section 1.4 Continuity

(page 87) 8.

g is continuous at x = −2, discontinuous at x = −1, 0, 1, and 2. It is left continuous at x = 0 and right continuous at x = 1. y (1, 2) 2 (−1, 1)

lim

x→−1−

-2

f (x) =

lim x = −1 6= 1

x→−1−

= f (−1) =

y = g(x)

lim x 2 =

x→−1+

lim

x→−1+

f (x).

1

9. -1

1

2

x

Fig. 1.4.1

10. 2.

4

x→0−

a) lim f (x 3 − x) = B (since x 3 − x < 0 if 0 < x < 1)

1.

3

g has removable discontinuities at x = −1 and x = 2. Redefine g(−1) = 1 and g(2) = 0 to make g continuous at those points.



1/x 2 if x 6= 0 is continuous everywhere ex0 if x = 0 cept at x = 0, where it is neither left nor right continuous since it does not have a real limit there.  2 if x ≤ 1 is continuous everywhere f (x) = x 0.987 if x > 1 except at x = 1, where it is left continuous but not right continuous because 0.987 6= 1. Close, as they say, but no cigar. f (x) =

31 Copyright © 2014 Pearson Canada Inc.

SECTION 1.4

(PAGE 87)

ADAMS and ESSEX: CALCULUS 8

11.

The least integer function ⌈x⌉ is continuous everywhere on R except at the integers, where it is left continuous but not right continuous.

12.

C(t) is discontinuous only at the integers. It is continuous on the left at the integers, but not on the right.

13.

14.

15.

16.

1 + t3

+ t 2)

(1 + t)(1 − t 1−t = = for 1 − t2 (1 + t)(1 − t) 1−t t 6= −1, we can define the function to be 3/2 at t = −1 to make it continuous there. The continuous extension is 1 − t + t2 . 1−t Since

t 2 − 5t + 6 (t − 2)(t − 3) t −2 = = for t 6= 3, t2 − t − 6 (t + 2)(t − 3) t +2 we can define the function to be 1/5 at t = 3 to make it t −2 continuous there. The continuous extension is . t +2

Let the numbers be x and y, where x ≥ 0, y ≥ 0, and x + y = 8. If S is the sum of their squares then S = x 2 + y 2 = x 2 + (8 − x)2

= 2x 2 − 16x + 64 = 2(x − 4)2 + 32.

Since 0 ≤ x ≤ 8, the maximum value of S occurs at x = 0 or x = 8, and is 64. The minimum value occurs at x = 4 and is 32.

+ t2

23.

Since T = 100 − 30x + 3x 2 = 3(x − 5)2 + 25, T will be minimum when x = 5. Five programmers should be assigned, and the project will be completed in 25 days.

24.

If x desks are shipped, the shipping cost per desk is

Since

Since √ √ √ x2 − 2 (x − 2)(x + 2) x+ 2 = √ √ = √ 2 x4 − 4 (x + 2)(x 2 + 2) √(x − 2)(x + 2)(x + 2) for x √ 6= 2, we can define the function to be 1/4 at x = 2 to make it continuous there. The continuous √ x+ 2 √ . (Note: cancelling the extension is (x + 2)(x 2 + 2) √ x + 2√factors provides a further continuous extension to x = − 2. limx→2+ f (x) = k − 4 and limx→2− f (x) = 4 = f (2). Thus f will be continuous at x = 2 if k − 4 = 4, that is, if k = 8.

18.

limx→3− g(x) = 3 − m and limx→3+ g(x) = 1 − 3m = g(3). Thus g will be continuous at x = 3 if 3 − m = 1 − 3m, that is, if m = −1.

19.

x 2 has no maximum value on −1 < x < 1; it takes all positive real values less than 1, but it does not take the value 1. It does have a minimum value, namely 0 taken on at x = 0.

21.

22.

x2 − 4 Since = x + 2 for x 6= 2, we can define the x −2 function to be 2 + 2 = 4 at x = 2 to make it continuous there. The continuous extension is x + 2.

17.

20.

Therefore P ≤ 16, so P is bounded. Clearly P = 16 if x = y = 4, so the largest value of P is 16.

The Max-Min Theorem says that a continuous function defined on a closed, finite interval must have maximum and minimum values. It does not say that other functions cannot have such values. The Heaviside function is not continuous on [−1, 1] (because it is discontinuous at x = 0), but it still has maximum and minimum values. Do not confuse a theorem with its converse. Let the numbers be x and y, where x ≥ 0, y ≥ 0, and x + y = 8. If P is the product of the numbers, then P = x y = x(8 − x) = 8x − x 2 = 16 − (x − 4)2 .

C=

245x − 30x 2 + x 3 = x 2 − 30x + 245 x = (x − 15)2 + 20.

This cost is minimized if x = 15. The manufacturer should send 15 desks in each shipment, and the shipping cost will then be $20 per desk.

25.

(x − 1)(x + 1) x2 − 1 = x x f = 0 at x = ±1. f is not defined at 0. f (x) > 0 on (−1, 0) and (1, ∞). f (x) < 0 on (−∞, −1) and (0, 1). f (x) =

26.

f (x) = x 2 + 4x + 3 = (x + 1)(x + 3) f (x) > 0 on (−∞, −3) and (−1, ∞) f (x) < 0 on (−3, −1).

27.

f (x) =

28.

x2 − 1 (x − 1)(x + 1) = x2 − 4 (x − 2)(x + 2) f = 0 at x = ±1. f is not defined at x = ±2. f (x) > 0 on (−∞, −2), (−1, 1), and (2, ∞). f (x) < 0 on (−2, −1) and (1, 2).

x2 + x − 2 (x + 2)(x − 1) = x3 x3 f (x) > 0 on (−2, 0) and (1, ∞) f (x) < 0 on (−∞, −2) and (0, 1). f (x) =

29.

f (x) = x 3 + x − 1, f (0) = −1, f (1) = 1. Since f is continuous and changes sign between 0 and 1, it must be zero at some point between 0 and 1 by IVT.

30.

f (x) = x 3 − 15x + 1 is continuous everywhere. f (−4) = −3, f (−3) = 19, f (1) = −13, f (4) = 5. Because of the sign changes f has a zero between −4 and −3, another zero between −3 and 1, and another between 1 and 4.

32 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

31.

32.

33.

SECTION 1.5

F(x) = (x − a)2 (x − b)2 + x. Without loss of generality, we can assume that a < b. Being a polynomial, F is continuous on [a, b]. Also F(a) = a and F(b) = b. Since a < 21 (a + b) < b, the Intermediate-Value Theorem guarantees that there is an x in (a, b) such that F(x) = (a + b)/2.

Section 1.5 The Formal Definition of Limit (page 92)

1.

lim f (x) = lim f (−y) = lim f (y) = f (0).

x→0−

y→0+

y→0+

The temperature should be kept between 12 ◦ C and 20 ◦ C.

2.

Since 1.2% of 8,000 is 96, we require the edge length x of the cube to satisfy 7904 ≤ x 3 ≤ 8096. It is sufficient that 19.920 ≤ x ≤ 20.079. The edge of the cube must be within 0.079 cm of 20 cm.

3.

3 − 0.02 ≤ 2x − 1 ≤ 3 + 0.02 3.98 ≤ 2x ≤ 4.02 1.99 ≤ x ≤ 2.01

4.

4 − 0.1 ≤ x 2 ≤ 4 + 0.1 1.9749 ≤ x ≤ 2.0024

Thus, f is continuous on the left at x = 0. Being continuous on both sides, it is therefore continuous.

34.

f odd ⇔ f (−x) = − f (x) f continuous on the right ⇔ lim f (x) = f (0) x→0+

Therefore, letting t = −x, we obtain lim f (x) = lim f (−t) = lim − f (t)

x→0−

t→0+

5.

t→0+

= − f (0) = f (−0) = f (0). Therefore f is continuous at 0 and f (0) = 0.

35.

max 1.593 at −0.831, min −0.756 at 0.629

36.

max 0.133 at x = 1.437; min −0.232 at x = −1.805

37.

max 10.333 at x = 3; min 4.762 at x = 1.260

38.

max 1.510 at x = 0.465; min 0 at x = 0 and x = 1

39.

root x = 0.682

40.

root x = 0.739

41.

roots x = −0.637 and x = 1.410

42.

roots x = −0.7244919590 and x = 1.220744085

43.

fsolve gives an approximation to the single real root to 10 significant figures; solve gives the three roots (including a complex conjugate pair) in exact form involving the  √ 1/3 quantity 108 + 12 69 ; evalf(solve) gives approximations to the three roots using 10 significant figures for the real and imaginary parts.

We require 39.9 ≤ L ≤ 40.1. Thus 39.9 ≤ 39.6 + 0.025T ≤ 40.1 0.3 ≤ 0.025T ≤ 0.5 12 ≤ T ≤ 20.

Let g(x) = f (x) − x. Since 0 ≤ f (x) ≤ 1 if 0 ≤ x ≤ 1, therefore, g(0) ≥ 0 and g(1) ≤ 0. If g(0) = 0 let c = 0, or if g(1) = 0 let c = 1. (In either case f (c) = c.) Otherwise, g(0) > 0 and g(1) < 0, and, by IVT, there exists c in (0, 1) such that g(c) = 0, i.e., f (c) = c. The domain of an even function is symmetric about the y-axis. Since f is continuous on the right at x = 0, therefore it must be defined on an interval [0, h] for some h > 0. Being even, f must therefore be defined on [−h, h]. If x = −y, then

(PAGE 92)

√ 1 − 0.1 ≤ x ≤ 1.1 0.81 ≤ x ≤ 1.21 1 ≤ −2 + 0.01 x 1 1 − ≥x ≥− 2.01 1.99 −0.5025 ≤ x ≤ −0.4975

6.

−2 − 0.01 ≤

7.

We need −0.03 ≤ (3x +1)−7 ≤ 0.03, which is equivalent to −0.01 ≤ x − 2 ≤ 0.01 Thus δ = 0.01 will do.

8.

We need −0.01 ≤

√

2x + 3 − 3 ≤ 0.01. Thus

√ 2.99 ≤ 2x + 3 ≤ 3.01 8.9401 ≤ 2x + 3 ≤ 9.0601 2.97005 ≤ x ≤ 3.03005 3 − 0.02995 ≤ x − 3 ≤ 0.03005. Here δ = 0.02995 will do.

9.

We need 8 − 0.2 ≤ x 3 ≤ 8.2, or 1.9832 ≤ x ≤ 2.0165. Thus, we need −0.0168 ≤ x − 2 ≤ 0.0165. Here δ = 0.0165 will do.

33 Copyright © 2014 Pearson Canada Inc.

SECTION 1.5

10.

11.

(PAGE 92)

ADAMS and ESSEX: CALCULUS 8

We need 1 − 0.05 ≤ 1/(x + 1) ≤ 1 + 0.05, or 1.0526 ≥ x + 1 ≥ 0.9524. This will occur if −0.0476 ≤ x ≤ 0.0526. In this case we can take δ = 0.0476.

This establishes the required limit.

18.

To be proved: lim (3x + 1) = 4. x→1

If |x +1| < 1, then −2 < x < 0, so −3 < x −1 < −1 and |x − 1| > 1. Ler δ = min(1, 2ǫ). If 0 < |x − (−1)| < δ then |x − 1| > 1 and |x + 1| < 2ǫ. Thus

To be proved: lim (5 − 2x) = 1. x→2

Proof: Let ǫ > 0 be given. Then |(5 − 2x) − 1| < ǫ holds if |2x −4| < ǫ, and so if |x −2| < δ = ǫ/2. This confirms the limit.

13.

14.

x→0

x −2 = 0. x→2 1 + x 2 Proof: Let ǫ > 0 be given. Then

19.

To be proved: lim

provided |x − 2| < δ = ǫ.

1 − 4x 2 = 2. To be proved: lim x→1/2 1 − 2x Proof: Let ǫ > 0 be given. Then if x 6= 1/2 we have

provided |x −

16.

provided |x − 1| < δ = ǫ. This completes the proof.

20.

21.

22.

x→1

1 1 1 − x |x − 1| − = x + 1 2 2(x + 1) = 2|x + 1| .

If |x − 1| < 1, then 0 < x < 2 and 1 < x + 1 < 3, so that |x + 1| > 1. Let δ = min(1, 2ǫ). If |x − 1| < δ, then 1 1 |x − 1| 2ǫ − x + 1 2 = 2|x + 1| < 2 = ǫ.

We say that limx→a− f (x) = L if the following condition holds: for every number ǫ > 0 there exists a number δ > 0, depending on ǫ, such that a−δ < x 0 be given. We have

ǫ × 19 = ǫ. 19

This completes the proof.

To be proved: lim

To be proved: lim

x→2

|x 3 − 8| = |x − 2||x 2 + 2x + 4| <

< δ = ǫ/2.

x 2 + 2x = −2. x→−2 x + 2 Proof: Let ǫ > 0 be given. For x 6= −2 we have

To be proved: lim x 3 = 8.

Proof: Let ǫ > 0 be given. We have |x 3 − 8| = |x − 2||x 2 + 2x + 4|. If |x − 2| < 1, then 1 < x < 3 and x 2 < 9. Therefore |x 2 + 2x + 4| ≤ 9 + 2 × 3 + 4 = 19. If |x − 2| < δ = min(1, ǫ/19), then

2 x + 2x x + 2 − (−2) = |x + 2| < ǫ

17.

x→1

x −1 √ ≤ |x − 1| < ǫ | x − 1| = √ x + 1

1 − 4x 2 = |(1 + 2x) − 2| = |2x − 1| = 2 x − 1 < ǫ − 2 1 − 2x 2 1 2|

This completes the required proof. √ To be proved: lim x = 1.

Proof: Let ǫ > 0 be given. We have

x −2 |x − 2| = − 0 ≤ |x − 2| < ǫ 1 + x2 1 + x2

15.

  x +1 1 |x + 1| 2ǫ − − = < = ǫ. x2 − 1 2 2|x − 1| 2

To be proved: lim x 2 = 0.

Let ǫ > 0 be given. √ Then |x 2 − 0| < ǫ holds if |x − 0| = |x| < δ = ǫ.

lim

x→−1

    x +1 1 1 1 |x + 1| − − = − − = . x2 − 1 2 x − 1 2 2|x − 1|

Proof: Let ǫ > 0 be given. Then |(3x + 1) − 4| < ǫ holds if 3|x −1| < ǫ, and so if |x −1| < δ = ǫ/3. This confirms the limit.

12.

x +1 1 =− . x2 − 1 2 Proof: Let ǫ > 0 be given. If x 6= −1, we have To be proved:

| f (x) − L| < ǫ.

We say that limx→−∞ f (x) = L if the following condition holds: for every number ǫ > 0 there exists a number R > 0, depending on ǫ, such that x < −R

23.

implies

implies

| f (x) − L| < ǫ.

We say that limx→a f (x) = −∞ if the following condition holds: for every number B > 0 there exists a number δ > 0, depending on B, such that 0 < |x − a| < δ

34 Copyright © 2014 Pearson Canada Inc.

implies

f (x) < −B.

INSTRUCTOR’S SOLUTIONS MANUAL

24.

SECTION 1.5

We say that limx→∞ f (x) = ∞ if the following condition holds: for every number B > 0 there exists a number R > 0, depending on B, such that x>R

implies

This implies that 3 < 2, a contradiction. Thus the original assumption that L 6= M must be incorrect. Therefore L = M.

f (x) > B.

32. 25.

We say that limx→a+ f (x) = −∞ if the following condition holds: for every number B > 0 there exists a number δ > 0, depending on R, such that a < x B.

33.

To be proved: limx→1−

1 = ∞. Proof: Let B > 0 x −1

29.

be given. We have

1 x2

+1

x→a

exists δ1 > 0 such that | f (x) − L| < ǫ/(2(1 + |M|)) if 0 < |x − a| < δ1 . Since lim g(x) = M, there exx→a

ists δ2 > 0 such that |g(x) − M| < ǫ/(2(1 + |L|)) if 0 < |x − a| < δ2 . By Exercise 32, there exists δ3 > 0 such that |g(x)| < 1 + |M| if 0 < |x − a| < δ3 . Let δ = min(δ1 , δ2 , δ3 ). If |x − a| < δ, then | f (x)g(x) − L M = | f (x)g(x) − Lg(x) + Lg(x) − L M| = |( f (x) − L)g(x) + L(g(x) − M)| ≤ |( f (x) − L)g(x)| + |L(g(x) − M)| = | f (x) − L||g(x)| + |L||g(x) − M| ǫ ǫ (1 + |M|) + |L| < 2(1 + |M|) 2(1 + |L|) ǫ ǫ ≤ + = ǫ. 2 2

= 0. Proof: Let ǫ > 0

1 √ 1 = √ 1 < R, where R = 1/ǫ. This completes the proof. √ To be proved: lim 0 be √x→∞ x = ∞. Proof: Let B > given. We have x > B if x > R where R = B 2 . This completes the proof. To be proved: if lim f (x) = L and lim f (x) = M, then x→a

x→a

L = M. Proof: Suppose L 6= M. Let ǫ = |L − M|/3. Then ǫ > 0. Since lim f (x) = L, there exists δ1 > 0 such that x→a

| f (x)−L| < ǫ if |x −a| < δ1 . Since lim f (x) = M, there

x→a

x→a

1 = −∞. Proof: Let B > 0 x −1

To be proved: limx→∞ √

x→a

Proof: Let ǫ > 0 be given. Since lim f (x) = L, there

1 < −B if 0 > x − 1 > −1/B, x −1 that is, if 1 − δ < x < 1, where δ = 1/B.. This completes the proof. be given. We have

To be proved: if lim f (x) = L and lim g(x) = M, then lim f (x)g(x) = L M.

1 be given. We have > B if 0 < x − 1 < 1/B, that x −1 is, if 1 < x < 1 + δ, where δ = 1/B. This completes the proof.

28.

x→a

|g(x)| = |(g(x) − M) + M| ≤ |G(x) − M| + |M| < 1 + |M|.

We say that limx→a− f (x) = ∞ if the following condition holds: for every number B > 0 there exists a number δ > 0, depending on B, such that implies

To be proved: if lim g(x) = M, then there exists δ > 0

such that if 0 < |x − a| < δ, then |g(x)| < 1 + |M|. Proof: Taking ǫ = 1 in the definition of limit, we obtain a number δ > 0 such that if 0 < |x − a| < δ, then |g(x) − M| < 1. It follows from this latter inequality that

f (x) < −B.

implies

a−δ < x 0 such that if 0 < |x − a| < δ, then |g(x)| > |M|/2. Proof: By the definition of limit, there exists δ > 0 such that if 0 < |x − a| < δ, then |g(x) − M| < |M|/2 (since |M|/2 is a positive number). This latter inequality implies that

x→a

exists δ2 > 0 such that | f (x) − M| < ǫ if |x − a| < δ2 . Let δ = min(δ1 , δ2 ). If |x − a| < δ, then

|M| = |g(x)+(M−g(x))| ≤ |g(x)|+|g(x)−M| < |g(x)|+

3ǫ = |L − M| = |( f (x) − M) + (L − f (x)| ≤ | f (x) − M| + | f (x) − L| < ǫ + ǫ = 2ǫ.

It follows that |g(x)| > |M| − (|M|/2) = |M|/2, as required.

|M| . 2

35 Copyright © 2014 Pearson Canada Inc.

SECTION 1.5

35.

(PAGE 92)

ADAMS and ESSEX: CALCULUS 8

Review Exercises 1 (page 93)

To be proved: if lim g(x) = M where M 6= 0, then x→a

1 1 = . g(x) M Proof: Let ǫ > 0 be given. Since lim g(x) = M 6= 0, lim

x→a

1.

The average rate of change of x 3 over [1, 3] is

x→a

there exists δ1 > 0 such that |g(x) − M| < ǫ|M|2 /2 if 0 < |x − a| < δ1 . By Exercise 34, there exists δ2 > 0 such that |g(x)| > |M|/2 if 0 < |x − a| < δ3 . Let δ = min(δ1 , δ2 ). If 0 < |x − a| < δ, then 1 1 |M − g(x)| ǫ|M|2 2 − < = ǫ. = g(x) M |M||g(x)| 2 |M|2

33 − 13 26 = = 13. 3−1 2

2.

The average rate of change of 1/x over [−2, −1] is (1/(−1)) − (1/(−2)) −1/2 1 = =− . −1 − (−2) 1 2

This completes the proof.

36.

To be proved: if lim f (x) = L and lim f (x) = M 6= 0, x→a

x→a

f (x) L then lim = . x→a g(x) M Proof: By Exercises 33 and 35 we have

3.

The rate of change of x 3 at x = 2 is lim

lim

x→a

37.

h→0

f (x) 1 1 L = lim f (x) × =L× = . x→a g(x) g(x) M M

To be proved: if f is continuous at L and lim g(x) = L,

h→0

4.

x→c

then lim f (g(x)) = f (L). x→c

To be proved: if f (x) ≤ g(x) ≤ h(x) in an open interval containing x = a (say, for a − δ1 < x < a + δ1 , where δ1 > 0), and if limx→a f (x) = limx→a h(x) = L, then also limx→a g(x) = L. Proof: Let ǫ > 0 be given. Since limx→a f (x) = L, there exists δ2 > 0 such that if 0 < |x − a| < δ2 , then | f (x) − L| < ǫ/3. Since limx→a h(x) = L, there exists δ3 > 0 such that if 0 < |x − a| < δ3 , then |h(x) − L| < ǫ/3. Let δ = min(δ1 , δ2 , δ3 ). If 0 < |x − a| < δ, then |g(x) − L| = |g(x) − f (x) + f (x) − L| ≤ |g(x) − f (x)| + | f (x) − L| ≤ |h(x) − f (x)| + | f (x) − L| = |h(x) − L + L − f (x)| + | f (x) − L| ≤ |h(x) − L| + | f (x) − L| + | f (x) − L| ǫ ǫ ǫ < + + = ǫ. 3 3 3 Thus limx→a g(x) = L.

The rate of change of 1/x at x = −3/2 is 1 − −(3/2) + h lim h→0 h

Proof: Let ǫ > 0 be given. Since f is continuous at L, there exists a number γ > 0 such that if |y − L| < γ , then | f (y) − f (L)| < ǫ. Since limx→c g(x) = L, there exists δ > 0 such that if 0 < |x − c| < δ, then |g(x) − L| < γ . Taking y = g(x), it follows that if 0 < |x − c| < δ, then | f (g(x)) − f (L)| < ǫ, so that limx→c f (g(x)) = f (L).

38.

(2 + h)3 − 23 8 + 12h + 6h 2 + h 3 − 8 = lim h→0 h h = lim (12 + 6h + h 2 ) = 12.

5.



1 −3/2



2 2 + = lim 2h − 3 3 h→0 h 2(3 + 2h − 3) = lim h→0 3(2h − 3)h 4 4 = lim =− . h→0 3(2h − 3) 9

lim (x 2 − 4x + 7) = 1 − 4 + 7 = 4

x→1

6.

x2 22 4 = =− x→2 1 − x 2 1 − 22 3

7.

x2 does not exist. The denominator approaches x→1 1 − x 2 0 (from both sides) while the numerator does not.

8.

lim

9.

lim

lim

x→2 x 2

(x − 2)(x + 2) x +2 x2 − 4 = lim = lim = −4 − 5x + 6 x→2 (x − 2)(x − 3) x→2 x − 3

x2 − 4 (x − 2)(x + 2) x +2 = lim = lim x→2 x→2 x − 2 − 4x + 4 (x − 2)2 does not exist. The denominator approaches 0 (from both sides) while the numerator does not. lim

x→2 x 2

36 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

10. 11. 12. 13.

lim

x→2− x 2

CHALLENGING PROBLEMS 1

x2 − 4 x +2 = lim = −∞ − 4x + 4 x→2− x − 2

27.

x2 − 4 x −2 = lim = −∞ + 4x + 4 x→−2+ x + 2 √ 2− x 1 4−x lim = lim =− √ x→4 x − 4 x→4 (2 + x)(x − 4) 4 √ √ x2 − 9 (x − 3)(x + 3)( x + 3) lim √ √ = lim x→3 x −3 x − 3 x→3 √ √ √ = lim (x + 3)( x + 3) = 12 3

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

−|x| ≤ x sin

x→−2+ x 2

lim √

h→0

lim

x→0+

lim

x→0

h

p

28.

29.

√

h( x + 3h + x) (x + 3h) − x √ √ √ x + 3h + x 2 x = lim = h→0 3 3

√ = lim h→0 x + 3h − x

p

√

√

x − x 2 is not de-

fined for x < 0. p √ lim x − x 2 does not exist because x − x 2 is not dex→1

fined for x > 1. p lim x − x 2 = 0

30. 31.

x→1−

lim

x→∞

1 − x2 (1/x 2 ) − 1 1 = lim =− 2 x→∞ −x −1 3 − (1/x) − (1/x ) 3

32.

3x 2

(2/x) + (100/x 2 ) 2x + 100 = lim =0 x→−∞ x→−∞ x 2 + 3 1 + (3/x 2 ) lim

x3

33.

− (1/x 2 )

lim

−1 x = lim = −∞ x 2 + 4 x→−∞ 1 + (4/x 2 )

lim

x4 x2 = lim =∞ − 4 x→∞ 1 − (4/x 2 )

x→−∞

34.

x→∞ x 2

lim √

x→0+

1 x − x2

=∞

35.

1 lim √ = √ =2 2 x→1/2 1/4 x−x 1

lim sin x does not exist; sin x takes the values −1 and 1

x→∞

lim

x→∞

cos x = 0 by the squeeze theorem, since x −

1 cos x 1 ≤ ≤ x x x

36.

37. 38.

for all x > 0

and limx→∞ (−1/x) = limx→∞ (1/x) = 0.

for all x 6= 0

1 does not exist; sin(1/x 2 ) takes the values −1 x2 and 1 in any interval (−δ, δ), where δ > 0, and limits, if they exist, must be unique. p lim [x + x 2 − 4x + 1] lim sin

x→0

x→−∞

x 2 − (x 2 − 4x + 1) √ x→−∞ x − x 2 − 4x + 1 4x − 1 p = lim x→−∞ x − |x| 1 − (4/x) + (1/x 2 ) x[4 − (1/x)] p = lim x→−∞ x + x 1 − (4/x) + (1/x 2 ) 4 − (1/x) p = 2. = lim x→−∞ 1 + 1 − (4/x) + (1/x 2 ) Note how we have used |x| = −x (in the second last line), because x → −∞. p lim [x + x 2 − 4x + 1] = ∞ + ∞ = ∞ = lim

x − x2 = 0

x − x 2 does not exist because

1 ≤ |x| x

and limx→0 (−|x|) = limx→0 |x| = 0.

in any interval (R, ∞), and limits, if they exist, must be unique.

26.

1 = 0 by the squeeze theorem, since x

lim

x→3

14.

lim x sin

x→0

(PAGE 94)

x→∞

f (x) = x 3 − 4x 2 + 1 is continuous on the whole real line and so is discontinuous nowhere. x is continuous everywhere on its domain, f (x) = x +1 which consists of all real numbers except x = −1. It is discontinuous nowhere.  2 if x > 2 f (x) = x is defined everywhere and disx if x ≤ 2 continuous at x = 2, where it is, however, left continuous since limx→2− f (x) = 2 = f (2).  2 if x > 1 f (x) = x is defined and continuous evx if x ≤ 1 erywhere, and so discontinuous nowhere. Observe that limx→1− f (x) = 1 = limx→1+ f (x). n 1 if x ≥ 1 f (x) = H (x − 1) = is defined everywhere 0 if x < 1 and discontinuous at x = 1 where it is, however, right continuous. n 1 if −3 ≤ x ≤ 3 f (x) = H (9 − x 2 ) = is defined 0 if x < −3 or x > 3 everywhere and discontinuous at x = ±3. It is right continuous at −3 and left continuous at 3. f (x) = |x|+|x +1| is defined and continuous everywhere. It is discontinuous nowhere. n |x|/|x + 1| if x 6= −1 f (x) = is defined everywhere 1 if x = −1 and discontinuous at x = −1 where it is neither left nor right continuous since limx→−1 f (x) = ∞, while f (−1) = 1.

37 Copyright © 2014 Pearson Canada Inc.

CHALLENGING PROBLEMS 1

Challenging Problems 1

(PAGE 94)

ADAMS and ESSEX: CALCULUS 8

(page 94)

5.

Use a − b = We have

1.

Let 0 < a < b. The average rate of change of [a, b] is b3 − a 3 = b2 + ab + a 2 . b−a

x3

The instantaneous rate of change of x 3 at x = c is

√ √ 1+a −1 − 1 + a , r− (a) = . a a a) lima→0 r− (a) does not exist. Observe that the right limit is −∞ and the left limit is ∞.

lim

6. r+ (a) =

p If c = (a 2 + ab + b2 )/3, then 3c2 = a 2 + ab + b2 , so the average rate ofpchange over [a, b] is the instantaneous rate of change at (a 2 + ab + b2 )/3. p 2 Claim: (a + ab + b2 )/3 > (a + b)/2. Proof: Since a 2 − 2ab + b2 = (a − b)2 > 0, we have 4a 2 + 4ab + 4b2 > 3a 2 + 6ab + 3b2

a 2 + ab + b2 a 2 + 2ab + b2 > = 3 4

s

2.



a+b 2

2

a+b a 2 + ab + b2 > . 3 2

For x near 0 we have |x − 1| = 1 − x and |x + 1| = x + 1. Thus

For x near 3 we have |5 − 2x| = 2x − 5, |x − 2| = x − 2, |x − 5| = 5 − x, and |3x − 7| = 3x − 7. Thus |5 − 2x| − |x − 2| 2x − 5 − (x − 2) lim = lim x→3 |x − 5| − |3x − 7| x→3 5 − x − (3x − 7) x −3 1 = lim =− . x→3 4(3 − x) 4

4.

Let y = x 1/6 . Then we have

a

r+ (a)

1 0.1 −0.1 0.01 −0.01 0.001 −0.001

0.41421 0.48810 0.51317 0.49876 0.50126 0.49988 0.50013

a→0

7.

√

1+a−1 a (1 + a) − 1 = lim √ a→0 a( 1 + a + 1) 1 1 = lim √ = . a→0 2 1+a+1

c) lim r+ (a) = lim

a→0

TRUE or FALSE

a) If limx→a f (x) exists  and limx→a  g(x) does not exist, then limx→a f (x) + g(x) does not exist.   TRUE, because if limx→a f (x) + g(x) were to exist then   lim g(x) = lim f (x) + g(x) − f (x) x→a x→a   = lim f (x) + g(x) − lim f (x) x→a

x 1/3 − 4 y2 − 4 = lim x→64 x 1/2 − 8 y→2 y 3 − 8 (y − 2)(y + 2) = lim y→2 (y − 2)(y 2 + 2y + 4) y+2 4 1 = lim 2 = = . y→2 y + 2y + 4 12 3 lim

−1 +

b) From the following table it appears that lima→0 r+ (a) = 1/2, the solution of the linear equation 2x − 1 = 0 which results from setting a = 0 in the quadratic equation ax 2 + 2x − 1 = 0.

x x 1 lim = lim =− . x→0 |x − 1| − |x + 1| x→0 (1 − x) − (x + 1) 2

3.

a 3 − b3 to handle the denominator. + ab + b2

√ 3+x −2 lim √ x→1 3 7 + x − 2 3+x −4 (7 + x)2/3 + 2(7 + x)1/3 + 4 = lim √ × x→1 3 + x + 2 (7 + x) − 8 (7 + x)2/3 + 2(7 + x)1/3 + 4 4+4+4 = lim = = 3. √ x→1 2+2 3+x +2

over

(c + h)3 − c3 3c2 h + 3ch 2 + h 3 = lim = 3c2 . h→0 h→0 h h

a2

x→a

would also exist.

b) If neither  limx→a f (x)  nor limx→a g(x) exists, then limx→a f (x) + g(x) does not exist. FALSE. Neither limx→0 1/x norlimx→0 (−1/x) exist, but limx→0 (1/x) + (−1/x) = limx→0 0 = 0 exists.

38 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 1

c) If f is continuous at a, then so is | f |. TRUE. For any two real numbers u and v we have

9.

|u| − |v| ≤ |u − v|.

This follows from

|u| = |u − v + v| ≤ |u − v| + |v|, and |v| = |v − u + u| ≤ |v − u| + |u| = |u − v| + |u|. Now we have | f (x)| − | f (a)| ≤ | f (x) − f (a)|

so the left side approaches zero whenever the right side does. This happens when x → a by the continuity of f at a. d) If | f | is continuous at a, then so  is f . −1 if x < 0 FALSE. The function f (x) = is 1 if x ≥ 0 discontinuous at x = 0, but | f (x)| = 1 everywhere, and so is continuous at x = 0. e) If f (x) < g(x) in an interval around a and if limx→a f (x) = L and limx→a g(x) = M both exist, then L < M.  x 2 if x 6= 0 and let FALSE. Let g(x) = 1 if x = 0 f (x) = −g(x). Then f (x) < g(x) for all x, but limx→0 f (x) = 0 = limx→0 g(x). (Note: under the given conditions, it is TRUE that L ≤ M, but not necessarily true that L < M.)

8.

a) To be proved: if f is a continuous function defined on a closed interval [a, b], then the range of f is a closed interval. Proof: By the Max-Min Theorem there exist numbers u and v in [a, b] such that f (u) ≤ f (x) ≤ f (v) for all x in [a, b]. By the Intermediate-Value Theorem, f (x) takes on all values between f (u) and f (v) at values of x between u and v, and hence at points of [a, b]. Thus the range of f is [ f (u), f (v)], a closed interval. b) If the domain of the continuous function f is an open interval, the range of f can be any interval (open, closed, half open, finite, or infinite).

10.

11.

(PAGE 94)

n x2 − 1 −1 if −1 < x < 1 = . 2 1 if x < −1 or x > 1 |x − 1| f is continuous wherever it is defined, that is at all points except x = ±1. f has left and right limits −1 and 1, respectively, at x = 1, and has left and right limits 1 and −1, respectively, at x = −1. It is not, however, discontinuous at any point, since −1 and 1 are not in its domain. f (x) =

1 1 1 = 1
=
. 1 2 1 1 2 x − x2 − − x + x 4 4 4 − x − 2 Observe that f (x) ≥ f (1/2) = 4 for all x in (0, 1). f (x) =

Suppose f is continuous on [0, 1] and f (0) = f (1).

a) To be proved: f (a) = f (a + 21 ) for some a in [0, 12 ]. Proof: If f (1/2) = f (0) we can take a = 0 and be done. If not, let g(x) = f (x + 21 ) − f (x). Then g(0) 6= 0 and g(1/2) = f (1) − f (1/2) = f (0) − f (1/2) = −g(0). Since g is continuous and has opposite signs at x = 0 and x = 1/2, the Intermediate-Value Theorem assures us that there exists a between 0 and 1/2 such that g(a) = 0, that is, f (a) = f (a + 21 ).

b) To be proved: if n > 2 is an integer, then f (a) = f (a + n1 ) for some a in [0, 1 − n1 ]. Proof: Let g(x) = f (x + n1 ) − f (x). Consider the numbers x = 0, x = 1/n, x = 2/n, . . . , x = (n − 1)/n. If g(x) = 0 for any of these numbers, then we can let a be that number. Otherwise, g(x) 6= 0 at any of these numbers. Suppose that the values of g at all these numbers has the same sign (say positive). Then we have 2 f (1) > f ( n−1 n ) > ··· > f (n) >

1 n

> f (0),

which is a contradiction, since f (0) = f (1). Therefore there exists j in the set {0, 1, 2, . . . , n − 1} such that g( j/n) and g(( j + 1)/n) have opposite sign. By the Intermediate-Value Theorem, g(a) = 0 for some a between j/n and ( j + 1)/n, which is what we had to prove.

39 Copyright © 2014 Pearson Canada Inc.

SECTION 2.1 (PAGE 100)

CHAPTER 2.

ADAMS and ESSEX: CALCULUS 8

DIFFERENTIATION

7. Slope of y =

√

x + 1 at x = 3 is √

√ 4+h −2 4+h +2 m = lim ·√ h→0 h 4+h +2 4+h−4 = lim
√ h→0 h h+h+2 1 1 = lim √ = . h→0 4 4+h +2

Section 2.1 Tangent Lines and Their Slopes (page 100) 1. Slope of y = 3x − 1 at (1, 2) is m = lim

h→0

3(1 + h) − 1 − (3 × 1 − 1) 3h = lim = 3. h→0 h h

The tangent line is y − 2 = 3(x − 1), or y = 3x − 1. (The tangent to a straight line at any point on it is the same straight line.)

Tangent line is y − 2 = 1 x

8. The slope of y = √ at x = 9 is

2. Since y = x/2 is a straight line, its tangent at any point

  1 1 1 √ − h→0 h 3 9+h √ √ 3− 9+h 3+ 9+h = lim √ · √ h→0 3h 9 + h 3+ 9+h 9−9−h = lim √ √ h→0 3h 9 + h(3 + 9 + h) 1 1 =− =− . 3(3)(6) 54

m = lim

(a, a/2) on it is the same line y = x/2.

3. Slope of y = 2x 2 − 5 at (2, 3) is 2(2 + h)2 − 5 − (2(22 ) − 5) h→0 h 8 + 8h + 2h 2 − 8 = lim h→0 h = lim (8 + 2h) = 8

m = lim

The tangent line at (9, 31 ) is y = 1 y = 12 − 54 x.

h→0

Tangent line is y − 3 = 8(x − 2) or y = 8x − 13.

9. Slope of y =

4. The slope of y = 6 − x − x 2 at x = −2 is 6 − (−2 + h) − (−2 + h)2 − 4 h→0 h 3h − h 2 = lim (3 − h) = 3. = lim h→0 h→0 h

The tangent line at (−2, 4) is y = 3x + 10.

5. Slope of y = x 3 + 8 at x = −2 is m = lim

= lim 12 − 6h + h

2

h→0

m = lim

h→0

x2 

10.

= 12

1 1 h h2 + 1

The tangent line at (0, 1) is y = 1.

h→0

−h = 0. h2 + 1

1 54 (x

− 9), or

2x at x = 2 is x +2

p 5 − (1 + h)2 − 2 h→0 h 5 − (1 + h)2 − 4  = lim p h→0 h 5 − (1 + h)2 + 2

m = lim

1 at (0, 1) is +1  − 1 = lim

−

1 Tangent line is y − 1 = (x − 2), 4 or x − 4y = −2. √ The slope of y = 5 − x 2 at x = 1 is

Tangent line is y − 0 = 12(x + 2) or y = 12x + 24.

6. The slope of y =

1 3

2(2 + h) −1 2 m = lim + h + 2 h→0 h 4 + 2h − 2 − h − 2 = lim h→0 h(2 + h + 2) 1 h = lim = . h→0 h(4 + h) 4

m = lim

(−2 + h)3 + 8 − (−8 + 8) h→0 h −8 + 12h − 6h 2 + h 3 + 8 − 0 = lim h→0 h  

1 (x − 3), or x − 4y = −5. 4

−2 − h 1 = lim p =− h→0 5 − (1 + h)2 + 2 2

The tangent line at (1, 2) is y = 2 − 21 (x − 1), or y = 52 − 21 x.

40 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.1 (PAGE 100)

If m = −3, then x0 = − 23 . The tangent line with slope m = −3 at (− 23 , 54 ) is y = 45 − 3(x + 32 ), that is, y = −3x − 13 4 .

11. Slope of y = x 2 at x = x0 is m = lim

h→0

(x0 + h)2 − x02 2x0 h + h 2 = lim = 2x0 . h→0 h h

19.

Tangent line is y − x02 = 2x0 (x − x0 ), or y = 2x0 x − x02 . 1 at (a, a1 ) is x   1 1 a−a−h 1 1 + = lim = − 2. m = lim h→0 h(a + h)(a) h→0 h a + h a a 1 1 1 The tangent line at (a, ) is y = − 2 (x − a), or a a a 2 x y = − 2. a a √ |0 + h| − 0 1 Since limh→0 = lim does not h→0 |h|sgn (h) h √ exist (and is not ∞ or −∞), the graph of f (x) = |x| has no tangent at x = 0.

a) Slope of y = x 3 at x = a is (a + h)3 − a 3 h→0 h a 3 + 3a 2 h + 3ah 2 + h 3 − a 3 = lim h→0 h = lim (3a 2 + 3ah + h 2 ) = 3a 2

m = lim

12. The slope of y =

13.

14. The slope of f (x) = (x − 1)4/3 at x = 1 is m = lim

h→0

(1 + h − 1)4/3 − 0 = lim h 1/3 = 0. h→0 h

The graph of f has a tangent line with slope 0 at x = 1. Since f (1) = 0, the tangent has equation y = 0

15. The slope of f (x) = (x + 2)3/5 at x = −2 is m = lim

h→0

(−2 + h + 2)3/5 − 0 = lim h −2/5 = ∞. h→0 h

The graph of f has vertical tangent x = −2 at x = −2.

16. The slope of f (x) = |x 2 − 1| at x = 1 is

17.

|(1 + h)2 − 1| − |1 − 1| |2h + h 2 | m = limh→0 = lim , h→0 h h which does not exist, and is not −∞ or ∞. The graph of f has no tangent at x = 1. √ x if x ≥ 0 √ If f (x) = , then − −x if x < 0 √ f (0 + h) − f (0) h lim = lim =∞ h→0+ h→0+ h h √ f (0 + h) − f (0) − −h lim = lim =∞ h→0− h→0− h h Thus the graph of f has a vertical tangent x = 0.

18. The slope of y = x 2 − 1 at x = x0 is

[(x0 + h)2 − 1] − (x02 − 1) h→0 h 2x0 h + h 2 = lim = 2x0 . h→0 h

m = lim

h→0

b) We have m = 3 if 3a 2 = 3, i.e., if a = ±1. Lines of slope 3 tangent to y = x 3 are y = 1 + 3(x − 1) and y = −1 + 3(x + 1), or y = 3x − 2 and y = 3x + 2.

20. The slope of y = x 3 − 3x at x = a is i 1h (a + h)3 − 3(a + h) − (a 3 − 3a) h→0 h i 1h 3 = lim a + 3a 2 h + 3ah 2 + h 3 − 3a − 3h − a 3 + 3a h→0 h = lim [3a 2 + 3ah + h 2 − 3] = 3a 2 − 3.

m = lim

h→0

At points where the tangent line is parallel to the x-axis, the slope is zero, so such points must satisfy 3a 2 − 3 = 0. Thus, a = ±1. Hence, the tangent line is parallel to the x-axis at the points (1, −2) and (−1, 2).

21. The slope of the curve y = x 3 − x + 1 at x = a is (a + h)3 − (a + h) + 1 − (a 3 − a + 1) h→0 h 3a 2 h + 3ah 2 + a 3 − h = lim h→0 h = lim (3a 2 + 3ah + h 2 − 1) = 3a 2 − 1.

m = lim

h→0

The tangent at x = a is parallel to the line y = 2x + 5 if 3a 2 − 1 = 2, that is, if a = ±1. The corresponding points on the curve are (−1, 1) and (1, 1).

22. The slope of the curve y = 1/x at x = a is 1 1 − a − (a + h) 1 a + h a m = lim = lim = − 2. h→0 h→0 ah(a + h) h a The tangent at x = a is perpendicular to the line y = 4x − 3 if −1/a 2 = −1/4, that is, if a = ±2. The corresponding points on the curve are (−2, −1/2) and (2, 1/2).

41 Copyright © 2014 Pearson Canada Inc.

SECTION 2.1 (PAGE 100)

ADAMS and ESSEX: CALCULUS 8

23. The slope of the curve y = x 2 at x = a is

27. Horizontal tangent at (−1/2, 5/4). No tangents at

(a + h)2 − a 2 m = lim = lim (2a + h) = 2a. h→0 h→0 h

(−1, 1) and (1, −1).

y 2

The normal at x = a has slope −1/(2a), and has equation 1 y − a = − (x − a), 2a 2

or

-3

This is the line x + y = k if 2a = 1, and so k = (1/2) + (1/2)2 = 3/4.

24. The curves y =

kx 2

The slope of y = m 1 = lim

h→0

1

x 1 + y = + a2 . 2a 2

and y = k(x at x = 1 is

kx 2

− 2)2

-2

-1

1

2

x

-1

intersect at (1, k).

-2

-3 Fig. 2.1.27

k(1 + h)2 − k = lim (2 + h)k = 2k. h→0 h

The slope of y = k(x − 2)2 at x = 1 is k(2 − (1 + h))2 − k m 2 = lim = lim (−2 + h)k = −2k. h→0 h→0 h

y = |x 2 − 1| − x

28. Horizontal tangent at (a, 2) and (−a, −2) for all a > 1. No tangents at (1, 2) and (−1, −2). y

The two curves intersect at right angles if 2k = −1/(−2k), that is, if 4k 2 = 1, which is satisfied if k = ±1/2.

2

y = |x + 1| − |x − 1|

1

25. Horizontal tangents at (0, 0), (3, 108), and (5, 0). y

(3, 108)

-3

-2

-1

100

1

2

x

-1

80

-2

60 -3 Fig. 2.1.28

40 y = x 3 (5 − x)2

20 -1

1

2

3

4

5

x

-20

29. Horizontal tangent at (0, −1). The tangents at (±1, 0) are vertical.

y Fig. 2.1.25

y = (x 2 − 1)1/3 2

26. Horizontal tangent at (−1, 8) and (2, −19). y

1

20 (−1, 8) 10 -2

-1

-3

y = 2x 3 − 3x 2 − 12x + 1 1

2

3

-2

-1

1

2

x

-1 -2

x

-10

-3 Fig. 2.1.29

-20

(2, −19)

-30

30. Horizontal tangent at (0, 1). No tangents at (−1, 0) and Fig. 2.1.26

(1, 0).

42 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.2 (PAGE 107)

y ((x 2

y=

2.

y

− 1)2 )1/3 2

1 y = g ′ (x) -2

-1

1

2

x

x

Fig. 2.1.30

31. The graph of the function f (x) = x 2/3 (see Figure 2.1.7

3.

y

in the text) has a cusp at the origin O, so does not have a tangent line there. However, the angle between O P and the positive y-axis does → 0 as P approaches 0 along the graph. Thus the answer is NO.

y = h ′ (x) x

32. The slope of P(x) at x = a is m = lim

h→0

P(a + h) − P(a) . h

Since P(a + h) = a0 + a1 h + a2 h 2 + · · · + an h n and P(a) = a0 , the slope is a0 + a1 h + a2 h 2 + · · · + an h n − a0 h→0 h = lim a1 + a2 h + · · · + an h n−1 = a1 .

m = lim

4.

y

h→0

Thus the line y = ℓ(x) = m(x − a) + b is tangent to y = P(x) at x = a if and only if m = a1 and b = a0 , that is, if and only if

x

P(x)−ℓ(x) = a2 (x − a)2 + a3 (x − a)3 + · · · + an (x − a)n h i = (x − a)2 a2 + a3 (x − a) + · · · + an (x − a)n−2

y = k ′ (x)

= (x − a)2 Q(x)

where Q is a polynomial.

Section 2.2 The Derivative 1.

(page 107)

5. Assuming the tick marks are spaced 1 unit apart, the function f is differentiable on the intervals (−2, −1), (−1, 1), and (1, 2).

y y = f ′ (x)

6. Assuming the tick marks are spaced 1 unit apart, the function g is differentiable on the intervals (−2, −1), (−1, 0), (0, 1), and (1, 2).

x

7.

y = f (x) has its minimum at x = 3/2 where f ′ (x) = 0

43 Copyright © 2014 Pearson Canada Inc.

SECTION 2.2 (PAGE 107)

ADAMS and ESSEX: CALCULUS 8

y

y y = f (x) = |x 3 − 1|

y = f (x) = 3x − x 2 − 1

x

x y y = f ′ (x)

y y = f ′ (x)

x x

Fig. 2.2.7

Fig. 2.2.9

10. 8.

y = f (x) has horizontal tangents at the points near 1/2 and 3/2 where f ′ (x) = 0 y

y = f (x) is constant on the intervals (−∞, −2), (−1, 1), and (2, ∞). It is not differentiable at x = ±2 and x = ±1. y y = f (x) = |x 2 − 1| − |x 2 − 4|

x x y = f (x) =

x3

− 3x 2

+ 2x + 1 y y = f ′ (x)

y

x x y = f ′ (x) Fig. 2.2.10 Fig. 2.2.8

11.

y = x 2 − 3x

(x + h)2 − 3(x + h) − (x 2 − 3x) h→0 h 2xh + h 2 − 3h = lim = 2x − 3 h→0 h d y = (2x − 3) d x y ′ = lim

9.

y = f (x) fails to be differentiable at x = −1, x = 0, and x = 1. It has horizontal tangents at two points, one between −1 and 0 and the other between 0 and 1.

44 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

12.

13.

14.

15.

16.

SECTION 2.2 (PAGE 107)

√ 2t + 1 √ √ 2 2 1 + 4(x + h) − 5(x + h) − (1 + 4x − 5x ′) 2(t + h) + 1 − 2t + 1 f ′ (x) = lim F (t) = lim h→0 h h→0 h 2t + 2h + 1 − 2t − 1 4h − 10xh − 5h 2 = lim √
√ = 4 − 10x = lim h→0 h 2(t + h) + 1 + 2t + 1 h→0 h 2 d f (x) = (4 − 10x) d x = lim √ √ h→0 2(t + h) + 1 + 2t + 1 1 = √ 2t + 1 1 f (x) = x 3 d F(t) = √ dt 3 3 2t + 1 (x + h) − x ′ f (x) = lim h→0 h 3x 2 h + 3xh 2 + h 3 = lim = 3x 2 h→0 h d f (x) = 3x 2 d x √ 18. f (x) = 43 2 − x √ 3 3√ ′ 4 2 − (x + h) − 4 2 − x f (x) = lim h→0 h   3 2−x −h −2+x 1 = lim √ √ s= h→0 4 h( 2 − (x + h) + 2 − x) 3 + 4t   ds 1 1 1 3 = lim − =− √ h→0 h 3 + 4(t + h) dt 3 + 4t 8 2−x 4 3 + 4t − 3 − 4t − 4h 3 = lim =− d f (x) = − √ dx h→0 h(3 + 4t)[3 + (4t + h)] (3 + 4t)2 8 2−x 4 ds = − dt (3 + 4t)2 f (x) = 1 + 4x − 5x 2

2−x g(x) = 2+x 2 − (x + h) 2 − x − 2+x ′ g (x) = lim 2 + x + h h→0 h (2 − x − h)(2 + x) − (2 + x + h)(2 − x) = lim h→0 h(2 + x + h)(2 + x) 4 =− (2 + x)2 4 dg(x) = − dx (2 + x)2

y = 31 x 3 − x i 1h1 y ′ = lim (x + h)3 − (x + h) − ( 31 x 3 − x) 3 h→0 h  1 2 = lim x h + xh 2 + 13 h 3 − h h→0 h = lim (x 2 + xh + 31 h 2 − 1) = x 2 − 1 h→0 2

d y = (x − 1) d x

17.

F(t) =

19.

y=x+

1 x

1 1 −x− x + h x y = lim h→0 h   x −x −h = lim 1 + h→0 h(x + h)x −1 1 = 1 + lim =1− 2 h→0 (x + h)x x   1 dy = 1 − 2 dx x ′

20.

x +h+

s 1+s   dz 1 s+h s = lim − h→0 h 1 + s + h ds 1+s (s + h)(1 + s) − s(1 + s + h) 1 = lim = h→0 h(1 + s)(1 + s + h) (1 + s)2 1 dz = ds (1 + s)2 z=

45 Copyright © 2014 Pearson Canada Inc.

SECTION 2.2 (PAGE 107)

21.

F(x) = √

ADAMS and ESSEX: CALCULUS 8

1 1 + x2

25. Since f (x) = x sgn x = |x|, for x 6= 0, f will become 1

continuous at x = 0 if we define f (0) = 0. However, f will still not be differentiable at x = 0 since |x| is not differentiable at x = 0.  2 Since g(x) = x 2 sgn x = x|x| = x 2 if x > 0 , g −x if x < 0 will become continuous and differentiable at x = 0 if we define g(0) = 0.

1

p −√ 1 + x2 1 + (x + h)2 F (x) = lim h→0 h 26. p √ 1 + x 2 − 1 + (x + h)2 = lim p √ h→0 h 1 + (x + h)2 1 + x 2 1 + x 2 − 1 − x 2 − 2hx − h 2 √ 27.  h(x) = |x 2 + 3x + 2| fails to be differentiable where = lim p p √ h→0 h 1 + (x + h)2 1 + x 2 1 + x 2 + 1 + (x + h)2 x 2 + 3x + 2 = 0, that is, at x = −2 and x = −1. Note: both of these are single zeros of x 2 + 3x + 2. If they −2x x = = − were higher order zeros (i.e. if (x + 2)n or (x + 1)n were 2(1 + x 2 )3/2 (1 + x 2 )3/2 a factor of x 2 + 3x + 2 for some integer n ≥ 2) then h x d F(x) = − d x would be differentiable at the corresponding point. (1 + x 2 )3/2 28. y = x 3 − 2x 1 y= 2 f (x) − f (1) f (x) − f (1) x x x   x − 1 x −1 1 1 1 − 2 y ′ = lim 0.9 0.71000 1.1 1.31000 2 h→0 h (x + h) x 0.99 0.97010 1.01 1.03010 x 2 − (x + h)2 2 0.999 0.99700 1.001 1.00300 = lim = − h→0 hx 2 (x + h)2 x3 0.9999 0.99970 1.0001 1.00030 2 dy = − 3 dx x ′

22.

23.

y= √

1+x

1 −√ 1+x +h 1+x y ′ (x) = lim h→0 h √ √ 1+x − 1+x +h = lim √ √ h→0 h 1 + x + h 1 + x 1+x −1−x −h = lim √
√ √ √ h→0 h 1 + x + h 1 + x 1+x + 1+x +h 1 = lim − √
√ √ √ h→0 1+x +h 1+x 1+x + 1+x +h 1 =− 2(1 + x)3/2 1 dy = − dx 2(1 + x)3/2 √

24.

(1 + h)3 − 2(1 + h) − (−1) d 3 = lim (x − 2x) h→0 dx h x=1 h + 3h 2 + h 3 = lim h→0 h = lim 1 + 3h + h 2 = 1

1 1

t2 − 3 t2 + 3   1 (t + h)2 − 3 t 2 − 3 f ′ (t) = lim − h→0 h (t + h)2 + 3 t 2 + 3 2 [(t + h) − 3](t 2 + 3) − (t 2 − 3)[(t + h)2 + 3] = lim h→0 h(t 2 + 3)[(t + h)2 + 3] 12th + 6h 2 12t = lim = 2 2 h→0 h(t + 3)[(t + h)2 + 3] (t + 3)2 12t d f (t) = 2 dt (t + 3)2 f (t) =

h→0

29.

f (x) = 1/x f (x) − f (2)

x

x −2 −0.26316 −0.25126 −0.25013 −0.25001

1.9 1.99 1.999 1.9999

x 2.1 2.01 2.001 2.0001

f (x) − f (2) x −2 −0.23810 −0.24876 −0.24988 −0.24999

1 −2 2 − (2 + h) 2 + h f ′ (2) = lim = lim h→0 h→0 h(2 + h)2 h 1 1 = lim − =− h→0 (2 + h)2 4

30. The slope of y = 5 + 4x − x 2 at x = 2 is 5 + 4(2 + h) − (2 + h)2 − 9 d y = lim d x x=2 h→0 h 2 −h = lim = 0. h→0 h Thus, the tangent line at x = 2 has the equation y = 9.

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INSTRUCTOR’S SOLUTIONS MANUAL

31.

√

x + 6. Slope at (3, 3) is √ 9+h −3 9+h−9 1 m = lim = lim √
= . h→0 h→0 h 6 h 9+h+3

y=

Tangent line is y − 3 =

32.

SECTION 2.2 (PAGE 107)

1 (x − 3), or x − 6y = −15. 6

t at t = −2 and y = −1 is The slope of y = 2 t −2   d y 1 −2 + h = lim − (−1) dt t=−2 h→0 h (−2 + h)2 − 2

−2 + h + [(−2 + h)2 − 2] 3 = lim =− . h→0 2 h[(−2 + h)2 − 2]

Thus, the tangent line has the equation y = −1 − 32 (t + 2), that is, y = − 23 t − 4.

33.

y=

2 t2 + t

34.

f ′ (x) = −17x −18 for x 6= 0

35.

g ′ (t) = 22t 21 for all t

37. 38. 39. 40. 41. 42. 43.

√

x at x = x0 is d y 1 = √ . d x x=x0 2 x0

Thus, the equation of the tangent line is 1 x + x0 √ y = x0 + √ (x − x0 ), that is, y = √ . 2 x0 2 x0

1 1 1 at x = a is − 2 = 2 . x x x=a a 1 Normal has slope a 2 , and equation y − = a 2 (x − a), a 1 or y = a 2 x − a 3 + a

45. Slope of y =

Slope at t = a is

2 2 − (a + h)2 + (a + h) a 2 + a m = lim h→0 h 2(a 2 + a − a 2 − 2ah − h 2 − a − h) = lim h→0 h[(a + h)2 + a + h](a 2 + a) −4a − 2h − 2 = lim h→0 [(a + h)2 + a + h](a 2 + a) 4a + 2 =− 2 (a + a)2 2 2(2a + 1) Tangent line is y = 2 − 2 (t − a) a +a (a + a)2

36.

44. The slope of y =

1 dy = x −2/3 for x 6= 0 dx 3 dy 1 = − x −4/3 for x 6= 0 dx 3 d −2.25 t = −2.25t −3.25 for t > 0 dt d 119/4 119 115/4 s = s for s > 0 ds 4 d √ 1 1 s = √ = . ds 6 2 s s=9 s=9   1 1 1 F(x) = , F ′ (x) = − 2 , F ′ = −16 x 4 x 2 1 f ′ (8) = − x −5/3 =− 3 48 x=8 d y 1 −3/4 1 = t = √ dt t=4 4 8 2 t=4

46. The intersection points of y = x 2 and x + 4y = 18 satisfy 4x 2 + x − 18 = 0 (4x + 9)(x − 2) = 0. Therefore x = − 94 or x = 2.

dy = 2x. dx 9 9 At x = − , m 1 = − . At x = 2, m 1 = 4. 4 2 The slope of x + 4y = 18, i.e. y = − 41 x + 18 4 , is m 2 = − 41 . Thus, at x = 2, the product of these slopes is (4)(− 14 ) = −1. So, the curve and line intersect at right angles at that point. The slope of y = x 2 is m 1 =

47. Let the point of tangency be (a, a 2 ). Slope of tangent is d 2 x = 2a d x x=a This is the slope from (a, a 2 ) to (1, −3), so a2 + 3 = 2a, and a−1 a 2 + 3 = 2a 2 − 2a

a 2 − 2a − 3 = 0 a = 3 or − 1

The two tangent lines are (for a = 3): y − 9 = 6(x − 3) or 6x − 9 (for a = −1): y − 1 = −2(x + 1) or y = −2x − 1

47 Copyright © 2014 Pearson Canada Inc.

SECTION 2.2 (PAGE 107)

ADAMS and ESSEX: CALCULUS 8

y

(a,a 2 )

y = x2

x (1,−3)

√

p 4a 2 − 4b = a ± a 2 − b. 2 √ If b < a 2 , i.e. a 2 − b > 0, then t = a ± a 2 − b has two real solutions. Therefore, there will be two distinct tangent passing through (a, b) with equations  lines √ y = b + 2 a ± a 2 − b (x − a). If b = a 2 , then t = a. There will be only one tangent line with slope 2a and equation y = b + 2a(x − a). If b > a 2 , then a 2 − b < 0. There will be no real solution for t. Thus, there will be no tangent line. Hence t =

2a ±

Fig. 2.2.47

48. The slope of y =

1 at x = a is x

51. Suppose f is odd: f (−x) = − f (x). Then f (−x + h) − f (−x) h→0 h f (x − h) − f (x) = lim − h→0 h

f ′ (−x) = lim

d y 1 = − 2. d x x=a a

1 1 If the slope is −2, then − 2 = −2, or a = ± √ . a 2 Therefore, the  equations  of the two straightlines are  √ √ 1 1 and y = − 2 − 2 x + √ , y = 2−2 x − √ 2 √ 2 or y = −2x ± 2 2. √

49. Let the point of tangency be (a, a)

d √ 1 Slope of tangent is x = √ 2 a x=a √ dx 1 a−0 Thus √ = , so a + 2 = 2a, and a = 2. a+2 2 a 1 The required slope is √ . 2 2

(let h = −k) = lim

k→0

f (x + k) − f (x) = f ′ (x) k

Thus f ′ is even. Now suppose f is even: f (−x) = f (x). Then f (−x + h) − f (−x) f ′ (−x) = lim h→0 h f (x − h) − f (x) = lim h→0 h f (x + k) − f (x) = lim k→0 −k = − f ′ (x) so f ′ is odd.

y

52. Let f (x) = x −n . Then

√ (a, a) √ y= x x

−2

Fig. 2.2.49

50. If a line is tangent to y = x 2 at (t, t 2 ), then its slope is

d y = 2t. If this line also passes through (a, b), then d x x=t its slope satisfies t2 − b = 2t, t −a

that is t 2 − 2at + b = 0.

(x + h)−n − x −n h→0 h   1 1 1 = lim − h→0 h (x + h)n xn n n x − (x + h) = lim h→0 hx n (x + h)n x − (x + h) = lim × h→0 hx n ((x + h)n   n−1 n−2 n−1 x +x (x + h) + · · · + (x + h)

f ′ (x) = lim

=−

48 Copyright © 2014 Pearson Canada Inc.

1 × nx n−1 = −nx −(n+1) . x 2n

INSTRUCTOR’S SOLUTIONS MANUAL

53.

SECTION 2.3 (PAGE 115)

f (x) = x 1/3

If f ′ (a+) is finite, call the half-line with equation y = f (a) + f ′ (a+)(x − a), (x ≥ a), the right tangent line to the graph of f at x = a. Similarly, if f ′ (a−) is finite, call the half-line y = f (a) + f ′ (a−)(x − a), (x ≤ a), the left tangent line. If f ′ (a+) = ∞ (or −∞), the right tangent line is the half-line x = a, y ≥ f (a) (or x = a, y ≤ f (a)). If f ′ (a−) = ∞ (or −∞), the right tangent line is the half-line x = a, y ≤ f (a) (or x = a, y ≥ f (a)). The graph has a tangent line at x = a if and only if f ′ (a+) = f ′ (a−). (This includes the possibility that both quantities may be +∞ or both may be −∞.) In this case the right and left tangents are two opposite halves of the same straight line. For f (x) = x 2/3 , f ′ (x) = 32 x −1/3 . At (0, 0), we have f ′ (0+) = +∞ and f ′ (0−) = −∞. In this case both left and right tangents are the positive y-axis, and the curve does not have a tangent line at the origin. For f (x) = |x|, we have

(x + h)1/3 − x 1/3 h→0 h (x + h)1/3 − x 1/3 = lim h→0 h (x + h)2/3 + (x + h)1/3 x 1/3 + x 2/3 × (x + h)2/3 + (x + h)1/3 x 1/3 + x 2/3 x +h−x = lim h→0 h[(x + h)2/3 + (x + h)1/3 x 1/3 + x 2/3 ] 1 = lim h→0 (x + h)2/3 + (x + h)1/3 x 1/3 + x 2/3 1 1 = 2/3 = x −2/3 3 3x

f ′ (x) = lim

54. Let f (x) = x 1/n . Then (x + h)1/n − x 1/n f (x) = lim h→0 h a−b = lim n a→b a − b n ′

f ′ (x) = sgn (x) =

(let x + h = a n , x = bn )

1 a n−1 + a n−2 b + a n−3 b2 + · · · + bn−1 1 1 = n−1 = x (1/n)−1 . nb n a→b

d n (x + h)n − x n x = lim h→0 dx h  1 n n n−1 n(n − 1) n−2 2 = lim x + x h+ x h h→0 h 1 1×2  n(n − 1)(n − 2) n−3 3 n n + x h +··· + h − x 1×2×3   n(n − 1) n−2 = lim nx n−1 + h x h h→0 1×2  n(n − 1)(n − 2) n−3 2 n−1 + x h +··· + h 1×2×3 = nx n−1

Section 2.3 (page 115)

y = 3x 2 − 5x − 7,

2.

y = 4x 1/2 −

3.

f (x) = Ax 2 + Bx + C,

4.

f (x) =

5.

z=

6.

y = x 45 − x −45

7.

g(t) = t 1/3 + 2t 1/4 + 3t 1/5 1 1 3 g ′ (t) = t −2/3 + t −3/4 + t −4/5 3 2 5 p 2 3 y = 3 t 2 − √ = 3t 2/3 − 2t −3/2 t3 dy = 2t −1/3 + 3t −5/2 dt

56. Let f (a + h) − f (a) h f (a + h) − f (a) ′ f (a−) = lim h→0− h h→0+

Differentiation Rules

1.

8.

f ′ (a+) = lim

if x > 0 if x < 0.

1 −1

At (0, 0), f ′ (0+) = 1, and f ′ (0−) = −1. In this case the right tangent is y = x, (x ≥ 0), and the left tangent is y = −x, (x ≤ 0). There is no tangent line.

= lim

55.

n

9.

5 , x

y ′ = 2x −1/2 + 5x −2

6 2 + 2 − 2, x3 x

s5 − s3 , 15

u=

y ′ = 6x − 5.

f ′ (x) = 2 Ax + B. f ′ (x) = −

18 4 − 3 x4 x

dz 1 1 = s 4 − s 2. dx 3 5 y ′ = 45x 44 + 45x −46

3 5/3 5 −3/5 x − x 5 3

du = x 2/3 + x −8/5 dx

49 Copyright © 2014 Pearson Canada Inc.

SECTION 2.3 (PAGE 115)

10.

11.

ADAMS and ESSEX: CALCULUS 8

F(x) = (3x − 2)(1 − 5x) F ′ (x) = 3(1 − 5x) + (3x − 2)(−5) = 13 − 30x

22.

  √ x2 1 x 5−x − = 5 x − x 3/2 − x 5/2 3 3 5 3√ 5 3/2 ′ y = √ − x− x 2 6 2 x y=

√

23. 12. 13.

14. 15.

16.

17.

18.

19.

20.

21.

1 , g(t) = 2t − 3

2 g ′ (t) = − (2t − 3)2

1 x 2 + 5x 1 2x + 5 y′ = − 2 (2x + 5) = − 2 2 (x + 5x) (x + 5x)2 y=

y=

4 , 3−x

y′ =

24.

4 (3 − x)2

π 2 − πt π π2 f ′ (t) = − (−π ) = 2 (2 − π t) (2 − π t)2 f (t) =

g(y) =

2 , 1 − y2

g ′ (y) =

25.

4y (1 − y 2 )2

1 − 4x 2 4 = x −3 − 3 x x 4x 2 − 3 ′ −4 −2 f (x) = −3x + 4x = x4 f (x) =

26.

√ u u −3 = u −1/2 − 3u −2 u2 √ 12 − u u 1 g ′ (u) = − u −3/2 + 6u −3 = 2 2u 3 g(u) =

27.

√ 2 + t + t2 = 2t −1/2 + t + t 3/2 √ t dy 1 3t 2 + t − 2 3√ −3/2 = −t + √ + t= √ dt 2 2 t 2t t y=

x −1 z = 2/3 = x 1/3 − x −2/3 x 1 2 x +2 dz = x −2/3 + x −5/3 = 5/3 dx 3 3 3x 3 − 4x 3 + 4x (3 + 4x)(−4) − (3 − 4x)(4) ′ f (x) = (3 + 4x)2 24 =− (3 + 4x)2 f (x) =

t 2 + 2t t2 − 1 (t 2 − 1)(2t + 2) − (t 2 + 2t)(2t) z′ = (t 2 − 1)2 2(t 2 + t + 1) =− (t 2 − 1)2 √ 1+ t s= √ 1− t √ √ 1 1 (1 − t) √ − (1 + t)(− √ ) ds 2 t 2 t = √ dt (1 − t)2 1 = √ √ t(1 − t)2 z=

x3 − 4 x +1 (x + 1)(3x 2 ) − (x 3 − 4)(1) ′ f (x) = (x + 1)2 3 2 2x + 3x + 4 = (x + 1)2 f (x) =

ax + b cx + d (cx + d)a − (ax + b)c f ′ (x) = (cx + d)2 ad − bc = (cx + d)2 f (x) =

t 2 + 7t − 8 t2 − t + 1 (t 2 − t + 1)(2t + 7) − (t 2 + 7t − 8)(2t − 1) F ′ (t) = (t 2 − t + 1)2 2 −8t + 18t − 1 = (t 2 − t + 1)2 F(t) =

f (x) = (1 + x)(1 + 2x)(1 + 3x)(1 + 4x) f ′ (x) = (1 + 2x)(1 + 3x)(1 + 4x) + 2(1 + x)(1 + 3x)(1 + 4x) + 3(1 + x)(1 + 2x)(1 + 4x) + 4(1 + x)(1 + 2x)(1 + 3x) OR f (x) = [(1 + x)(1 + 4x)] [(1 + 2x)(1 + 3x)] = (1 + 5x + 4x 2 )(1 + 5x + 6x 2 )

= 1 + 10x + 25x 2 + 10x 2 (1 + 5x) + 24x 4 = 1 + 10x + 35x 2 + 50x 3 + 24x 4

f (x) = 10 + 70x + 150x 2 + 96x 3 ′

28.

f (r ) = (r −2 + r −3 − 4)(r 2 + r 3 + 1)

f ′ (r ) = (−2r −3 − 3r −4 )(r 2 + r 3 + 1) or

+ (r −2 + r −3 − 4)(2r + 3r 2 )

f (r ) = −2 + r −1 + r −2 + r −3 + r − 4r 2 − 4r 3

f ′ (r ) = −r −2 − 2r −3 − 3r −4 + 1 − 8r − 12r 2

50 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

29.

30.

√ y = (x 2 + 4)( x + 1)(5x 2/3 − 2) √ y ′ = 2x( x + 1)(5x 2/3 − 2) 1 + √ (x 2 + 4)(5x 2/3 − 2) 2 x √ 10 −1/3 2 x (x + 4)( x + 1) + 3 y= = y′ =

= =

31.

32.

33.

34.

SECTION 2.3 (PAGE 115)

y=

35.

(x 2 + 1)(x 3 + 2) (x 2 + 2)(x 3 + 1) x 5 + x 3 + 2x 2 + 2 x 5 + 2x 3 + x 2 + 2 (x 5 + 2x 3 + x 2 + 2)(5x 4 + 3x 2 + 4x) (x 5 + 2x 3 + x 2 + 2)2 (x 5 + x 3 + 2x 2 + 2)(5x 4 + 6x 2 + 2x) − (x 5 + 2x 3 + x 2 + 2)2 7 6 2x − 3x − 3x 4 − 6x 2 + 4x (x 5 + 2x 3 + x 2 + 2)2 7 2x − 3x 6 − 3x 4 − 6x 2 + 4x (x 2 + 2)2 (x 3 + 1)2 x

=

36.

d dx



37.

d dx



38.

3x 2 + x + 2x + 1

√ ( x − 1)(2 − x)(1 − x 2 ) √ x(3 + 2x)   2 − x − 2x 2 + x 3 1 · = 1− √ 3 + 2x x     1 1 2 − x − 2x 2 + x 3 f ′ (x) = x −3/2 + 1− √ 2 3 + 2x x (3 + 2x)(−1 − 4x + 3x 2 ) − (2 − x − 2x 2 + x 3 )(2) × (3 + 2x)2 (2 − x)(1 − x 2 ) = 2x 3/2 (3 + 2x)   1 4x 3 + 5x 2 − 12x − 7 + 1− √ (3 + 2x)2 x

d dx

d dx





 f (x)(2x) − x 2 f ′ (x) x 2 = f (x) x=2 [ f (x)]2 x=2 ′ 4 f (2) − 4 f (2) 4 = = − = −1 [ f (2)]2 4  f (x) x 2 f ′ (x) − 2x f (x) = x 2 x=2 x4 x=2 ′ 4 f (2) − 4 f (2) 4 1 = = = 16 16 4

   x2 − 4 d 8 | = 1 − x=−2 2 2 x +4 dx x + 4 x=−2 8 = 2 (2x) 2 (x + 4) x=−2

6x 2

f (x) =

 f (x) 2 x + f (x) x=2 (x 2 + f (x)) f ′ (x) − f (x)(2x + f ′ (x)) = (x 2 + f (x))2 x=2 18 − 14 1 (4 + f (2)) f ′ (2) − f (2)(4 + f ′ (2)) = = = 2 2 9 (4 + f (2)) 6

32 1 =− =− 64 2

1 3x + 1 (6x 2 + 2x + 1)(6x + 1) − (3x 2 + x)(12x + 2) y′ = (6x 2 + 2x + 1)2 6x + 1 = (6x 2 + 2x + 1)2 2x +

   d  2 = 2x f (x) + x 2 f ′ (x) x f (x) dx x=2 x=2 = 4 f (2) + 4 f ′ (2) = 20

√  t (1 + t) 5−t t=4   3/2 d t +t = dt 5−t

d dt



t=4

(5 − t)(1 + 32 t 1/2 ) − (t + t 3/2 )(−1) = (5 − t)2 t=4 (1)(4) − (12)(−1) = 16 = (1)2

39.

f (x) =

√

x x +1

√ 1 (x + 1) √ − x(1) 2 x f ′ (x) = (x + 1)2 √ 3 √ − 2 1 2 2 f ′ (2) = =− √ 9 18 2

40.

d [(1 + t)(1 + 2t)(1 + 3t)(1 + 4t)] dt t=0 = (1)(1 + 2t)(1 + 3t)(1 + 4t) + (1 + t)(2)(1 + 3t)(1 + 4t)+ (1 + t)(1 + 2t)(3)(1 + 4t) + (1 + t)(1 + 2t)(1 + 3t)(4) t=0

= 1 + 2 + 3 + 4 = 10

41.

y=

2 2 √ , y′ = −  √ 2 3−4 x 3−4 x



4 − √ 2 x



8 =4 (−1)2 2 Tangent line has the equation y = −2 + 4(x − 1) or y = 4x − 6

Slope of tangent at (1, −2) is m =

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SECTION 2.3 (PAGE 115)

ADAMS and ESSEX: CALCULUS 8

y

x +1 we calculate x −1

42. For y =

(x − 1)(1) − (x + 1)(1) 2 =− . (x − 1)2 (x − 1)2

y′ =

y= 

b

At x = 2 we have y = 3 and y ′ = −2. Thus, the equation of the tangent line is y = 3 − 2(x − 2), or y = −2x + 7. The normal line is y = 3 + 12 (x − 2), or y = 21 x + 2.

43.

y=x+

1 1 ′ , y =1− 2 x x

For horizontal tangent: 0 = y ′ = 1 −

44. If y =

x 2 (4

−

x 2 ),

then

y ′ = 2x(4 − x 2 ) + x 2 (−2x) = 8x − 4x 3 = 4x(2 − x 2 ). The slope of a horizontal line must be zero, so √ 4x(2 − x 2 ) = 0, which implies√that x = 0 or x = ± 2. At x = 0, y = 0 and at x = ± 2, y = 4. Hence, there are two horizontal lines that are tangent to the curve. Their equations are y = 0 and y = 4.

45.

1 2x + 1 , y′ = − 2 x2 + x + 1 (x + x + 1)2 For horizon2x + 1 tal tangent we want 0 = y ′ = − 2 . Thus (x + x + 1)2 1 2x + 1 = 0 and x = − 2   1 4 The tangent is horizontal only at − , . 2 3 y=

46. If y =

x +1 , then x +2 y′ =

(x + 2)(1) − (x + 1)(1) 1 = . (x + 2)2 (x + 2)2

In order to be parallel to y = 4x, the tangent line must have slope equal to 4, i.e., 1 = 4, (x + 2)2

or (x + 2)2 = 41 .

Hence x + 2 = ± 21 , and x = − 23 or − 52 . At x = − 32 , y = −1, and at x = − 25 , y = 3. Hence, the tangent is parallel to y = 4x at the points 
− 23 , −1 and − 52 , 3 .

47. Let the point of tangency be (a, a1 ). The slope of the

b − a1 1 2 = . Thus b − a1 = a1 and a = . 2 0−a b a b2 b2 Tangent has slope − so has equation y = b − x. 4 4

1 a, a

 x

1 so x 2 = 1 and x2

x = ±1 The tangent is horizontal at (1, 2) and at (−1, −2)

1 x

Fig. 2.3.47 1 x the intersection point. The slope of y = x 2 at x = 1 is 1 = 2. The slope of y = √ at x = 1 is 2x x

48. Since √ = y = x 2 ⇒ x 5/2 = 1, therefore x = 1 at

x=1

1 d y 1 −3/2 =− . = − x d x x=1 2 2 x=1


The product of the slopes is (2) − 21 = −1. Hence, the two curves intersect at right angles.

49. The tangent to y = x 3 at (a, a 3 ) has equation

y = a 3 + 3a 2 (x − a), or y = 3a 2 x − 2a 3 . This line passes through (2, 8) if 8 = 6a 2 − 2a 3 or, equivalently, if a 3 − 3a 2 + 4 = 0. Since (2, 8) lies on y = x 3 , a = 2 must be a solution of this equation. In fact it must be a double root; (a − 2)2 must be a factor of a 3 − 3a 2 + 4. Dividing by this factor, we find that the other factor is a + 1, that is, a 3 − 3a 2 + 4 = (a − 2)2 (a + 1). The two tangent lines to y = x 3 passing through (2, 8) correspond to a = 2 and a = −1, so their equations are y = 12x − 16 and y = 3x + 2.

50. The tangent to y = x 2 /(x − 1) at (a, a 2 /(a − 1)) has slope m=

(x − 1)2x − x 2 (1) a 2 − 2a = . 2 (x − 1) (a − 1)2 x=a

The equation of the tangent is y−

a 2 − 2a a2 = (x − a). a−1 (a − 1)2

This line passes through (2, 0) provided

tangent is −

52 Copyright © 2014 Pearson Canada Inc.

0−

a2 a 2 − 2a = (2 − a), a−1 (a − 1)2

INSTRUCTOR’S SOLUTIONS MANUAL

51.

52.

SECTION 2.4 (PAGE 120)

or, upon simplification, 3a 2 − 4a = 0. Thus we can have either a = 0 or a = 4/3. There are two tangents through (2, 0). Their equations are y = 0 and y = −8x + 16. √ √ d p f (x + h) − f (x) f (x) = lim h→0 dx h f (x + h) − f (x) 1 √ = lim √ h→0 h f (x + h) + f (x) f ′ (x) = √ 2 f (x) d p 2 2x x x +1 = √ = √ 2 2 dx 2 x +1 x +1  3 f (x) = |x 3 | = x 3 if x ≥ 0 . Therefore f is differen−x if x < 0 tiable everywhere except possibly at x = 0, However,

54. To be proved: ( f 1 f 2 · · · f n )′ = f 1′ f 2 · · · f n + f 1 f 2′ · · · f n + · · · + f 1 f 2 · · · f n′ Proof: The case n = 2 is just the Product Rule. Assume the formula holds for n = k for some integer k > 2. Using the Product Rule and this hypothesis we calculate ( f 1 f 2 · · · f k f k+1 )′ = [( f 1 f 2 · · · f k ) f k+1 ]′ ′ = ( f 1 f 2 · · · f k )′ f k+1 + ( f 1 f 2 · · · f k ) f k+1

= ( f 1′ f 2 · · · f k + f 1 f 2′ · · · f k + · · · + f 1 f 2 · · · f k′ ) f k+1 ′ + ( f 1 f 2 · · · f k ) f k+1

= f 1′ f 2 · · · f k f k+1 + f 1 f 2′ · · · f k f k+1 + · · · ′ + f 1 f 2 · · · f k′ f k+1 + f 1 f 2 · · · f k f k+1

f (0 + h) − f (0) = lim h 2 = 0 h→0+ h f (0 + h) − f (0) = lim (−h 2 ) = 0. lim h→0− h→0− h lim

so the formula is also true for n = k + 1. The formula is therefore for all integers n ≥ 2 by induction.

h→0+

Thus

f ′ (0)

exists and equals 0. We have f ′ (x) =



3x 2 −3x 2

Section 2.4 1.

if x ≥ 0 if x < 0.

2.

d n/2 n x = x (n/2)−1 for n = 1, 2, 3, . . . . dx 2 Proof: It is already known that the case n = 1 is true: the derivative of x 1/2 is (1/2)x −1/2 . Assume that the formula is valid for n = k for some positive integer k:

53. To be proved:

3.

4. 5.

d k/2 k x = x (k/2)−1 . dx 2 Then, by the Product Rule and this hypothesis,

6. d 1/2 k/2 d (k+1)/2 x = x x dx dx 1 −1/2 k/2 k 1/2 (k/2)−1 k + 1 (k+1)/2−1 = x x + x x = x . 7. 2 2 2 Thus the formula is also true for n = k + 1. Therefore it is true for all positive integers n by induction. For negative n = −m (where m > 0) we have d n/2 d 1 x = dx d x x m/2 −1 m = m x (m/2)−1 x 2 m n = − x −(m/2)−1 = x (n/2)−1 . 2 2

8.

The Chain Rule

(page 120)

y = (2x + 3)6 , y ′ = 6(2x + 3)5 2 = 12(2x + 3)5  x 99 y = 1− 3     x 98 1 x 98 ′ y = 99 1 − − = −33 1 − 3 3 3 f (x) = (4 − x 2 )10

f ′ (x) = 10(4 − x 2 )9 (−2x) = −20x(4 − x 2 )9 dy d p −6x 3x = 1 − 3x 2 = √ = −√ dx dx 2 1 − 3x 2 1 − 3x 2   3 −10 F(t) = 2 + t     3 −11 −3 3 −11 30 ′ F (t) = −10 2 + = 2 2+ t t t2 t z = (1 + x 2/3 )3/2

z ′ = 23 (1 + x 2/3 )1/2 ( 32 x −1/3 ) = x −1/3 (1 + x 2/3 )1/2 3 5 − 4x 3 12 y′ = − (−4) = (5 − 4x)2 (5 − 4x)2 y=

y = (1 − 2t 2 )−3/2

y ′ = − 32 (1 − 2t 2 )−5/2 (−4t) = 6t (1 − 2t 2 )−5/2

9. 10.

y = |1 − x 2 |,

y ′ = −2xsgn (1 − x 2 ) =

f (t) = |2 + t 3 | f ′ (t) = [sgn (2 + t 3 )](3t 2 ) =

2x 3 − 2x |1 − x 2 |

3t 2 (2 + t 3 ) |2 + t 3 |

53 Copyright © 2014 Pearson Canada Inc.

SECTION 2.4 (PAGE 120)

11.

12.

ADAMS and ESSEX: CALCULUS 8

17.

y = 4x + |4x − 1| y ′ = 4 + 4(sgn (4x − 1))  8 if x > 41 = 0 if x < 14

y

y=|2+t 3 |

3 1/3

y = (2 + |x| )

y ′ = 31 (2 + |x|3 )−2/3 (3|x|2 )sgn (x)   x 2 3 −2/3 = |x| (2 + |x| ) = x|x|(2 + |x|3 )−2/3 |x|

13.

18. y

slope 8

1 √ 2 + 3x + 4   1 3 ′ y = − √ 2 √ 2 3x + 4 2 + 3x + 4 y=

y=4x+|4x−1| slope 0



3 =− √  2 √ 2 3x + 4 2 + 3x + 4

14.

4 x −2 3 r 3  x −2 f ′ (x) = 4 1 + 3 r r  3 x 2 1+ = 3 x −2 

f (x) = 1 +

r

1 2

r

−2 3

3 x −2 3

!  1 3

19.

d 1/4 d x = dx dx

20.

d 3/4 x dx

21.

d 3/2 x dx

22. 23.

 z= u+

15.



1 u−1

−5/3

−8/3  1−

24. 

dz 5 1 1 =− u+ du 3 u−1 (u − 1)2   −8/3 5 1 1 =− 1− u + 3 u−1 (u − 1)2

16.

t

−21/3

√ x5 3 + x6 y= (4 + x 2 )3

25. 26.

!# " p 1 3x 5 5 2 3 4 6 y = √ (4 + x ) 5x 3 + x + x (4 + x 2 )6 3 + x6  h i p − x 5 3 + x 6 3(4 + x 2 )2 (2x) h i (4 + x 2 ) 5x 4 (3 + x 6 ) + 3x 10 − x 5 (3 + x 6 )(6x) = √ (4 + x 2 )4 3 + x 6 60x 4 − 3x 6 + 32x 10 + 2x 12 = √ (4 + x 2 )4 3 + x 6 ′

27. 28.



1 4 ,1

x

q √

1 1 1 x = p√ × √ = x −3/4 4 2 x 2 x   q √ √ d 1 x 3 = x x= p √ x+ √ = x −1/4 dx 4 2 x 2 x x d p 3 3 1 = x = √ (3x 2 ) = x 1/2 dx 2 2 x3

d f (2t + 3) = 2 f ′ (2t + 3) dt d f (5x − x 2 ) = (5 − 2x) f ′ (5x − x 2 ) dx   3   2     d 2 2 −2 ′ 2 f =3 f f dx x x x x2     2 2 2 2 = − 2 f′ f x x x

2 f ′ (x) d p f ′ (x) 3 + 2 f (x) = √ = √ dx 2 3 + 2 f (x) 3 + 2 f (x) √ √ d 2 f ( 3 + 2t ) = f ′ ( 3 + 2t) √ dt 2 3 + 2t √ 1 = √ f ′ ( 3 + 2t ) 3 + 2t √ √ d 1 f (3 + 2 x) = √ f ′ (3 + 2 x) dx x    d f 2 f 3 f (x) dt      = f ′ 2 f 3 f (x) · 2 f ′ 3 f (x) · 3 f ′ (x)      = 6 f ′ (x) f ′ 3 f (x) f ′ 2 f 3 f (x)

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INSTRUCTOR’S SOLUTIONS MANUAL

29.

30.

31. 32.

33.

 d  f 2 − 3 f (4 − 5t) dx    = f ′ 2 − 3 f (4 − 5t) −3 f ′ (4 − 5t) (−5)   = 15 f ′ (4 − 5t) f ′ 2 − 3 f (4 − 5t)

! √ d x 2 − 1 dx x 2 + 1 x=−2 p x − x 2 − 1(2x) (x 2 + 1) √ x2 − 1 = (x 2 + 1)2 x=−2   √ 2 (5) − √ − 3(−4) 2 3 = = √ 25 25 3 d√ 3 3 = √ 3t − 7 = √ dt 2 3t − 7 2 2 t=3 t=3

2

x=−2

=

=− 2 (1 + x ) x 2 3 x=−1 The tangent√line at (−1, 23/2 ) has equation y = 23/2 − 2(x + 1).

x=b/a

7 = 8a(ax + b)

40.

3 3 5 = − (9−5/2 )(−5) = − (x 2 −x+3)−5/2 (2x−1) 2 2 162 x=−2 1 The tangent line at (−2, ) has equation 27 1 5 y= + (x + 2). 27 162 Given that f (x) = (x − a)m (x − b)n then f ′ (x) = m(x − a)m−1 (x − b)n + n(x − a)m (x − b)n−1

F(x) = (1 + x)(2 + x) (3 + x) (4 + x)

= (x − a)m−1 (x − b)n−1 (mx − mb + nx − na).

If x 6= a and x 6= b, then f ′ (x) = 0 if and only if

2(1 + x)(2 + x)(3 + x)3 (4 + x)4 +

3(1 + x)(2 + x)2 (3 + x)2 (4 + x)4 +

mx − mb + nx − na = 0,

4(1 + x)(2 + x)2 (3 + x)3 (4 + x)3 2

4

2

3

which is equivalent to 3

3(1)(2 )(3 )(4 ) + 4(1)(2 )(3 )(4 )

35.

= 4(22 · 33 · 44 ) = 110, 592   −1/2 −6 y = x + (3x)5 − 2

√

x=

n m a+ b. m +n m +n

This point lies lies between a and b.

  −1/2 −7 y ′ = −6 x + (3x)5 − 2  −3/2   1 × 1− (3x)5 − 2 5(3x)4 3 2   −3/2  15 = −6 1 − (3x)4 (3x)5 − 2 2   −1/2 −7 × x + (3x)5 − 2

36. The slope of y =

b and a

39. Slope of y = 1/(x 2 − x + 3)3/2 at x = −2 is

F ′ (x) = (2 + x)2 (3 + x)3 (4 + x)4 +

2

= 1024ab7 .

y = (2b)8 = 256b8 is  b 8 7 y = 256b +1024ab x − , or y = 210 ab7 x −3×28 b8 . a

4

F ′ (0) = (22 )(33 )(44 ) + 2(1)(2)(33 )(44 )+

x=b/a

b is a

The equation of the tangent line at x =

17 (12) = 102 2

3

37. Slope of y = (1+ x 2/3 )3/2  at x = −1 is √ 3 2 2/3 1/2 −1/3

d y dx

f (x) = √ 2x + 1 1 1 ′ f (4) = − =− 3/2 (2x + 1) x=4 27

y = (x 3 + 9)17/2 17 3 y ′ = (x + 9)15/2 3x 2 2

Thus, the equation of the tangent line at (2, 3) is y = 3 + 43 (x − 2), or y = 43 x + 31 .

38. The slope of y = (ax + b)8 at x =

1

x=−2

34.

SECTION 2.5 (PAGE 125)

1 + 2x 2 at x = 2 is

d y 4x 4 = = . √ 2 d x x=2 3 2 1 + 2x x=2

41.

x(x 4 + 2x 2 − 2)/(x 2 + 1)5/2

42. 4(7x 4 − 49x 2 + 54)/x 7 43. 857, 592 44. 5/8

45. The Chain Rule does not enable you to calculate the derivatives of |x|2 and |x 2 | at x = 0 directly as a composition of two functions, one of which is |x|, because |x| is not differentiable at x = 0. However, |x|2 = x 2 and |x 2 | = x 2 , so both functions are differentiable at x = 0 and have derivative 0 there.

46. It may happen that k = g(x + h) − g(x) = 0 for values

of h arbitrarily close to 0 so that the division by k in the “proof” is not justified.

55 Copyright © 2014 Pearson Canada Inc.

SECTION 2.5 (PAGE 125)

ADAMS and ESSEX: CALCULUS 8

Section 2.5 Derivatives of Trigonometric Functions (page 125)

23.

d d 1 cos x csc x = = − 2 = − csc x cot x dx d x sin x sin x

24.

1.

d d cos x − cos2 x − sin2 x cot x = = = −csc2 x dx d x sin x sin2 x

25.

2. 3.

y = cos 3x,

y ′ = −3 sin 3x

4.

x y = sin , 5

5.

y = tan π x,

6.

y = sec ax,

7.

y = cot(4 − 3x),

y′ =

1 x cos . 5 5

f (x) = cos(s − r x),

12. 13. 14.

29.

y ′ = 3 csc2 (4 − 3x)

9.

11.

28.

y ′ = a sec ax tan ax.

d π −x 1 π−x sin = − cos dx 3 3 3

10.

27.

y ′ = π sec2 π x

8.

y = sin(Ax + B),

f ′ (x) = r sin(s − r x)

′

26.

30.

y = A cos(Ax + B)

d sin(π x 2 ) = 2π x cos(π x 2 ) dx √ √ 1 d cos( x) = − √ sin( x) dx 2 x √ − sin x y = 1 + cos x, y ′ = √ 2 1 + cos x d sin(2 cos x) = cos(2 cos x)(−2 sin x) dx = −2 sin x cos(2 cos x)

15.

f (x) = cos(x + sin x) f ′ (x) = −(1 + cos x) sin(x + sin x)

16.

g(θ ) = tan(θ sin θ )

31. 32.

33.

34.

g ′ (θ ) = (sin θ + θ cos θ ) sec2 (θ sin θ )

17. u = sin3 (π x/2), 18. 19.

20.

21. 22.

y = sec(1/x),

u′ =

3π cos(π x/2) sin2 (π x/2) 2

y ′ = −(1/x 2 ) sec(1/x) tan(1/x)

1 sin 2at) 2 ′ F (t) = a cos at cos at − a sin at sin at ( = a cos 2at) F(t) = sin at cos at

(=

35. 36.

sin aθ cos bθ a cos bθ cos aθ + b sin aθ sin bθ G ′ (θ ) = . cos2 bθ  d  sin(2x) − cos(2x) = 2 cos(2x) + 2 sin(2x) dx d d (cos2 x − sin2 x) = cos(2x) dx dx = −2 sin(2x) = −4 sin x cos x G(θ ) =

d (tan x + cot x) = sec2 x − csc2 x dx d (sec x − csc x) = sec x tan x + csc x cot x dx d (tan x − x) = sec2 x − 1 = tan2 x dx d d tan(3x) cot(3x) = (1) = 0 dx dx d (t cos t − sin t) = cos t − t sin t − cos t = −t sin t dt d (t sin t + cos t) = sin t + t cos t − sin t = t cos t dt d sin x (1 + cos x)(cos x) − sin(x)(− sin x) = d x 1 + cos x (1 + cos x)2 cos x + 1 1 = = (1 + cos x)2 1 + cos x d cos x (1 + sin x)(− sin x) − cos(x)(cos x) = d x 1 + sin x (1 + sin x)2 −1 − sin x − 1 = = 2 (1 + sin x) 1 + sin x

d 2 x cos(3x) = 2x cos(3x) − 3x 2 sin(3x) dx p g(t) = (sin t)/t t cos t − sin t 1 × g ′ (t) = √ t2 2 (sin t)/t t cos t − sin t = √ 2t 3/2 sin t v = sec(x 2 ) tan(x 2 )

v ′ = 2x sec(x 2 ) tan2 (x 2 ) + 2x sec3 (x 2 ) √ sin x z= √ 1 + cos x √ √ √ √ √ √ (1 + cos x)(cos x/2 x) − (sin x)(− sin x/2 x) ′ z = √ (1 + cos x)2 √ 1 + cos x 1 = √ √ 2 = √ √ 2 x(1 + cos x) 2 x(1 + cos x)

d sin(cos(tan t)) = −(sec2 t)(sin(tan t)) cos(cos(tan t)) dt f (s) = cos(s + cos(s + cos s)) f ′ (s) = −[sin(s + cos(s + cos s))] × [1 − (sin(s + cos s))(1 − sin s)]

37. Differentiate both sides of sin(2x) = 2 sin x cos x and divide by 2 to get cos(2x) = cos2 x − sin2 x.

38. Differentiate both sides of cos(2x) = cos2 x − sin2 x and divide by −2 to get sin(2x) = 2 sin x cos x.

39. Slope of y = sin x at (π, 0) is cos π = −1. Therefore

the tangent and normal lines to y = sin x at (π, 0) have equations y = −(x − π ) and y = x − π , respectively.

56 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.5 (PAGE 125)

40. The slope of y = tan(2x) at (0, 0) is 2 sec2 (0) = 2.

41.

Therefore the tangent and normal lines to y = tan(2x) at (0, 0) have equations y = 2x and y = −x/2, respectively. √ The√slope of y = 2 cos(x/4) at (π, 1) is −( 2/4) sin(π/4) =√−1/4. Therefore the tangent and normal lines to y = 2 cos(x/4) at (π, 1) have equations y = 1 − (x − π )/4 and y = 1 + 4(x − π ), respectively.

42. The slope of y =√cos2 x at (π/3, 1/4) is

− sin(2π/3) = − 3/2. Therefore the tangent and normal lines to y = tan(2x) at (0, 0) have equations √ y = (1/4) − ( 3/2)(x − (π/3)) and √ y = (1/4) + (2/ 3)(x − (π/3)), respectively.  πx  = sin(x ◦ ) = sin is 180  πx  π ′ y = cos . At x = 45 the tangent line has 180 180 equation π 1 y= √ + √ (x − 45). 2 180 2  xπ  For y = sec (x ◦ ) = sec we have 180

43. Slope of y

44.

 xπ   xπ  π dy = sec tan . dx 180 180 180

√ π √ π 3 At x = 60 the slope is (2 3) = . 180 90 90 Thus, the normal line has slope − √ and has equation π 3 90 y = 2 − √ (x − 60). π 3

45. The slope of y = tan x at x = a is sec2 a. The tan-

gent there is parallel to y = 2x if sec2 a = 2, or √ cos a = ±1/ 2. The only solutions in (−π/2, π/2) are a = ±π/4. The corresponding points on the graph are (π/4, 1) and (−π/4, 1).

46. The slope of y = tan(2x) at x = a is 2 sec2 (2a). The

tangent there is normal to y = −x/8 if 2 sec2 (2a) = 8, or cos(2a) = ±1/2. The only solutions in (−π/4, π/4) are a = ±π/6. The corresponding points on the graph are √ √ (π/6, 3) and (−π/6, − 3).

47.

d sin x = cos x = 0 at odd multiples of π/2. dx d cos x = − sin x = 0 at multiples of π . dx d sec x = sec x tan x = 0 at multiples of π . dx d csc x = − csc x cot x = 0 at odd multiples of π/2. dx Thus each of these functions has horizontal tangents at infinitely many points on its graph.

48.

d tan x = sec2 x = 0 nowhere. dx d cot x = − csc2 x = 0 nowhere. dx Thus neither of these functions has a horizontal tangent.

49.

y = x + sin x has a horizontal tangent at x = π because d y/d x = 1 + cos x = 0 there.

50.

y = 2x + sin x has no horizontal tangents because d y/d x = 2 + cos x ≥ 1 everywhere.

51.

y = x + 2 sin x has horizontal tangents at x = 2π/3 and x = 4π/3 because d y/d x = 1 + 2 cos x = 0 at those points.

52.

y = x + 2 cos x has horizontal tangents at x = π/6 and x = 5π/6 because d y/d x = 1 − 2 sin x = 0 at those points.

53. 54. 55.

56.

57. 58.

lim

x→0

tan(2x) sin(2x) 2 = lim =1×2=2 x→0 x 2x cos(2x)

lim sec(1 + cos x) = sec(1 − 1) = sec 0 = 1

x→π

 x 2 cos x = 12 × 1 = 1 x→0 sin x

lim x 2 csc x cot x = lim

x→0

lim cos

x→0

lim

h→0



π − π cos2 x x2



= lim cos π x→0

 sin x 2 x

1 − cos h 2 sin2 (h/2) 1 = lim = lim 2 2 h→0 h→0 h h 2



= cos π = −1

sin(h/2) h/2

2

=

f will be differentiable at x = 0 if 2 sin 0 + 3 cos 0 = b, and d = a. (2 sin x + 3 cos x) dx x=0

Thus we need b = 3 and a = 2.

59. There are infinitely many lines through the origin that are tangent to y = cos x. The two with largest slope are shown in the figure. y

π

−π

2π

x

y = cos x Fig. 2.5.59

57 Copyright © 2014 Pearson Canada Inc.

1 2

SECTION 2.5 (PAGE 125)

ADAMS and ESSEX: CALCULUS 8

The tangent to y = cos x at x = a has equation y = cos a − (sin a)(x − a). This line passes through the origin if cos a = −a sin a. We use a calculator with a “solve” function to find solutions of this equation near a = −π and a = 2π as suggested in the figure. The solutions are a ≈ −2.798386 and a ≈ 6.121250. The slopes of the corresponding tangents are given by − sin a, so they are 0.336508 and 0.161228 to six decimal places.

3.

y ′′ = 36(x − 1)−4

y ′′′ = −144(x − 1)−5

4.

60. 1 61. 62.

√ − 2π + 3(2π 3/2 − 4π + 3)/π a) As suggested by the figure in the problem, the square of the length of chord AP is (1 − cos θ )2 + (0 − sin θ )2 , and the square of the length of arc AP is θ 2 . Hence

0 ≤ |1 − cos θ | < |θ |,

0 ≤ | sin θ | < |θ |.

′

7.

8.

h→0

lim sin(θ0 + h) = lim (sin θ0 cos h + cos θ0 sin h) = sin θ0 . h→0

9. This says that cosine and sine are continuous at any point θ0 .

Section 2.6 Higher-Order Derivatives (page 130) 11.

12.

y ′′ = 168(3 − 2x)5

y ′′′ = −1680(3 − 2x)4 1 x 1 ′ y = 2x + 2 x y = x2 −

x −1 x +1 2 y′ = (x + 1)2

y ′′ = 2 − y ′′′ =

6 x4

2 x3

y ′′ = 90x 8 + 112x 6

y ′′′ = 720x 7 + 672x 5

4 (x + 1)3 12 y ′′′ = (x + 1)4 y ′′ = −

y=

y = tan x

y ′′ = 2 sec2 x tan x

y ′′′ = 2 sec4 x + 4 sec2 x tan2 x

y = sec x y ′ = sec x tan x y = cos(x 2 )

y ′ = −2x sin(x 2 )

y = (3 − 2x)7

y ′ = −14(3 − 2x)6

2.

7

√ y = (x 2 + 3) x = x 5/2 + 3x 1/2 3 5 y ′ = x 3/2 + x −1/2 2 2 15 1/2 3 −3/2 ′′ x − x y = 4 4 15 −1/2 9 −5/2 ′′′ y = x + x 8 8

y ′ = sec2 x

10.

1.

9

y = 10x + 16x

lim cos(θ0 + h) = lim (cos θ0 cos h − sin θ0 sin h) = cos θ0 h→0

y ′′′

y = x 10 + 2x 8

From the first of these, limθ →0 cos θ = 1.

h→0

a2 4(ax + b)3/2 3a 3 = 8(ax + b)5/2

y ′′ = −

ax + b a y′ = √ 2 ax + b

6.

θ →0

b) Using the result of (a) and the addition formulas for cosine and sine we obtain

√

y = x 1/3 − x −1/3 1 1 y ′ = x −2/3 + x −4/3 3 3 2 −5/3 4 −7/3 ′′ y =− x − x 9 9 10 −8/3 28 −10/3 ′′′ y = x + x 27 27

Since limθ →0 |θ | = 0, the squeeze theorem implies that lim 1 − cos θ = 0, lim sin θ = 0. θ →0

y=

5.

(1 + cos θ )2 + sin2 θ < θ 2 , and, since squares cannot be negative, each term in the sum on the left is less than θ 2 . Therefore

6 = 6(x − 1)−2 (x − 1)2 y ′ = −12(x − 1)−3 y=

y ′′ = sec x tan2 x + sec3 x

y ′′′ = sec x tan3 x + 5 sec3 x tan x y ′′ = −2 sin(x 2 ) − 4x 2 cos(x 2 )

y ′′′ = −12x cos(x 2 ) + 8x 3 sin(x 2 )

sin x x cos x sin x ′ − 2 y = x x 2 ) sin x (2 − x 2 cos x y ′′ = − x3 x2 2 ) cos x 2 − 2) sin x (6 − x 3(x y ′′′ = + x3 x4 y=

58 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

13.

SECTION 2.6 (PAGE 130)

1 = x −1 x f ′ (x) = −x −2

Proof: Evidently, the above formula holds for n = 2, 3 and 4. Assume that it holds for n = k, i.e.

f (x) =

f ′′ (x) = 2x −3

f (k) (x) = (−1)k−1

f ′′′ (x) = −3!x −4

f (4) (x) = 4!x −5 Guess: f (n) (x) = (−1)n n!x −(n+1) (∗) Proof: (*) is valid for n = 1 (and 2, 3, 4). −(k+1) for some k ≥ 1 Assume f (k) (x) = (−1)k k!x   (k+1) k Then f (x) = (−1) k! −(k + 1) x −(k+1)−1

Then d (k) f (x) dx   −(2k − 1) −[(2k−1)/2]−1 k−1 1 · 3 · 5 · · · (2k − 3) = (−1) · x 2k 2 1 · 3 · 5 · · · (2k − 3)[2(k + 1) − 3] −[2(k+1)−1]/2 = (−1)(k+1)−1 x . 2k+1

f (k+1) (x) =

= (−1)k+1 (k + 1)!x −((k+1)+1) which is (*) for n = k + 1. Therefore, (*) holds for n = 1, 2, 3, . . . by induction.

14.

1 = x −2 x2 f ′ (x) = −2x −3 f ′′ (x) = −2(−3)x −4 = 3!x −4 f (3) (x) = −2(−3)(−4)x −5 = −4!x −5 Conjecture: f (x) =

f (n) (x) = (−1)n (n + 1)!x −(n+2)

Thus, the formula is also true for n = k + 1. Hence, it is true for n ≥ 2 by induction.

17.

for n = 1, 2, 3, . . .

f ′′′ (x) = −3!b3 (a + bx)−4 Guess: f (n) (x) = (−1)n n!bn (a + bx)−(n+1) (∗) Proof: (*) holds for n = 1, 2, 3 Assume (*) holds for n = k: f (k) (x) = (−1)k k!bk (a + bx)−(k+1) Then   f (k+1) (x) = (−1)k k!bk −(k + 1) (a + bx)−(k+1)−1 (b)

= (−1)k+1 (k + 2)!x −[(k+1)+2] .

Thus, the formula is also true for n = k + 1. Hence it is true for n = 1, 2, 3, . . . by induction. 1 = (2 − x)−1 2−x f ′ (x) = +(2 − x)−2 f (x) =

f ′′ (x) = 2(2 − x)−3

16.

= (−1)k+1 (k + 1)!bk+1 (a + bx)((k+1)+1) So (*) holds for n = k + 1 if it holds for n = k. Therefore, (*) holds for n = 1, 2, 3, 4, . . . by induction.

18.

f ′′′ (x) = +3!(2 − x)−4 Guess: f (n) (x) = n!(2 − x)−(n+1) (∗) Proof: (*) holds for n = 1, 2, 3. Assume f (k) (x) = k!(2 − x)−(k+1) (i.e., (*) holds for n = k)   Then f (k+1) (x) = k! −(k + 1)(2 − x)−(k+1)−1 (−1)

1 = (a + bx)−1 a + bx f ′ (x) = −b(a + bx)−2 f (x) =

f ′′ (x) = 2b2 (a + bx)−3

Proof: Evidently, the above formula holds for n = 1, 2 and 3. Assume it holds for n = k, i.e., f (k) (x) = (−1)k (k + 1)!x −(k+2) . Then d (k) f (x) f (k+1) (x) = dx = (−1)k (k + 1)![(−1)(k + 2)]x −(k+2)−1

15.

1 · 3 · 5 · · · (2k − 3) −(2k−1)/2 x . 2k

f (x) = x 2/3 f ′ (x) = 32 x −1/3 f ′′ (x) = 32 (− 13 )x −4/3 f ′′′ (x) = 32 (− 13 )(− 43 )x −7/3 Conjecture: 1 · 4 · 7 · · · · (3n − 5) −(3n−2)/3 f (n) (x) = 2(−1)n−1 x for 3n n ≥ 2. Proof: Evidently, the above formula holds for n = 2 and 3. Assume that it holds for n = k, i.e.

= (k + 1)!(2 − x)−((k+1)+1) . 1 · 4 · 7 · · · · (3k − 5) −(3k−2)/3 Thus (*) holds for n = k + 1 if it holds for k. f (k) (x) = 2(−1)k−1 x . Therefore, (*) holds for n = 1, 2, 3, . . . by induction. 3k √ f (x) = x = x 1/2 Then, f ′ (x) = 12 x −1/2 f ′′ (x) = 12 (− 21 )x −3/2 d (k) f (x) f (k+1) (x) = f ′′′ (x) = 12 (− 12 )(− 23 )x −5/2 dx   −(3k − 2) −[(3k−2)/3]−1 1 · 4 · 7 · · · · (3k − 5) f (4) (x) = 12 (− 12 )(− 32 )(− 25 )x −7/2 = 2(−1)k−1 · x Conjecture: 3k 3 1 · 3 · 5 · · · (2n − 3) −(2n−1)/2 (k+1)−1 1 · 4 · 7 · · · · (3k − 5)[3(k + 1) − 5] −[3(k+1)−2]/3 x . f (n) (x) = (−1)n−1 x (n ≥ 2). = 2(−1) 3( k + 1) 2n

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Thus, the formula is also true for n = k + 1. Hence, it is true for n ≥ 2 by induction.

19.

If x 6= 0 we have d sgn x = 0 dx

f (x) = cos(ax) f ′ (x) = −a sin(ax)

f ′′′ (x) = a 3 sin(ax)

f (4) (x) = a 4 cos(ax) = a 4 f (x) It follows that f (n) (x) = a 4 f (n−4) (x) for n ≥ 4, and if if if if

n n n n

f

= 4k = 4k + 1 (k = 0, 1, 2, . . .) = 4k + 2 = 4k + 3

f (x) = x cos x f ′ (x) = cos x − x sin x f ′′ (x) = −2 sin x − x cos x f ′′′ (x) = −3 cos x + x sin x

 n sin x + x cos x   n cos x − x sin x (n) f (x) =  −n sin x − x cos x  −n cos x + x sin x

if if if if

n n n n

= 4k = 4k + 1 = 4k + 2 = 4k + 3

Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula.

21.

f (x) = x sin(ax) f ′ (x) = sin(ax) + ax cos(ax)

f ′′ (x) = 2a cos(ax) − a 2 x sin(ax)

f ′′′ (x) = −3a 2 sin(ax) − a 3 x cos(ax)

f 4) (x) = −4a 3 cos(ax) + a 4 x sin(ax) This suggests the formula  −na n−1 cos(ax) + a n x sin(ax)   n−1 n (n) f (x) = na n−1 sin(ax) + a nx cos(ax)  cos(ax) − a x sin(ax)  na −na n−1 sin(ax) − a n x cos(ax)

22.

if if if if

n n n n

= 4k = 4k + 1 = 4k + 2 = 4k + 3

for k = 0, 1, 2, . . .. Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula. d 1 = |x|−1 . Recall that |x| = sgn x, so f (x) = |x| dx f ′ (x) = −|x|−2 sgn x.

(x) = −3!|x|−4 sgn x

The pattern suggests that f (n) (x) =

23.

f (4) (x) = 4 sin x + x cos x This suggests the formula (for k = 0, 1, 2, . . .)

f ′′ (x) = 2|x|−3 (sgn x)2 = 2|x|−3

(3)

f (4) (x) = 4!|x|−5 .

Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula.

20.

(sgn x)2 = 1.

Thus we can calculate successive derivatives of f using the product rule where necessary, but will get only one nonzero term in each case:

f ′′ (x) = −a 2 cos(ax)

 n a cos(ax)   n −a sin(ax) (n) f (x) = n cos(ax)   −a n a sin(ax)

and



−n!|x|−(n+1) sgn x n!|x|−(n+1)

if n is odd if n is even

Differentiating this formula leads to the same formula with n replaced by n + 1 so the formula is valid for all n ≥ 1 by induction. √ f (x) = 1 − 3x = (1 − 3x)1/2 1 f ′ (x) = (−3)(1 − 3x)−1/2 2  1 1 f ′′ (x) = − (−3)2 (1 − 3x)−3/2 2 2    1 3 1 − − (−3)3 (1 − 3x)−5/2 f ′′′ (x) = 2 2 2     1 3 5 1 − − − (−3)4 (1 − 3x)7/2 f (4) (x) = 2 2 2 2 1 × 3 × 5 × · · · × (2n − 3) n Guess: f (n) (x) = − 3 2n −(2n−1)/2 (1 − 3x) (∗) Proof: (*) is valid for n = 2, 3, 4, (but not n = 1) Assume (*) holds for n = k for some integer k ≥ 2 1 × 3 × 5 × . . . × (2k − 3) k 3 i.e., f (k) (x) = − 2k −(2k−1)/2 (1 − 3x) 1 × 3 × 5 × · · · × (2k − 3) k (k+1) Then f (x) = − 3 2k   2(k − 1) − (1 − 3x)−(2k−1)/2−1 (−3) 2   1 × 3 × 5 × · · · 2(k + 1) − 1 =− 3k+1 2k+1 (1 − 3x)−(2(k+1)−1)/2 Thus (*) holds for n = k + 1 if it holds for n = k. Therefore, (*) holds for n = 2, 3, 4, . . . by induction.

24. If y = tan(kx), then y ′ = k sec2 (kx) and y ′′ = 2k 2 sec2 (kx)tan(kx)

= 2k 2 (1 + tan2 (kx)) tan(kx) = 2k 2 y(1 + y 2 ).

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25. If y = sec(kx), then y ′ = k sec(kx) tan(kx) and

29.

d ( f g)′′ dx d = [ f ′′ g + 2 f ′ g ′ + f g ′′ ] dx = f (3) g + f ′′ g ′ + 2 f ′′ g ′ + 2 f ′ g ′′ + f ′ g ′′ + f g (3)

( f g)(3) =

y ′′ = k 2 (sec2 (kx) tan2 (kx) + sec3 (kx))

= k 2 y(2 sec2 (kx) − 1) = k 2 y(2y 2 − 1). ( f g)(4)

26. To be proved: if f (x) = sin(ax + b), then f (n) (x) =



(−1)k a n sin(ax + b) (−1)k a n cos(ax + b)

if n = 2k if n = 2k + 1

for k = 0, 1, 2, . . . Proof: The formula works for k = 0 (n = 2 × 0 = 0 and n = 2 × 0 + 1 = 1): 

f (0) (x) = f (x) = (−1)0 a 0 sin(ax + b) = sin(ax + b) f (1) (x) = f ′ (x) = (−1)0 a 1 cos(ax + b) = a cos(ax + b)

Now assume the formula holds for some k ≥ 0. If n = 2(k + 1), then f (n) (x) =

d (n−1) d (2k+1) f (x) = f (x) dx dx   d = (−1)k a 2k+1 cos(ax + b) dx = (−1)k+1 a 2k+2 sin(ax + b)

and if n = 2(k + 1) + 1 = 2k + 3, then  d (−1)k+1 a 2k+2 sin(ax + b) dx = (−1)k+1 a 2k+3 cos(ax + b).

f (n) (x) =

Thus the formula also holds for k + 1. Therefore it holds for all positive integers k by induction.

27. If y = tan x, then y ′ = sec2 x = 1 + tan2 x = 1 + y 2 = P2 (y), where P2 is a polynomial of degree 2. Assume that y (n) = Pn+1 (y) where Pn+1 is a polynomial of degree n + 1. The derivative of any polynomial is a polynomial of one lower degree, so y (n+1) =

dy d Pn+1 (y) = Pn (y) = Pn (y)(1+y 2 ) = Pn+2 (y), dx dx

a polynomial of degree n + 2. By induction, (d/d x)n tan x = Pn+1 (tan x), a polynomial of degree n + 1 in tan x.

28.

( f g)′′ = ( f ′ g + f g ′ ) = f ′′ g + f ′ g ′ + f ′ g ′ + f g ′′ = f ′′ g + 2 f ′ g ′ + f g ′′

= f (3) g + 3 f ′′ g ′ + 3 f ′ g ′′ + f g (3) . d = ( f g)(3) dx d = [ f (3) g + 3 f ′′ g ′ + 3 f ′ g ′′ + f g (3) ] dx = f (4) g + f (3) g ′ + 3 f (3) g ′ + 3 f ′′ g ′′ + 3 f ′′ g ′′ + 3 f ′ g (3) + f ′ g (3) + f g (4)

= f (4) g + 4 f (3) g ′ + 6 f ′′ g ′′ + 4 f ′ g (3) + f g (4) . n! f (n−2) g ′′ ( f g)(n) = f (n) g + n f (n − 1)g ′ + 2!(n − 2)! n! + f (n−3) g (3) + · · · + n f ′ g (n−1) + f g (n) 3!(n − 3)! n X n! f (n−k) g (k) . = k!(n − k)! k=0

Section 2.7 Using Differentials and Derivatives (page 136) 0.01 1 d x = − 2 = −0.0025. x2 2 If x = 2.01, then y ≈ 0.5 − 0.0025 = 0.4975.

1. 1y ≈ d y = −

3 dx

2. 1 f (x) ≈ d f (x) = √

=

2 3x + 1 f (1.08) ≈ f (1) + 0.06 = 2.06.

3 (0.08) = 0.06 4

πt π 1 1 π sin dt − (1) =− . 4 4 4 10π 40   1 1 1 h 2+ ≈ h(2) − =− . 10π 40 40

3. 1h(t) ≈ dh(t) = −

s  1 1 sec2 ds = (2)(−0.04) = −0.04. 4 4 4 If s = π − 0.06, then u ≈ 1 − 0.04 ≈ 0.96.

4. 1u ≈ du =

5. If y = x 2 , then 1y ≈ d y = 2x d x. If d x = (2/100)x, then 1y ≈ (4/100)x 2 = (4/100)y, so y increases by about 4%.

6. If y = 1/x, then 1y ≈ d y = (−1/x 2 ) d x. If

d x = (2/100)x, then 1y ≈ (−2/100)/x = (−2/100)y, so y decreases by about 2%.

7. If y = 1/x 2 , then 1y ≈ d y = (−2/x 3 ) d x. If

d x = (2/100)x, then 1y ≈ (−4/100)/x 2 = (−4/100)y, so y decreases by about 4%.

8. If y = x 3 , then 1y ≈ d y = 3x 2 d x. If d x = (2/100)x, then 1y ≈ (6/100)x 3 = (6/100)y, so y increases by about 6%.

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√

√ x, then d y ≈ d y = (1/2 √ x) d x. If 1x = (2/100)x, then 1y ≈ (1/100) x = (1/100)y, so y increases by about 1%.

9. If y =

21. Volume in tank is V (t) = 350(20 − t)2 L at t min. a) At t = 5, water volume is changing at rate d V = −10, 500. = −700(20 − t) dt t=5 t=5

10. If y = x −2/3 , then 1y ≈ d y = (−2/3)x −5/3 d x. If

d x = (2/100)x, then 1y ≈ (−4/300)x 2/3 = (−4/300)y, so y decreases by about 1.33%.

Water is draining out at 10,500 L/min at that time. At t = 15, water volume is changing at rate d V = −3, 500. = −700(20 − t) dt t=15 t=15

11. If V = 43 πr 3 , then 1V ≈ d V = 4πr 2 dr . If r increases by 2%, then dr = 2r/100 and 1V ≈ 8πr 3 /100. Therefore 1V /V ≈ 6/100. The volume increases by about 6%.

Water is draining out at 3,500 L/min at that time.

12. If V is the volume and x is the edge length of the cube then V = x 3 . Thus 1V ≈ d V = 3x 2 1x. If 1V = −(6/100)V , then −6x 3 /100 ≈ 3x 2 d x, so d x ≈ −(2/100)x. The edge of the cube decreases by about 2%.

b) Average rate of change between t = 5 and t = 15 is V (15) − V (5) 350 × (25 − 225) = = −7, 000. 15 − 5 10

13. Rate change of Area A with respect to side s, where dA A= is = 2s. When s = 4 ft, the area is changing ds 2 at rate 8 ft /ft. √ √ If A = s 2 , then s = A and ds/d A = 1/(2 A). If A = 16 m2 , then the side is changing at rate ds/d A = 1/8 m/m2 .

The average rate of draining is 7,000 L/min over that interval.

s 2,

14.

15. The diameter D and area A of a circle are related by √

D = 2 A/π . The rate of change of diameter with re√ spect to area is d D/d A = 1/(π A) units per square unit.

16. Since A = π D 2 /4, the rate of change of area with re-

17.

spect to diameter is d A/d D = π D/2 square units per unit. 4 Rate of change of V = πr 3 with respect to radius r is 3 dV 2 = 4πr . When r = 2 m, this rate of change is 16π dr m3 /m.

18. Let A be the area of a square, s be its side length and L be its diagonal. Then, L 2 = s 2 + s 2 = 2s 2 and dA = L. Thus, the rate of change of A = s 2 = 21 L 2 , so dL the area of a square with respect to its diagonal L is L.

19. If the radius of the circle is r then C = 2πr and A = πr 2 .

r

√ √ A = 2 π A. π Rate of √ change of C with respect to A is dC π 1 = √ = . dA r A

Thus C = 2π

F kr 4 = = 0.025r. 3 40kr 40kr 3 The flow rate will increase by 10% if the radius is increased by about 2.5%. 1r ≈

23.

F = k/r 2 implies that d F/dr = −2k/r 3 . Since d F/dr = 1 pound/mi when r = 4, 000 mi, we have 2k = 4, 0003 . If r = 8, 000, we have d F/dr = −(4, 000/8, 000)3 = −1/8. At r = 8, 000 mi F decreases with respect to r at a rate of 1/8 pounds/mi.

24. If price = $ p, then revenue is $R = 4, 000 p − 10 p2 . a) Sensitivity of R to p is d R/d p = 4, 000 − 20 p. If p = 100, 200, and 300, this sensitivity is 2,000 $/$, 0 $/$, and −2, 000 $/$ respectively. b) The distributor should charge $200. This maximizes the revenue.

25. Cost is $C(x) = 8, 000 + 400x − 0.5x 2 if x units are manufactured.

a) Marginal cost if x = 100 is C ′ (100) = 400 − 100 = $300. b) C(101) − C(100) = 43, 299.50 − 43, 000 = $299.50 which is approximately C ′ (100). where P(x) = 8x − 0.005x 2 − 1, 000.

ds Then V = ⇒s = and = 13 V −2/3 . Hence, dV the rate of change of the side length of a cube with respect to its volume V is 31 V −2/3 . V 1/3

then

26. Daily profit if production is x sheets per day is $P(x)

20. Let s be the side length and V be the volume of a cube. s3

22. Flow rate F = kr 4 , so 1F ≈ 4kr 3 1r . If 1F = F/10,

a) Marginal profit P ′ (x) = 8 − 0.01x. This is positive if x < 800 and negative if x > 800. b) To maximize daily profit, production should be 800 sheets/day.

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80, 000 n2 + 4n + n 100 dC 80, 000 n =− + 4 + . dn n2 50 dC = −2. Thus, the marginal cost of (a) n = 100, dn production is −$2.

27. C =

dC 82 = ≈ 9.11. Thus, the marginal cost dn 9 of production is approximately $9.11.

(b) n = 300,

28. Daily profit P = 13x − C x = 13x − 10x − 20 −

x2 1000

x2 1000 Graph of P is a parabola opening downward. P will be maximum where the slope is zero: = 3x − 20 −

0=

dP 2x =3− so x = 1500 dx 1000

Should extract 1500 tonnes of ore per day to maximize profit.

29. One of the components comprising C(x) is usually a fixed cost, $S, for setting up the manufacturing operation. On a per item basis, this fixed cost $S/x, decreases as the number x of items produced increases, especially when x is small. However, for large x other components of the total cost may increase on a per unit basis, for instance labour costs when overtime is required or maintenance costs for machinery when it is over used. C(x) Let the average cost be A(x) = . The minimal avx erage cost occurs at point where the graph of A(x) has a horizontal tangent: xC ′ (x) − C(x) dA = 0= . dx x2 C(x) = A(x). x ′ Thus the marginal cost C (x) equals the average cost at the minimizing value of x. Hence, xC ′ (x) − C(x) = 0 ⇒ C ′ (x) =

30. If y = C p−r , then the elasticity of y is −

p dy p = − −r (−r )C p−r−1 = r. y dp Cp

Section 2.8 The Mean-Value Theorem (page 143) 1.

2

f (x) = x ,

′

f (x) = 2x

b2 − a 2 f (b) − f (a) = b+a = b−a b−a b+a ′ = f (c) = 2c ⇒ c = 2

2. If f (x) =

1 1 , and f ′ (x) = − 2 then x x

f (2) − f (1) 1 1 1 = − 1 = − = − 2 = f ′ (c) 2−1 2 2 c where c =

3.

√ 2 lies between 1 and 2.

f (x) = x 3 − 3x + 1, f ′ (x) = 3x 2 − 3, a = −2, b = 2 f (b) − f (a) f (2) − f (−2) = b−a 4 8 − 6 + 1 − (−8 + 6 + 1) = 4 4 = =1 4 ′ 2 f (c) = 3c − 3 2 3c2 − 3 = 1 ⇒ 3c2 = 4 ⇒ c = ± √ 3 (Both points will be in (−2, 2).)

4. If f (x) = cos x + (x 2 /2), then f ′ (x) = x − sin x > 0

for x > 0. By the MVT, if x > 0, then f (x) − f (0) = f ′ (c)(x − 0) for some c > 0, so f (x) > f (0) = 1. Thus cos x + (x 2 /2) > 1 and cos x > 1 − (x 2 /2) for x > 0. Since both sides of the inequality are even functions, it must hold for x < 0 as well.

5. Let f (x) = tan x. If 0 < x < π/2, then by the MVT f (x) − f (0) = f ′ (c)(x − 0) for some c in (0, π/2). Thus tan x = x sec2 c > x, since secc > 1.

6. Let f (x) = (1 + x)r − 1 − r x where r > 1.

Then f ′ (x) = r (1 + x)r−1 − r . If −1 ≤ x < 0 then f ′ (x) < 0; if x > 0, then f ′ (x) > 0. Thus f (x) > f (0) = 0 if −1 ≤ x < 0 or x > 0. Thus (1 + x)r > 1 + r x if −1 ≤ x < 0 or x > 0.

7. Let f (x) = (1 + x)r where 0 < r < 1. Thus,

f ′ (x) = r (1 + x)r−1 . By the Mean-Value Theorem, for x ≥ −1, and x 6= 0, f (x) − f (0) = f ′ (c) x −0 (1 − x)r − 1 ⇒ = r (1 + c)r−1 x

for some c between 0 and x. Thus, (1 + x)r = 1 + r x(1 + c)r−1 . If −1 ≤ x < 0, then c < 0 and 0 < 1 + c < 1. Hence (1 + c)r−1 > 1 r−1

r x(1 + c)

< rx

(since r − 1 < 0), (since x < 0).

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Hence, (1 + x)r < 1 + r x. If x > 0, then

17.

f (x) = x 3 (5 − x)2

f ′ (x) = 3x 2 (5 − x)2 + 2x 3 (5 − x)(−1)

c>0 1+c >1

= x 2 (5 − x)(15 − 5x)

= 5x 2 (5 − x)(3 − x) > 0 if x < 0, 0 < x < 3, or x > 5 f ′ (x) < 0 if 3 < x < 5 f is increasing on (−∞, 3) and (5, ∞). f is decreasing on (3, 5).

(1 + c)r−1 < 1

f ′ (x)

r x(1 + c)r−1 < r x. Hence, (1 + x)r < 1 + r x in this case also. Hence, (1 + x)r < 1 + r x for either −1 ≤ x < 0 or x > 0.

8. If f (x) = x 3 − 12x + 1, then f ′ (x) = 3(x 2 − 4).

18. If f (x) = x − 2 sin x, then f ′ (x) = 1 − 2 cos x = 0 at

x = ±π/3 + 2nπ for n = 0, ±1, ±2, . . .. f is decreasing on (−π/3 + 2nπ, π + 2nπ ). f is increasing on (π/3 + 2nπ, −π/3 + 2(n + 1)π ) for integers n.

The critical points of f are x = ±2. f is increasing on (−∞, −2) and (2, ∞) where f ′ (x) > 0, and is decreasing on (−2, 2) where f ′ (x) < 0.

9. If f (x) = x 2 − 4, then f ′ (x) = 2x. The critical point of

f is x = 0. f is increasing on (0, ∞) and decreasing on (−∞, 0).

19. If f (x) = x + sin x, then f ′ (x) = 1 + cos x ≥ 0

f ′ (x) = 0 only at isolated points x = ±π, ±3π, .... Hence f is increasing everywhere.

10. If y = 1 − x − x 5 , then y ′ = −1 − 5x 4 < 0 for all x. Thus

20. If f (x) = x + 2 sin x, then f ′ (x) = 1 + 2 cos x > 0

11. If y = x 3 + 6x 2 , then y ′ = 3x 2 + 12x = 3x(x + 4).

21.

12. If f (x) = x 2 + 2x + 2 then f ′ (x) = 2x + 2 = 2(x + 1).

22. There is no guarantee that the MVT applications for f

y has no critical points and is decreasing on the whole real line.

if cos x > −1/2. Thus f is increasing on the intervals (−(4π/3) + 2nπ, (4π/3) + 2nπ ) where n is any integer.

The critical points of y are x = 0 and x = −4. y is increasing on (−∞, −4) and (0, ∞) where y ′ > 0, and is decreasing on (−4, 0) where y ′ < 0.

Evidently, f ′ (x) > 0 if x > −1 and f ′ (x) < 0 if x < −1. Therefore, f is increasing on (−1, ∞) and decreasing on (−∞, −1).

13.

f (x) = x 3 − 4x + 1 f ′ (x) = 3x 2 − 4

and g yield the same c.

23. CPs x = 0.535898 and x = 7.464102 24. CPs x = −1.366025 and x = 0.366025 25. CPs x = −0.518784 and x = 0

2 f ′ (x) > 0 if |x| > √ 3 2 ′ f (x) < 0 if |x| < √ 3

26. CP x = 0.521350 27. If x1 < x2 < . . . < xn belong to I , and f (xi ) = 0,

2 2 f is increasing on (−∞, − √ ) and ( √ , ∞). 3 3 2 2 f is decreasing on (− √ , √ ). 3 3

14. If f (x) = x 3 + 4x + 1, then f ′ (x) = 3x 2 + 4. Since

f ′ (x) > 0 for all real x, hence f (x) is increasing on the whole real line, i.e., on (−∞, ∞).

15.

(1 ≤ i ≤ n), then there exists yi in (xi , xi+1 ) such that f ′ (yi ) = 0, (1 ≤ i ≤ n − 1) by MVT.

28. For x 6= 0, we have f ′ (x) = 2x sin(1/x) − cos(1/x) which has no limit as x → 0. However, f ′ (0) = limh→0 f (h)/ h = limh→0 h sin(1/ h) = 0 does exist even though f ′ cannot be continuous at 0.

29. If f ′ exists on [a, b] and f ′ (a) 6= f ′ (b), let us assume,

f (x) = (x 2 − 4)2 f ′ (x) = 2x2(x 2 − 4) = 4x(x − 2)(x + 2) f ′ (x) > 0 if x > 2 or −2 < x < 0 f ′ (x) < 0 if x < −2 or 0 < x < 2 f is increasing on (−2, 0) and (2, ∞). f is decreasing on (−∞, −2) and (0, 2). 1 −2x then f ′ (x) = 2 . Evidently, +1 (x + 1)2 ′ ′ f (x) > 0 if x < 0 and f (x) < 0 if x > 0. Therefore, f is increasing on (−∞, 0) and decreasing on (0, ∞).

16. If f (x) =

f (x) = x 3 is increasing on (−∞, 0) and (0, ∞) because f ′ (x) = 3x 2 > 0 there. But f (x1 ) < f (0) = 0 < f (x2 ) whenever x1 < 0 < x2 , so f is also increasing on intervals containing the origin.

x2

30.

without loss of generality, that f ′ (a) > k > f ′ (b). If g(x) = f (x) − kx on [a, b], then g is continuous on [a, b] because f , having a derivative, must be continuous there. By the Max-Min Theorem, g must have a maximum value (and a minimum value) on that interval. Suppose the maximum value occurs at c. Since g ′ (a) > 0 we must have c > a; since g ′ (b) < 0 we must have c < b. By Theorem 14, we must have g ′ (c) = 0 and so f ′ (c) = k. Thus f ′ takes on the (arbitrary) intermediate value k.  2 f (x) = x + 2x sin(1/x) if x 6= 0 0 if x = 0.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 2.9 (PAGE 148)

f (0 + h) − f (0) h→0 h h + 2h 2 sin(1/ h) = lim h→0 h = lim (1 + 2h sin(1/ h) = 1,

Then, by MVT applied to f ′ on the interval [r, s], for some b in (r, s),

a) f ′ (0) = lim

f ′ (s) − f ′ (r ) 1−0 = s −r s −r 1 1 > = s −r 2

f ′′ (b) =

h→0

because |2h sin(1/ h)| ≤ 2|h| → 0 as h → 0. b) For x 6= 0, we have

since s − r < 2.

f ′ (x) = 1 + 4x sin(1/x) − 2 cos(1/x).

c) Since f ′′ (x) exists on [0, 2], therefore f ′ (x) is continuous there. Since f ′ (r ) = 0 and f ′ (s) = 1, and since 0 < 17 < 1, the Intermediate-Value Theorem assures us that f ′ (c) = 71 for some c between r and s.

There are numbers x arbitrarily close to 0 where f ′ (x) = −1; namely, the numbers x = ±1/(2nπ ), where n = 1, 2, 3, . . . . Since f ′ (x) is continuous at every x 6= 0, it is negative in a small interval about every such number. Thus f cannot be increasing on any interval containing x = 0.

Section 2.9 (page 148)

31. Let a, b, and c be three points in I where f vanishes; that is, f (a) = f (b) = f (c) = 0. Suppose a < b < c. By the Mean-Value Theorem, there exist points r in (a, b) and s in (b, c) such that f ′ (r ) = f ′ (s) = 0. By the Mean-Value Theorem applied to f ′ on [r, s], there is some point t in (r, s) (and therefore in I ) such that f ′′ (t) = 0.

32. If f (n) exists on interval I and f vanishes at n + 1 dis-

tinct points of I , then f (n) vanishes at at least one point of I . Proof: True for n = 2 by Exercise 8. Assume true for n = k. (Induction hypothesis) Suppose n = k + 1, i.e., f vanishes at k + 2 points of I and f (k+1) exists. By Exercise 7, f ′ vanishes at k + 1 points of I . By the induction hypothesis, f (k+1) = ( f ′ )(k) vanishes at a point of I so the statement is true for n = k + 1. Therefore the statement is true for all n ≥ 2 by induction. (case n = 1 is just MVT.)

33. Given that f (0) = f (1) = 0 and f (2) = 1: a) By MVT, f ′ (a) =

1.

x y − x + 2y = 1 Differentiate with respect to x: y + x y ′ − 1 + 2y ′ = 0 1− y Thus y ′ = 2+x

2.

x 3 + y3 = 1 3x 2 + 3y 2 y ′ = 0, so y ′ = −

x 2 + x y = y3 Differentiate with respect to x: 2x + y + x y ′ = 3y 2 y ′ 2x + y y′ = 2 3y − x

4.

x 3 y + x y5 = 2 3x 2 y + x 3 y ′ + y 5 + 5x y 4 y ′ = 0 −3x 2 y − y 5 y′ = 3 x + 5x y 4

5.

x 2 y 3 = 2x − y 2x y 3 + 3x 2 y 2 y ′ = 2 − y ′ 2 − 2x y 3 y′ = 2 2 3x y + 1

6.

x 2 + 4(y − 1)2 = 4 2x + 8(y − 1)y ′ = 0, so y ′ =

b) By MVT, for some r in (0, 1), f ′ (r ) =

f (1) − f (0) 0−0 = = 0. 1−0 1−0

Also, for some s in (1, 2), f ′ (s) =

f (2) − f (1) 1−0 = = 1. 2−1 2−1

x2 . y2

3.

1−0 1 f (2) − f (0) = = 2−0 2−0 2

for some a in (0, 2).

Implicit Differentiation

7.

x 4(1 − y)

x−y x2 x2 + y = +1 = x+y y y Thus x y − y 2 = x 3 + x 2 y + x y + y 2, or x 3 + x 2 y + 2y 2 = 0 Differentiate with respect to x: 3x 2 + 2x y + x 2 y ′ + 4yy ′ = 0 3x 2 + 2x y y′ = − 2 x + 4y

65 Copyright © 2014 Pearson Canada Inc.

SECTION 2.9 (PAGE 148)

8.

ADAMS and ESSEX: CALCULUS 8

√ x x + y = 8 − xy 1 √ x+y+x √ (1 + y ′ ) = −y − x y ′ 2 x+y √ 2(x + y) + x(1 + y ′ ) √ = −2 x + y(y + x y ′ ) 3x + 2y + 2y x + y y′ = − √ x + 2x x + y

15.

16.

9. 2x 2 + 3y 2 = 5 4x + 6yy ′ = 0

At (1, 1): 4 + 6y ′ = 0, y ′ = −

2 3

2 Tangent line: y − 1 = − (x − 1) or 2x + 3y = 5 3

10.

11.

12.

x 2 y 3 − x 3 y 2 = 12 2x y 3 + 3x 2 y 2 y ′ − 3x 2 y 2 − 2x 3 yy ′ = 0 At (−1, 2): −16 + 12y ′ − 12 + 4y ′ = 0, so the slope is 12 + 16 28 7 y′ = = = . 12 + 4 16 4 Thus, the equation of the tangent line is y = 2 + 74 (x + 1), or 7x − 4y + 15 = 0. x  y 3 =2 + y x 4 4 x + y = 2x 3 y 4x 3 + 4y 3 y ′ = 6x 2 y + 2x 3 y ′ at (−1, −1): −4 − 4y ′ = −6 − 2y ′ 2y ′ = 2, y ′ = 1 Tangent line: y + 1 = 1(x + 1) or y = x.

17.

y2 x −1 (x − 1)2yy ′ − y 2 (1) ′ 1 + 2y = (x − 1)2 At (2, −1) we have 1 + 2y ′ = −2y ′ − 1 so y ′ = − 12 . Thus, the equation of the tangent is y = −1 − 12 (x − 2), or x + 2y = 0.

19.

14. tan(x y 2 ) = (2/π )x y

(sec2 (x y 2 ))(y 2 + 2x yy ′ ) = (2/π )(y + x y ′ ). At (−π, 1/2): 2((1/4) − π y ′ ) = (1/π ) − 2y ′ , so y ′ = (π − 2)/(4π(π − 1)). The tangent has equation y=

1 π −2 + (x + π ). 2 4π(π − 1)

2 + 8(y ′ )2 + 8yy ′′ = 0.

−2 − 8(y ′ )2 1 x2 −4y 2 − x 2 1 =− − = = − 3. 3 8y 4y 16y 16y 3 4y

x 3 − y2 + y3 = x

1 − 3x 2 3y 2 − 2y ′ 2 ′′ ′ 2 6x − 2(y ) − 2yy + 6y(y ) + 3y 2 y ′′ = 0 (1 − 3x 2 )2 (2 − 6y) − 6x ′ 2 (2 − 6y)(y ) − 6x (3y 2 − 2y)2 y ′′ = = 3y 2 − 2y 3y 2 − 2y 2 2 (2 − 6y)(1 − 3x ) 6x = − 2 (3y 2 − 2y)3 3y − 2y

3x 2 − 2yy ′ + 3y 2 y ′ = 1 ⇒ y ′ =

√

π 4  x− . 4−π 4

x 2 + 4y 2 = 4, 2x + 8yy ′ = 0, −x and Thus, y ′ = 4y y ′′ =

y′

y =1−

y−1 1−x

y ′ + y ′ + x y ′′ = y ′′ 2y ′ 2(y − 1) Therefore, y ′′ = = 1−x (1 − x)2

13. 2x + y −√ 2 sin(x y) = π/2

2 + − 2 cos(x y)(y + x y ′ ) = 0 At (π/4, 1): 2 + y ′ − (1 + (π/4)y ′ ) = 0, so y ′ = −4/(4 − π ). The tangent has equation

xy = x + y y + x y′ = 1 + y′ ⇒ y′ =

18.

x + 2y + 1 =

x sin(x y − y 2 ) = x 2 − 1 sin(x y − y 2 ) + x(cos(x y − y 2 ))(y + x y ′ − 2yy ′) = 2x. At (1, 1): 0+(1)(1)(1− y ′ ) = 2, so y ′ = −1. The tangent has equation y = 1 − (x − 1), or y = 2 − x.  π y  x2 17 cos − = x y 2 h  π y i π(x y ′ − y) 2x y − x 2 y ′ = . − sin 2 x √ x y2 3 π(3y ′ − 1) At (3, 1): − = 6 − 9y ′ , 2 √ 9 √ so y ′ = (108 − 3π )/(162 − 3 3π ). The tangent has equation √ 108 − 3π y =1+ √ (x − 3). 162 − 3 3π

20.

x 3 − 3x y + y 3 = 1 3x 2 − 3y − 3x y ′ + 3y 2 y ′ = 0 6x − 3y ′ − 3y ′ − 3x y ′′ + 6y(y ′ )2 + 3y 2 y ′′ = 0 Thus

y − x2 y2 − x −2x + 2y ′ − 2y(y ′ )2 y ′′ = y2 − x     2  2 y − x2 y − x2 = 2 −x + − y y −x y2 − x y2 − x   2 −2x y 4x y = 2 = . y − x (y 2 − x)2 (x − y 2 )3 y′ =

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INSTRUCTOR’S SOLUTIONS MANUAL

21.

22.

SECTION 2.9 (PAGE 148)

x 2 + y 2 = a2 x 2x + 2yy ′ = 0 so x + yy ′ = 0 and y ′ = − y 1 + y ′ y ′ + yy ′′ = 0 so x2 1+ 2 ′ )2 1 + (y y =− y ′′ = − y y y2 + x 2 a2 =− =− 3 3 y y Ax 2 + By 2 = C

2 Ax + 2Byy ′ = 0 ⇒ y ′ = −

Similarly, the slope of the hyperbola (x, y) satisfies 2x 2y − 2 y ′ = 0, 2 A B

− A − B(y ′ )2 = By

Ax By

x2 b2 + B 2 A2 a 2 = · . y2 B 2 b2 a 2 − A2 2 2 2 2 Since a − b = A + B , therefore B 2 + b2 = a 2 − A2 , x2 A2 a 2 and 2 = 2 2 . Thus, the product of the slope of the y B b two curves at (x, y) is

−A − B



Ax By

2

By AC − A(By 2 + Ax 2 ) = − 2 3. = 2 3 B y B y

−

29. If z = tan(x/2), then 1 = sec2 (x/2)

25. Maple gives the value −26.

27.

855, 000 . 371, 293

2 −1 sec2 (x/2) 2 1 − z2 = − 1 = 1 + z2 1 + z2 2 tan(x/2) 2z sin x = 2 sin(x/2) cos(x/2) = = . 1 + z2 1 + tan2 (x/2)

cos x = 2 cos2 (x/2) − 1 =

x 2y Hyperbola: 2x 2 − 2y 2 = 1 4x − 4yy ′ = 0 ′ = x Slope of hyperbola: y H y  x 2 + 2y 2 = 2 At intersection points 2x 2 − 2y 2 = 1 1 3x 2 = 3 so x 2 = 1, y 2 = 2 x x x2 ′ ′ Thus y E y H = − = − 2 = −1 2y y 2y Therefore the curves intersect at right angles. Slope of ellipse: y E′ = −

x2 y2 + 2 = 1 is found from 2 a b

2x 2y + 2 y ′ = 0, a2 b

1 dx 1 + tan2 (x/2) d x 1 + z2 d x = = . 2 dz 2 dz 2 dz

Thus d x/dz = 2/(1 + z 2 ). Also

Ellipse: x 2 + 2y 2 = 2 2x + 4yy ′ = 0

28. The slope of the ellipse

b 2 B 2 A2 a 2 b2 x B 2 x · 2 = − 2 2 · 2 2 = −1. 2 a y A y a A B b

Therefore, the curves intersect at right angles.

206 . Maple gives the slope as 55

26. Maple gives the value −

B2x . A2 y

Thus,

23. Maple gives 0 for the value. 24.

or y ′ =

If the point (x, y) is an intersection of the two curves, then x2 y2 x2 y2 + 2 = 2 − 2 2 b A B  a  1 1 1 1 2 2 − + x = y . A2 a2 B2 b2

2 A + 2B(y ′ )2 + 2Byy ′′ = 0. Thus,

y ′′ =

x2 y2 − 2 = 1 at 2 A B

b2 x i.e. y = − 2 . a y ′

30.

x−y x = + 1 ⇔ x y − y2 = x 2 + x y + x y + y2 x+y y ⇔ x 2 + 2y 2 + x y = 0 Differentiate with respect to x: 2x + 4yy ′ + y + x y ′ = 0

⇒

y′ = −

2x + y . 4y + x

However, since x 2 + 2y 2 + x y = 0 can be written x + xy +

1 2 7 2 y 7 y + y = 0, or (x + )2 + y 2 = 0, 4 4 2 4

the only solution is x = 0, y = 0, and these values do not satisfy the original equation. There are no points on the given curve.

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SECTION 2.9 (PAGE 148)

ADAMS and ESSEX: CALCULUS 8

Section 2.10 Antiderivatives and Initial-Value Problems (page 154) 1.

Z

2.

Z

3.

Z

√

4.

Z

x 12 d x =

5.

Z

x3 dx =

6.

Z

7.

5 d x = 5x + C 2

x dx =

1 3 3x

21.

+C

2 x d x = x 3/2 + C 3 1 13 13 x

+C

1 4 x +C 4

23.

x2 (x + cos x) d x = + sin x + C 2 Z Z tan x cos x d x = sin x d x = − cos x + C 1 + cos3 x dx = cos2 x

Z

9.

Z

(a 2 − x 2 ) d x = a 2 x −

10.

Z

(A + Bx + C x 2 ) d x = Ax +

11.

Z

13. 14.

Z

(sec2 x+cos x) d x = tan x+sin x+C

B 2 C 3 x + x +K 2 3

16. 17. 18. 19.

√

2x + 3 d x =

2x sin(x 2 ) d x = − cos(x 2 ) + C

d p 2 x x +1= √ , therefore 2 dx x +1 Z p 2x √ d x = 2 x 2 + 1 + C. x2 + 1 Z Z tan2 x d x = (sec2 x − 1) d x = tan x − x + C

26.

28.

1 (2x + 3)3/2 + C 3

29.

Z

Z

1 1 sin(2x) d x = − cos(2x) + C 2 4 Z Z 1 + cos(2x) x sin(2x) cos2 x d x = dx = + +C 2 2 4 Z Z 1 − cos(2x) x sin(2x) sin2 x d x = dx = − +C 2 2 4 ( 1 y ′ = x − 2 ⇒ y = x 2 − 2x + C 2 y(0) = 3 ⇒ 3 = 0 + C therefore C = 3 1 Thus y = x 2 − 2x + 3 for all x. 2 Given that  y ′ = x −2 − x −3 y(−1) = 0, Z then y = (x −2 − x −3 ) d x = −x −1 + 21 x −2 + C sin x cos x d x =

and 0 = y(−1) = −(−1)−1 + 12 (−1)−2 + C so C = − 23 . 1 1 3 Hence, y(x) = − + 2 − which is valid on the x 2x 2 interval (−∞, 0).  √ y ′ = 3 x ⇒ y = 2x 3/2 + C y(4) = 1 ⇒ 1 = 16 + C so C = −15 Thus y = 2x 3/2 − 15 for x > 0.

30. Given that

= 105t + 35t 3 + 21t 5 + 15t 7 + C Z 1 cos(2x) d x = sin(2x) + C 2 Z x  x  sin d x = −2 cos +C 2 2 Z dx 1 =− +C (1 + x)2 1+x Z sec(1 − x) tan(1 − x) d x = − sec(1 − x) + C Z

25.

1/3

= 105(t + 31 t 3 + 51 t 5 + 71 t 7 ) + C

15.

24.

27.

1 3 x +C 3

4 9 (2x + 3x d x = x 3/2 + x 4/3 + C 3 4 Z Z 6(x − 1) d x = (6x −1/3 − 6x −4/3 ) d x x 4/3 = 9x 2/3 + 18x −1/3 + C  Z  3 x2 1 4 1 3 1 2 x − + x − 1 dx = x − x + x −x +C 3 2 12 6 2 Z 105 (1 + t 2 + t 4 + t 6 ) dt 1/2

Z

d √ 1 , therefore x +1= √ dx 2 x +1 Z √ 4 d x = 8 x + 1 + C. √ x +1

22. Since

8.

12.

20. Since

then y =

Z

 x 1/3 d x =

Hence, y(x) = line.

y ′ = x 1/3 y(0) = 5,

3 4/3 4x

3 4/3 +5 4x

+ C and 5 = y(0) = C.

which is valid on the whole real

31. Since y ′ = Ax 2 + Bx + C we have

A 3 B 2 x + x + C x + D. Since y(1) = 1, therefore 3 2 A B A B 1 = y(1) = + + C + D. Thus D = 1 − − − C, 3 2 3 2 and A B y = (x 3 − 1) + (x 2 − 1) + C(x − 1) + 1 for all x 3 2 y=

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INSTRUCTOR’S SOLUTIONS MANUAL

32. Given that

then y =

33.



Z

SECTION 2.10 (PAGE 154)

then y ′ =

y ′ = x −9/7 y(1) = −4,

Also, −4 = y(1) = − 72 + C, so C = − 12 . Hence, y = − 27 x −2/7 − 12 , which is valid in the interval (0, ∞).  ′ y = cos x For , we have y(π/6) = 2 Z

1 π + C = + C H⇒ 6 2 3 (for all x). y = sin x + 2



34. For

C=

3 2

Z

1 sin(2x) d x = − cos(2x) + C 2 1 1 1 1 = − cos π + C = + C H⇒ C = 2 2 2  1 y= 1 − cos(2x) (for all x). 2

36.



y′

1 4 Since y ′ (0) = 0, therefore 0 = 0 − 0 + C1 , and 1 y ′ = x 4 − x. 4 1 5 1 2 Thus y = x − x + C2 . 20 2 Since y(0) = 8, we have 8 = 0 − 0 + C2 . 1 5 1 2 Hence y = x − x + 8 for all x. 20 2

39. Since y ′′ = x 3 − 1, therefore y ′ = x 4 − x + C1 .

40. Given that

we have y ′ =

sec2

= x , we have y(0) = 1 Z y = sec2 x d x = tan x + C

For

y=

y′ =

1 = tan π + C = C H⇒ C = 1 y = tan x + 1 (for π/2 < x < 3π/2).

Since y ′ (0) = 5, therefore 5 = 0 + C1 , and y ′ = 2x + 5. Thus y = x 2 + 5x + C2 . Since y(0) = −3, therefore −3 = 0 + 0 + C2 , and C2 = −3. Finally, y = x 2 + 5x − 3, for all x.

38. Given that

y=

  y ′′ = x −4 y ′ (1) = 2  y(1) = 1,

5x 2 − 3x −1/2 d x =

5 3 −6+ 5 3 1/2 + 19 , and 3 x − 6x 3

Z 

1/2 5 3 3 x − 6x

+

19 3



5 3 3x

− 6x 1/2 + C.

C so that C =

dx =

5 4 12 x

19 . Thus, 3

− 4x 3/2 +

19 3 x

+ D.

5 11 Finally, 0 = y(1) = 12 − 4 + 19 3 + D so that D = − 4 . 5 4 19 11 Hence, y(x) = 12 x − 4x 3/2 + 3 x − 4 .

sec2 x d x = tan x + C

37. Since y ′′ = 2, therefore y ′ = 2x + C1 .

Z

  y ′′ = 5x 2 − 3x −1/2 y ′ (1) = 2  y(1) = 0,

Also, 2 = y ′ (1) =

1 = tan 0 + C = C H⇒ C = 1 y = tan x + 1 (for −π/2 < x < π/2).  ′ 2 y = sec x For , we have y(π ) = 1 Z

Z   − 13 x −3 + 37 d x = 16 x −2 + 73 x + D,

and 1 = y(1) = 16 + 73 + D, so that D = − 23 . Hence, y(x) = 61 x −2 + 37 x − 32 , which is valid in the interval (0, ∞).

y ′ = sin(2x) , we have y(π/2) = 1 y=

35.

y=

cos x d x = sin x + C

2 = sin

x −4 d x = − 31 x −3 + C.

Since 2 = y ′ (1) = − 13 + C, therefore C = 37 , and y ′ = − 31 x −3 + 73 . Thus

x −9/7 d x = − 72 x −2/7 + C.

y=

Z

41.

  y ′′ = cos x For y(0) = 0 we have  ′ y (0) = 1 y′ =

Z

cos x d x = sin x + C1

1 = sin 0 + C1 H⇒ C1 = 1 Z y = (sin x + 1) d x = − cos x + x + C2 0 = − cos 0 + 0 + C2 y = 1 + x − cos x.

H⇒

C2 = 1

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SECTION 2.10 (PAGE 154)

42.

ADAMS and ESSEX: CALCULUS 8

  y ′′ = x + sin x we have For y(0) = 2  ′ y (0) = 0

B 7 , so B = 18, A = − . 2 2 7 Thus y = − x 1/2 + 18x −1/2 (for x > 0). 2 Hence 9 =

x2 − cos x + C1 2 0 = 0 − cos 0 + C1 H⇒ C1 = 1  Z  2 x x3 y= − cos x + 1 d x = − sin x + x + C2 2 6 2 = 0 − sin 0 + 0 + C2 H⇒ C2 = 2

y′ =

y=

Z

(x + sin x) d x =

46. Consider

Let y = x r , y ′ = r x r−1 , y ′′ = r (r − 1)x r−2 . Substituting these expressions into the differential equation we obtain

x3 − sin x + x + 2. 6

x 2 [r (r − 1)x r−2 ] − 6x r = 0 [r (r − 1) − 6]x r = 0.

B 2B B . Then y ′ = A − 2 , and y ′′ = 3 . x x x Thus, for all x 6= 0,

43. Let y = Ax +

x 2 y ′′ + x y ′ − y =

  x 2 y ′′ − 6y = 0 y(1) = 1  ′ y (1) = 1.

Since this equation must hold for all x > 0, we must have r (r − 1) − 6 = 0

2B B B + Ax − − Ax − = 0. x x x

r2 − r − 6 = 0 (r − 3)(r + 2) = 0.

We will also have y(1) = 2 and y ′ (1) = 4 provided A + B = 2,

and

There are two roots: r1 = −2, and r2 = 3. Thus the differential equation has solutions of the form y = Ax −2 + Bx 3 . Then y ′ = −2 Ax −3 + 3Bx 2 . Since 1 = y(1) = A + B and 1 = y ′ (1) = −2 A + 3B, therefore A = 25 and B = 35 . Hence, y = 25 x −2 + 35 x 3 .

A − B = 4.

These equations have solution A = 3, B = −1, so the initial value problem has solution y = 3x − (1/x).

44. Let r1 and r2 be distinct rational roots of the equation ar (r − 1) + br + c = 0 (x > 0) Let y = Ax r1 + Bx r2 Then y ′ = Ar 1 x r1 −1 + Br2 x r2 −1 , and y ′′ = Ar 1 (r1 − 1)x r1 −2 + Br2 (r2 − 1)x r2 −2 . Thus ax 2 y ′′ + bx y ′ + cy = ax 2 (Ar 1 (r1 − 1)x r1 −2 + Br2 (r2 − 1)x r2 −2

+ bx(Ar 1 x r1 −1 + Br2 x r2 −1 ) + c(Ax r1 + Bx r2 )   = A ar1 (r1 − 1) + br1 + c x r1  + B(ar2 (r2 − 1) + br2 + c x r2

45.

= 0x r1 + 0x r2 ≡ 0 (x > 0)   4x 2 y ′′ + 4x y ′ − y = 0 (∗) ⇒ a = 4, b = 4, c = −1 y(4) = 2  ′ y (4) = −2 Auxilary Equation: 4r (r − 1) + 4r − 1 = 0 4r 2 − 1 = 0 1 r =± 2 By #31, y = Ax 1/2 + Bx −1/2 solves (∗) for x > 0. A B Now y ′ = x −1/2 − x −3/2 2 2 Substitute the initial conditions: B 2 A B −2 = − 4 16 2 = 2A +

⇒1 = A +

Section 2.11 (page 160) 1.

Velocity and Acceleration

dx dv = 2t − 4, a = =2 dt dt a) particle is moving: to the right for t > 2

x = t 2 − 4t + 3, v =

b) to the left for t < 2 c) particle is always accelerating to the right d) never accelerating to the left e) particle is speeding up for t > 2 f) slowing down for t < 2 g) the acceleration is 2 at all times h) average velocity over 0 ≤ t ≤ 4 is

B 4

⇒−8= A−

B . 4

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16 − 16 + 3 − 3 x(4) − x(0) = =0 4−0 4

INSTRUCTOR’S SOLUTIONS MANUAL

2.

SECTION 2.11 (PAGE 160)

x = 4 + 5t − t 2 , v = 5 − 2t, a = −2.

d) The point √ is accelerating √ to the left if a < 0, i.e., for t < − 3 or 0 < t < 3.

a) The point is moving to the right if v > 0, i.e., when t < 25 .

e) The particle is speeding √ up if v and a have the same sign, i.e.,√for t < − 3, or −1 < t < 0 or 1 < t < 3.

b) The point is moving to the left if v < 0, i.e., when t > 52 .

f) The particle is slowing √ down if v and a have opposite √ sign, i.e., for − 3 < t < −1, or 0 < t < 1 or t > 3. −2(−2) 1 g) v = 0 at t = ±1. At t = −1, a = = . (2)3 2 2(−2) 1 At t = 1, a = =− . (2)3 2

c) The point is accelerating to the right if a > 0, but a = −2 at all t; hence, the point never accelerates to the right. d) The point is accelerating to the left if a < 0, i.e., for all t. e) The particle is speeding up if v and a have the same sign, i.e., for t > 25 .

h) The average velocity over [0, 4] is 4 −0 x(4) − x(0) 1 = 17 = . 4 4 17

f) The particle is slowing down if v and a have opposite sign, i.e., for t < 52 . g) Since a = −2 at all t, a = −2 at t =

3.

5 2

when v = 0.

5.

h) The average velocity over [0, 4] is x(4) − x(0) 8−4 = = 1. 4 4 dv dx = 3t 2 − 4, a = = 6t x = t 3 − 4t + 1, v = dt dt √ a) particle√moving: to the right for t < −2/ 3 or t > 2/ 3, √ √ b) to the left for −2/ 3 < t < 2/ 3 c) particle is accelerating: to the right for t > 0 d) to the left for t < 0 √ e) particle √ is speeding up for t > 2/ 3 or for −2/ 3 < t < 0 √ f) particle is slowing down for t < −2/ 3 or for √ 0 < t < 2/ 3 √ g) velocity is zero √ at t = ±2/ 3. Acceleration at these times is ±12/ 3. h) average velocity on [0, 4] is 43 − 4 × 4 + 1 − 1 = 12 4−0

4.

6. Given that y = 100 − 2t − 4.9t 2 , the time t at which the ball reaches the ground is the positive root of the equation y = 0, i.e., 100 − 2t − 4.9t 2 = 0, namely, t=

t2

a) The point is moving to the right if v > 0, i.e., when 1 − t 2 > 0, or −1 < t < 1.

b) The point is moving to the left if v < 0, i.e., when t < −1 or t > 1. c) The point is accelerating to the right if a > 0, i.e., when√2t (t 2 −√3) > 0, that is, when t > 3 or − 3 < t < 0.

−2 +

√ 4 + 4(4.9)(100) ≈ 4.318 s. 9.8

−100 = −23.16 m/s. 4.318 Since −23.159 = v = −2 − 9.8t, then t ≃ 2.159 s. The average velocity of the ball is

7.

t (t 2 + 1)(1) − (t)(2t) 1 − t2 , v= = 2 , 2 2 +1 (t + 1) (t + 1)2 (t 2 + 1)2 (−2t) − (1 − t 2 )(2)(t 2 + 1)(2t) 2t (t 2 − 3) a= = . (t 2 + 1)4 (t 2 + 1)3 x=

y = 9.8t − 4.9t 2 metres (t in seconds) dy velocity v = = 9.8 − 9.8t dt dv = −9.8 acceleration a = dt The acceleration is 9.8 m/s2 downward at all times. Ball is at maximum height when v = 0, i.e., at t = 1. Thus maximum height is y = 9.8 − 4.9 = 4.9 metres. t=1 Ball strikes the ground when y = 0, (t > 0), i.e., 0 = t (9.8 − 4.9t) so t = 2. Velocity at t = 2 is 9.8 − 9.8(2) = −9.8 m/s. Ball strikes the ground travelling at 9.8 m/s (downward).

D = t 2 , D in metres, t in seconds dD = 2t velocity v = dt Aircraft becomes airborne if 200, 000 500 v = 200 km/h = = m/s. 3600 9 Time for aircraft to become airborne is t =

250 s, that 9

is, about 27.8 s. Distance travelled during takeoff run is t 2 ≈ 771.6 metres.

8. Let y(t) be the height of the projectile t seconds after it is fired upward from ground level with initial speed v 0 . Then y ′′ (t) = −9.8, y ′ (0) = v 0 , y(0) = 0.

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SECTION 2.11 (PAGE 160)

ADAMS and ESSEX: CALCULUS 8

13. Let x(t) be the distance travelled by the train in

Two antidifferentiations give

the t seconds after the brakes are applied. Since d 2 x/dt 2 = −1/6 m/s2 and since the initial speed is v 0 = 60 km/h = 100/6 m/s, we have

y = −4.9t 2 + v 0 t = t (v 0 − 4.9t). Since the projectile returns to the ground at t = 10 s, we have y(10) = 0, so v 0 = 49 m/s. On Mars, the acceleration of gravity is 3.72 m/s2 rather than 9.8 m/s2 , so the height of the projectile would be

x(t) = −

The speed of the train at time t is v(t) = −(t/6) + (100/6) m/s, so it takes the train 100 s to come to a stop. In that time it travels x(100) = −1002 /12 + 1002 /6 = 1002 /12 ≈ 833 metres.

y = −1.86t 2 + v 0 t = t (49 − 1.86t). The time taken to fall back to ground level on Mars would be t = 49/1.86 ≈ 26.3 s.

9. The height of the ball after t seconds is y(t) = −(g/2)t 2 + v 0 t m if its initial speed was v 0 m/s. Maximum height h occurs when d y/dt = 0, that is, at t = v 0 /g. Hence h=−

14.

v2 g v 02 v0 · 2 + v0 · = 0. 2 g g 2g

An initial speed of 2v 0 means the maximum height will be 4v 02 /2g = 4h. To get a maximum height of 2h an √ initial speed of 2v 0 is required.

10. To get to 3h metres above Mars, the ball would have to be thrown upward with speed vM =

q p p 6g M h = 6g M v 02 /(2g) = v 0 3g M /g.

Since g M = 3.72 and g = 9.80, we have v M ≈ 1.067v 0 m/s.

11. If the cliff is h ft high, then the height of the rock t seconds after it falls is y = h −√16t 2 ft. The rock hits the ground (y = 0) at time √ t = h/16 s. Its speed√at that time is v = −32t = −8 h = −160 ft/s. Thus h = 20, and the cliff is h = 400 ft high.

12. If the cliff is h ft high, then the height of the rock t seconds after it is thrown down is y = h − 32t − 16t 2 ft. The rock hits the ground (y = 0) at time t=

−32 +

√ 1√ 322 + 64h = −1 + 16 + h s. 32 4

Its speed at that time is √ v = −32 − 32t = −8 16 + h = −160 ft/s. Solving this equation for h gives the height of the cliff as 384 ft.

1 2 100 t + t. 12 6

15.

x = At 2 + Bt + C, v = 2 At + B. The average velocity over [t1 , t2 ] is x(t2 ) − x(t1 ) t2 − t1 At22 + Bt1 + C − At12 − Bt1 − C = t2 − t1 A(t22 − t12 ) + B(t2 − t1 ) = (t2 − t1 ) A(t2 + t1 )(t2 − t1 ) + B(t2 − t1 ) = (t2 − t1 ) = A(t2 + t1 ) + B. The velocity  at the midpoint of [t1 , t2 ] is  instantaneous   t2 + t1 t2 + t1 v = 2A + B = A(t2 + t1 ) + B. 2 2 Hence, the average velocity over the interval is equal to the instantaneous velocity at the midpoint.   t2 0≤t ≤2 s = 4t − 4 2 →∞ h |h| |h|

as h → 0. Therefore f ′ (0) does not exist.

6. Given that f ′ (0) = k, f (0) 6= 0, and f (x + y) = f (x) f (y), we have f (0) = f (0+0) = f (0) f (0)

H⇒

f (0) = 0 or

f (0) = 1.

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CHALLENGING PROBLEMS 2 (PAGE 162)

ADAMS and ESSEX: CALCULUS 8

If a 6= 0, the x-axis is another tangent to y = x 3 that passes through (a, 0).

Thus f (0) = 1. f (x + h) − f (x) h f (x) f (h) − f (x) = lim = f (x) f ′ (0) = k f (x). h→0 h

The number of tangents to y = x 3 that pass through (x0 , y0 ) is

f ′ (x) = lim

h→0

7. Given that

g ′ (0)

three, if x0 6= 0 and y0 is between 0 and x03 ; two, if x0 6= 0 and either y0 = 0 or y0 = x03 ;

= k and g(x + y) = g(x) + g(y), then

one, otherwise.

a) g(0) = g(0 + 0) = g(0) + g(0). Thus g(0) = 0.

g(x + h) − g(x) b) g (x) = lim h→0 h g(x) + g(h) − g(x) g(h) − g(0) = lim = lim h→0 h→0 h h = g ′ (0) = k.

This is the number of distinct real solutions b of the cubic equation 2b3 − 3b2 x0 + y0 = 0, which states that the tangent to y = x 3 at (b, b3 ) passes through (x0 , y0 ).

′

h ′ (x)

10. By symmetry, any line tangent to both curves must pass through the origin. y = x 2 + 4x + 1

g ′ (x) − k

c) If h(x) = g(x) − kx, then = =0 for all x. Thus h(x) is constant for all x. Since h(0) = g(0) − 0 = 0, we have h(x) = 0 for all x, and g(x) = kx.

8.

f (x + k) − f (x) (let k = −h) k f (x − h) − f (x) f (x) − f (x − h) = lim = lim . h→0 h→0 −h h   1 ′ f ′ (x) = f (x) + f ′ (x) 2 1 f (x + h) − f (x) lim = 2 h→0 h  f (x) − f (x − h) + lim h→0 h f (x + h) − f (x − h) . = lim h→0 2h b) The change of variables used in the first part of (a) shows that

y

a) f ′ (x) = lim

x

k→0

lim

h→0

f (x + h) − f (x) h

and

lim

h→0

f (x) − f (x − h) h

y = −x 2 + 4x − 1 Fig. C-2.10 The tangent to y = x 2 + 4x + 1 at x = a has equation y = a 2 + 4a + 1 + (2a + 4)(x − a) = (2a + 4)x − (a 2 − 1),

11. The slope of y = x 2 at x = a is 2a.

are always equal if either exists. c) If f (x) = |x|, then f ′ (0) does not exist, but lim

h→0

|h| − |h| 0 f (0 + h) − f (0 − h) = lim = lim = 0. 12. h→0 h→0 h 2h h

9. The tangent to y = x 3 at x = 3a/2 has equation y=

27a 3 27 + 2 8 4a



x−

3a 2



.

This line passes through (a, 0) because 27a 3 27 + 2 8 4a



a−

3a 2



= 0.

which passes through the origin if a = ±1. The two common tangents are y = 6x and y = 2x.

The slope of the line from (0, b) to (a, a 2 ) is (a 2 − b)/a. This line is normal to y = x 2 if either a = 0 or 2a((a 2 − b)/a) = −1, that is, if a = 0 or 2a 2 = 2b − 1. There are three real solutions for a if b > 1/2 and only one (a = 0) if b ≤ 1/2.

The point Q = (a, a 2 ) on y = x 2 that is closest to P = (3, 0) is such that P Q is normal to y = x 2 at Q. Since P Q has slope a 2 /(a − 3) and y = x 2 has slope 2a at Q, we require 1 a2 =− , a−3 2a

which simplifies to 2a 3 + a − 3 = 0. Observe that a = 1 is a solution of this cubic equation. Since the slope of y = 2x 3 + x − 3 is 6x 2 + 1, which is always positive, the cubic equation can have only one real solution. Thus Q = (1, 1) is the point on y = x 2 that is closest √ to P. The distance from P to the curve is |P Q| = 5 units.

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INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 2 (PAGE 162)

13. The curve y = x 2 has slope m = 2a at (a, a 2 ). The

There is a single common tangent, and since the points of tangency on the two curves coincide, the two curves are tangent to each other.

tangent there has equation

y = a 2 + m(x − a) = mx −

m2 . 4

CASE IV. If A 6= 1 and B 2 − 4(A − 1)C < 0, there are no real solutions for b, so there can be no common tangents.

The curve y = Ax 2 + Bx + C has slope m = 2 Aa + B at (a, Aa 2 + Ba + C). Thus a = (m − B)/(2 A), and the tangent has equation

CASE V. If A 6= 1 and B 2 − 4(A − 1)C > 0, there are two distinct real solutions for b, and hence two common tangent lines. y y

y = Aa 2 + Ba + C + m(x − a)

(m − B)2 B(m − B) m(m − B) + +C − 4A 2A 2A (m − B)2 (m − B)2 = mx + C + − 4A 2A = mx + f (m),

= mx +

x

x

one common two common tangent tangents

where f (m) = C − (m − B)2 /(4 A).

tangent curves y

14. Parabola y = x 2 has tangent y = 2ax − a 2 at (a, a 2 ). Parabola y = Ax 2 + Bx + C has tangent

y

y = (2 Ab + B)x − Ab2 + C at (b, Ab2 + Bb + C). These two tangents coincide if (∗)

2 Ab + B = 2a 2

x

no common tangent

2

Ab − C = a .

The two curves have one (or more) common tangents if (∗) has real solutions for a and b. Eliminating a between the two equations leads to

Fig. C-2.14

15.

a) The tangent to y = x 3 at (a, a 3 ) has equation

(2 Ab + B)2 = 4 Ab2 − 4C,

y = 3a 2 x − 2a 3 .

or, on simplification, 4 A(A − 1)b2 + 4 ABb + (B 2 + 4C) = 0.

For intersections of this line with y = x 3 we solve x 3 − 3a 2 x + 2a 3 = 0

(x − a)2 (x + 2a) = 0.

This quadratic equation in b has discriminant D = 16 A2 B 2 −16 A(A−1)(B 2 +4C) = 16 A(B 2 −4(A−1)C). There are five cases to consider: CASE I. If A = 1, B 6= 0, then (∗) gives b=−

B 2 + 4C , 4B

a=

B 2 − 4C . 4B

There is a single common tangent in this case. CASE II. If A = 1, B = 0, then (∗) forces C = 0, which is not allowed. There is no common tangent in this case. CASE III. If A 6= 1 but B 2 = 4(A − 1)C, then b=

−B = a. 2(A − 1)

x

The tangent also intersects y = x 3 at (b, b3 ), where b = −2a.

b) The slope of y = x 3 at x = −2a is 3(−2a)2 = 12a 2 , which is four times the slope at x = a. c) If the tangent to y = x 3 at x = a were also tangent at x = b, then the slope at b would be four times that at a and the slope at a would be four times that at b. This is clearly impossible.

d) No line can be tangent to the graph of a cubic polynomial P(x) at two distinct points a and b, because if there was such a double tangent y = L(x), then (x − a)2 (x − b)2 would be a factor of the cubic polynomial P(x) − L(x), and cubic polynomials do not have factors that are 4th degree polynomials.

79 Copyright © 2014 Pearson Canada Inc.

CHALLENGING PROBLEMS 2 (PAGE 162)

16.

ADAMS and ESSEX: CALCULUS 8

a) y = x 4 − 2x 2 has horizontal tangents at points x satisfying 4x 3 − 4x = 0, that is, at x = 0 and x = ±1. The horizontal tangents are y = 0 and y = −1. Note that y = −1 is a double tangent; it is tangent at the two points (±1, −1).

These two latter roots are equal (and hence correspond to a double tangent) if the expression under the square root is 0, that is, if 8a 2 p2 + 4abp + 4ac − b2 = 0.

b) The tangent to y = x 4 − 2x 2 at x = a has equation

This quadratic has two real solutions for p provided its discriminant is positive, that is, provided

y = a 4 − 2a 2 + (4a 3 − 4a)(x − a)

16a 2 b2 − 4(8a 2 )(4ac − b2 ) > 0.

= 4a(a 2 − 1)x − 3a 4 + 2a 2 .

This condition simplifies to

Similarly, the tangent at x = b has equation

3b2 > 8ac.

y = 4b(b2 − 1)x − 3b4 + 2b2 .

For example, for y = x 4 −2x 2 +x −1, we have a = 1, b = 0, and c = −2, so 3b2 = 0 > −16 = 8ac, and the curve has a double tangent.

These tangents are the same line (and hence a double tangent) if

b) From the discussion above, the second point of tangency is

4a(a 2 − 1) = 4b(b2 − 1)

− 3a 4 + 2a 2 = −3b4 + 2b2 .

q=

The second equation says that either a 2 = b2 or 3(a 2 + b2 ) = 2; the first equation says that a 3 − b3 = a − b, or, equivalently, a 2 + ab + b2 = 1. If a 2 = b2 , then a = −b (a = b is not allowed). Thus a 2 = b2 = 1 and the two points are (±1, −1) as discovered in part (a). If a 2 +b2 = 2/3, then ab = 1/3. This is not possible since it implies that

The slope of P Q is f (q) − f ( p) b3 − 4abc + 8a 2 d = . q−p 8a 2 Calculating f ′ (( p + q)/2) leads to the same expression, so the double tangent P Q is parallel to the tangent at the point horizontally midway between P and Q.

0 = a 2 + b2 − 2ab = (a − b)2 > 0.

c) The inflection points are the real zeros of

Thus y = −1 is the only double tangent to y = x 4 − 2x 2 .

f ′′ (x) = 2(6ax 2 + 3bx + c).

c) If y = Ax + B is a double tangent to y = x 4 − 2x 2 + x, then y = (A − 1)x + B is a double tangent to y = x 4 − 2x 2 . By (b) we must have A − 1 = 0 and B = −1. Thus the only double tangent to y = x 4 − 2x 2 + x is y = x − 1.

17.

This equation has distinct real roots provided 9b2 > 24ac, that is, 3b2 > 8ac. The roots are √ −3b − 9b2 − 24ac r= √12a −3b + 9b2 − 24ac . s= 12a

a) The tangent to y = f (x) = ax 4 + bx 3 + cx 2 + d x + e

The slope of the line joining these inflection points is f (s) − f (r ) b3 − 4abc + 8a 2 d = , s −r 8a 2

at x = p has equation y = (4ap3 +3bp2 +2cp +d)x −3ap4 −2bp3 −cp2 +e. This line meets y = f (x) at x = p (a double root), and x=

−2ap − b ±

p b2 − 4ac − 4abp − 8a 2 p2 . 2a

−2ap − b b = −p − . 2a 2a

18.

so this line is also parallel to the double tangent.  dn nπ  n a) Claim: cos(ax) = a cos ax + . dxn 2 Proof: For n = 1 we have  d π cos(ax) = −a sin(ax) = a cos ax + , dx 2

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INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 2 (PAGE 162)

so the formula above is true for n = 1. Assume it is true for n = k, where k is a positive integer. Then    d k+1 d kπ k cos(ax) = a cos ax + d x k+1 dx 2    kπ k = a −a sin ax + 2   (k + 1)π k+1 . =a cos ax + 2 Thus the formula holds for n = 1, 2, 3, . . . by induction.  dn nπ  b) Claim: sin(ax) = a n sin ax + . n dx 2 Proof: For n = 1 we have  π d sin(ax) = a cos(ax) = a sin ax + , dx 2 so the formula above is true for n = 1. Assume it is true for n = k, where k is a positive integer. Then    d kπ d k+1 k sin(ax) = a sin ax + d x k+1 dx 2    kπ = a k a cos ax + 2   (k + 1)π k+1 =a sin ax + . 2 Thus the formula holds for n = 1, 2, 3, . . . by induction. c) Note that d (cos4 x + sin4 x) = −4 cos3 x sin x + 4 sin3 x cos x dx = −4 sin x cos x(cos2 − sin2 x) = −2 sin(2x) cos(2x)  π = − sin(4x) = cos 4x + . 2

19.

v (m/s) (3, 39.2)

40 30 20 10

t (s) 2

-10

6

8

10

12

14

(15, −1)

-20 -30 -40 Fig. C-2.19

(12, −49)

a) The fuel lasted for 3 seconds. b) Maximum height was reached at t = 7 s. c) The parachute was deployed at t = 12 s. d) The upward acceleration in [0, 3] was 39.2/3 ≈ 13.07 m/s2 . e) The maximum height achieved by the rocket is the distance it fell from t = 7 to t = 15. This is the area under the t-axis and above the graph of v on that interval, that is, 12 − 7 49 + 1 (49) + (15 − 12) = 197.5 m. 2 2 f) During the time interval [0, 7], the rocket rose a distance equal to the area under the velocity graph and above the t-axis, that is, 1 (7 − 0)(39.2) = 137.2 m. 2

It now follows from part (a) that  nπ  dn . (cos4 x + sin4 x) = 4n−1 cos 4x + n dx 2

4

Therefore the height of the tower was 197.5 − 137.2 = 60.3 m.

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SECTION 3.1 (PAGE 169)

CHAPTER 3. TIONS

TRANSCENDENTAL FUNC-

Section 3.1 Inverse Functions 1.

2.

3.

4.

5.

ADAMS and ESSEX: CALCULUS 8

f (x) = 2x − 1. If f (x1 ) = f (x2 ), then 2x1 − 1 = 2x2 − 1. Thus 2(x1 − x2 ) = 0 and x1 = x2 . Hence, f is one-toone. Let y = f −1 (x). Thus x = f (y) = 2y − 1, so y = 21 (x + 1). Thus f −1 (x) = 21 (x + 1). D( f ) = R( f −1 ) = (−∞, ∞). R( f ) = D( f −1 ) = (−∞, ∞).

8.

9.

√ f (x) = x − 1 p p f (x1 ) = f (x2 ) ⇔ x1 − 1 = x2 − 1, (x1 , x2 ≥ 1) ⇔ x1 − 1 = x2 − 1 = 0 ⇔ x1 = x2 Thus f is one-to-one.√ Let y = f −1 (x). Then x = f (y) = y − 1, and y = 1 + x 2 . Thus f −1 (x) = 1 + x 2 , (x ≥ 0). D( f ) = R( f −1 ) = [1, ∞), R( f ) = D( f −1 ) = [0, ∞). √ f (x) = − x − 1 for x ≥ 1. √ √ If f (x1 ) = f (x2 ), then − x1 − 1 = − x2 − 1 and x1 − 1 = x2 − 1. Thus x1 = x2 and f is one-to-one. √ Let y = f −1 (x). Then x = f (y) = − y − 1 so 2 2 −1 x = y − 1 and y = x + 1. Thus, f (x) = x 2 + 1. D( f ) = R( f −1 ) = [1, ∞). R( f ) = D( f −1 ) = (−∞, 0].

10.

f (x) = x 3 f (x1 ) = f (x2 ) ⇔ x13 = x23

⇒ (x1 − x2 )(x12 + x1 x2 + x22 ) = 0 ⇒ x1 = x2 Thus f is one-to-one. Let y = f −1 (x). Then x = f (y) = y 3 so y = x 1/3 . Thus f −1 (x) = x 1/3 . D( f ) = D( f −1 ) = R = R( f ) = R( f −1 ).

6.

f (x1 ) = f (x2 ) ⇔ x12 = x22 , (x1 ≤ 0, x2 ≤ 0) ⇔ x1 = x2 Thus f is one-to-one. Let y = f −1 (x). 2 Then x = f (y) = 0). √ y (y ≤−1 √ therefore y = − x and f (x) = − x. −1 D( f ) = (−∞, 0] = R( f ), D( f −1 ) = [0, ∞) = R( f ).

(page 169)

f (x) = x − 1 f (x1 ) = f (x2 ) ⇒ x1 − 1 = x2 − 1 ⇒ x1 = x2 . Thus f is one-to-one. Let y = f −1 (x). Then x = f (y) = y − 1 and y = x + 1. Thus f −1 (x) = x + 1. D( f ) = D( f −1 ) = R = R( f ) = R( f −1 ).

√ f (x)√ = 1 + 3 √ x. If f (x1 ) = f (x2 ), then 1 + 3 x 1 = 1 + 3 x 2 so x1 = x2 . Thus, f is one-toone. √ Let y = f −1 (x) so that x = f (y) = 1 + 3 y. Thus 3 −1 3 y = (x − 1) and f (x) = (x − 1) . D( f ) = R( f −1 ) = (−∞, ∞). R( f ) = D( f −1 ) = (−∞, ∞).

f (x) = x 2 , (x ≤ 0)

7.

11.

f (x) = (1 − 2x)3 . If f (x1 ) = f (x2 ), then (1 − 2x1 )3 = (1 − 2x2 )3 and x1 = x2 . Thus, f is one-toone. 3 so Let y = f −1√(x). Then x = f (y) = (1 − 2y) √ 1 1 −1 3 3 y = 2 (1 − x). Thus, f (x) = 2 (1 − x). D( f ) = R( f −1 ) = (−∞, ∞). R( f ) = D( f −1 ) = (−∞, ∞). 1 . D( f ) = {x : x 6= −1} = R( f −1 ). x +1 1 1 f (x1 ) = f (x2 ) ⇔ = x1 + 1 x2 + 1 ⇔ x2 + 1 = x1 + 1 ⇔ x2 = x1 Thus f is one-to-one; Let y = f −1 (x). 1 Then x = f (y) = y+1 1 1 so y + 1 = and y = f −1 (x) = − 1. x x D( f −1 ) = {x : x 6= 0} = R( f ). f (x) =

x1 x2 x . If f (x1 ) = f (x2 ), then = . 1+x 1 + x1 1 + x2 Hence x1 (1 + x2 ) = x2 (1 + x1 ) and, on simplification, x1 = x2 . Thus, f is one-to-one. y Let y = f −1 (x). Then x = f (y) = and 1+y x = f −1 (x). x(1 + y) = y. Thus y = 1−x D( f ) = R( f −1 ) = (−∞, −1) ∪ (−1, ∞). R( f ) = D( f −1 ) = (−∞, 1) ∪ (1, ∞). f (x) =

1 − 2x . D( f ) = {x : x 6= −1} = R( f −1 ) 1+x 1 − 2x1 1 − 2x2 f (x1 ) = f (x2 ) ⇔ = 1 + x1 1 + x2 ⇔ 1 + x2 − 2x1 − 2x1 x2 = 1 + x1 − 2x2 − 2x1 x2 ⇔ 3x2 = 3x1 ⇔ x1 = x2 Thus f is one-to-one. Let y = f −1 (x). 1 − 2y Then x = f (y) = 1+ y so x + x y = 1 − 2y 1−x and f −1 (x) = y = . 2+x −1 D( f ) = {x : x 6= −2} = R( f ). f (x) =

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INSTRUCTOR’S SOLUTIONS MANUAL

12.

SECTION 3.1 (PAGE 169)

x . If f (x1 ) = f (x2 ), then x2 + 1 x1 x2 q = q . (∗) 2 x1 + 1 x22 + 1 f (x) = √

Thus x12 (x22 + 1) = x22 (x12 + 1) and x12 = x22 . From (*), x1 and x2 must have the same sign. Hence, x1 = x2 and f is one-to-one. y Let y = f −1 (x). Then x = f (y) = p , and 2 y +1 x2 x 2 (y 2 + 1) = y 2 . Hence y 2 = . Since f (y) and y 1 − x2 x have the same sign, we must have y = √ , so 1 − x2 x f −1 (x) = √ . 1 − x2 D( f ) = R( f −1 ) = (−∞, ∞). R( f ) = D( f −1 ) = (−1, 1).

13.

1 + f (x) . Let y = s −1 (x). 1 − f (x) 1 + f (y) . Solving for f (y) we obtain Then x = s(y) = 1 − f (y)   x −1 x −1 f (y) = . Hence s −1 (x) = y = f −1 . x +1 x +1

20. s(x) =

21.

g(x) = f (x) − 2 Let y = g −1 (x). Then x = g(y) = f (y) − 2, so f (y) = x + 2 and g −1 (x) = y = f −1 (x + 2).

f (x) = x 2 + 1 if x ≥ 0, and f (x) = x + 1 if x < 0. If f (x1 ) = f (x2 ) then if x1 ≥ 0 and x2 ≥ 0 then x12 + 1 = x22 + 1 so x1 = x2 ; if x1 ≥ 0 and x2 < 0 then x12 + 1 = x2 + 1 so x2 = x12 (not possible); if x1 < 0 and x2 ≥ 0 then x1 = x22 (not possible); if x1 < 0 and x2 < 0 then x1 + 1 = x2 + 1 so x1 = x2 . Therefore f is one-to-one. Let y = f −1 (x). Then y 2 + 1 if y ≥ 0 x = f (y) = y + 1  if y < 0. √ x − 1 if x ≥ 1 Thus f −1 (x) = y = x −1 if x < 1. y

14. h(x) = f (2x). Let y = h −1 (x). Then x = h(y) = f (2y) and 2y = f −1 (x). Thus h −1 (x) = y =

1 2

1

f −1 (x).

y = f (x)

15. k(x) = −3 f (x). Let y = k −1 (x). Then

x x = k(y) = −3 f (y), so f (y) = − and 3  x k −1 (x) = y = f −1 − . 3

x

16. m(x) = f (x − 2). Let y = m −1 (x). Then

x = m(y) = f (y − 2), and y − 2 = f −1 (x). Hence m −1 (x) = y = f −1 (x) + 2.

17.

1 p(x) = . Let y = p−1 (x). 1 + f (x) 1 1 Then x = p(y) = so f (y) = − 1, 1 + f(y) x  1 −1 . and p−1 (x) = y = f −1 x f (x) − 3 Let y = q −1 (x). Then 2 f (y) − 3 x = q(y) = and f (y) = 2x + 3. Hence 2 −1 −1 q (x) = y = f (2x + 3).

Fig. 3.1.21

22.

18. q(x) =

23. If x1 and x2 are both positive or both negative, and h(x1 ) = h(x2 ), then x12 = x22 so x1 = x2 . If x1 and x2 have opposite sign, then h(x1 ) and h(x2 ) are on opposite sides of 1, so cannot be equal. Hence h is one-to-one.  2+1 y if y ≥ 0 If y = h −1 (x), then x = h(y) = 2 + 1 if y < 0 . If −y √ √ y ≥ 0, then y = √x − 1. If y < 0, then y = 1 − x. Thus h −1 (x) = √ x − 1 if x ≥ 1 1 − x if x < 1

19. r (x) = 1 − 2 f (3 − 4x)

Let y = r −1 (x). Then x = r (y) = 1 − 2 f (3 − 4y). f (3 − 4y) =

1−x 2

3 − 4y = f −1

1−x 2



   1 1−x and r −1 (x) = y = 3 − f −1 . 4 2

g(x) = x 3 if x ≥ 0, and g(x) = x 1/3 if x < 0. Suppose f (x1 ) = f (x2 ). If x1 ≥ 0 and x2 ≥ 0 then x13 = x23 so x1 = x2 . Similarly, x1 = x2 if both are negative. If x1 and x2 have opposite sign, then so do g(x1 ) and g(x2 ). Therefore g is one-to-one. Let y = g −1 (x). Then y3 if y ≥ 0 x = g(y) = y 1/3 if y < 0. 1/3 if x ≥ 0 Thus g −1 (x) = y = x 3 x if x < 0.

24.

y = f −1 (x) ⇔ x = f (y) = y 3 + y. To find y = f −1 (2) we solve y 3 + y = 2 for y. Evidently y = 1 is the only solution, so f −1 (2) = 1.

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SECTION 3.1 (PAGE 169)

25.

ADAMS and ESSEX: CALCULUS 8

g(x) = 1 if x 3 + x = 10, that is, if x = 2. Thus g −1 (1) = 2.

26. h(x) = −3 if x|x| = −4, that is, if x = −2. Thus h −1 (−3) = −2.

31.

32.

27. If y = f −1 (x) then x = f (y).

dy dy 1 1 so = ′ = =y 1 dx dx f (y) y (since f ′ (x) = 1/x).

Thus 1 = f ′ (y)

28.

4x 3 , then +1

dx dy = (2 + cos y) dx d x 1 d y −1 ′ = ≈ 0.36036. (g ) (2) = d x x=2 2 + cos y

1=

33. If f (x) = x sec x, then f ′ (x) = sec x + x sec x tan x ≥ 1

x2

for x in (−π/2, π/2). Thus f is increasing, and so oneto-one on that interval. Moreover, lim x→−(π/2)+ f (x) = −∞ and lim x→(π/2)+ f (x) = ∞, so, being continuous, f has range (−∞, ∞), and so f −1 has domain (−∞, ∞). Since f (0) = 0, we have f −1 (0) = 0, and

(x 2 + 1)(12x 2 ) − 4x 3 (2x) 4x 2 (x 2 + 3) f (x) = = . (x 2 + 1)2 (x 2 + 1)2 ′

Since f ′ (x) > 0 for all x, except x = 0, f must be oneto-one and so it has an inverse. 4y 3 If y = f −1 (x), then x = f (y) = 2 , and y +1 1 = f ′ (y) =

(y 2 + 1)(12y 2 y ′ ) − 4y 3 (2yy ′ ) . (y 2 + 1)2

(y 2 + 1)2 . Since f (1) = 2, therefore 4y 4 + 12y 2 f −1 (2) = 1 and

Thus y ′ =



−1

f

′

√

(y 2 + 1)2 1 (2) = 4 = . 4 4y + 12y 2 y=1

30. If f (x) = x 3p+ x 2 and y = f −1 (x), then x = f (y) = y 3 + y 2 , so,

q 2yy ′ 1 = y′ 3 + y2 + y p 2 3 + y2

⇒

y′ =

p 3 + y2 . 3 + 2y 2

Since f (−1) = −2 implies that f −1 (−2) = −1, we have 

f

−1

′

g(x) = 2x + sin x ⇒ g ′ (x) = 2 + cos x ≥ 1 for all x. Therefore g is increasing, and so one-to-one and invertible on the whole real line.

y = g −1 (x) ⇔ x = g(y) = 2y + sin y. For y = g −1 (2), we need to solve 2y + sin y − 2 = 0. The root is between 0 and 1; to five decimal places g −1 (2) = y ≈ 0.68404. Also

f (x) = 1 + 2x 3 Let y = f −1 (x). Thus x = f (y) = 1 + 2y 3 . dy 1 1 dy so ( f −1 )′ (x) = = 2 = 1 = 6y 2 −1 dx dx 6y 6[ f (x)]2

29. If f (x) =

√ y = f −1 (2) ⇔ 2 = f (y) = y 2 /(1 + y). We must solve √ 2 2 + 2 y = y for y. There is a root between 2 and 3: f −1 (2) ≈ 2.23362 to 5 decimal places.

p

2 3 + y 2 (−2) = = . 3 + 2y 2 y=−1 5

√ Note: f (x) = x 3 + x 2 = −2 ⇒ x 2 (3 + x 2 ) = 4 ⇒ x 4 + 3x 2 − 4 = 0 ⇒ (x 2 + 4)(x 2 − 1) = 0. Since (x 2 + 4) = 0 has no real solution, therefore x 2 − 1 = 0 and x = 1 or −1. Since it is given that f (x) = −2, therefore x must be −1.

( f −1 )′ (0) =

1 1 = ′ = 1. f ′ ( f −1 (0) f (0)

34. If y = ( f ◦ g)−1 (x), then x = f ◦ g(y) = f (g(y)). Thus

35.

g(y) = f −1 (x) and y = g −1 ( f −1 (x)) = g −1 ◦ f −1 (x). That is, ( f ◦ g)−1 = g −1 ◦ f −1 . x −a f (x) = bx − c y−a Let y = f −1 (x). Then x = f (y) = and by − c cx − a . We have bx y − cx = y − a so y = bx − 1 x −a cx − a f −1 (x) = f (x) if = . Evidently it is bx − c bx − 1 necessary and sufficient that c = 1. a and b may have any values.

36. Let f (x) be an even function. Then f (x) = f (−x).

Hence, f is not one-to-one and it is not invertible. Therefore, it cannot be self-inverse. An odd function g(x) may be self-inverse if its graph is symmetric about the line x = y. Examples are g(x) = x and g(x) = 1/x.

37. No. A function that is one-to-one on a single interval need not be either increasing consider the function defined  x f (x) = −x

or decreasing. For example, on [0, 2] by if 0 ≤ x ≤ 1 if 1 < x ≤ 2.

It is one-to-one but neither increasing nor decreasing on all of [0, 2].

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SECTION 3.2 (PAGE 173)

38. First we consider the case where the domain of f is a closed interval. Suppose that f is one-to-one and continuous on [a, b], and that f (a) < f (b). We show that f must be increasing on [a, b]. Suppose not. Then there are numbers x1 and x2 with a ≤ x1 < x2 ≤ b and f (x1 ) > f (x2 ). If f (x1 ) > f (a), let u be a number such that u < f (x1 ), f (x2 ) < u, and f (a) < u. By the Intermediate-Value Theorem there exist numbers c1 in (a, x1 ) and c2 in (x1 , x2 ) such that f (c1 ) = u = f (c2 ), contradicting the one-to-oneness of f . A similar contradiction arises if f (x1 ) ≤ f (a) because, in this case, f (x2 ) < f (b) and we can find c1 in (x1 , x2 ) and c2 in (x2 , b) such that f (c1 ) = f (c2 ). Thus f must be increasing on [a, b]. A similar argument shows that if f (a) > f (b), then f must be decreasing on [a, b].

12.

= 1 + logx 2 = 1 +

13. (log4 16)(log4 2) = 2 × 14.

1 =1 2

log15 75 + log15 3 = log15 225 = 2

15. log6 9 + log6 4 = log6 36 = 2  42 · 32 24 · 34 −2 = log3 (3 ) = −2 

16.

2 log3 12 − 4 log3 6 = log3

17.

loga (x 4 + 3x 2 + 2) + loga (x 4 + 5x 2 + 6) p − 4 loga x 2 + 2     = loga (x 2 + 2)(x 2 + 1) + loga (x 2 + 2)(x 2 + 3) − 2 log1 (x 2 + 2)

= loga (x 2 + 1) + loga (x 2 + 3)

= loga (x 4 + 4x 2 + 3)

18. Section 3.2 Exponential and Logarithmic Functions (page 173) √ 33 √ = 33−5/2 = 31/2 = 3 35

1 log2 x

(since 152 = 225)

Finally, if the interval I where f is defined is not necessarily closed, the same argument shows that if [a, b] is a subinterval of I on which f is increasing (or decreasing), then f must also be increasing (or decreasing) on any intervals of either of the forms [x1 , b] or [a, x2 ], where x1 and x2 are in I and x1 ≤ a < b ≤ x2 . So f must be increasing (or decreasing) on the whole of I .

1.

  logx x(log y y 2 ) = log x (2x) = log x x + logx 2

19.

logπ (1 − cos x) + logπ (1 + cos x) − 2 logπ sin x   (1 − cos x)(1 + cos x) sin2 x = logπ = log π sin2 x sin2 x = logπ 1 = 0 √ √ y = 3 √2 , log10 y = 2 log10 3, y = 10 2 log10 3 ≈ 4.72880

2. 21/2 81/2 = 21/2 23/2 = 22 = 4

20. log3 5 = (log10 5)/(log10 3 ≈ 1.46497

3. (x −3 )−2 = x 6

21. 22x = 5x+1 , 2x log10 2 = (x + 1) log10 5,

4.

( 21 )x 4x/2

2x = x =1 2

22.

5. log5 125 = log5 53 = 3 6. If log4 ( 18 ) = y then 4 y =

1 8,

or 22y = 2−3 . Thus 1 3 2y = −3 and log4 ( 8 ) = y = − 2 . 2x

7. log1/3 3 3/2

8. 4

=8

 −2x 1 = log1/3 = −2x 3 ⇒

9. 10− log10 (1/x) = 10. Since loga

log4 8 =

3 2

⇒

2

3/2

=2

√ =2 2

1 =x 1/x

x 1/(loga x)

x 1/(loga x) = a 1 = a.




1 = loga x = 1, therefore loga x

11. (loga b)(logb a) = loga a = 1

23. log x 3 = 5, (log10 3)/(log10 x) = 5,

log10 x = (log10 3)/5, x = 10(log10 3)/5 ≈ 1.24573

24. log3 x = 5, (log10 x)/(log10 3) = 5, 25.

log4 8

x = (log10 5)/(2 log10 2 − log10 5) ≈ −7.21257 √ √ x 2 = 3, 2 log x = log10 3, √ 10 x = 10(log10 3)/ 2 ≈ 2.17458

log10 x = 5 log10 3, x = 105 log10 3 = 35 = 243   1 1 Let u = loga then a u = = x −1 . Hence, a −u = x x x and u = −  logax. 1 Thus, loga = − loga x. x

26. Let loga x = u, loga y = v.

Then x = a u , y = a v . x au Thus = v = a u−v y  a x and loga = u − v = loga x − loga y. y

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SECTION 3.2 (PAGE 173)

ADAMS and ESSEX: CALCULUS 8

27. Let u = loga (x y ), then a u = x y and a u/y = x. u Therefore = loga x, or u = y loga x. y Thus, loga (x y ) = y loga x.

1.

28. Let logb x = u, logb a = v.

Thus bu = x and bv = a. Therefore x = bu = bv(u/v) = a u/v u logb x and loga x = = . v logb a

29.

log4 (x + 4) − 2 log4 (x + 1) =

1 2

1 2

log3 x. Now

2 log3 x + log9 x = 10 log3 x x

5/2

5/2

= 10

= 10

Since lim x→∞ log2 x = ∞, therefore lim x→∞ logx 2 = 0.

34. Note that log x 2 = 1/ log2 x. Since lim x→1− log2 x = lim x→1− logx 2 = −∞.

35.

f (x) = a x and f ′ (0) = lim

h→0

6.

7. 3 ln 4 − 4 ln 3 = ln √

11.

2x+1 = 3x (x + 1) ln 2 = x ln 3 ln 2 ln 2 x= = ln 3 − ln 2 ln(3/2)

12.

3x = 91−x ⇒ 3x = 32(1−x) ⇒

0+, therefore

13. 0−, therefore ah − 1 = k. Thus h

a x+h − a x h→0 h a x ah − a x = lim h→0 h ah − 1 x = a lim = a x f ′ (0) = a x k = k f (x). h→0 h

y = f −1 (x) ⇒ x = f (y) = a y dx dy ⇒1= = ka y dx dx dy 1 1 ⇒ = y = . dx ka kx Thus ( f −1 )′ (x) = 1/(kx).

43 64 = ln 34 81

8. 4 ln x + 6 ln(x 1/3 ) = 2 ln x + 2 ln x = 4 ln x

−∞, therefore

f ′ (x) = lim

36.

1 = ln e−3x = −3x e3x  2 e2 ln cos x + ln esin x = cos2 x + sin2 x = 1

5. ln

32. Note that log x (1/2) = − log x 2 = −1/ log2 x.

=

7 6



10. ln(x 2 + 6x + 9) = ln[(x + 3)2 ] = 2 ln(x + 3)

31. Note that log x 2 = 1/ log2 x.

Since lim x→1+ log2 x lim x→1+ logx 2 = ∞.

=



= 3 , so x = (310 )2/5 = 34 = 81

33. Note that log x 2 = 1/ log2 x.

2 3

9. 2 ln x + 5 ln(x − 2) = ln x 2 (x − 2)5

10

Since lim x→0+ log2 x = lim x→0+ logx (1/2) = 0.

+

4. e(3 ln 9)/2 = 93/2 = 27

30. First observe that log9 x = log3 x/ log3 9 = log3 x + log3 x

1 2

3. e5 ln x = x 5

x +4 1 = 2 (x + 1)2 x +4 = 41/2 = 2 (x + 1)2 2x 2 + 3x − 2 = 0 but we need x + 1 > 0, so x = 1/2.

1/2

√ e3 √ = e3−5/2 = e1/2 = e 5 e

2. ln(e1/2 e2/3 ) =

log4

2

Section 3.3 The Natural Logarithm and Exponential (page 181)

14.

x = 2(1 − x)

⇒

x=

2 3

1 5 = x+3 2x 8 −x ln 2 = ln 5 − (x + 3) ln 8 = ln 5 − (3x + 9) ln 2 2x ln 2 = ln 5 − 9 ln 2 ln 5 − 9 ln 2 x= 2 ln 2 2x

2 −3

2

= 4x = 22x ⇒ x 2 − 3 = 2x

x − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0 Hence, x = −1 or 3.

15. ln(x/(2 − x)) is defined if x/(2 − x) > 0, that is, if 0 < x < 2. The domain is the interval (0, 2).

16. ln(x 2 − x − 2) = ln[(x − 2)(x + 1)] is defined if

(x − 2)(x + 1) > 0, that is, if x < −1 or x > 2. The domain is the union (−∞, −1) ∪ (2, ∞).

17. ln(2x − 5) > ln(7 − 2x) holds if 2x − 5 > 0, 7 − 2x > 0, and 2x − 5 > 7 − 2x, that is, if x > 5/2, x < 7/2, and 4x > 12 (i.e., x > 3). The solution set is the interval (3, 7/2).

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.3 (PAGE 181)

2 2 18. ln(x 2 − 2) ≤ ln x holds √ if x > 2, x > 0, and x − 2 ≤ x.

x2

Thus we need x > 2 and − x − 2 ≤ 0. This latter inequality says that (x − 2)(x + 1) ≤ 0, so it holds for −1 √ ≤ x ≤ 2. The solution set of the given inequality is ( 2, 2].

19. 20. 21.

y = xe x − x, y ′ = e x + xe x − 1 x y = 2x = xe−2x e y ′ = e−2x − 2xe−2x 2 x/2

y=x e

′

,

y = 2xe

x/2

+

3 3x − 2 3 y′ = 3x − 2 ex y′ = 1 + ex

y = ln(3x − 2)

24.

y = ln |3x − 2|,

25.

y = ln(1 + e x )

26.

f (x) = e x ,

f ′ (x) = (2x)e x

27.

y=

y′ =

28.

x = e3t ln t,

29.

y = e(e ) ,

30.

y=

31.

y = e x sin x,

32.

y = e−x cos x,

y ′ = e x (sin x + cos x)

33.

y = ln ln x

y′ =

34.

y = x ln x − x   1 y ′ = ln x + x − 1 = ln x x

35.

36. 37.

2

e x + e−x , 2

44.

2

e x − e−x 2

x

1 ex =1− , 1 + ex 1 + ex

x

y′ =

45. ex (1 + e x )2

y ′ = −e−x cos x − e−x sin x

39.

46.

47.

48.

y = 52x+1

g(x) = t x x t ,

,

40. h(t) = t x − x t ,

y ′ = (2x − 3)(ln 2)2(x

x

√

=e 

x ln x

√

x ln x

2 −3x+8)

g ′ (x) = t x x t ln t + t x+1 x t−1 h ′ (t) = xt x−1 − x t ln x

    1 2 ln x 1 ln x =− . x x x

y = ln | sec x + tan x|

y

y y′

x2 2 x2 2x y ′ = 2x ln x + − = 2x ln x x 2 cos x y = ln | sin x|, y′ = = cot x sin x

2 −3x+8)

√

√  ln x x √ + x 2 x    √ 1 1 ln x + 1 =x x √ x 2  ln x 1 , let u = ln x. Then x = eu and Given that y = x  u 1 2 y= = (e−u )u = e−u . Hence, u e

y′

y = x 2 ln x −

y = 2(x

y=x

sec x tan x + sec2 x sec x + tan x = sec x p = ln |x + x 2 − a 2 | 2x 1+ √ 2√ x 2 − a 2 = √ 1 = x + x 2 − a2 x 2 − a2 p = ln( x 2 + a 2 − x) x −1 √ 2 2 = √x + a x 2 + a2 − x 1 = −√ 2 x + a2

y′ =

1 x ln x

y ′ = 2(52x+1 ) ln 5 = (2 ln 5)52x+1

38.

b (bs + c) ln a

dy d y du 2 = · = (−2ue−u ) dx du d x

dx 1 = 3e3t ln t + e3t dt t y ′ = e x e(e ) = e x+e

ln(bs + c) ln a

ln(2x + 3) g(x) = log x (2x + 3) = ln x     2 1 ln x − [ln(2x + 3)] 2x + 3 x g ′ (x) = (ln x)2 2x ln x − (2x + 3) ln(2x + 3) = x(2x + 3)(ln x)2 y′ = e

y′ =

23.

x

43.

1 2 x/2 2x e

f (s) = loga (bs + c) = f ′ (s) =

42.

y = e5x , y ′ = 5e5x

= (1 − 2x)e−2x

22.

41.

y = (cos x)x − x cos x = e x ln cos x − e(cos x)(ln x)     1 ′ x ln cos x y =e ln cos x + x (− sin x) cos x   1 − e(cos x)(ln x) − sin x ln x + cos x x = (cos x)x (ln cos x − x tan x)   1 − x cos x − sin x ln x + cos x x

87 Copyright © 2014 Pearson Canada Inc.

SECTION 3.3 (PAGE 181)

ADAMS and ESSEX: CALCULUS 8

f (x) = xeax f ′ (x) = eax (1 + ax)

49.

Thus ln x =

f ′′ (x) = eax (2a + a 2 x) ax

′′′

2

55.

3

f (x) = e (3a + a x) .. . f (n) (x) = eax (na n−1 + a n x)

50. Since d (ax 2 + bx + c)e x = (2ax + b)e x + (ax 2 + bx + c)e x dx = [ax 2 + (2a + b)x + (b + c)]e x

56.

= [Ax 2 + Bx + C]e x .

Thus, differentiating (ax 2 + bx + c)e x produces another function of the same type with different constants. Any number of differentiations will do likewise.

51.

y=e

57.

x2

y ′ = 2xe x

2

2

2

y ′′ = 2e x + 4x 2 e x = 2(1 + 2x 2 )e x 2

2

2

y ′′′ = 2(4x)e x + 2(1 + 2x 2 )2xe x = 4(3x + 2x 3 )e x

y (4) = 4(3 + 6x 2 )e 2

x2

+ 4(3x + 2x 3 )2xe

= 4(3 + 12x + 4x 4 )e x

52.

f (x) = ln(2x + 1) 2

′′

f ′ (x) = 2(2x + 1)−1

f (4) (x) = −(3!)24 (2x + 1)−4

f ′′′ (x) = (2)23 (2x + 1)−3

Thus, if n = 1, 2, 3, . . . we have f (n) (x) = (−1)n−1 (n − 1)!2n (2x + 1)−n .

53.

a)

f (x) = (x x )x = x (x

2)

ln f (x) = x 2 ln x 1 ′ f = 2x ln x + x f b)

f ′ = xx

2 +1

xx

(2 ln x + 1)

g(x) = x ln g = x x ln x xx 1 ′ g = x x (1 + ln x) ln x + ′ g x   ′ xx x 1 g =x x + ln x + (ln x)2 x Evidently g grows more rapidly than does f as x grows large.

54. Given that

x xx

..

= a where a > 0, then ln a = x

xx

..

.

ln x = a ln x.

(x 2 − 1)(x 2 − 2)(x 2 − 3) (x 2 + 1)(x 2 + 2)(x 2 + 3) 3×2×1 1 f (2) = = , f (1) = 0 5×6×7 35 ln f (x) = ln(x 2 − 1) + ln(x 2 − 2) + ln(x 2 − 3) f (x) =

2

58. Since y = x 2 e−x , then 2

2

2

y ′ = 2xe−x − 2x 3 e−x = 2x(1 − x)(1 + x)e−x . The tangent is horizontal at (0, 0) and

59.

.

f (x) = (x − 1)(x − 2)(x − 3)(x − 4) ln f (x) = ln(x − 1) + ln(x − 2) + ln(x − 3) + ln(x − 4) 1 1 1 1 1 f ′ (x) = + + + f (x) x −1 x −2 x −3 x −4   1 1 1 1 ′ f (x) = f (x) + + + x −1 x −2 x −3 x −4 √ 1 + x(1 − x)1/3 F(x) = (1 + 5x)4/5 1 ln F(x) = 2 ln(1 + x) + 31 ln(1 − x) − 54 ln(1 + 5x) F ′ (x) 1 1 4 = − − F(x) 2(1 + x) 3(1 − x) (1 + 5x)     23 1 1 4 1 1 ′ − − − −4 = − F (0) = F(0) = (1) 2 3 1 2 3 6

− ln(x 2 + 1) − ln(x 2 + 2) − ln(x 2 + 3) 1 2x 2x 2x f ′ (x) = 2 + 2 + 2 f (x) x −1 x −2 x −3 2x 2x 2x − 2 − − 2 x + 1  x2 + 2 x +3 1 1 1 + 2 + 2 f ′ (x) = 2x f (x) 2 x −1 x −2 x −3  1 1 1 − 2 − 2 − 2 x +1 x +2 x +3   4 1 1 1 1 1 1 ′ + + − − − f (2) = 35 3 2 1 5 6 7 4 139 556 = × = 35 105 3675 Since f (x) = (x 2 − 1)g(x) where g(1) 6= 0, then f ′ (x) = 2xg(x) + (x 2 − 1)g ′ (x) and 1 (−1)(−2) = . f ′ (1) = 2g(1) + 0 = 2 × 2×3×4 6

x2

2

−2

f (x) = (−1)2 (2x + 1)

2

1 ln a = ln a 1/a , so x = a 1/a . a

  1 . ±1, e

f (x) = xe−x

f ′ (x) = e−x (1 − x),

C.P. x = 1, f (1) =

f ′ (x) > 0 if x < 1 ( f increasing) f ′ (x) < 0 if x > 1 ( f decreasing)

88 Copyright © 2014 Pearson Canada Inc.

1 e

INSTRUCTOR’S SOLUTIONS MANUAL

y (1,1/e) y

SECTION 3.3 (PAGE 181)

63. Let the point of tangency be (a, 2a ). Slope of the tangent

= x e−x

is x

2a − 0 d x = 2a ln 2. = 2 a−1 d x x=a

1 1 , a =1+ . ln 2 ln 2 a 1+(1/ ln 2) So the slope is 2 ln 2 = 2 ln 2 = 2e ln 2. 1 1/ ln 2 ln 2 = 1 ⇒ 21/ ln 2 = e) (Note: ln 2 = ln 2 The tangent line has equation y = 2e ln 2(x − 1). Thus a − 1 =

Fig. 3.3.59 1 = 4 then x = 14 and x y = ln 41 = − ln 4. The tangent line of slope 4 is y = − ln 4 + 4(x − 14 ), i.e., y = 4x − 1 − ln 4.

60. Since y = ln x and y ′ =

64. The tangent line to y = a x which passes through the origin is tangent at the point (b, a b ) where

61. Let the point of tangency be (a, ea ).

ab − 0 d x = a b ln a. = a b−0 dx x=b

Tangent line has slope ea − 0 d x = ea . = e a−0 dx x=a

1 = ln a, so a b = a 1/ ln a = e. The line y = x will b intersect y = a x provided the slope of this tangent line e does not exceed 1, i.e., provided ≤ 1, or e ln a ≤ 1. b 1/e Thus we need a ≤ e . Thus

Therefore, a = 1 and line has slope e. The line has equation y = ex. y

y

(a,ea )

(b, a b )

y = ax

y = ex

x

x

Fig. 3.3.64

Fig. 3.3.61 1 1 = . The x x=a a line from (0, 0) to (a, ln a) is tangent to y = ln x if

62. The slope of y = ln x at x = a is y ′ =

x 1 =x+ y y   x y y − x y′ 1 xy ′ e (y + x y ) ln + e xy = 1 − 2 y′ 2 y x y y   1 we have At e, e   1 1 e + ey ′ 2 + e 2 (e − e3 y ′ ) = 1 − e2 y ′ e e 2 + 2e2 y ′ + 1 − e2 y ′ = 1 − e2 y ′ . 1 Thus the slope is y ′ = − 2 . e

65. e xy ln

ln a − 0 1 = a−0 a

i.e., if ln a = 1, or a = e. Thus, the line is y =

x . e

y

(a, ln a) x

y = ln x Fig. 3.3.62

66.

xe y + y − 2x = ln 2 ⇒ e y + xe y y ′ + y ′ − 2 = 0. At (1, ln 2), 2 + 2y ′ + y ′ − 2 = 0 ⇒ y ′ = 0. Therefore, the tangent line is y = ln 2.

89 Copyright © 2014 Pearson Canada Inc.

SECTION 3.3 (PAGE 181)

67.

68.

ADAMS and ESSEX: CALCULUS 8

f (x) = Ax cos ln x + Bx sin ln x f ′ (x) = A cos ln x − A sin ln x + B sin ln x + B cos ln x = (A + B) cos ln x + (B − A) sin ln x 1 If A = B = then f ′ (x) = cos ln x. Z 2 1 1 Therefore cos ln x d x = x cos ln x + x sin ln x + C. 2 2 1 1 If B = , A = − then f ′ (x) = sin ln x. 2 Z 2 1 1 Therefore sin ln x d x = x sin ln x − x cos ln x + C. 2 2 F A,B (x) = Ae x cos x + Be x sin x d F A,B (x) dx = Ae x cos x − Ae x sin x + Be x sin x + Be x cos x = (A + B)e x cos x + (B − A)e x sin x = F A+B,B−A (x)

1 + ln x = C (constant). Taking x = 1, we x 1 get C = ln 1 + ln 1 = 0. Thus ln = − ln x. x   x 1 1 ln = ln x = ln x + ln = ln x − ln y. y y y Therefore ln

72. 73.

d r x r−1 r r r [ln(x r ) − r ln x] = − = − = 0. dx xr x x x r Therefore ln(x ) − r ln x = C (constant). Taking x = 1, we get C = ln 1 − r ln 1 = 0 − 0 = 0. Thus ln(x r ) = r ln x.

74. Let x > 0, and F(x) be the area bounded by y = t 2 , the t-axis, t = 0 and t = x. For h > 0, F(x + h) − F(x) is the shaded area in the following figure. y

d F A,B (x) = F A+B,B−A (x) we have dx 2 d d F A,B (x) = F A+B,B−A (x) = F2B,−2 A (x) dx2 dx d3 x d3 d e cos x = F1,0 (x) = F0,−2 (x) 3 dx dx3 dx = F−2,−2 (x) = −2e x cos x − 2e x sin x

69. Since a) b)

70.

d (Aeax cos bx + Beax sin bx) dx = Aaeax cos bx − Abeax sin bx + Baeax sin bx + Bbeax cos bx = (Aa + Bb)eax cos bx + (Ba − Ab)eax sin bx. (a) If Aa + Bb = 1 and Ba − Ab = 0, then A = b . Thus and B = 2 a + b2

a a 2 + b2

Z

eax sin bx d x  1  ax = 2 ae sin bx − beax cos bx + C. 2 a +b

71.

d 1 1 ln + ln x = dx x 1/x

−1 x2



+

t

hx 2 < F(x + h) − F(x) < h(x + h)2 . Hence, the Newton quotient for F(x) satisfies x2 <

−b (b) If Aa + Bb = 0 and Ba − Ab = 1, then A = 2 a + b2 a and B = 2 . Thus a + b2



x +h

Comparing this area with that of the two rectangles, we see that

eax cos bx d x  1  ax ax = 2 ae cos bx + be sin bx + C. a + b2



x Fig. 3.3.74

Z



y = t2

1 1 1 = − + = 0. x x x

F(x + h) − F(x) < (x + h)2 . h

Letting h approach 0 from the right (by the Squeeze Theorem applied to one-sided limits) lim

h→0+

F(x + h) − F(x) = x 2. h

If h < 0 and 0 < x + h < x, then (x + h)2 <

F(x + h) − F(x) < x 2, h

so similarly, lim

h→0−

F(x + h) − F(x) = x 2. h

Combining these two limits, we obtain d F(x + h) − F(x) F(x) = lim = x 2. h→0 dx h

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INSTRUCTOR’S SOLUTIONS MANUAL

Therefore F(x) =

Z

x2 dx =

F(0) = C = 0, therefore F(x) the area of the region is F(2) =

75.

SECTION 3.4 (PAGE 189)

1 3 3 x + C. Since = 13 x 3 . For x = 8 3 square units.

Section 3.4 2,

1.

a) The shaded area A in part (i) of the figure is less than the area of the rectangle (actually a square) with base from t = 1 to t = 2 and height 1/1 = 1. Since ln 2 = A < 1, we have 2 < e1 = e; i.e., e > 2. (i) (ii) y y

2. 3. 4. 5.

A

6.

A1

A2 t

7.

b) If f (t) = 1/t, then f ′ (t) = −1/t 2 and f ′′ (t) = 2/t 3 > 0 for t > 0. Thus f ′ (t) is an increasing function of t for t > 0, and so the graph of f (t) bends upward away from any of its tangent lines. (This kind of argument will be explored further in Chapter 5.)

8.

1

2

1

t

2

3

Fig. 3.3.75

c) The tangent to y = 1/t at t = 2 has slope −1/4. Its equation is y=

1 1 − (x − 2) 2 4

or y = 1 −

x . 4

The tangent to y = 1/t at t = 3 has slope −1/9. Its equation is y=

1 1 − (x − 3) 3 9

or y =

x3 = 0 (exponential wins) ex

lim x −3 e x = lim

ex =∞ x3

x→∞

A1 =

1 2



3 1 + 4 2



=

5 . 8

The trapezoid bounded by x = 2, x = 3, y = 0, and y = (2/3) − (x/9) has area 1 A2 = 2



4 1 + 9 3



7 = . 18

5 7 73 + = > 1. 8 18 72 1 Thus 3 > e = e. Combining this with the result of (a) we conclude that 2 < e < 3.

e) ln 3 > A1 + A2 =

x→∞

x→∞

lim

x→∞

lim

x→∞

x→∞

2e x

−3 2 − 3e−x 2−0 = lim = =2 x→∞ 1 + 5e −x ex + 5 1+0

x − 2e−x 1 − 2/(xe x ) 1−0 = lim = =1 −x x→∞ x + 3e 1 + 3/(xe x ) 1+0

lim x ln x = 0 (power wins)

x→0+

lim

x→0+

ln x = −∞ x

lim x(ln |x|)2 = 0

x→0

lim

x→∞

(ln x)3 =0 √ x

(power wins)

9. Let N (t) be the number of bacteria present after t hours. Then N (0) = 100, N (1) = 200. dN Since = k N we have N (t) = N (0)ekt = 100ekt . dt k Thus 200 =  and k = ln 2.  100e 5 = 100e(5/2) ln 2 ≈ 565.685. Finally, N 2 There will be approximately 566 bacteria present after another 1 21 hours.

10. Let y(t) be the number of kg undissolved after t hours. Thus, y(0) = 50 and y(5) = 20. Since y ′ (t) = ky(t), therefore y(t) = y(0)ekt = 50ekt . Then 20 = y(5) = 50e5k ⇒ k =

2 x − . 3 9

d) The trapezoid bounded by x = 1, x = 2, y = 0, and y = 1 − (x/4) has area

(page 189)

lim x 3 e−x = lim

y=1/t

y=1/t

Growth and Decay

1 5

ln 52 .

If 90% of the sugar is dissolved at time T then 5 = y(T ) = 50ekT , so T =

1 1 5 ln(0.1) ln = ≈ 12.56. k 10 ln(0.4)

Hence, 90% of the sugar will dissolved in about 12.56 hours.

11. Let P(t) be the percentage undecayed after t years. Thus P(0) = 100, P(15) = 70. dP Since = k P, we have P(t) = P(0)ekt = 100ekt . dt 1 Thus 70 = P(15) = 100e15k so k = ln(0.7). 15 The half-life T satisfies if 50 = P(T ) = 100ekT , so 1 15 ln(0.5) T = ln(0.5) = ≈ 29.15. k ln(0.7) The half-life is about 29.15 years.

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SECTION 3.4 (PAGE 189)

ADAMS and ESSEX: CALCULUS 8

12. Let P(t) be the percentage remaining after t years. Thus P ′ (t) = k P(t) and P(t) = P(0)ekt = 100ekt . Then, 50 = P(1690) = 100e

1690k

b) P(1000) = 100e1000k ≈ 66.36, i.e., about 66.36% remains after 1000 years.

13. Let P(t) be the percentage of the initial amount remaining after t years. Then P(t) = 100ekt and 99.57 = P(1) = 100ek . Thus k = ln(0.9957). The half-life T satisfies 50 = P(T ) = 100ekT , 1 ln(0.5) so T = ln(0.5) = ≈ 160.85. k ln(0.995) The half-life is about 160.85 years.

14. Let N (t) be the number of bacteria in the culture t days after the culture was set up. Thus N (3) = 3N (0) and N (7) = 10 × 106 . Since N (t) = N (0)ekt , we have 3N (0) = N (3) = N (0)e3k ⇒ k = 10 = N (7) = N (0)e

7k

1000ekt = 2000 1 5 ln 2 ⇒ t = ln 2 = = 8.55. k ln( 32 )

1 1 ⇒k= ln ≈ 0.0004101. 1690 2

a) P(100) = 100e100k ≈ 95.98, i.e., about 95.98% remains after 100 years.

7

a) Let t be the time such that y(t) = 2000, i.e.,

1 3

ln 3.

⇒ N (0) = 107 e−(7/3) ln 3 ≈ 770400.

There were approximately 770,000 bacteria in the culture initially. (Note that we are approximating a discrete quantity (number of bacteria) by a continuous quantity N (t) in this exercise.)

15. Let W (t) be the weight t days after birth.

Hence, the doubling time for the investment is about 8.55 years. b) Let r % be the effective annual rate of interest; then r ) = y(1) = 1000ek 100 ⇒r = 100(ek − 1) = 100[exp ( 15 ln 32 ) − 1] 1000(1 +

= 8.447.

The effective annual rate of interest is about 8.45%.

19. Let the purchasing power of the dollar be P(t) cents after t years. Then P(0) = 100 and P(t) = 100ekt . Now 91 = P(1) = 100ek so k = ln(0.91). If 25 = P(t) = 100kt then 1 ln(0.25) t = ln(0.25) = ≈ 14.7. k ln(0.91) The purchasing power will decrease to $0.25 in about 14.7 years.

20. Let i % be the effectiverate, thenan original investment

i in one year. Let r % 100 be the nominal rate per annum compounded n times per year, then an original investment of $ A will grow to

of $ A will grow to $ A 1 +

Thus W (0) = 4000 and W (t) = 4000ekt .

1 ln(1.1). Also 4400 = W (14) = is k = 14 Five days after birth, the baby weighs W (5) = 4000e(5/14) ln(1.1) ≈ 4138.50 ≈ 4139 grams.



r $A 1 + 100n

4000e14k ,

16. Since I ′ (t) = k I (t) ⇒ I (t) = I (0)ekt = 40ekt , 1 15 3 15 = I (0.01) = 40e0.01k ⇒ k = ln = 100 ln , 0.01 40 8 thus,

n

in one year, if compounding is performed n times per year. For i = 9.5 and n = 12, we have   12 r 9.5 = $A 1 + $A 1 + 100 1200  √  12 ⇒r = 1200 1.095 − 1 = 9.1098. 

The nominal rate of interest is about 9.1098%. 

3 I (t) = 40 exp 100t ln 8



 100t 3 = 40 . 8

17. $P invested at 4% compounded continuously grows to $P(e0.04 )7 = $Pe0.28 in 7 years. This will be $10,000 if $P = $10, 000e−0.28 = $7, 557.84.

18. Let y(t) be the value of the investment after t years. Thus y(0) = 1000 and y(5) = 1500. Since y(t) = 1000ekt and 1500 = y(5) = 1000e5k , therefore, k = 15 ln 32 .

21. Let x(t) be the number of rabbits on the island t years after they were introduced. Thus x(0) = 1,000, x(3) = 3,500, and x(7) = 3,000. For t < 5 we have d x/dt = k1 x, so x(t) = x(0)ek1 t = 1,000ek1 t

x(2) = 1,000e2k1 = 3,500 H⇒ e2k1 = 3.5  5/2 = 1,000(3.5)5/2 x(5) = 1,000e5k1 = 1,000 e2k1

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≈ 22,918.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.4 (PAGE 189)

For t > 5 we have d x/dt = k2 x, so that x(t) = x(5)ek2 (t−5) x(7) = x(5)e2k2 = 3,000 x(10) = x(5)35k2

H⇒

 5/2 = x(5) e2k2

≈ 142.

3,000 22,918   3,000 5/2 ≈ 22,918 22,918 e2k2 ≈

so there are approximately 142 rabbits left after 10 years.

22. Let N (t) be the number of rats on the island t months after the initial population was released and before the first cull. Thus N (0) = R and N (3) = 2R. Since N (t) = Rekt , we have e3k = 2, so ek = 21/3 . Hence N (5) = Re5k = 25/3 R. After the first 1,000 rats are killed the number remaining is 25/3 R − 1,000. If this number is less than R, the number at the end of succeeding 5-year periods will decline. The minimum value of R for which this won’t happen must satisfy 25/3 R −1,000 = R, that is, R = 1,000/(25/3 −1) ≈ 459.8. Thus R = 460 rats should be brought to the island initially.

23.

f ′ (x) = a + b f (x). a) If u(x) = a + b f (x), then u ′ (x) = b f ′ (x) = b[a + b f (x)] = bu(x). This equation for u is the equation of exponential growth/decay. Thus bx

u(x) = C1 e ,  a 1 C1 ebx − a = Cebx − . f (x) = b b b) If

 a  −bt a x(t) = x(0) − e + b  b a −bt = 1−e . b c) We will have x(t) = 12 (a/b) if 1 − e−bt = 21 , that is, if e−bt = 12 , or −bt = ln(1/2) = − ln 2. The time required to attain half the limiting concentration is t = (ln 2)/b.

25. Let T (t) be the reading t minutes after the Thermometer is moved outdoors. Thus T (0) = 72, T (1) = 48. dT By Newton’s law of cooling, = k(T − 20). dt dV If V (t) = T (t) − 20, then = kV , so dt kt kt V (t) = V (0)e = 52e . Also 28 = V (1) = 52ek , so k = ln(7/13). Thus V (5) = 52e5 ln(7/13) ≈ 2.354. At t = 5 the thermometer reads about T (5) = 20 + 2.354 = 22.35 ◦ C.

26. Let T (t) be the temperature of the object t minutes after its temperature was 45 ◦ C. Thus T (0) = 45 and dT T (40) = 20. Also = k(T + 5). Let dt u(t) = T (t) + 5, so u(0) = 50, u(40) = 25, and dT du = = k(T + 5) = ku. Thus, dt dt

dy = a + by and y(0) = y0 , then, from part (a), dx y = Cebx −

a , b

y0 = Ce0 −

u(t) = 50ekt ,

25 = u(40) = 50e40k , 1 25 1 1 ⇒k = ln = ln . 40 50 40 2

a . b

Thus C = y0 + (a/b), and a  bx a y = y0 + e − . b b 

24.

b) The differential equation for x(t) resembles that of Exercise 21(b), except that y(x) is replaced by x(t), and b is replaced by −b. Using the result of Exercise 21(b), we obtain, since x(0) = 0,

dx = a − bx(t). dt This says that x(t) is increasing if it is less than a/b and decreasing if it is greater than a/b. Thus, the limiting concentration is a/b.

We wish to know t such that T (t) = 0, i.e., u(t) = 5, hence 5 = u(t) = 50ekt   5 40 ln 50   = 132.88 min. t= 1 ln 2

a) The concentration x(t) satisfies

Hence, it will take about (132.88 − 40) = 92.88 minutes more to cool to 0 ◦ C.

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SECTION 3.4 (PAGE 189)

ADAMS and ESSEX: CALCULUS 8

27. Let T (t) be the temperature of the body t minutes after it was 5 ◦ . Thus T (0) = 5, T (4) = 10. Room temperature = 20◦ . dT By Newton’s law of cooling (warming) = k(T − 20). dt dV If V (t) = T (t) − 20 then = kV , dt kt kt so V (t) = V (0)e = −15e .   2 1 4k . Also −10 = V (4) = −15e , so k = ln 4 3 ◦ kt If T (t) = 15 , then −5  = V(t) = −15e 1   ln 1 1 3 so t = ln = 4   ≈ 10.838. 2 k 3 ln 3 It will take a further 6.84 minutes to warm to 15 ◦ C.

31. The solution y=

of the logistic equation is valid on any interval containing t = 0 and not containing any point where the denominator is zero. The denominator is zero if y0 = (y0 − L)e−kt , that is, if   1 y0 ∗ t = t = − ln . k y0 − L Assuming k and L are positive, but y0 is negative, we have t ∗ > 0. The solution is therefore valid on (−∞, t ∗ ). The solution approaches −∞ as t → t ∗ −.

32.

28. By the solution given for the logistic equation, we have y1 =

L y0 , y0 + (L − y0 )e−k

y2 =

L y0 y0 + (L − y0 )e−2k

(L − y1 )y0 y1 (L − y0 )

2

=

(L − y2 )y0 y2 (L − y0 )

Thus 200(1 + M) = L = 10, 000, so M = 49. Also 1, 000(1 + 49e−k ) = L = 10, 000, so e−k = 9/49 and k = ln(49/9) ≈ 1.695.

33.

Now simplify: y0 y2 (L − y1 )2 = y12 (L − y0 )(L − y2 ) y0 y2 L 2 −2y1 y0 y2 L+y0 y12 y2 = y12 L 2 −y12(y0 +y2 )L+y0 y12 y2 Assuming L 6= 0,

29.

y12 (y0 + y2 ) − 2y0 y1 y2 y12 − y0 y2

  k L 2 kL dy =− y− + , dt L 2 4

dy L is greatest when y = . dt 2 L y0 The solution y = is valid on the y0 + (L − y0 )e−kt largest interval containing t = 0 on which the denominator does not vanish. If y0 > L then y0 + (L − y0 )e−kt = 0 if 1 y0 t = t ∗ = − ln . k y0 − L Then the solution is valid on (t ∗ , ∞). limt→t ∗ + y(t) = ∞. thus

L 10, 000 = ≈ 7671 cases 1 + Me−3k 1 + 49(9/49)3 Lk Me−3k ≈ 3, 028 cases/week. y ′ (3) = (1 + Me−3k )2 y(3) =

Section 3.5 The Inverse Trigonometric Functions (page 197)

.

If y0 = 3, y1 = 5, y2 = 6, then 45 25(9) − 180 = ≈ 6.429. L= 25 − 18 7 The rate of growth of y in the logistic equation is   dy y = ky 1 − . dt L Since

30.

L=

L 1 + Me−kt L 200 = y(0) = 1+M L 1, 000 = y(1) = 1 + Me−k 10, 000 = lim y(t) = L y(t) =

t→∞

Thus y1 (L − y0 )e−k = (L − y1 )y0 , and y2 (L − y0 )e−2k = (L − y2 )y0 . Square the first equation and thus eliminate e−k : 

L y0 y0 + (L − y0 )e−kt

1. sin 2. 3. 4.

√

π 3 = 2 3   1 2π cos−1 − = 2 3 π tan−1 (−1) = − 4 √ π −1 sec 2= 4 −1

5. sin(sin−1 0.7) = 0.7 6.

7.

p 1 − sin2 ( arcsin 0.7) √ √ = 1 − 0.49 = 0.51   √ 2π π tan−1 tan = tan−1 (− 3) = − 3 3 cos(sin−1 0.7) =

8. sin−1 (cos 40◦ ) = 90◦ − cos−1 (cos 40◦ ) = 50◦ 9.

  π   cos−1 sin(−0.2) = − sin−1 sin(−0.2) 2 π = + 0.2 2

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INSTRUCTOR’S SOLUTIONS MANUAL

10.

11.

SECTION 3.5 (PAGE 197)

  q sin cos−1 (− 13 ) = 1 − cos2 ( arccos (− 13 ) √ √ q 8 2 2 1 = 1− 9 = = 3 3   1 1   cos tan−1 = 1 2 −1 sec tan 2 2 1 = s   = √5 1 1 + tan2 tan−1 2

21.

y′ = − s

13.

sin(cos

22.

15. 16.

17.

18.

23.

19.

20.

24.

1 1 = √ sec(tan−1 x) 1 + x2 p tan( arctan x) = x ⇒ sec( arctan x) = 1 + x 2 1 ⇒ cos( arctan x) = √ 1 + x2 x ⇒ sin( arctan x) = √ 1 + x2

25.

26.

sin(cos−1 x) cos(cos−1 x √ 1 − x2 (by # 13) = x

tan(cos−1 x) =

27.

y = tan−1 (ax + b),

y′ =

a . 1 + (ax + b)2

(assuming) a > 0).

f (x) = x sin−1 x

f (t) = t tan−1 t

t 1 + t2

u = z 2 sec−1 (1 + z 2 )

F(x) = (1 + x 2 ) tan−1 x

F ′ (x) = 2x tan−1 x + 1 a  y = sin−1 (|x| > |a|) x h 1 ai a y′ = r  a 2 − x 2 = − |x|√x 2 − a 2 1− x G(x) =

sin−1 x sin−1 (2x)

1 2 − sin−1 x √ 2 1 − x 1 − 4x 2 G ′ (x) =  2 sin−1 (2x) √ √ 1 − 4x 2 sin−1 (2x) − 2 1 − x 2 sin−1 x =  2 √ √ 1 − x 2 1 − 4x 2 sin−1 (2x) sin−1 (2x) √

√

1 x2 − 1 1− 2 = x |x| p ⇒ tan(sec−1 x) = x 2 − 1 sgn x √ x2 − 1 if x ≥ 1 √ = − x 2 − 1 if x ≤ −1   2x − 1 y = sin−1 3 2 1 ′ y =s 2 3  2x − 1 1− 3 2 =p 9 − (4x 2 − 4x + 1) 1 =√ 2 + x − x2 1 ⇒ sin(sec−1 x) = x

−1

a 2 − (x − b)2

du z 2 (2z) p = 2z sec−1 (1 + z 2 ) + dz (1 + z 2 ) (1 + z 2 )2 − 1 2z 2 sgn (z) = 2z sec−1 (1 + z 2 ) + √ (1 + z 2 ) z 2 + 2

q
p 1 − sin2 sin−1 x = 1 − x 2

r

1 a

a2

f ′ (t) = tan−1 t +

cos(tan−1 x) =

cos(sec−1 x) =

= p

(x

− b)2

x f ′ (x) = sin−1 x + √ . 1 − x2

p x) = 1 − cos2 (cos−1 x) p = 1 − x2

14. cos(sin−1 x) =

x −b a 1

1−

12. tan(tan−1 200) = 200 −1

y = cos−1

28.

H (t) =

sin−1 t sin  t

sin t ′

H (t) = =

29.

√

1



− sin−1 t cos t 1 − t2 sin2 t

1 √ − csc t cot t sin−1 t (sin t) 1 − t 2

f (x) = (sin−1 x 2 )1/2 2x 1 f ′ (x) = (sin−1 x 2 )−1/2 √ 2 1 − x4 x = √ √ 4 1 − x sin−1 x 2

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SECTION 3.5 (PAGE 197)

30.

31.



 a √ a2 + x 2   −1/2  a2 a 2 2 −3/2 y′ = − 1 − 2 (a + x ) (2x) − a + x2 2 asgn (x) = 2 a + x2 y = cos−1

y=

p

x a a

=√

x a2 − x 2

a−x

a2

−

x2

+s

=

r

1−

x2

vals (−∞, −1] and [1, ∞), the fact that the derivative of sec−1 is positive wherever defined does not imply that sec−1 is increasing over its whole domain, only that it is increasing on each of those intervals taken independently. In fact, sec−1 (−1) = π > 0 = sec−1 (1) even though −1 < 1.

1 a

a2

a−x a+x

37. (a > 0)

  p x y = a cos−1 1 − (a > 0) − 2ax − x 2 a  −1/2     1 2a − 2x x 2 − − √ y ′ = −a 1 − 1 − a a 2 2ax − x 2 x =√ 2ax − x 2

d 1 d csc−1 x = sin−1 dx dx x  1 1 − 2 = r x 1 1− 2 x 1 =− √ |x| x 2 − 1 y

(1,π/2)

x



 2x πx = 2 y y 1 2y − 2x y ′ y 2 − 2x yy ′ =π 2 2 y y4 4x 1+ 2 y 1 4 − 2y ′ 4 − 4y ′ At (1, 2) =π 2 4 16 π −2 ′ ′ ′ 8 − 4y = 4π − 4π y ⇒ y = π −1 π −2 At (1, 2) the slope is π −1

33. tan−1

36. Since the domain of sec−1 consists of two disjoint inter-

a 2 − x 2 + a sin−1

y′ = − √

32.

ADAMS and ESSEX: CALCULUS 8

34. If y = sin−1 x, then y ′ = √

1

y = csc−1 x (−1,−π/2)

Fig. 3.5.37

38. cot−1 x = arctan (1/x); . If the slope is 2

1 −√ x2 3 . Thus the equations then √ = 2 so that x = ± 2 2 1−x of the two tangent√lines are √     π 3 π 3 y = +2 x − and y = − + 2 x + . 3 2 3 2

d cot−1 x = dx

1

35.

d 1 sin−1 x = √ > 0 on (−1, 1). dx 1 − x2 Therefore, sin−1 is increasing. d 1 tan−1 x = > 0 on (−∞, ∞). dx 1 + x2 Therefore tan−1 is increasing. d 1 cos−1 x = − √ < 0 on (−1, 1). dx 1 − x2 Therefore cos−1 is decreasing.

1

−1 1 =− 1 x2 1 + x2 1+ 2 x y

π/2

y = cot−1 x x −π/2

Fig. 3.5.38 Remark: the domain of cot−1 can be extended to include 0 by defining, say, cot−1 0 = π/2. This will make cot−1 right-continuous (but not continuous) at x = 0. It is also possible to define cot−1 in such a way that it is continuous on the whole real line, but we would then lose the identity cot−1 x = tan−1 (1/x), which we prefer to maintain for calculation purposes.

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INSTRUCTOR’S SOLUTIONS MANUAL

39.

SECTION 3.5 (PAGE 197)

  d d 1 (tan−1 x + cot−1 x) = tan−1 x + tan−1 dx dx x   1 1 1 + = − 2 = 0 if x 6= 0 1 1 + x2 x 1+ 2 x Thus tan−1 x + cot−1 x = C1 (const. for x > 0) π π At x = 1 we have + = C1 4 4 π Thus tan−1 x + cot−1 x = for x > 0. 2 −1 −1 Also tan x + cot x = C2 for (x < 0). π π At x = −1, we get − − = C2 . 4 4 π Thus tan−1 x + cot−1 x = − for x < 0. 2

42.

d 1 sin−1 (cos x) = √ (− sin x) dx 1 − cos2 x n −1 if sin x > 0 = 1 if sin x < 0

sin−1 (cos x) is continuous everywhere and differentiable everywhere except at x = nπ for integers n. y y = sin−1 (cos x) π/2 −π

40. If g(x) = tan(tan−1 x) then

π x

Fig. 3.5.42

sec2 (tan−1 x) g ′ (x) = 1 + x2 1 + x2 1 + [tan(tan−1 x)]2 = = 1. = 2 1+x 1 + x2

43.

If h(x) = tan−1 (tan x) then h is periodic with period π , and sec2 x h ′ (x) = =1 1 + tan2 x

tan−1 (tan x) is continuous and differentiable everywhere except at x = (2n + 1)π/2 for integers n. It is not defined at those points. y y = tan−1 (tan x) π/2

provided that x 6= (k + 21 )π where k is an integer. h(x) is π not defined at odd multiples of . 2 y

1 d (sec2 x) = 1 except at odd tan−1 (tan x) = dx 1 + tan2 x multiples of π/2.

y

π

−π

(π/2,π/2)

x

y=tan(tan−1 x) π

−π

x

x

Fig. 3.5.43

y=tan−1 (tan x)

Fig. 3.5.40(a)

41.

Fig. 3.5.40(b)

−1 d cos−1 (cos x) = √ (− sin x) dx 1 − cos2 x n 1 if sin x > 0 = −1 if sin x < 0

44.

cos−1 (cos x)

is continuous everywhere and differentiable everywhere except at x = nπ for integers n. y y = cos−1 (cos x) π

d 1 tan−1 (cot x) = (− csc2 x) = −1 except at dx 1 + cot2 x integer multiples of π . tan−1 (cot x) is continuous and differentiable everywhere except at x = nπ for integers n. It is not defined at those points. y y = tan−1 (cot x) π/2 π

−π π

−π

x

x

Fig. 3.5.41

Fig. 3.5.44

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SECTION 3.5 (PAGE 197)

45. If |x| < 1 and y = tan−1 √ and tan y = √

x

ADAMS and ESSEX: CALCULUS 8

x 1 − x2

Thus sec y = x and y = sec−1√ x. x2 − 1 If x ≤ −1 and y = π − sin−1 , then x and sec y < 0. Therefore

, then y > 0 ⇔ x > 0

1 − x2 x2 1 sec2 y = 1 + = 1 − x2 1 − x2 sin2 y = 1 − cos2 y = 1 − (1 − x 2 ) = x 2 sin y = x.

−1

sin y = sin π − sin cos2 y = 1 −

x Thus y = sin−1 x and sin−1 x = tan−1 √ . 1 − x2 An alternative method of proof involves showing that the derivative of the left side minus the right side is 0, and both sides are 0 at x = 0. √ x 2 − 1, then tan y = x 2 − 1 −1 x. and sec y = x, so that y = sec√ If x ≤ −1 and y = π − tan−1 x 2 − 1, then π2 < y < 3π 2 , so sec y < 0. Therefore

46. If x ≥ 1 and y = tan−1

√

tan y = tan(π − tan−1

p

x

49.

p x 2 − 1) = − x 2 − 1

 π  3π f (x) = tan−1 1 − − = x→−∞ 2 4   x −1 3π Thus tan−1 − tan−1 x = on (−∞, −1). x +1 4

50. Since f (x) = x − tan−1 (tan x) then

x2 1 = 1 + x2 1 + x2 tan2 y = sec2 y − 1 = 1 + x 2 − 1 = x 2 tan y = x. x 1 + x2

x2 − 1 , then 0 ≤ y < x

.

x2 − 1 x x2 − 1 1 2 cos y = 1 − = 2 2 x x sec2 y = x 2 .

π 2

sec2 x =1−1=0 1 + tan2 x

if x 6= −(k + 21 )π where k is an integer. Thus, f is π constant on intervals not containing odd multiples of . 2 f (0) = 0 but f (π ) = π −0 = π . There is no contraπ diction here because f ′ is not defined, so f is not 2 constant on the interval containing 0 and π .

51.

sin y =

1 x2 − 1 = 2 x2 x

f ′ (x) = 1 −

√

x2 − 1 x

Evaluate the limit as x → −∞:

cos2 y = 1 − sin2 y = 1 −

48. If x ≥ 1 and y = sin−1

√

f ′ (x) ≡ 0 on (−∞, −1)   x −1 −1 − tan−1 x = C on Thus f (x) = tan x +1 (−∞, −1).

x sin y = √ 1 + x2

√

=

lim

, then y > 0 ⇔ x > 0 and

Thus y = tan−1 x and tan−1 x = sin−1 √

!

3π 2

because both x and sec y are negative. Thus y = sec−1 x in this case also.

because both x and sec y are negative. Thus y = sec−1 x in this case also.

1 + x2

x2 − 1 x

≤y<

sec2 y = x 2 sec y = x,

sec2 y = 1 + (x 2 − 1) = x 2 sec y = x,

47. If y = sin−1 √

√

π 2

and

f (x) = x − sin−1 (sin x) (−π ≤ x ≤ π ) 1 f ′ (x) = 1 − √ cos x 1 − sin2 x cos x =1− | cos x|  π π  0 if − < x < 2 2 =  2 if −π < x < − π or π < x < π 2 2 π Note: f is not differentiable at ± . 2

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 3.6 (PAGE 203)

y

2.

(π,π )

cosh x cosh y + sinh x sinh y

= 41 [(e x + e−x )(e y + e−y ) + (e x − e−x )(e y − e−y )] = 41 (2e x+y + 2e−x−y ) = 21 (e x+y + e−(x+y) )

−π/2

= cosh(x + y). sinh x cosh y + cosh x sinh y

x

π/2

= 14 [(e x − e−x )(e y + e−y ) + (e x + e−x )(e y − e−y )] = 21 (e x+y − e−(x+y) ) = sinh(x + y).

y = f (x)

cosh(x − y) = cosh[x + (−y)] = cosh x cosh(−y) + sinh x sinh(−y) = cosh x cosh y − sinh x sinh y. sinh(x − y) = sinh[x + (−y)] = sinh x cosh(−y) + cosh x sinh(−y) = sinh x cosh y − cosh x sinh y.

(−π,−π )

Fig. 3.5.51 1 ⇒ y = tan−1 x + C 1 + x2 y(0) = C = 1

52.

y′ =

53.

Thus, y = tan−1 x + 1.  1 x 1   y′ = ⇒ y = tan−1 + C 9 + x2 3 3 π 1 −1   y(3) = 2 C =2− 2 = tan 1 + C 3 12 1 x π Thus y = tan−1 + 2 − . 3 3 12

54.

y′ = √

55.

sinh(x ± y) cosh(x ± y) sinh x cosh y ± cosh x sinh y = cosh x cosh y ± sinh x sinh y tanh x ± tanh y = 1 ± tanh x tanh y

3.

tanh(x ± y) =

4.

y = coth x =

1

⇒ y = sin−1 x + C 1 − x2 y( 21 ) = sin−1 ( 21 ) + C = 1 π π ⇒ +C =1⇒C =1− . 6 6 π Thus, y = sin−1 x + 1 − . 6 ( 4 x y′ = √ ⇒ y = 4sin−1 + C 2 5 25 − x y(0) = 0 0 =0+C ⇒C = 0 x Thus y = 4sin−1 . 5

y

1

d d 1 sech x = dx d x cosh x 1 =− sinh x = − sech x tanh x cosh2 x d d 1 csch x = dx d x sinh x 1 =− cosh x = − csch x coth x sinh2 x d d cosh coth x = dx d x sinh x sinh2 x − cosh2 x 1 = =− = − csch 2 x sinh2 x sinh2 x

y = sech x = y

y = coth x

1 x

−1

Section 3.6 Hyperbolic Functions (page 203) 1.

e x + e−x e x − e−x

Fig. 3.6.4(a) y = csch x =

2 e x − e−x

2 e x + e−x

y = sech x x

Fig. 3.6.4(b)

y

y = csch x x

Fig. 3.6.4

99 Copyright © 2014 Pearson Canada Inc.

SECTION 3.6 (PAGE 203)

5.

d d sinh−1 x = ln(x + dx dx 1 = √ x2 + 1

p

ADAMS and ESSEX: CALCULUS 8

x 1+ √ 2+1 x x 2 + 1) = √ x + x2 + 1

x 1+ √ p 2−1 d d x −1 cosh x = ln(x + x 2 − 1) = √ dx dx x + x2 − 1 1 = √ 2 x −1   d d 1 1+x −1 tanh x = ln dx dx 2 1−x 1 1 − x 1 − x − (1 + x)(−1) 1 = = 21+x (1 − x)2 1 − x2 R dx = sinh−1 x + C √ x2 + 1 R dx √ = cosh−1 x + C (x > 1) x2 − 1 R dx = tanh−1 x + C (−1 < x < 1) 1 − x2

6. Let y = sinh−1 Thus,

x  a

⇔ x = a sinh y ⇒ 1 = a(cosh y)

  1 1 1 x2 − 1 x− = 2 2 x 2x   2+1 1 ln x 1 1 x b) cosh ln x = (e + e− ln x ) = x+ = 2 2 x 2x sinh ln x x2 − 1 c) tanh ln x = = 2 cosh ln x x +1 cosh ln x + sinh ln x x 2 + 1 + (x 2 − 1) d) = 2 = x2 cosh ln x − sinh ln x (x + 1) − (x 2 − 1)   x +1 1 The domain of coth−1 x = ln consists of all 2 x −1 x +1 > 0. x satisfying |x| > 1. For such x, we have x −1 Since this fraction takes very large values for x close to 1 and values close to 0 for x close to −1, the range of coth−1 x consists of a ll real numbers except 0.

7. a) sinh ln x = (eln x − e− ln x ) =

8.

d d coth−1 x = tanh−1 dx dx 1 = 1 − (1/x)2

dy . dx

y

y = coth−1 x

x  d 1 sinh−1 = dx a a cosh y 1 1 = q = √ 2 a + x2 a 1 + sinh2 y Z dx x √ (a > 0) = sinh−1 + C. a a2 + x 2 x Let y = cosh−1 ⇔ x = a Cosh y = a cosh y a dy for y ≥ 0, x ≥ a. We have 1 = a(sinh y) . Thus, dx 1 d x cosh−1 = dx a a sinh y 1 1 = q =√ 2 − a2 2 x a cosh y − 1 Z dx x √ (a > 0, x ≥ a) = cosh−1 + C. a x 2 − a2 Let y =

Thus,

tanh−1

1 x −1 −1 = 2 . x2 x −1

−1

x

1

Fig. 3.6.8

9. Since sech −1 x = cosh−1 (1/x) is defined in terms of

the restricted function Cosh, its domain consists of the reciprocals of numbers in [1, ∞), and is therefore the interval (0, 1]. The range of sech −1 is the domain of Cosh, that is, [0, ∞). Also, d 1 d sech −1 x = cosh−1 dx dx x  1 −1 −1 = s  = √ . 2 2 x x 1 − x2 1 −1 x y

x dy ⇔ x = a tanh y ⇒ 1 = a(sech2 y) . a dx

d x 1 tanh−1 = dx a a sech2 y a a = = 2 2 − a 2 tanh2 x a − x2 a Z dx 1 x = tanh−1 + C. a2 − x 2 a a

100 Copyright © 2014 Pearson Canada Inc.

y = Sech−1 x

1

Fig. 3.6.9

x

INSTRUCTOR’S SOLUTIONS MANUAL

10.

SECTION 3.7 (PAGE 210)

csch −1 has domain and range consisting of all real numbers x except x = 0. We have

where A = 12 e−ka (L + M) and B = 12 eka (L − M).

13.

d d 1 csch−1 x = sinh−1 dx dx x  1 −1 −1 = s  2 x 2 = |x|√x 2 + 1 . 1 1+ x

y ′′ − k 2 y = 0 ⇒ y = h L ,M (x) = L cosh k(x − a) + M sinh k(x − a) y(a) = y0 ⇒ y0 = L + 0 ⇒ L = y0 , v0 y ′ (a) = v 0 ⇒ v 0 = 0 + Mk ⇒ M = k Therefore y = h y0 ,v0 /k (x) = y0 cosh k(x − a) + (v 0 /k) sinh k(x − a).

Section 3.7 Second-Order Linear DEs with Constant Coefficients (page 210)

y

y = csch−1 x

y ′′ + 7y ′ + 10y = 0

1.

auxiliary eqn r 2 + 7r + 10 = 0 (r + 5)(r + 2) = 0

x

y = Ae−5t + Be−2t

11.

y ′′ − 2y ′ − 3y = 0

2.

Fig. 3.6.10

auxiliary eqn r 2 − 2r − 3 = 0

f A,B (x) = Aekx + Be−kx

y = Ae−t + Be3t

′ f A,B (x) = k Aekx − k Be−kx

y ′′ + 2y ′ = 0

3.

′′ f A,B (x) = k 2 Aekx + k 2 Be−kx ′′ − k 2 f Thus f A,B A,B = 0

auxiliary eqn r 2 + 2r = 0

4.

′′ gC,D (x) = k 2 C cosh kx + k 2 D sinh kx ′′ Thus gC,D − k 2 gC,D = 0 cosh kx + sinh kx = ekx cosh kx − sinh kx = e−kx Thus f A,B (x) = (A + B) cosh kx + (A − B) sinh kx, that is, f A,B (x) = g A+B,A−B (x), and c D gC,D (x) = (ekx + e−kx ) + (ekx − e−kx ), 2 2 that is gC,D (x) = f (C+D)/2,(C−D)/2 (x).

12. Since

4r 2 − 4r − 3 = 0 ⇒ (2r + 1)(2r − 3) = 0

Thus, r1 = − 21 , r2 = 32 , and y = Ae−(1/2)t + Be(3/2)t . y ′′ + 8y ′ + 16y = 0

5.

auxiliary eqn r 2 + 8r + 16 = 0 y = Ae

6.

h ′′L ,M (x) = Lk 2 cosh k(x − a) + Mk 2 sinh k(x − a) = k 2 h L ,M (x)

+ Bte

−4t

⇒ r = −4, −4

y ′′ − 2y ′ + y = 0

y ′′ − 6y ′ + 10y = 0

auxiliary eqn r 2 − 6r + 10 = 0

⇒ r = 3±i

y = Ae3t cos t + Be3t sin t

8.

9y ′′ + 6y ′ + y = 0

9r 2 + 6r + 1 = 0 ⇒ (3r + 1)2 = 0

Thus, r = − 31 , − 13 , and y = Ae−(1/3)t + Bte−(1/3)t .

= 0 and

h L ,M (x)  M  L  kx−ka = e + e−kx+ka + ekx−ka − e−kx+ka 2 2     M −ka kx L ka M L −ka e + e e + e − eka e−kx = 2 2 2 2 kx −kx = Ae + Be = f A,B (x)

−4t

r 2 − 2r + 1 = 0 ⇒ (r − 1)2 = 0 Thus, r = 1, 1, and y = Aet + Btet .

h L ,M (x) = L cosh k(x − a) + M sinh k(x − a)

hence, h L ,M (x) is a solution of

⇒ r = 0, −2

4y ′′ − 4y ′ − 3y = 0

7.

− k2 y

⇒ r = −1, r = 3

y = A + Be−2t

gC,D (x) = C cosh kx + D sinh kx ′ gC,D (x) = kC cosh kx + k D sinh kx

y ′′

⇒ r = −5, −2

y ′′ + 2y ′ + 5y = 0

9.

auxiliary eqn r 2 + 2r + 5 = 0 ⇒ r = −1 ± 2i y = Ae−t cos 2t + Be−t sin 2t

10. For y ′′ − 4y ′ + 5y = 0 the auxiliary equation is

r 2 − 4r + 5 = 0, which has roots r = 2 ± i . Thus, the general solution of the DE is y = Ae2t cos t + Be2t sin t.

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11. For y ′′ + 2y ′ + 3y = 0 the auxiliary equation is√ 2 r + 2r + 3 = 0, which has solutions r = −1 ± the general solution of the given√equation is √ y = Ae−t cos( 2t) + Be−t sin( 2t).

2i . Thus

12. Given that y ′′ + y ′ + y = 0, hence r 2 + r + 1 = 0. Since

16. The auxiliary equation r 2 − (2 + ǫ)r + (1 + ǫ) factors

to (r − 1 − ǫ)(r − 1) = 0 and so has roots r = 1 + ǫ and r = 1. Thus the DE y ′′ − (2 + ǫ)y ′ + (1 + ǫ)y = 0 has general solution y = Ae(1+ǫ)t + Bet . The function e(1+ǫ)t − et yǫ (t) = is of this form with A = −B = 1/ǫ. ǫ We have, substituting ǫ = h/t,

a = 1, b = 1 and c = 1, the discriminant is D = b2 − 4ac = −3 < 0 and −(b/2a) = − 21 and √ ω = 3/2. Thus, the general solution is √ √    3 3 y = Ae−(1/2)t cos t + Be−(1/2)t sin t . 2 2

13.

  2y ′′ + 5y ′ − 3y = 0 y(0) = 1  ′ y (0) = 0 The DE has auxiliary equation 2r 2 + 5y − 3 = 0, with roots r = 12 and r = −3. Thus y = Aet/2 + Be−3t . A Now 1 = y(0) = A + B, and 0 = y ′ (0) = − 3B. 2 Thus B = 1/7 and A = 6/7. The solution is 6 1 y = et/2 + e−3t . 7 7

r 2 + 10r + 25 = 0 ⇒ (r + 5)2 = 0 ⇒ r = −5. Thus, y = Ae

+ Bte

ǫ→0

which is, along with et , a solution of the CASE II DE y ′′ − 2y ′ + y = 0.

17. Given that a > 0, b > 0 and c > 0: Case 1: If D = b2 − 4ac > 0 then the two roots are √ −b ± b2 − 4ac r1,2 = . 2a Since b2 − 4ac < b2 p ± b2 − 4ac < b p −b ± b2 − 4ac < 0

therefore r1 and r2 are negative. The general solution is

14. Given that y ′′ + 10y ′ + 25y = 0, hence −5t

−5t

y(t) = Aer1 t + Ber2 t .

If t → ∞, then er1 t → 0 and er2 t → 0. Thus, lim y(t) = 0. t→∞

Case 2: If D = b2 − 4ac = 0 then the two equal roots r1 = r2 = −b/(2a) are negative. The general solution is

y ′ = −5e−5t (A + Bt) + Be−5t .

y(t) = Aer1 t + Bter2 t .

Since 0 = y(1) = Ae ′

−5

2 = y (1) = −5e −2e5

+ Be

−5

−5

(A + B) + Be

−5

,

2e5 .

we have A = and B = Thus, y = −2e5 e−5t + 2te5 e−5t = 2(t − 1)e−5(t−1) .

15.

e(1+ǫ)t − et ǫ→0 ǫ et+h − et = t lim h→0  h d t =t e = t et dt

lim yǫ (t) = lim

  y ′′ + 4y ′ + 5y = 0 y(0) = 2  ′ y (0) = 0 The auxiliary equation for the DE is r 2 + 4r + 5 = 0, which has roots r = −2 ± i . Thus y = Ae−2t cos t + Be−2t sin t

y ′ = (−2 Ae−2t + Be−2t ) cos t − (Ae−2t + 2Be−2t ) sin t.

Now 2 = y(0) = A ⇒ A = 2, and 2 = y ′ (0) = −2 A + B ⇒ B = 6. Therefore y = e−2t (2 cos t + 6 sin t).

If t → ∞, then er1 t → 0 and er2 t → 0 at a faster rate than Bt → ∞. Thus, lim y(t) = 0. t→∞

Case 3: If D = b2 − 4ac < 0 then the general solution is

y = Ae−(b/2a)t cos(ωt) + Be−(b/2a)t sin(ωt) √ 4ac − b2 where ω = . If t → ∞, then the amplitude of 2a both terms Ae−(b/2a)t → 0 and Be−(b/2a)t → 0. Thus, lim y(t) = 0. t→∞

18. The auxiliary equation ar 2 + br + c = 0 has roots

√ √ −b − D −b + D r1 = , r2 = , 2a 2a where D = √ b2 − 4ac. Note that a(r 2 − r1 ) = D = −(2ar1 + b). If y = er1 t u, then y ′ = er1 t (u ′ + r1 u), and y ′′ = er1 t (u ′′ + 2r1 u ′ + r12 u). Substituting these expressions into the DE ay ′′ + by ′ + cy = 0, and simplifying, we obtain er1 t (au ′′ + 2ar1 u ′ + bu ′ ) = 0,

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or, more simply, u ′′ − (r2 − r1 )u ′ = 0. Putting v = u ′ reduces this equation to first order:

Substituting into ay ′′ + by ′ + cy = 0 leads to   0 = ekt au ′′ + (2ka + b)u ′ + (ak 2 + bk + c)u   = ekt au ′′ + 0 + ((b2 /(4a) − (b2 /(2a) + c)u   = a ekt u ′′ + ω2 u .

v ′ = (r2 − r1 )v, which has general solution v = Ce(r2 −r1 )t . Hence u=

Z

Ce(r2 −r1 )t dt = Be(r2 −r1 )t + A,

Thus u satisfies u ′′ + ω2 u = 0, which has general solution u = A cos(ωt) + B sin(ωt)

and y = er1 t u = Aer1 t + Ber2 t .

by the previous problem. Therefore ay ′′ + by ′ + cy = 0 has general solution

19. If y = A cos ωt + B sin ωt then y ′′ + ω2 y = − Aω2 cos ωt − Bω2 sin ωt

y = Aekt cos(ωt) + Bekt sin(ωt).

+ ω2 (A cos ωt + B sin ωt) = 0

for all t. So y is a solution of (†).

20. If f (t) is any solution of (†) then f ′′ (t) = −ω2 f (t) for

24. Because y ′′ + 4y = 0, therefore y = A cos 2t + B sin 2t. Now

all t. Thus,

y(0) = 2 ⇒ A = 2,

y ′ (0) = −5 ⇒ B = − 52 .

2 i 2  d h 2 ω f (t) + f ′ (t) dt = 2ω2 f (t) f ′ (t) + 2 f ′ (t) f ′′ (t)

= 2ω2 f (t) f ′ (t) − 2ω2 f (t) f ′ (t) = 0

 2  2 for all t. Thus, ω2 f (t) + f ′ (t) is constant. (This can be interpreted as a conservation of energy statement.)

21. If g(t) satisfies (†) and also g(0) = g ′ (0) = 0, then by Exercise 20,

25.

 2  2 ω2 g(t) + g ′ (t)  2  2 = ω2 g(0) + g ′ (0) = 0. Since a sum of squares cannot vanish unless each term vanishes, g(t) = 0 for all t.

22. If f (t) is any solution of (†), let g(t) = f (t) − A cos ωt − B sin ωt where A = f (0) and Bω = f ′ (0). Then g is also solution of (†). Also g(0) = f (0) − A = 0 and g ′ (0) = f ′ (0) − Bω = 0. Thus, g(t) = 0 for all t by Exercise 24, and therefore f (x) = A cos ωt + B sin ωt. Thus, it is proved that every solution of (†) is of this form. 4ac − b2 b and ω2 = which is 2a 4a 2 positive for Case III. If y = ekt u, then

23. We are given that k = − ′

y =e

kt





′

u + ku



26.

Thus, y = 2 cos 2t − 52 sin 2t. circular frequency = ω = 2, frequency = 1 ω = ≈ 0.318 2π π 2π period = = π ≈ 3.14 ω q amplitude = (2)2 + (− 25 )2 ≃ 3.20   y ′′ + 100y = 0 y(0) = 0  ′ y (0) = 3 y = A cos(10t) + B sin(10t) A = y(0) = 0, 10B = y ′ (0) = 3 3 y= sin(10t) 10     y = A cos ω(t − c) + B sin ω(t − c)

(easy to calculate y ′′ + ω2 y = 0)  y = A cos(ωt) cos(ωc) + sin(ωt) sin(ωc)   + B sin(ωt) cos(ωc) − cos(ωt) sin(ωc)   = A cos(ωc) − B sin(ωc) cos ωt   + A sin(ωc) + B cos(ωc) sin ωt = A cos ωt + B sin ωt where A = A cos(ωc) − B sin(ωc) and B = A sin(ωc) + B cos(ωc)

27. For y ′′ + y = 0, we have y = A sin t + B cos t. Since,

 y ′′ = ekt u ′′ + 2ku ′ + k 2 u .

y(2) = 3 = A sin 2 + B cos 2 y ′ (2) = −4 = A cos 2 − B sin 2,

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31. Using the addition identities for cosine and sine,

therefore A = 3 sin 2 − 4 cos 2 B = 4 sin 2 + 3 cos 2.

y = ekt [A cos ω(t − t0 )B sin ω(t − t0 )]

= ekt [A cos ωt cos ωt0 + A sin ωt sin ωt0 + B sin ωt cos ωt0 − B cos ωt sin ωt0 ]

Thus,

= ekt [A1 cos ωt + B1 sin ωt],

y = (3 sin 2 − 4 cos 2) sin t + (4 sin 2 + 3 cos 2) cos t = 3 cos(t − 2) − 4 sin(t − 2).

28.

  y ′′ + ω2 y = 0 y(a) = A  ′ y (a) = B   B   y = A cos ω(t − a) + sin ω(t − a) ω

where A1 = A cos ωt0 − B sin ωt0 and B1 = A sin ωt0 + B cos ωt0 . Under the conditions of this problem we know that ekt cos ωt and ekt sin ωt are independent solutions of ay ′′ + by ′ + cy = 0, so our function y must also be a solution, and, since it involves two arbitrary constants, it is a general solution.

32. Expanding the hyperbolic functions in terms of exponentials, y = ekt [A cosh ω(t − t0 )B sinh ω(t − t0 )]  A kt A ω(t−t0 ) =e e + e−ω(t−t0 ) 2 2  B ω(t−t0 ) B −ω(t−t0 ) + e − e 2 2 = A1 e(k+ω)t + B1 e(k−ω)t

29. From Example 9, the spring constant is k = 9 × 104 gm/sec2 . For a frequency of 10 Hz (i.e., a circular √ frequency ω = 20π rad/sec.), a mass m satisfying k/m = 20π should be used. So, m=

k 9 × 104 = = 22.8 gm. 2 400π 400π 2

The motion is determined by   y ′′ + 400π 2 y = 0 y(0) = −1  ′ y (0) = 2

therefore, y = A cos 20π t + B sin 20π t and y(0) = −1 ⇒ A = −1 2 1 y ′ (0) = 2 ⇒ B = = . 20π 10π

33.

The DE has auxiliary equation r 2 + 2r + 5 = 0 with roots r = −1 ± 2i . By the second previous problem, a general solution can be expressed in the form y = e−t [A cos 2(t − 3) + B sin 2(t − 3)] for which

1 sin 20π t, with y in cm 10π and t in second, gives the displacement at time t. The r 1 amplitude is (−1)2 + ( )2 ≈ 1.0005 cm. 10π

Thus, y = − cos 20π t +

k ω , ω2 = (k = spring const, m = mass) 2π m Since the spring does not change, ω2 m = k (constant) For m = 400 gm, ω = 2π(24) (frequency = 24 Hz) 4π 2 (24)2 (400) If m = 900 gm, then ω2 = 900 2π × 24 × 2 so ω = = 32π . 3 32π Thus frequency = = 16 Hz 2π 4π 2 (24)2 400 For m = 100 gm, ω = 100 ω so ω = 96π and frequency = = 48 Hz. 2π

where A1 = (A/2)e−ωt0 + (B/2)e−ωt0 and B1 = (A/2)eωt0 − (B/2)eωt0 . Under the conditions of this problem we know that Rr = k ± ω are the two real roots of the auxiliary equation ar 2 +br +c = 0, so e(k±ω)t are independent solutions of ay ′′ + by ′ + cy = 0, and our function y must also be a solution. Since it involves two arbitrary constants, it is a general solution.   y ′′ + 2y ′ + 5y = 0 y(3) = 2  ′ y (3) = 0

y ′ = −e−t [A cos 2(t − 3) + B sin 2(t − 3)] + e−t [−2 A sin 2(t − 3) + 2B cos 2(t − 3)]. The initial conditions give

30. Frequency =

2 = y(3) = e−3 A

0 = y ′ (3) = −e−3 (A + 2B) Thus A = 2e3 and B = − A/2 = −e3 . The IVP has solution y = e3−t [2 cos 2(t − 3) − sin 2(t − 3)].

34.

  y ′′ + 4y ′ + 3y = 0 y(3) = 1  ′ y (3) = 0

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The DE has auxiliary equation r 2 + 4r + 3 = 0 with roots r = −2 + 1 = −1 and r = −2 − 1 = −3 (i.e. k ± ω, where k = −2 and ω = 1). By the second previous problem, a general solution can be expressed in the form y = e−2t [A cosh(t − 3) + B sinh(t − 3)] for which

Let v = x + (1/5). Then v ′′ = x ′′ = −(x + 1/5) = −v, v(π ) = −(3/5) + (1/5) = −(2/5), and v ′ (π ) = x ′ (π ) = 0. Thius initial-value problem has solution v(t) = −(2/5) cos(t − π ) = (2/5) cos t, so that x(t) = (2/5) cos t − (1/5) and x ′ (t) = −(2/5) sin t. These formulas remain valid for t ≥ π until t = 2π when x ′ becomes 0 again. We have x(2π ) = (2/5) − (1/5) = 1/5 and x ′ (2π ) = 0.

y ′ = −2e−2t [A cosh(t − 3) + B sinh(t − 3)]

+ e−2t [A sinh(t − 3) + B cosh(t − 3)].

The conditions for stopping the motion are met at t = 2π ; the mass remains at rest thereafter. Thus  1 4   5 cos t + 5 if 0 ≤ t ≤ π 2 x(t) = 5 cos t − 15 if π < t ≤ 2π  1 if t > 2π 5

The initial conditions give 1 = y(3) = e−6 A

0 = y ′ (3) = −e−6 (−2 A + B) Thus A = e6 and B = 2 A = 2e6 . The IVP has solution y = e6−2t [cosh(t − 3) + 2 sinh(t − 3)].

Review Exercises 3 (page 211) 1.

35. Let u(x) = c − k 2 y(x). Then u(0) = c − k 2 a.

Also u ′ (x) = −k 2 y ′ (x), so u ′ (0) = −k 2 b. We have

f (x) = 3x + x 3 ⇒ f ′ (x) = 3(1 + x 2 ) > 0 for all x, so f is increasing and therefore one-to-one and invertible. Since f (0) = 0, therefore f −1 (0) = 0, and 1 1 1 d −1 = ′ −1 ( f )(x) = ′ = . dx f ( f (0)) f (0) 3 x=0

  u ′′ (x) = −k 2 y ′′ (x) = −k 2 c − k 2 y(x) = −k 2 u(x) This IVP for the equation of simple harmonic motion has solution

2.

u(x) = (c − k 2 a) cos(kx) − kb sin(kx) so that  1  y(x) = 2 c − u(x) k   c = 2 c − (c − k 2 a) cos(kx) + kb sin(kx) k c b = 2 (1 − cos(kx) + a cos(kx) + sin(kx). k k

( f −1 )′ (2) =

3.

Let u = x − (1/5). Then u ′′ = x ′′ = −(x − 1/5) = −u, u(0) = 4/5, and u ′ (0) = x ′ (0) = 0. This simple harmonic motion initial-value problem has solution u(t) = (4/5) cos t. Thus x(t) = (4/5) cos t + (1/4) and x ′ (t) = u ′ (t) = −(4/5) sin t. These formulas remain valid until t = π when x ′ (t) becomes 0 again. Note that x(π ) = −(4/5) + (1/5) = −(3/5). Since x(π ) < −(1/5), the motion for t > π will be governed by x ′′ = −x − (1/5) until such time t > π when x ′ (t) = 0 again.

lim

x→±∞

f (x) = lim

x→±∞

1 1 1 = ′ = . f ′ ( f −1 (2)) f (π/4) 8 x ex

2

= 0.

2

4. Observe f ′ (x) = e−x (1 − 2x 2 ) is positive if x 2 < 1/2

36. Since x ′ (0) = 0 and x(0) = 1 > 1/5, the motion will be

governed by x ′′ = −x + (1/5) until such time t > 0 when x ′ (t) = 0 again.

f (x) = sec2 x tan x ⇒ f ′ (x) = 2 sec2 x tan2 x + sec4 x > 0 for x in (−π/2, π/2), so f is increasing and therefore one-to-one and invertible there. The domain of f −1 is (−∞, ∞), the range of f . Since f (π/4) = 2, therefore f −1 (2) = π/4, and

5. 6.

2 and is√negative √ if x > 1/2. Thus f is increasing √ on 2, 1/ 2) and is decreasing on (−∞, −1/ 2) and (−1/ √ on (1/ 2, ∞). √ √ The max√and min values √ of f are 1/ 2e (at x = 1/ 2) and −1/ 2e (at x = −1/ 2).

y = e−x sin x, (0 ≤ x ≤ 2π ) has a horizontal tangent where dy 0= = e−x (cos x − sin x). dx This occurs if tan x = 1, √ so x = π/4 or x = 5π/4. √ The points are (π/4, e−π/4 / 2) and (5π/4, −e−5π/4 / 2).

7. If f ′ (x) = x for all x, then

d f (x) f ′ (x) − x f (x) = = 0. 2 d x e x /2 e x 2 /2

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2

Thus f (x)/e x /2 = C (constant) for all x. Since f (2) = 3, we have C = 3/e2 and 2 2 f (x) = (3/e2 )e x /2 = 3e(x /2)−2 .

12. If f (x) = (ln x)/x, then f ′ (x) = (1 − ln x)/x 2 . Thus

f ′ (x) > 0 if ln x < 1 (i.e., x < e) and f ′ (x) < 0 if ln x > 1 (i.e., x > e). Since f is increasing to the left of e and decreasing to the right, it has a maximum value f (e) = 1/e at x = e. Thus, if x > 0 and x 6= e, then

8. Let the length, radius, and volume of the clay cylinder at time t be ℓ, r , and V , respectively. Then V = πr 2 ℓ, and

ln x 1 < . x e

dV dr dℓ = 2πr ℓ + πr 2 . dt dt dt

Putting x = π we obtain (ln π )/π < 1/e. Thus

Since d V /dt = 0 and dℓ/dt = kℓ for some constant k > 0, we have dr 2πr ℓ = −kπr 2 ℓ, dt

⇒

dr kr =− . dt 2

ln(π e ) = e ln π < π = π ln e = ln eπ , and π e < eπ follows because ln is increasing.

13.

That is, r is decreasing at a rate proportional to itself.

9.

1T ≈

dT 100 ln 2 1r = − 1r. dr r2

If r = 13.863% and 1r = −0.5%, then 1T ≈ −

100 ln 2 (−0.5) ≈ 0.1803 years. 13.8632

The doubling time will increase by about 66 days.

10.

ah − 1 a 0+h − a 0 d x = ln a. a) lim = lim = a h→0 h→0 h h d x x=0   Putting h = 1/n, we get lim n a 1/n − 1 = ln a. n→∞

b) Using the technique described in the exercise, we calculate   10 210 21/2 − 1 ≈ 0.69338183   11 211 21/2 − 1 ≈ 0.69326449

2  2 d  f (x) = f ′ (x) dx  2 ⇒ 2 f (x) f ′ (x) = f ′ (x)

This line passes through the origin if 0 = a a [1−a(1+ln a)], that is, if (1+ln a)a = 1. Observe that a = 1 solves this equation. Therefore the slope of the line is 11 (1 + ln 1) = 1, and the line is y = x.

14.

a)

ln x ln 2 = is satisfied if x = 2 or x = 4 (because x 2 ln 4 = 2 ln 2).

b) The line y = mx through the origin intersects the curve y = ln x at (b, ln b) if m = (ln b)/b. The same line intersects y = ln x at a different point (x, ln x) if (ln x)/x = m = (ln b)/b. This equation will have only one solution x = b if the line y = mx intersects the curve y = ln x only once, at x = b, that is, if the line is tangent to the curve at x = b. In this case m is the slope of y = ln x at x = b, so ln b 1 =m= . b b Thus ln b = 1, and b = e.

15. Let the rate be r %. The interest paid by account A is 1, 000(r/100) = 10r . The interest paid by account B is 1, 000(er/100 − 1). This is $10 more than account A pays, so 1, 000(er/100 − 1) = 10r + 10. A TI-85 solve routine gives r ≈ 13.8165%.

Thus ln 2 ≈ 0.693.

11.

y = a a + a a (1 + ln a)(x − a).

a) An investment of $P at r % compounded continuously grows to $Per T /100 in T years. This will be $2P provided er T /100 = 2, that is, r T = 100 ln 2. If T = 5, then r = 20 ln 2 ≈ 13.86%. b) Since the doubling time is T = 100 ln 2/r , we have

y = x x = e x ln x ⇒ y ′ = x x (1 + ln x). The tangent to y = x x at x = a has equation

16. If y = cos−1 x, then x = cos y and 0 ≤ y ≤ π . Thus

⇒ f ′ (x) = 0 or f ′ (x) = 2 f (x). Since f (x) is given to be nonconstant, we have f ′ (x) = 2 f (x). Thus f (x) = f (0)e2x = e2x .

r √ q 1 1 − x2 2 tan y = sgn x sec y − 1 = sgn x −1= . x2 x √ Thus cos−1 x = tan−1 (( 1 − x 2 )/x). Since cot x = 1/ tan x, cot−1 x = tan−1 (1/x).

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1 π 1 = − cos−1 x 2 x p π 1 − (1/x)2 −1 = − tan 2 1/x p π −1 x 2 − 1. = − sgn xtan 2

Thus

csc−1 x = sin−1

17.

 60 − R 2 40 − R = e20k = 96 − R 96 − R (60 − R)2 = (96 − R)(40 − R) 

3600 − 120R + R 2 = 3840 − 136R + R 2 16R = 240 R = 15.

π cos−1 x = − sin−1 x. 2

If y = cot−1 x, then x = cot y and 0 < y < π/2. Thus q p csc y = sgn x 1 + cot2 y = sgn x 1 + x 2 sgn x sin y = √ . 1 + x2 sgn x 1 Thus cot−1 x = sin−1 √ = sgn xsin−1 √ . 2 1+x 1 + x2 1 csc−1 x = sin−1 . x

Room temperature is 15◦ .

20. Let f (x) = e x − 1 − x. Then f (0) = 0 and by the MVT, f (x) f (x) − f (0) = = f ′ (c) = ec − 1 x x −0 for some c between 0 and x. If x > 0, then c > 0, and f ′ (c) > 0. If x < 0, then c < 0, and f ′ (c) < 0. In either case f (x) = x f ′ (c) > 0, which is what we were asked to show.

21. Suppose that for some positive integer k, the inequality

18. Let T (t) be the temperature of the milk t minutes after it

ex > 1 + x +

is removed from the refrigerator. Let U (t) = T (t) − 20. By Newton’s law, ′

U (t) = kU (t)

⇒

holds for all x > 0. This is certainly true for k = 1, as shown in the previous exercise. Apply the MVT to

kt

U (t) = U (0)e .

g(t) = et − 1 − t −

Now T (0) = 5 ⇒ U (0) = −15 and T (12) = 12 ⇒ U (12) = −8. Thus

k=

1 12

g(x) − g(0) g(x) = = g ′ (c) x x −0

ln(8/15).

for some c in (0, x). Since x and g ′ (c) are both positive, so is g(x). This completes the induction and shows the desired inequality holds for x > 0 for all positive integers k.

If T (s) = 18, then U (s) = −2, so −2 = −15esk . Thus sk = ln(2/15), and s=

t2 t k+1 −··· − 2! (k + 1)!

on the interval (0, x) (where x > 0) to obtain

− 8 = U (12) = U (0)e12k = −15e12k

e12k = 8/15,

x2 xk + ··· + 2! k!

ln(2/15) ln(2/15) = 12 ≈ 38.46. k ln(8/15)

Challenging Problems 3 (page 212) It will take another 38.46 − 12 = 26.46 min for the milk to warm up to 18◦ .

19. Let R be the temperature of the room, Let T (t) be the temperature of the water t minutes after it is brought into the room. Let U (t) = T (t) − R. Then U ′ (t) = kU (t)

⇒

U (t) = U (0)ekt .

We have T (0) = 96 ⇒ U (0) = 96 − R

T (10) = 60 ⇒ U (10) = 60 − R ⇒ 60 − R = (96 − R)e10k

T (20) = 40 ⇒ U (20) = 40 − R ⇒ 40 − R = (96 − R)e20k .

1.

a) (d/d x)x x = x x (1 + ln x) > 0 if ln x > −1, that is, if x > e−1 . Thus x x is increasing on [e−1 , ∞).

b) Being increasing on [e−1 , ∞), f (x) = x x is invertible on that interval. Let g = f −1 . If y = x x , then x = g(y). Note that y → ∞ if and only if x → ∞. We have ln y = x ln x ln(ln y) = ln x + ln(ln x) g(y) ln(ln y) x(ln x + ln(ln x)) lim = lim y→∞ x→∞ ln y x ln x   ln(ln x) = lim 1 + . x→∞ ln x

107 Copyright © 2014 Pearson Canada Inc.

CHALLENGING PROBLEMS 3 (PAGE 212)

ADAMS and ESSEX: CALCULUS 8

√ Now ln x 0) dt √ a) Let u = 2t gk. If v(t) =

r

gk

, then y(0) = y0

√ √ g 1 2 gke2t gk − √ k k 1 + e2t gk √ r g 1 − e2t gk = √ = v(t). k 1 + e2t gk r

Thus y(t) gives the height of the object at time t during its fall.   dp dy If p = e−bt y, then = e−bt − by . dt dt  dp p  The DE = kp 1 − −bt therefore transforms to dt e M  p  dy = by + kpebt 1 − −bt dt e M  y ky 2 = Ky 1 − = (b + k)y − , M L b+k M. This is a standard k Logistic equation with solution (as obtained in Section 3.4) given by where K = b + k and L =

y=

L y0 , y0 + (L − y0 )e−K t

where y0 = y(0) = p(0) = p0 . Converting this solution back in terms of the function p(t), we obtain L p0 e−bt p0 + (L − p0 )e−(b+k)t (b + k)M p0   = . bt p0 ke + (b + k)M − kp0 e−kt

g 1 − eu , then k 1 + eu

r

√ g 1 − e2t gk Thus v(t) = √ . k 1 + e2t gk √ r r g e−2t gk − 1 g b) lim v(t) = lim =− √ t→∞ t→∞ k e−2t gk + 1 k r

√

p(t) =

g (1 + eu )(−eu ) − (1 − eu )eu p 2 gk k (1 + eu )2 −4geu = (1 + eu )2   (1 − eu )2 2 kv − g = g −1 (1 + eu )2 u −4ge dv . = = (1 + eu )2 dt dv = dt

g 1 1 + e2t t − ln k k 2

dy = dt

4.

a) Let u(t) = −g − kv(t). Then

3.

and

g(y) ln(ln y) = 1 + 0 = 1. ln y

dv = −g − kv. dt

c)

c) If y(t) = y0 +

r

Since p represents a percentage, we must have (b + k)M/k < 100. If k = 10, b = 1, M = 90, and p0 = 1, then b+k M = 99 < 100. The numerator of the final expresk sion for p(t) given above is a constant. Therefore p(t) will be largest when the derivative of the denominator,   f (t) = p0 kebt + (b + k)M − kp0 e−kt = 10et + 980e−10t is zero. Since f ′ (t) = 10et − 9, 800e−10t , this will happen at t = ln(980)/11. The value of p at this t is approximately 48.1. Thus the maximum percentage of potential clients who will adopt the technology is about 48.1%.

108 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.1 (PAGE 218)

CHAPTER 4. MORE APPLICATIONS OF DIFFERENTIATION

7.

V =

4 3 dV dr πr , so = 4πr 2 . 3 dt dt

When r = 30 cm and d V /dt = 20 cm3 /s, we have

Section 4.1 Related Rates

(page 218)

20 = 4π(30)2

1. If the side and area of the square at time t are x and A,

20 1 dr = = . dt 3600π 180π

respectively, then A = x 2 , so dA dx = 2x . dt dt

The radius is increasing at 1/(180π ) cm/s.

8. The volume V of the ball is given by

If x = 8 cm and d x/dt = 2 cm/min, then the area is increasing at rate d A/dt = 32 cm2 /min.

4 4π V = πr 3 = 3 3

2. As in Exercise 1, d A/dt = 2x d x/dt. If d A/dt = −2 ft2 /s and x = 8 ft, then d x/dt = −2/(16). The side length is decreasing at 1/8 ft/s. pact be r and A respectively. Then A =

πr 2 .

4. Let A and r denote the area and radius of the circle. A = πr 2 ⇒ r = ⇒

dr = dt

1 √ 2 Aπ



dA . dt

dA dr 1 = −2, and A = 100, = − √ . The dt dt 10 π 1 radius is decreasing at the rate √ cm/min when the 10 π area is 100 cm2 . d A/dt = 1/3 km2 /h, √ = 1/(6πr ) km/h, or √ then (a) dr/dt (b) dr/dt = 1/(6π A/π ) = 1/(6 π A) km/h

6. Let the length, width, and area be l, w, and A at time t.

9. The volume V , surface area S, and edge length x of a cube are related by V = x 3 and S = 6x 2 , so that

10. Let V , r and h denote the volume, radius and height of the cylinder at time t. Thus, V = πr 2 h and dr dV dh = 2πr h + πr 2 . dt dt dt If V = 60,

dV dr = 2, r = 5, = 1, then dt dt V 12 60 = = 2 πr  25π 5π  dV dh 1 dr = − 2πr h dt dt πr 2 dt   1 12 22 = 2 − 10π =− . 25π 5π 25π h=

dA dw = 3, = 0, we have dt dt dl dl 48 ⇒ =− = −4 dt dt 12

The length is decreasing at 4 m/s.

dS dx = 12x . dt dt

If V = 64 cm3 and d V /dt = 2 cm3 /s, then x = 4 cm and d x/dt = 2/(3 × 16) = 1/24 cm/s. Therefore, d S/dt = 12(4)(1/24) = 2. The surface area is increasing at 2 cm2 /s.

dA dw dl =l +w dt dt dt

0 = 16 × 3 + 12

π 3 D , 6

The volume is decreasing at about 28.3 cm3 /h.

5. For A = πr 2 , we have d A/dt = 2πr dr/dt. If

When l = 16, w = 12,

=

dV π = (36)(−0.5) = −9π ≈ −28.3. dt 2

dV dx = 3x 2 , dt dt

When

Thus A = lw.

3

When D = 6 cm, d D/dt = −.5 cm/h. At that time

dr If r = 20 cm and = 4 cm/s, then dt dA = 40π(4) = 160π . dt The area is increasing at 160π cm2 /s.

A π 

D 2

dV π dD = D2 . dt 2 dt

We have

dA dr = 2πr . dt dt

r



where D = 2r is the diameter of the ball. We have

3. Let the radius and area of the ripple t seconds after im-

Then

dr dt

The height is decreasing at the rate

22 cm/min. 25π

109 Copyright © 2014 Pearson Canada Inc.

SECTION 4.1 (PAGE 218)

ADAMS and ESSEX: CALCULUS 8

11. Let the length, width, depth, and volume at time t be l, w, h and V respectively. Thus V = lwh, and dV dl dw dh = wh + lh + lw . dt dt dt dt If l = 6 cm, w =5cm, h= 4cm, dw = −2cm/s, then dt

The √ distance from the origin is increasing at a rate of 2/ 5.

16. From the figure, x 2 + k 2 = s 2 . Thus

dh dl = = 1m/s, and dt dt

dV = 20 − 48 + 30 = 2. dt The volume is increasing at a rate of 2 cm3 /s.

12. Let the length, width and area at time t be x, y and A respectively. Thus A = x y and

x

√ When angle PC A = 45◦ , x = k and s = 2k. The radar gun indicates√that ds/dt = 100 km/h. Thus d x/dt = 100 2k/k ≈ 141. The car is travelling at about 141 km/h. A C x

dA dy dx =x +y . dt dt dt If

dx ds =s . dt dt

k

dA dx = 5, = 10, x = 20, y = 16, then dt dt

s

P

dy dy 31 5 = 20 + 16(10) ⇒ =− . dt dt 4 Thus, the width is decreasing at

13.

Fig. 4.1.16

31 m/s. 4

dx dx dy = 2x . If x = −2 and = −3, y = x 2 . Thus dt dt dt dy = −4(−3) = 12. y is increasing at rate 12. then dt

14. Since x 2 y 3 = 72, then 2x y 3

dx dy dy 2y d x + 3x 2 y 2 =0⇒ =− . dt dt dt 3x dt

dx dy 8 = 2, then = − . Hence, the dt dt 9 8 vertical velocity is − units/s. 9 We have dy dx xy = t ⇒ x +y =1 dt dt dy dx y = tx2 ⇒ = x 2 + 2xt dt dt If x = 3, y = 2,

15.

17. We continue the notation of Exercise 16. If d x/dt √ = 90

km/h, and √ angle PC A = 30◦ ,√then s = 2k, x = 3k, and ds/dt = ( 3k/2k)(90) = 45 3 = 77.94. The radar gun will read about 78 km/h.

18. Let the distances x and y be as shown at time t. Thus x 2 + y 2 = 25 and 2x

dy dx + 2y = 0. dt dt

dx 1 4 dy = and y = 3, then x = 4 and + 3 = 0 so dt 3 3 dt dy 4 =− . dt 9 4 The top of the ladder is slipping down at a rate of 9 m/s. If

At t = 2 we have x y = 2, y = 2x 2 ⇒ 2x 3 = 2 ⇒ x = 1, y = 2. dx dx dy dy +2 = 1, and 1 + 4 = . Thus dt dt dt dt dx dx dy So 1 + 6 =1⇒ =0⇒ = 1 ⇒. dt dt dt p Distance D from origin satisfies D = x 2 + y 2 . So   dD 1 dx dy = p 2x + 2y dt dt dt 2 x 2 + y2  1  2 = √ 1(0) + 2(1) = √ . 5 5

110 Copyright © 2014 Pearson Canada Inc.

5 m

y

x 1/3 m/s

Fig. 4.1.18

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.1 (PAGE 218)

19. Let x and y be the distances shown in the following fig-

We are given that d x/dt = 2 ft/s, so d y/dt = 24/13 ft/s when x = 12 ft. Now the similar triangles in the figure show that s+y s = , 6 15 so that s = 2y/3. Hence ds/dt = 48/39. The woman’s shadow is changing at rate 48/39 ft/s when she is 12 ft from the point on the path nearest the lamppost.

ure. From similar triangles: x x+y 2y dx 2 dy = ⇒x= ⇒ = . 2 5 3 dt 3 dt Since

dy 1 = − , then dt 2 dx 1 d 1 1 5 = − and (x + y) = − − = − . dt 3 dt 2 3 6

Hence, the man’s shadow is decreasing at 13 m/s and the shadow of his head is moving towards the lamppost at a rate of 56 m/s.

21.

x2 C = 10, 000 + 3x + 8, 000   dC x dx = 3+ . dt 4, 000 dt If dC/dt = 600 when x = 12, 000, then d x/dt = 100. The production is increasing at a rate of 100 tons per day.

22. Let x, y be distances travelled by A and B from their positions at 1:00 pm in t hours. dx dy Thus = 16 km/h, = 20 km/h. dt dt Let s be the distance between A and B at time t. Thus s 2 = x 2 + (25 + y)2

5 m 2 m y

x

2s

Fig. 4.1.19

ds dx dy = 2x + 2(25 + y) dt dt dt


At 1:30 t = 21 we have x = 8, y = 10, √ √ s = 82 + 352 = 1289 so

20.

√

1289

ds = 8 × 16 + 35 × 20 = 828 dt

ds 828 = √ ≈ 23.06. At 1:30, the ships are dt 1289 separating at about 23.06 km/h.

15

and

6

y 5

x

s

A

16 km/h

pos. of

A

at 1:00 p.m.

B

at 1:00 p.m.

x 25 km

s

pos. of

Fig. 4.1.20 y

Refer to the figure. s, y, and x are, respectively, the length of the woman’s shadow, the distances from the woman to the lamppost, and the distances from the woman to the point on the path nearest the lamppost. From one of triangles in the figure we have y 2 = x 2 + 25. If x = 12, then y = 13. Moreover, 2y

dy dx = 2x . dt dt

20 km/h

B

Fig. 4.1.22

23. Let θ and ω be the angles that the minute hand and hour hand made with the vertical t minutes after 3 o’clock. Then dθ π = rad/min dt 30 dω π = rad/min. dt 360

111 Copyright © 2014 Pearson Canada Inc.

SECTION 4.1 (PAGE 218)

Since θ = 0 and ω = π θ= t 30

ADAMS and ESSEX: CALCULUS 8

π at t = 0, therefore 2 and

25. Let V , r and h be the volume, radius and height of the cone. Since h = r , therefore

π π ω= t+ . 360 2

V = 31 πr 2 h = 13 π h 3 dV dh dh 1 dV = π h2 ⇒ = . dt dt dt π h 2 dt

At the first time after 3 o’clock when the hands of the clock are together, i.e., θ = ω,

4 Thus, the hands will be together at 16 11 minutes after 3 o’clock.

12 θ

dV 1 dV 1 = and h = 3, then = . Hence, the dt 2 dt 18π 1 height of the pile is increasing at m/min. 18π Let r , h, and V be the top radius, depth, and volume of 10 r and the water in the tank at time t. Then = h 8 1 2 π 25 3 V = πr h = h . We have 3 3 16

If

π π π 180 ⇒ t= t+ ⇒t = . 30 360 2 11

26.

ω

9

1 π 25 2 dh dh 16 = 3h ⇒ = . 10 3 16 dt dt 250π h 2

3

When h = 4 m, we have

6

dh 1 = . dt 250π

The water level is rising at a rate of

Fig. 4.1.23

24. Let y be the height of balloon t seconds after release.

depth is 4 m.

Then y = 5t m. Let θ be angle of elevation at B of balloon at time t. Then tan θ = y/100. Thus

10 m

r

1 dy 5 1 dθ = = = sec θ dt 100 dt 100 20   dθ 1 1 + tan2 θ = dt 20   y 2  dθ 1 1+ = . 100 dt 20 2

dθ 1 dθ 1 When y = 200 we have 5 = so = . dt 20 dt 100 The angle of elevation of balloon at B is increasing at a 1 rate of rad/s. 100

1 m/min when 250π

8 m h

Fig. 4.1.26

27. Let r and h be the radius and height of the water in the tank at time t. By similar triangles, r 10 5 = ⇒ r = h. h 8 4 The volume of water in the tank at time t is V =

1 2 25π 3 πr h = h . 3 48

Thus, y

dV 25π 2 dh dh 16 d V = h ⇒ = . dt 16 dt dt 25π h 2 dt If

θ B

100 m

Fig. 4.1.24

A

dV 1 h3 = − and h = 4, then dt 10 1000   dh 16 1 43 9 = − = . 2 dt (25π )(4) 10 1000 6250π

112 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.1 (PAGE 218)

9 m/min 6250π when the water is 4 m deep. The maximum depth occurs dh when = 0, i.e., dt

x

Hence, the depth of water is increasing at

16 25π h 2



h3 1 − 10 1000



30 m

1 h3 − =0 10 1000 √ 3 ⇒ h = 100.

s

=0⇒

Thus, √ the maximum depth the water in the tank can get is 3 100 ≈ 4.64 m.

28. Let r , h, and V be the top radius, depth, and volume of the water in the tank at time t. Then 3 1 r = = h 9 3 π 3 1 h V = πr 2 h = 3 27 dV π dh = h2 . dt 9 dt If

10 m/min

Fig. 4.1.29

30. Let P, x, and y be your position, height above centre, and horizontal distance from centre at time t. Let θ be the angle shown. Then y = 10 sin θ , and x = 10 cos θ . We have dy dθ = 10 cos θ , dt dt

dθ = 1 rpm = 2π rad/min. dt

6 dy 6 , so = 10 × × 12π . 10 dt 10 You are rising or falling at a rate of 12π m/min at the time in question. When x = 6, then cos θ =

dh 2 = 20 cm/h = m/h when h = 6 m, then dt 10

P

dV π 2 4π = × 36 × = ≈ 2.51 m3 /h. dt 9 10 5

10 m

m3 /h,

Since water is coming in at a rate of 10 it must be leaking out at a rate of 10 − 2.51 ≈ 7.49 m3 /h.

y

θ C

x

3 m

r 9 m

Fig. 4.1.30 h

31. Let x and y denote the distances of the two aircraft east and north of the airport respectively at time t as shown in the following diagram. Also let the distance between the two aircraft be s, then s 2 = x 2 + y 2 . Thus, Fig. 4.1.28

2s

29. Let x and s be the distance as shown. Then s 2 = x 2 + 302 and 2s

dx dy = −200 and = 150 when x = 144 and dt √ dt y = 60, we have s = 1442 + 602 = 156, and

Since

ds dx ds x dx = 2x ⇒ = . dt dt dt s dt

√ dx = 10, s = 402 + 302 = 50, then dt ds 40 = (10) = 8. Hence, one must let out line at 8 dt 50 m/min. When x = 40,

ds dx dy = 2x + 2y . dt dt dt

1 ds = [144(−200) + 60(150)] ≈ −126.9. dt 156 Thus, the distance between the aircraft is decreasing at about 126.9 km/h.

113 Copyright © 2014 Pearson Canada Inc.

SECTION 4.1 (PAGE 218)

ADAMS and ESSEX: CALCULUS 8

34. Let x and y be the distances travelled from the intersec-

150 km/h

tion point by the boat and car respectively in t minutes. Then

s y

1000 1000 dx = 20 × = m/min dt 60 3 dy 1000 4000 = 80 × = m/min dt 60 3 The distance s between the boat and car satisfy

x airport

200 km/h

Fig. 4.1.31

32.

ds dx dy =x +y . dt dt dt 1000 4000 After one minute, x = , y= so s ≈ 1374. m. 3 3 Thus ds 1000 1000 4000 4000 1374.5 = + ≈ 1, 888, 889. dt 3 3 3 3 ds Hence ≈ 1374.2 m/min ≈ 82.45 km/h after 1 minute. dt s 2 = x 2 + y 2 + 202 ,

1 0.6 0.4 x y 3 dP 0.6 −0.4 0.4 d x 0.4 0.6 −0.6 d y = x y + x y . dt 3 dt 3 dt P=

If d P/dt = 0, x = 40, d x/dt = 1, and y = 10, 000, then dy 6y 0.4 y 0.6 d x 6y d x = − 0.4 =− = −375. dt x 4x 0.6 dt 4x dt

s

The daily expenses are decreasing at $375 per day. y 20 m

Car

33. Let the position of the ant be (x, y) and the position of its shadow be (0, s). By similar triangles,

x s

y 3y s−y = ⇒s= . x 3−x 3−x

Boat

Then, dy dx 3(3 − x) + 3y ds dt dt . = dt (3 − x)2 If the ant is at (1, 2) and ds = dt

1 dy 1 dx = , = − , then dt 3 dt 4

3(2)(− 14 ) + 3(2)( 13 ) 4

Hence, the ant’s shadow is moving at along the y-axis.

1 8

=

Fig. 4.1.34

35. Let h and b (measured in metres) be the depth and the surface width of the water in the trough at time t. We have √ 2 h = tan 60◦ = 3 ⇒ b = √ h. 1 ( 2 b) 3

1 . 8

units/s upwards

y S

Thus, the volume of the water is   1 10 V = hb (10) = √ h 2 , 2 3 and

y

ant

dV 1 = and h = 0.2 metres, then dt 4 √   √ dh 3 1 3 = = . dt 20(0.2) 4 16 √ 3 Hence, the water level is rising at m/min. 16

If

x Fig. 4.1.33

3 lamp

x

√ 20 dh dh dV 3 dV =√ h ⇒ = . dt dt 20h dt 3 dt

114 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.1 (PAGE 218)

60◦ b/2

10 m

b/2

30 cm

h

√

y

30◦

3 m

32 +x 2

x s

Fig. 4.1.35

Fig. 4.1.37 a) By similar triangles,

36. Let V and h be the volume and depth of water in the

dy dx −30x dx dy = = . 2 3/2 dt d x dt (9 + x ) dt dx 1 If x = 4 and = , then dt 5   dy −30(4) 1 24 = . =− dt (9 + 16)3/2 5 125

pool at time t. If h ≤ 2, then x 20 = = 10, h 2

so V =

1 xh8 = 40h 2 . 2

If 2 ≤ h ≤ 3, then V = 160 + 160(h − 2). a) If h = 2.5m, then −1 =

dV dh = 160 . dt dt

So surface of water is dropping at a rate of m/min.

Thus,

y 3 30 = √ ⇒y= √ . 2 2 10 3 +x 9 + x2

Hence, the free top end of the ladder is moving vertically downward at 24/125 m/s.

1 160

b) By similar triangles, √

dV dh dh b) If h = 1m, then −1 = = 80h = 80 . dt dt dt So surface of water is dropping at a rate of 1 m/min. 80 20 8 3

1

x 32

Then,

+

x2

=

10x s ⇒s= √ . 10 9 + x2

ds d x ds = dt d x dt   √ 2x ( 9 + x 2 )(10) − (10x) √ 2 9 + x2 dx = 2 (9 + x ) dt 90 dx . = (9 + x 2 )3/2 dt dx 1 = , then dt 5   90 18 ds 1 = = . dt (9 + 16)3/2 5 125

If x = 4 and

x h

This is the rate of change of the length of the horizontal projection of the ladder. The free top end of the ladder is moving horizontally to the right at rate dx ds 1 18 7 − = − = m/s. dt dt 5 125 125

Fig. 4.1.36

38. Let x, y, and s be distances shown at time t. Then

37. Let the various distances be as shown in the figure.

s 2 = x 2 + 16, ds dx s =x , dt dt

(15 − s)2 = y 2 + 16 ds dy − (15 − s) =y . dt dt

115 Copyright © 2014 Pearson Canada Inc.

SECTION 4.1 (PAGE 218)

ADAMS and ESSEX: CALCULUS 8

dx 1 When x = 3 and = , then s = 5 and dt 2 √ √ 2 = 84. y = 102 − 4  3 1 3 ds = so = Also dt 5 2 10

40. Let y be height of ball t seconds after it drops. Thus

dy 10 3 3 = −√ = − √ ≈ 0.327. dt 84 10 84 Crate B is moving toward Q at a rate of 0.327 m/s.

d2 y dy = −9.8, |t=0 = 0, y|t=0 = 20, and dt 2 dt dy = −9.8t. dt

y = −4.9t 2 + 20,

Let s be distance of shadow of ball from base of pole. s s − 10 = . y 20 200 20s − 200 = sy, s = 20 − y ds dy ds =y +s . 20 dt dt dt

By similar triangles,

P

15−s B

4

y

x

A

Q

Fig. 4.1.38

39. Let θ be the angle of elevation, and x and y the horizontal and vertical distances from the launch site. We have

tan θ =

y x

dx dy −y x dθ sec2 θ = dt 2 dt . dt x

⇒

dy = −9.8, y = 15.1, dt ds 200 4.9 = (−9.8). dt 4.9 The shadow is moving at a rate of 81.63 m/s after one second.

a) At t = 1, we have

s

At the instant in question

b) As the r ball hits the ground, y =r0, s = 10, 20 dy 20 t = , and = −9.8 , so 4.9 dt 4.9 ds dy 20 = 0 + 10 . dt dt r 20 Now y = 0 implies that t = . Thus 4.9 r 1 ds 20 = − (9.8) ≈ −9.90. dt 2 4.9 The shadow is moving at about 9.90 m/s when the ball hits the ground. 10 m

√ dy dx = 4 cos 30◦ = 2 3, = 4 sin 30◦ = 2, dt dt x = 50 km, y = 100 km. Thus tan θ =

100 = 2, sec2 θ = 1 + tan2 θ = 5, and 50

20−y

20 m

√ √ dθ 1 50(2) − 100(2 3) 1−2 3 = = ≈ −0.0197. dt 5 (50)2 125

y

s−10

10

Therefore, the angle of elevation is decreasing at about 0.0197 rad/s.

s

Fig. 4.1.40

4 km/s 30◦ y

41. Let y(t) be the height of the rocket t seconds after it blasts off. We have

θ x

d2 y = 10, dt 2

Fig. 4.1.39

116 Copyright © 2014 Pearson Canada Inc.

dy =y=0 dt

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.2 (PAGE 227)

at t = 0. Hence y = 5t 2 , (y in metres, t in seconds). Now dθ d y/dt y tan θ = , so sec2 θ = , and 2000 dt 2000   y 2  dθ 10t t = = 1+ 2000 dt 2000 200 t 1 dθ = · dt 200 25t 4 1+ 20002 t 1 800t = · . = 200 4002 + t 4 t4 1+ 4002 At t = 10, we have

Starting with x0 = 1.5, get x4 = x5 = 1.73205080757.

9.

f (x) = x 3 + 2x − 1, f ′ (x) = 3x 2 + 2. Newton’s formula xn+1 = g(xn ), where g(x) = x −

2x 3 + 1 x 3 + 2x − 1 = 2 . 2 3x + 2 3x + 2

Starting with x0 = 0.5, get x3 = x4 = 0.45339765152.

10.

dθ 8000 = ≈ 0.047 rad/s. dt 4002 + 1002

f (x) = x 3 + 2x 2 − 2, f ′ (x) = 3x 2 + 4x. Newton’s formula xn+1 = g(xn ), where g(x) = x −

2x 3 + 2x 2 + 2 x 3 + 2x 2 − 2 = . 3x 2 + 4x 3x 2 + 4x

Starting with x0 = 1.5, get x5 = x6 = 0.839286755214.

y θ

11.

2 km

Fig. 4.1.41

Section 4.2 Finding Roots of Equations (page 227)

g(x) = x −

2. To solve 1 + xn+1 = 1 +

1 4

1 4

sin x = x, start with x0 = 1 and iterate sin xn . x5 and x6 round to 1.23613.

12.

3. To solve cos(x/3) = x, start with x0 = 0.9 and iterate xn+1 = cos(xn /3). x4 and x5 round to 0.95025.

xn+1 = (xn + 9)1/3 . x4 and x5 round to 2.24004.

xn+1 = 1/(2 + xn2 ). x6 and x7 round to 0.45340.

6. To solve x 3 + 10x − 10 = 0, start with x0 = 1 and iterate 7.

x2

x7 and x8 round to 0.92170.

f ′ (x)

f (x) = − 2, = 2x. Newton’s formula xn+1 = g(xn ), where g(x) = x −

x2 − 2 x2 + 2 = . 2x 2x

13.

f (x) = sin x − 1 + x, f ′ (x) = cos x + 1. Newton’s formula is xn+1 = g(xn ), where

Starting with x0 = 1.5, get x3 = x4 = 1.41421356237.

8.

f (x) = x 2 − 3, f ′ (x) = 2x. Newton’s formula xn+1 = g(xn ), where

x2 − 3 x2 + 3 g(x) = x − = . 2x 2x

x 3 + 3x 2 − 1 2x 3 + 3x 2 + 1 = . 2 3x + 6x 3x 2 + 6x

Because f (−3) = −1, f (−2) = 3, f (−1) = 1, f (0) = −1, f (1) = 3, there are roots between −3 and −2, between −1 and 0, and between 0 and 1. Starting with x0 = −2.5, get x5 = x6 = −2.87938524157. Starting with x0 = −0.5, get x4 = x5 = −0.652703644666. Starting with x0 = 0.5, get x4 = x5 = 0.532088886328.

5. To solve 1/(2 + x 2 ) = x, start with x0 = 0.5 and iterate 1 3 10 x n .

f (x) = x 3 + 3x 2 − 1, f ′ (x) = 3x 2 + 6x. Newton’s formula xn+1 = g(xn ), where g(x) = x −

4. To solve (x + 9)1/3 = x, start with x0 = 2 and iterate

xn+1 = 1 −

3x 4 − 8x 2 − 16 x 4 − 8x 2 − x + 16 = . 3 4x − 16x − 1 4x 3 − 16x − 1

Starting with x0 = 1.5, get x4 = x5 = 1.64809536561. Starting with x0 = 2.5, get x5 = x6 = 2.35239264766.

1 2 round to 0.35173.

1. Iterate xn+1 = e−xn starting with x + 0 = 0.3. Both x10 and x11

f (x) = x 4 − 8x 2 − x + 16, f ′ (x) = 4x 3 − 16x − 1. Newton’s formula xn+1 = g(xn ), where

g(x) = x −

sin x − 1 + x . cos x + 1

The graphs of sin x and 1−x suggest a root near x = 0.5. Starting with x0 = 0.5, get x3 = x4 = 0.510973429389.

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SECTION 4.2 (PAGE 227)

ADAMS and ESSEX: CALCULUS 8

y

y y =1−x y = sin x

y=x π

0.5

1.0

1.5x

x

Fig. 4.2.13

14.

y = tan x

f (x) = x 2 − cos x, f ′ (x) = 2x + sin x. Newton’s formula is xn+1 = g(xn ), where

Fig. 4.2.15

16. A graphing calculator shows that the equation

x 2 − cos x g(x) = x − . 2x + sin x

√ (1 + x 2 ) x − 1 = 0

The graphs of cos x and x 2 , suggest a root near x = ±0.8. Starting with x0 = 0.8, get x3 = x4 = 0.824132312303. The other root is the negative of this one, because cos x and x 2 are both even functions. y

18.

y = x2

(1 + x 2 ) cos x − 2x sin x (1 + x 2 )2 2 0 = (1 + x ) cos x − 2x sin x 0 = f ′ (x) =

y = cos x -1.5 -1.0 -0.5 0.5 1.0 1.5x Fig. 4.2.14

19.

15. Since tan x takes all real values between any two consecutive odd multiples of π/2, its graph intersects y = x infinitely often. Thus, tan x = x has infinitely many solutions. The one between π/2 and 3π/2 is close to 3π/2, so start with x0 = 4.5. Newton’s formula here is xn+1 = xn −

has a root near x = 0.6. Use of a solve routine or Newton’s Method gives x = 0.56984029099806. sin x . Since | f (x)| ≤ 1/(1 + x 2 ) → 0 Let f (x) = 1 + x2 as x → ±∞ and f (0) = 0, the maximum and minimum values of f will occur at the two critical points of f that are closest to the origin on the right and left, respectively. For CP:

tan xn − xn . sec2 xn − 1

We get x3 = x4 = 4.49340945791.

with 0 < x < π for the maximum and −π < x < 0 for the minimum. Solving this equation using a solve routine or Newton’s Method starting, say, with x0 = 1.5, we get x = ±0.79801699184239. The corresponding max and min values of f are ±0.437414158279. cos x Let f (x) = . Note that f is an even function, and 1 + x2 that f has maximum value 1 at x = 0. (Clearly f (0) = 1 and | f (x)| < 1 if x 6= 0.) The minimum value will occur at the critical points closest to but not equal to 0. For CP: (1 + x 2 )(− sin x) − 2x cos x (1 + x 2 )2 2 0 = (1 + x ) sin x + 2x cos x. 0 = f ′ (x) =

The first CP to the right of zero is between π/2 and 3π/2, so start with x = 2.5, say, and get x = 2.5437321475261. The minimum value is f (x) = −0.110639672192.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.2 (PAGE 227)

20. For x 2 = 0 we have xn+1 = xn − (xn2 /(2xn )) = xn /2.

It follows that

If x0 = 1, then x1 = 1/2, x2 = 1/4, x3 = 1/8.

2 2xn xn2 + 1 1 − yn = 4yn2 = 4yn (1 − yn ). yn

a) xn = 1/2n , by induction.

yn+1 =

b) xn approximates the root x = 0 to within 0.0001 provided 2n > 10, 000. We need n ≥ 14 to ensure this. c) To ensure that xn2 is within 0.0001 of 0 we need (1/2n )2 < 0.0001, that is, 22n > 10, 000. We need n ≥ 7.

21.

d) Convergence of Newton approximations to the root x = 0 of x 2 = 0 is slower than usual because the derivative 2x of x 2 is zero at the root. √ x if x ≥ 0 , f (x) = √ −x if x 0 f ′ (x) = . −1/(2 −x) if x < 0 The Newton’s Method formula says that xn+1 = xn −

22. Newton’s Method formula for f (x) = x 1/3 is = xn − 3xn = −2xn .

23. Newton’s Method formula for f (x) = x 2/3 is

1 , we have 1 + xn2

xn

−1/3 (2/3)xn

= xn − 32 xn = − 12 xn .

If x0 = 1, then x1 = −1/2, x2 = 1/4, x3 = −1/8, x4 = 1/16, and, in general, xn = (−1/2)n . The successive approximations oscillate around the root x = 0, but still converge to it (though more slowly than is usual for Newton’s Method). xn2 − 1 , we have 2xn

2 1 + xn+1 =1+



xn2 − 1 2xn

2

=



xn2 + 1 2xn

2xn d xn . (1 + xn2 )2

Hence d xn = −

(1 + xn2 )2 2 sin(u n ) cos(u n ) 2n du 0 . 2xn

Since the values of xn are assumed to neither converge nor diverge, the exponential factor 2n will dominate for large n (because f is), and since a ≤ f (x) ≤ b whenever a ≤ x ≤ b (by condition (i)), we know that g(a) ≥ 0 and g(b) ≤ 0. By the Intermediate-Value Theorem there exists r in [a, b] such that g(r ) = 0, that is, such that f (r ) = r . The fixed point r is unique because if there were two such fixed points, say r1 and r2 , then condition (ii) would imply that |r1 − r2 | = | f (r1 ) − f (r2 )| ≤ K |r1 − r2 |,

2/3

24. Since xn+1 =

b) Since yn =

26. Let g(x) = f (x) − x for a ≤ x ≤ b. g is continuous

If x0 = 1, then x1 = −2, x2 = 4, x3 = −8, x4 = 16, and, in general, xn = (−2)n . The successive “approximations” oscillate ever more widely, diverging from the root at x = 0.

xn+1 = xn −

a) Since sin2 (u n+1 ) = 4 sin2 (u n )(1−sin2 (u n )) = 4 sin2 (u n ) cos2 (u n ) = sin2 (2u n ), we have u n+1 = 2u n . Thus u n+1 = 2n u 0 . It follows that d yn = 2 sin(u n ) cos(u n ) 2n du 0 .

d yn = −

If x0 = a, then x1 = −a, x2 = a, and, in general, xn = (−1)n a. The approximations oscillate back and forth between two numbers. If one observed that successive approximations were oscillating back and forth between two values a and b, one should try their average, (a + b)/2, as a new starting guess. It may even turn out to be the root!

xn+1 = xn −

25. Let y j = sin2 (u j ).

f (xn ) = xn − 2xn = −xn . f ′ (xn )

1/3 xn −2/3 (1/3)xn



2

.

which is impossible if r1 6= r2 and K < 1.

27. We are given that there is a constant K satisfying 0 < K < 1, such that | f (u) − f (v)| ≤ K |u − v| holds whenever u and v are in [a, b]. Pick any x0 in [a, b], and let x1 = f (x0 ), x2 = f (x1 ), and, in general, xn+1 = f (xn ). Let r be the fixed point of f in [a, b] found in Exercise 24. Thus f (r ) = r . We have |x1 − r | = | f (x0 ) − f (r )| ≤ K |x0 − r |

|x2 − r | = | f (x1 ) − f (r )| ≤ K |x1 − r | ≤ K 2 |x0 − r |,

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SECTION 4.2 (PAGE 227)

ADAMS and ESSEX: CALCULUS 8

and, in general, by induction

8.

|xn − r | ≤ K n |x0 − r |. Since K < 1, limn→∞ K n = 0, so limn→∞ xn = r . The iterates converge to the fixed point as claimed in Theorem 6.

Section 4.3 Indeterminate Forms (page 233) 9. 1.

2.

3.

4.

5.

6.

7.

  0 0 3 3 = = lim x→0 4 sec2 4x 4   ln(2x − 3) 0 lim x→2 x 2 − 4 0   2 1 2x − 3 = = . 2x 2   sin ax 0 lim x→0 sin bx 0 a cos ax a = lim = x→0 b cos bx b   0 1 − cos ax lim x→0 1 − cos bx 0   a sin ax 0 = lim x→ 0 b sin bx 0 a2 a 2 cos ax = lim 2 = 2. x→0 b cos bx b 3x lim x→0 tan 4x

sin−1

10.

11.

12.

13.   0 0

x lim x→0 tan−1 x 1 + x2 = lim √ =1 x→0 1 − x2

lim x cot x [0 × ∞]  x  = lim cos x x→0 sin x   x 0 = 1 × lim x→0 sin x 0 1 = lim =1 x→0 cos x

  sin2 t 0 t→π t − π 0 2 sin t cos t =0 = lim t→π 1 lim

  10x − e x 0 x→0 x 0 10x ln 10 − e x = lim = ln 10 − 1. x→0 1 lim

  0 cos 3x lim x→π/2 π − 2x 0 −3 sin 3x 3 3 = lim = (−1) = − x→π/2 −2 2 2   0 ln(ex) − 1 x→1 sin π x 0 1 1 x = lim =− . x→1 π cos(π x) π lim

lim x sin

x→∞

1 x

sin = lim

[∞ × 0] 1 x

  0 0

1 x 1 1 − 2 cos 1 x x = lim = lim cos = 1. 1 x→∞ x→∞ x − 2 x x→∞

  0 x 1/3 − 1 lim x→1 x 2/3 − 1 0 1 −2/3 ( )x 1 = . = lim 32 x→1 ( )x −1/3 2 3 x→0

  1 − cos x 0 2 x→0 ln(1 + x ) 0 sin x  = lim  2x x→0 1 + x2 sin x = lim (1 + x 2 ) lim x→0 x→0 2x cos x 1 = lim = . x→0 2 2 lim

14.

  x − sin x 0 x→0 x3 0   0 1 − cos x = lim 2 x→0 3x 0   sin x 0 = lim x→0 6x 0 cos x 1 = lim = . x→0 6 6 lim

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INSTRUCTOR’S SOLUTIONS MANUAL

15.

16.

17.

18.

19.

20.

21.

  x − sin x 0 x→0 x − tan x 0   1 − cos x 0 = lim x→0 1 − sec2 x 0 1 − cos x 2 = lim (cos x) 2 x→0 cos x − 1 cos x − 1 = −1 × lim x→0 (cos x − 1)(cos x + 1) 1 =− 2 lim

  2 − x 2 − 2 cos x 0 x→0 0 x4   −2x + 2 sin x 0 = lim x→0 4x 3 0 x − sin x 1 = − lim 3 x→0 2  x 1 1 1 =− (by Exercise 14). =− 2 6 12

SECTION 4.3 (PAGE 233)

22.

23.

24.

  sin2 x 0 x→0+ tan x − x 0   0 2 sin x cos x = lim x→0+ sec2 x − 1 0 cos x =∞ = 2 × 1 × lim x→0+ 2 sec2 x tan x

lim

t→(π/2)−



 1 1 − at (∞ − ∞) t→0 t te   0 eat − 1 = lim t→0 te at 0 aeat = lim at =a t→0 e + ate at lim

√

ln x x ln x = lim −1/2 x→0+ x→0+ x   1 x  = 0, = lim  1 x→0+ x −3/2 − 2

Since lim

lim

hence lim x

√

  0 0

x

x→0+

= lim e

√

x ln x

x→0+

= e0 = 1.

2

25. Let y = (csc x)sin x .

  ln sin r 0 lim r→π/2 cos r 0  cos r  sin r = 0. = lim r→π/2 − sin r lim

[∞ − ∞]   0 0

1 − sin t cos t − cos t = lim = 0. t→(π/2)− − sin t =

lim

t→π/2

(sec t − tan t)

lim

t→(π/2)−

Then ln y = sin2 x ln(csc x) h∞i ln(csc x) lim ln y = lim 2 x→0+ x→0+ csc x ∞ csc x cot x − csc x = lim x→0+ −2 csc2 x cot x 1 = 0. = lim x→0+ 2 csc2 x 2x sin Thus lim x→0+ (csc x) = e0 = 1.

sin t 2 = t π

  cos−1 x 0 x→1− x − 1 0   1 − √ 1 − x 2 = −∞. = lim x→1− 1 lim

lim x(2 tan−1 x − π )

x→∞

2 tan−1

x −π 1 x 2 . 1 = lim − 2 x→∞ 1 + x 2 x 2x 2 = lim − = −2 x→∞ 1 + x 2 = lim

x→∞

[0 × ∞]   0 0

26.

 1 x − [∞ − ∞] x→1+ x − 1 ln x   x ln x − x + 1 0 = lim x→1+ (x − 1)(ln x) 0   ln x 0 = lim 1 x→1+ 0 ln x + 1 − x 1 x = lim 1 x→1+ 1 + 2 x x x 1 = lim = = . x→1+ x +1 2 lim



121 Copyright © 2014 Pearson Canada Inc.

SECTION 4.3 (PAGE 233)

27.

28.

ADAMS and ESSEX: CALCULUS 8

  3 sin t − sin 3t 0 t→0 3 tan t − tan 3t 0   3(cos t − cos 3t) 0 = lim t→0 3(sec2 t − sec2 3t) 0 cos t − cos 3t = lim t→0 cos2 3t − cos2 t cos2 t cos2 3t cos 3t − cos t = − lim t→0 cos2 3t − cos2 t 1 1 =− = − lim t→0 cos 3t + cos t 2  sin x 1/x 2 Let y = . x  sin x    ln 0 x lim ln y = lim x→0 x→0 x2 0  x   x cos x − sin x  sin x x2 = lim x→0 2x   0 x cos x − sin x = lim 2 x→0 2x sin x 0 −x sin x = lim x→0 4x sin x + 2x 2 cos x   − sin x 0 = lim x→0 4 sin x + 2x cos x 0 1 − cos x = lim =− . x→0 6 cos x − 2x sin x 6  sin x 1/x 2 −1/6 Thus, lim =e . x→0 x

31.

lim

32.

33.

34.

2

ln(cos 2t) . We have t2   ln(cos 2t) 0 2 t→0 t 0   −2 tan 2t 0 = lim t→0 2t 0 2 sec2 2t = − lim = −2. t→0 1

lim ln y = lim

t→0

Let y = (1 + tan x)1/x .   0 ln(1 + tan x) lim ln y = lim x→0 x→0 x 0 sec2 x = lim = 1. x→0 1 + tan x Thus, lim (1 + tan x)1/x = e. x→0

29. Let y = (cos 2t)1/t . Then ln y =

h∞i ln sin π x x→1− csc π x ∞ π cos π x sin π x = lim x→1− −π csc π x cot π x −π = lim tan π x = 0 π x→1− lim

  f (x + h) − 2 f (x) + f (x − h) 0 h→0 0 h2   f ′ (x + h) − f ′ (x − h) 0 = lim h→0 2h 0 f ′′ (x + h) + f ′′ (x − h) = lim h→0 2 2 f (x) ′′ = = f (x) 2 lim

f (x + 3h) − 3 f (x + h) + 3 f (x − h) − f (x − 3h) h3 ′ ′ 3 f (x + 3h) − 3 f (x + h) − 3 f ′ (x − h) + 3 f ′ (x − 3h) = lim h→0 3h 2 ′′ ′′ 3 f (x + 3h) − f (x + h) + f ′′ (x − h) − 3 f ′′ (x − 3h) = lim h→0 2h 9 f ′′′ (x + 3h) − f ′′′ (x + h) − f ′′′ (x − h) + 9 f ′′′ (x − 3h) = lim h→0 2 =8 f ′′′ (x). lim

h→0

35. Suppose that f and g are continuous on [a, b] and differentiable on (a, b) and g(x) 6= 0 there. Let a < x < t < b, and apply the Generalized Mean-Value Theorem; there exists c in (x, t) such that

2

30.

Therefore limt→0 (cos 2t)1/t = e−2 . csc x h ∞ i − lim x→0+ ln x ∞ − csc x cot x h ∞ i = lim − 1 x→0+ ∞ x   −x cos x 0 = lim x→0+ sin2 x 0   1 = − lim cos x lim x→0+ x→0+ 2 sin x cos x = −∞.

⇒ ⇒ ⇒ ⇒ ⇒

f (x) − f (t) f ′ (c) = ′ g(x) − g(t) g (c)    f ′ (c) f (x) − f (t) g(x) = ′ g(x) g(x) − g(t) g (c)   f (x) f (t) f ′ (c) g(x) − g(t) − = ′ g(x) g(x) g (c) g(x) f (x) f ′ (c) g(t) f ′ (c) f (t) = ′ − + ′ g(x) g (c) g(x) g (c) g(x)   f (x) f ′ (c) 1 f ′ (c) = ′ + f (t) − g(t) ′ g(x) g (c) g(x) g (c)   f (x) f ′ (c) 1 f ′ (c) −L = ′ −L+ f (t) − g(t) ′ . g(x) g (c) g(x) g (c)

122 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.4 (PAGE 239)

Since |m + n| ≤ |m| + |n|, therefore, ′ ′   f (c) f (c) f (x) 1 g(x) − L ≤ g ′ (c) − L + |g(x)| | f (t)|+|g(t)| g ′ (c) .

Now suppose that ǫ is an arbitrary small positive number. Since limc→a+ f ′ (c)/g ′ (c) = L, and since a < x < c < t, we can choose t sufficiently close to a to ensure that

In particular,

′ f (c) ǫ < . − L g ′ (c) 2

8.

f (x) = x 3 + x − 4 on (a, b) Since f ′ (x) = 3x 2 + 1 > 0 for all x, therefore f is increasing. Since (a, b) is open, f has no max or min values.

9.

f (x) = x 5 + x 3 + 2x on (a, b] f ′ (x) = 5x 4 + 3x 2 + 2 > 0 for all x. f has no min value, but has abs max value b5 + b3 + 2b at x = b.

10.

′ f (c) ǫ g ′ (c) < |L| + 2 .

11.

Since lim x→a+ |g(x)| = ∞, we can choose x between a and t sufficiently close to a to ensure that  1 h ǫ i ǫ | f (t)| + |g(t)| |L| + < . |g(x)| 2 2

12.

It follows that

Thus lim x→a+

1 on [2, 3] x −1 1 abs min 2 at x = 3, abs max 1 at x = 2. f (x) =

f ′ (x) = sgn (x − 1). No CP; SP x = 1, f (1) = 0. Max value of f is 3 at x = −2; min value is 0 at x = 1.

f (x) = L. g(x)

14. Let f (x) = |x 2 − x − 2| = |(x − 2)(x + 1)| on [−3, 3]: f (−3) = 10, f (3) = 4. f ′ (x) = (2x − 1)sgn (x 2 − x − 2). CP x = 1/2; SP x = −1, and x = 2. f (1/2) = 9/4, f (−1) = 0, f (2) = 0. Max value of f is 10 at x = −3; min value is 0 at x = −1 or x = 2.

(page 239)

1.

f (x) = x + 2 on [−1, 1] f ′ (x) = 1 so f is increasing. f has absolute minimum 1 at x = −1 and absolute maximum 3 at x = 1.

2.

f (x) = x + 2 on (−∞, 0] abs max 2 at x = 0, no min.

3.

f (x) = x + 2 on [−1, 1) f has absolute minimum 1 at x = −1 and has no absolute maximum.

4.

f (x) = x 2 − 1 no max, abs min −1 at x = 0.

5.

f (x) = x 2 − 1 on [−2, 3] f has abs min −1 at x = 0, abs max 8 at x = 3, and local max 3 at x = −2.

6.

f (x) = x 2 − 1 on (2, 3) no max or min values.

7.

1 on (0, 1) x −1 1 f ′ (x) = − < 0 on (0, 1) (x − 1)2 f has no max or min values. f (x) =

13. Let f (x) = |x − 1| on [−2, 2]: f (−2) = 3, f (2) = 1.

ǫ f (x) < + ǫ = ǫ. − L 2 2 g(x)

Section 4.4 Extreme Values

1 −1 . Since f ′ (x) = < 0 for all x in x −1 (x − 1)2 the domain of f , therefore f has no max or min values. f (x) =

1 2x , f ′ (x) = − 2 x2 + 1 (x + 1)2 f has abs max value 1 at x = 0; f has no min values.

15.

f (x) =

16.

f (x) = (x + 2)(2/3) no max, abs min 0 at x = −2.

17.

f (x) = (x − 2)1/3 , f ′ (x) =

1 (x − 2)−2/3 > 0 3 f has no max or min values. y

2

f (x) = x 3 + x − 4 on [a, b] f ′ (x) = 3x 2 + 1 > 0 for all x. Therefore f has abs min a 3 + a − 4 at x = a and abs max b3 + b − 4 at x = b.

x

y = (x − 2)1/3 Fig. 4.4.17

123 Copyright © 2014 Pearson Canada Inc.

SECTION 4.4 (PAGE 239)

18.

ADAMS and ESSEX: CALCULUS 8

f (x) = x 2 + 2x, f ′ (x) = 2x + 2 = 2(x + 1) Critical point: x = −1. f (x) → ∞ as x → ±∞.

20.

f (x) = (x 2 − 4)2 , f ′ (x) = 4x(x 2 − 4) = 4x(x + 2)(x − 2) Critical points: x = 0, ±2. f (x) → ∞ as x → ±∞. CP CP CP f′ − −2 + 0 − +2 + −−−−−−−−−−|−−−−−−−−−−|−−−−−−−−−− | −−−−−−→x abs loc abs f ց min ր max ց min ր

CP f′ − −1 + −−−−−−−−−−−−−|−−−−−−−−− →x abs f ց ր min

Hence, f (x) has abs min 0 at x = ±2 and loc max 16 at x = 0.

Hence, f (x) has no max value, and the abs min is −1 at x = −1.

y

y

y = (x 2 − 4)2

16

y = x 2 + 2x

x x

Fig. 4.4.20

Fig. 4.4.18

21.

19.

2

−2

(−1,−1)

f (x) = x 3 (x − 1)2 f ′ (x) = 3x 2 (x − 1)2 + 2x 3 (x − 1)

= x 2 (x − 1)(5x − 3) 3 CP x = 0, , 1 5

x3

f (x) = − 3x − 2 f ′ (x) = 3x 2 − 3 = 3(x − 1)(x + 1) CP CP + −1 − 1 + −−−−−−−−−−−− | −−−−−−−−−−−|−−−−−−−−→x loc loc f ր ց ր max min f′

f has no absolute extrema.

CP CP CP 3 f′ + 0 + − 1 + 5 −−−−−−−−−|−−−−−−−−−− | −−−−−−−−−|−−−−−−→x loc loc f ր ր max ց min ր f has no absolute extrema.

y

y

−1

x



3 108 5 , 55

 1

y = x 3 − 3x − 2

y = x 3 (x − 1)2

(1,−4)

Fig. 4.4.19

Fig. 4.4.21

124 Copyright © 2014 Pearson Canada Inc.

x

INSTRUCTOR’S SOLUTIONS MANUAL

22.

SECTION 4.4 (PAGE 239)

f (x) = x 2 (x − 1)2 , f ′ (x) = 2x(x − 1)2 + 2x 2 (x − 1) = 2x(2x − 1)(x − 1) Critical points: x = 0, 12 and 1. f (x) → ∞ as x → ±∞.

24.

1 16

at x =

1 2

x2

CP CP f′ − −1 + +1 − −−−−−−−−−−−− | −−−−−−−−−−−|−−−−−−−−→x abs abs f ց ր ց min max

CP CP CP 1 − 0 + − 1 + 2 −−−−−−−−−−|−−−−−−−−−−|−−−−−−−−−− | −−−−−−→x abs loc abs f ց min ր max ց min ր f′

Hence, f (x) has loc max x = 0 and x = 1.

x 1 − x2 , f ′ (x) = 2 +1 (x + 1)2 Critical point: x = ±1. f (x) → 0 as x → ±∞. f (x) =

Hence, f has abs max x = −1.

and abs min 0 at

y

1 2

at x = 1 and abs min − 21 at y (1,0.5)

y=

(−1,−0.5)

y = x 2 (x − 1)2




x

1

x2

CP f′ − 0 + −−−−−−−−−−− | −−−−−−→x abs f ց min ր

f (x) = x(x 2 − 1)2

f ′ (x) = (x 2 − 1)2 + 2x(x 2 − 1)2x

y

= (x 2 − 1)(x 2 − 1 + 4x 2 )

y=1

= (x 2 − 1)(5x 2 − 1) √ √ = (x − 1)(x + 1)( 5x − 1)( 5x + 1)

y=

CP CP CP CP f ′ + −1 − − √1 + √1 − 1 + 5 5 −−−−−−−−|−−−−−−−−|−−−−−−−−|−−−−−−−− | −−−−→x loc ց loc ր loc ց loc ր f ր max min max min √ √ f (±1) = 0, f (±1/ 5) = ±16/25 5

√ −1/ 5

√ 1/ 5

y = x(x 2 − 1)2 Fig. 4.4.23

1

x

x2 x2 + 1 x

Fig. 4.4.25

26.

y

−1

x

x2 1 =1− 2 0) x x − ln x 1 − ln x f ′ (x) = x 2 = x x2 f (x) → −∞ as x → 0+ (vertical asymptote), f (x) → 0 as x → ∞ (horizontal asymptote). ASY CP f′ 0 + e − −−−−− |−−−−−−−− | −−−−→x abs ց f ր max f (x) =

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SECTION 4.4 (PAGE 239)



y

1 e, e

ADAMS and ESSEX: CALCULUS 8



38.

π 3π 5π f ′ (x) = sgn (x) cos |x| = 0 at x = ± , ± , ± , ... 2 2 2 0 is a singular point. Since f (x) is an even function, its graph is symmetric about the origin.

x

y=

ln x x

f (x) = sin |x|

CP CP SP CP CP π π 3π 3π − − + − − 0 + − + 2 2 2 2|−−−−→x −−−−−−−−− | −−−−−−−− |−−−−−−−− |−−−−−−−− |−−−−−−−− abs ր abs ց loc ր abs ց abs ր f ց min max min max min f′

Fig. 4.4.35

36. Since f (x) = |x + 1|, f ′ (x) = sgn (x + 1) =



π Hence, f has abs max 1 at x = ±(4k + 1) and abs min 2 π −1 at x = ±(4k + 3) where k = 0, 1, 2, . . . and loc 2 min 0 at x = 0.

1, if x > −1; −1, if x < −1.

−1 is a singular point; f has no max but has abs min 0 at x = −1. f (x) → ∞ as x → ±∞.

y

y

1 π

−π

x

y = sin |x|

y = |x + 1|

Fig. 4.4.38

39.

−1

x

Fig. 4.4.36

37.

f (x) = | sin x| (2n + 1)π , SP = ±nπ CP: x = ± 2 f has abs max 1 at all CP. f has abs min 0 at all SP. y = | sin x|

y

f (x) = |x 2 − 1|

f ′ (x) = 2xsgn (x 2 − 1) CP: x = 0 SP: x = ±1

π

−π

x

2π

Fig. 4.4.39

SP CP SP − −1 + 0 − 1 + −−−−−−−−− | −−−−−−−−− | −−−−−−−−|−−−−−→x abs ր loc ց abs ր f ց min max min f′

40.

f (x) = (x − 1)2/3 − (x + 1)2/3 f ′ (x) = 23 (x − 1)−1/3 − 32 (x + 1)−1/3 Singular point at x = ±1. For critical points: (x − 1)−1/3 = (x + 1)−1/3 ⇒ x − 1 = x + 1 ⇒ 2 = 0, so there are no critical points.

y

SP SP f′ + −1 − +1 + −−−−−−−−−−−− | −−−−−−−−−−−|−−−−−−−−→x abs abs f ր ց ր max min

y = |x 2 − 1|

Hence, f has abs max 22/3 at x = −1 and abs min −22/3 at x = 1.

1

(−1,22/3 )

y

y = (x − 1)2/3 − (x + 1)2/3 1

−1

x

x (1,−22/3 )

Fig. 4.4.37

Fig. 4.4.40

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INSTRUCTOR’S SOLUTIONS MANUAL

41.

SECTION 4.5 (PAGE 244)

√ f (x) = x/ x 2 + 1. Since √ f ′ (x) =

46.

2x x2 + 1 − x √ 1 2 x2 + 1 = > 0, x2 + 1 (x 2 + 1)3/2

for all x, f cannot have any maximum or minimum value.

42.

√ f (x) = x/ x 4 + 1. f is continuous on R, and lim x→±∞ f (x) = 0. Since f (1) > 0 and f (−1) < 0, f must have both maximum and minimum values. √ f ′ (x) =

x4

47. If it exists, an absolute max value is the maximum of the set of all the local max values. Hence, if a function has an absolute max value, it must have one or more local max values. On the other hand, if a function has a local max value, it may or may not have an absolute max value. Since a local max value, say f (x0 ) at the point x0 , is defined such that it is the max within some interval |x − x0 | < h where h > 0, the function may have greater values, and may even approach ∞ outside this interval. There is no absolute max value in this latter case.

4x 3 +1−x √ 4 2 x4 + 1 = 1 − x . x4 + 1 (x 4 + 1)3/2

√ √ CP x = ±1. f (±1)√= ±1/ 2. f has max value 1/ 2 and min value −1/ 2.   y 1 1, √ 2



1 −1,− √ 2

43.



y= √

x

x

48. No. f (x) = −x 2 has abs max value 0, but g(x) = | f (x)| = x 2 has no abs max value.

x4 + 1

Fig. 4.4.42

49.

p −2x 2(2 − x 2 ) 4 − x2 + x √ = √ . 2 4 − x2 4 − x2

√ √ maximum value 2 CP x =√± 2. f (± 2) = ±2. f has √ at x = 2 and min value −2 at x = − 2.

44.

√ f (x) = x 2 / 4 − x 2 is continuous on (−2, 2), and lim x→−2+ f (x) = limx→2− f (x) = ∞. Thus f can have no maximum value, but will have a minimum value. √ −2x 2x 4 − x 2 − x 2 √ 3 2 4 − x 2 = 8x − x . f ′ (x) = 2 2 3/2 4−x (4 − x ) √ √ CP x = 0, x = ± 8. f (0) = 0, and ± 8 is not in the domain of f . f has minimum value 0 at x = 0.

45.

f (x) = 1/[x sin x] is continuous on (0, π ), and lim x→0+ f (x) = ∞ = limx→π − f (x). Thus f can have no maximum value, but will have a minimum value. Since f is differentiable on (0, π ), the minimum value must occur at a CP in that interval.

f (x) =

(

x sin

1 x

if x > 0

0 if x < 0 | f (x)| ≤ |x| if x > 0 so lim x→0+ f (x) = 0 = f (0). 1 Therefore f is continuous at x = 0. Clearly x sin x is continuous at x > 0. Therefore f is continuous on [0, ∞). Given any h > 0 there exists x1 in (0, h) and x2 in (0, h) such that f (x1 ) > 0 = f (0) and f (x2 ) < 0 = f (0). Therefore f cannot be a local max or min value at 0. 1 Specifically, let positive integer n satisfy 2nπ > h 1 1 and let x1 = , x = . 2 π 3π 2nπ + 2nπ + 2 2 Then f (x1 ) = x1 > 0 and f (x2 ) < 0.

√ f (x) = x 4 − x 2 is continuous on [−2, 2], and f (±2) = 0. f ′ (x) =

f (x) = (sin x)/x is continuous and differentiable on R except at x = 0 where it is undefined. Since lim x→0 f (x) = 1 (Theorem 8 of Section 2.5), and | f (x)| < 1 for all x 6= 0 (because | sin x| < |x|), f cannot have a maximum value. Since lim x→±∞ f (x) = 0 and since f (x) < 0 at some points, f must have a minimum value occurring at a critical point. In fact, since | f (x)| ≤ 1/|x| for x 6= 0 and f is even, the minimum value will occur at the two critical points closest to x = 0. (See Figure 2.20 In Section 2.5 of the text.)

Section 4.5 (page 244)

Concavity and Inflections

√

1 1 x, f ′ (x) = √ , f ′′ (x) = − x −3/2 4 2 x f ′′ (x) < 0 for all x > 0. f is concave down on (0, ∞).

1.

f (x) =

2.

f (x) = 2x − x 2 , f ′ (x) = 2 − 2x, f ′′ (x) = −2 < 0. Thus, f is concave down on (−∞, ∞).

3.

f (x) = x 2 + 2x + 3, f ′ (x) = 2x + 2, f ′′ (x) = 2 > 0. f is concave up on (−∞, ∞).

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SECTION 4.5 (PAGE 244)

4.

ADAMS and ESSEX: CALCULUS 8

f (x) = x − x 3 , f ′ (x) = 1 − 3x 2 , f ′′ (x) = −6x.

10.

f ′′ + 0 − −−−−−−−−−−−−|−−−−−−−−→x f ⌣ infl ⌢

5.

f ′′ (x) = 60(x − x 3 ) = 60x(1 − x)(1 + x). f ′′ + −1 − 0 + 1 − −−−−−−−−−−|−−−−−−−−− | −−−−−−−−|−−−−−−→x f ⌣ infl ⌢ infl ⌣ infl ⌢

6.

7.

11.

f (x) = sin x, f ′ (x) = cos x, f ′′ (x) = − sin x. f is concave down on intervals (2nπ, (2n + 1)π ) and concave up on intervals ((2n − 1)π, 2nπ ), where n ranges over the integers. Points x = nπ are inflection points.

12.

f (x) = cos 3x, f ′ (x) = −3 sin 3x, f ′′ (x) = −9 cos 3x.
π Inflection points: x = n + 21 for n = 0, ±1, ±2, .... 3   4n + 1 4n + 3 f is concave up on π, π and concave 6   6 4n + 3 4n + 5 down on π, π . 6 6

f (x) = 10x 3 + 3x 5 , f ′ (x) = 30x 2 + 15x 4 , f ′′ (x) = 60x + 60x 3 = 60x(1 + x 2 ). f ′′ − 0 + −−−−−−−−−−−−|−−−−−−−−→x f ⌢ infl ⌣

13.

f ′′ (x) = −12 + 12x 2 = 12(x − 1)(x + 1). f ′′ + −1 − 1 + −−−−−−−−−−−− | −−−−−−−−−− | −−−−−−−→x f ⌣ infl ⌢ infl ⌣ f (x) = (2 + 2x − x 2 )2 , 2

14.

f (x) = x − 2 sin x, f ′ (x) = 1 − 2 cos x, f ′′ (x) = 2 sin x. Inflection points: x = nπ  for n = 0, ±1, ±2,  .... f is concave down on (2n+1)π, (2n+2)π and concave   up on (2n)π, (2n + 1)π .

15.

f (x) = tan−1 x, f ′ (x) =

f ′ (x) = 2(2 + 2x − x 2 )(2 − 2x),

f (x) = 2(2 − 2x) + 2(2 + 2x − x 2 )(−2) = 12x(x − 2). ′′

f ′′ (x) =

f ′′ + 0 − 2 + −−−−−−−−−−−− | −−−−−−−−−− | −−−−−−−→x f ⌣ infl ⌢ infl ⌣

9.

f (x) = x + sin 2x, f ′ (x) = 1 + 2 cos 2x, f ′′ (x) = −4 sin 2x.

  (2n − 1)π , nπ , and conf is concave up on intervals 2   nπ (2n + 1)π . Points are cave down on intervals nπ, 2 2 inflection points.

f (x) = (3 − x 2 )2 ,

f ′ (x) = −4x(3 − x 2 ) = −12x + 4x 3 ,

8.

x2

f ′′ − −3 + 0 − 3 + −−−−−−−−−|−−−−−−−−− | −−−−−−−−|−−−−−→x f ⌢ infl ⌣ infl ⌢ infl ⌣

f (x) = 10x 3 − 3x 5 ,

f ′ (x) = 30x 2 − 15x 4 ,

x 3 − x2 , f ′ (x) = 2 , +3 (x + 3)2 2x(x 2 − 9) f ′′ (x) = . (x 2 + 3)3 f (x) =

−2x . (1 + x 2 )2

1 , 1 + x2

f ′′ + 0 − −−−−−−−−−−−−|−−−−−−−−→x f ⌣ infl ⌢

f (x) = (x 2 − 4)3 ,

f ′ (x) = 6x(x 2 − 4)2 ,

f ′′ (x) = 6(x 2 − 4)2 + 24x 2 (x 2 − 4)

16.

= 6(x 2 − 4)(5x 2 − 4).

f (x) = xe x , f ′ (x) = e x (1 + x), f ′′ (x) = e x (2 + x).

f ′′ + −2 − − √2 + √2 − 2 + 5 5 −−−−−−−−− | −−−−−−−− | −−−−−−−− | −−−−−−− | −−−−→x f ⌣ infl ⌢ infl ⌣ infl ⌢ infl ⌣

130 Copyright © 2014 Pearson Canada Inc.

f ′′ − −2 + −−−−−−−−−−−−|−−−−−−−−→x f ⌢ infl ⌣

INSTRUCTOR’S SOLUTIONS MANUAL

17.

18.

19.

2

SECTION 4.5 (PAGE 244)

2

f (x) = e−x , f ′ (x) = −2xe−x , 2 f ′′ (x) = e−x (4x 2 − 2).

23. According to Definition 4.3.1 and the subsequent discussion, f (x) = ax + b has no concavity and therefore no inflections.

f ′′ + − √1 − √1 + 2 2 −−−−−−−−−−|−−−−−−−−−|−−−−−→x f ⌣ infl ⌢ infl ⌣

24.

2 − ln(x 2 ) ln(x 2 ) ′ , f (x) = , x x2 2) −6 + 2 ln(x f ′′ (x) = . x3 f has inflection point at x = ±e3/2 and f is undefined at x = 0. f is concave up on (−e3/2 , 0) and (e3/2 , ∞); and concave down on (−∞, −e3/2 ) and (0, e3/2 ).

25.

f (x) =

2x , 1 + x2 2 (1 + x )(2) − 2x(2x) 2(1 − x 2 ) f ′′ (x) = = . (1 + x 2 )2 (1 + x 2 )2 f (x) = ln(1 + x 2 ),

20.

f (x) = (ln x)2 , f ′ (x) = f ′′ (x) =

26.

27.

2 ln x, x

2(1 − ln x) for all x > 0. x2

x3 25 − 4x 2 + 12x − , 3 3 f ′ (x) = x 2 − 8x + 12, f ′′ (x) = 2x − 8 = 2(x − 4).

f (x) = x 3 +

1 x

f (x) = (x − 1)1/3 + (x + 1)1/3 , f ′ (x) = 31 [(x − 1)−2/3 + (x + 1)−2/3 ], f ′′ (x) = − 29 [(x − 1)−5/3 + (x + 1)−5/3 ]. f (x) = 0 ⇔ x − 1 = −(x + 1) ⇔ x = 0. Thus, f has inflection point at x = 0. f ′′ (x) is undefined at x = ±1. f is defined at ±1 and x = ±1 are also inflection points. f is concave up on (−∞, −1) and (0, 1); and down on (−1, 0) and (1, ∞).

1 3x 4 − 1 = , x2 x2

1 . CP: x = ± √ 4 3

2 f ′′ (x) = 6x + 3 . x     1 ′′ −1 > 0, f √ < 0. f ′′ √ 4 4 3 3 −1 1 Therefore f has a loc min at √ and a loc max at √ . 4 4 3 3

28.

f (x) =

f ′′ − 4 + −−−−−−−−−−−−|−−−−−−−−→x f ⌢ infl ⌣

22.

4 ′ 4 , f (x) = 1 − 2 , f ′′ (x) = 8x −3 . x x The critical points are x = 2, f ′′ (2) > 0 ⇒ local min; x = −2, f ′′ (−2) < 0 ⇒ local max. f (x) = x +

f ′ (x) = 3x 2 −

f ′′ 0 + e − −−−−−− |−−−−−−−−−− | −−−−−−−→x f ⌣ infl ⌢

21.

f (x) = x(x − 2)2 + 1 = x 3 − 4x 2 + 4x + 1

f ′ (x) = 3x 2 − 8x + 4 = (x − 2)(3x − 2) 2 CP: x = 2, x = 3   2 ′′ ′′ ′′ f (x) = 6x − 8, f (2) = 4 > 0, f = −4 < 0. 3 Therefore, f has a loc min at x = 2 and a loc max at 2 x= . 3

f ′ (x) =

f ′′ − −1 + 1 − −−−−−−−−−− | −−−−−−−−− | −−−−−→x f ⌢ infl ⌣ infl ⌢

f (x) = 3x 3 − 36x − 3, f ′ (x) = 9(x 2 − 4), f ′′ (x) = 18x. The critical points are x = 2, f ′′ (2) > 0 ⇒ local min; x = −2, f ′′ (−2) < 0 ⇒ local max.

29.

x 1 − x ln 2 , f ′ (x) = , 2x 2x ln 2(x ln 2 − 2) f ′′ (x) = . 2x The critical point  is  1 1 ′′ , f < 0 ⇒ local max. x= ln 2 ln 2 f (x) =

x 1 + x2 (1 + x 2 ) − x2x 1 − x2 f ′ (x) = = (1 + x 2 )2 (1 + x 2 )2 CP: x = ±1 (1 + x)2 (−2x) − (1 − x 2 )2(1 + x 2 )2x f ′′ (x) = (1 + x 2 )4 3 −2x − 2x − 4x + 4x 3 −6x + 2x 3 = = (1 + x 2 )3 (1 + x 2 )3 1 ′′ 1 ′′ f (1) = − , f (−1) = . 2 2 f has a loc max at 1 and a loc min at −1. f (x) =

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SECTION 4.5 (PAGE 244)

30.

31.

ADAMS and ESSEX: CALCULUS 8

f (x) = xe x , f ′ (x) = e x (1 + x), f ′′ (x) = e x (2 + x). The critical point is x = −1. f ′′ (−1) > 0, ⇒ local min. f (x) = x ln x, f ′ (x) = 1 + ln x,

CP: x =

1 e

1 1 , f ′′ ( ) = e > 0. x e 1 f has a loc min at . e f ′′ (x) =

32.

33.

f (x) = (x 2 − 4)2 , f ′ (x) = 4x 3 − 16x, f ′′ (x) = 12x 2 − 16. The critical points are x = 0, f ′′ (0) < 0 ⇒ local max; x = 2, f ′′ (2) > 0 ⇒ local min; x = −2, f ′′ (−2) > 0 ⇒ local min. f (x) = (x 2 − 4)3

f ′ (x) = 6x(x 2 − 4)2 CP: x = 0, x = ±2 f ′′ (x) = 6(x 2 − 4)2 + 24x 2 (x 2 − 4)

= 6(x 2 − 4)(5x 2 − 4) f (0) > 0, f ′′ (±2) = 0. f has a loc min at x = 0. Second derivative test yields no direct information about ±2. However, since f ′′ has opposite signs on opposite sides of the points 2 and −2, each of these points is an inflection point of f , and therefore f cannot have a local maximum or minimum value at either. ′′

34.

f (x) = (x 2 − 3)e x ,

f ′ (x) = (x 2 + 2x − 3)e x = (x + 3)(x − 1)e x ,

f ′′ (x) = (x 2 + 4x − 1)e x . The critical points are x = −3, f ′′ (−3) < 0 ⇒ local max; x = 1, f ′′ (1) > 0 ⇒ local min.

35.

f (x) = x 2 e−2x

2

f ′ (x) = e−2x (2x − 4x 3 ) = 2(x − 2x 3 )e−2x 1 CP: x = 0, x = ± √ 2 2 f ′′ (x) = e−2x (2 − 20x 2 + 16x 4 )   1 4 f ′′ (0) > 0, f ′′ ± √ = − < 0. e 2 Therefore, f has a loc (and abs) min value at 0, and loc 1 (and abs) max values at ± √ . 2

36. Since f (x) =



x2 −x 2

if x ≥ 0 if x < 0,

2x if x ≥ 0 = 2|x| −2x if x < 0 n 2 if x > 0 = 2sgn x. f ′′ (x) = −2 if x < 0

f ′ (x) =

n

f ′ (x) = 0 if x = 0. Thus, x = 0 is a critical point of f . It is also an inflection point since the conditions of Definition 3 are satisfied. f ′′ (0) does not exist. If a the graph of a function has a tangent line, vertical or not, at x0 , and has opposite concavity on opposite sides of x0 , the x0 is an inflection point of f , whether or not f ′′ (x0 ) even exists.

37. Suppose f is concave up (i.e., f ′′ (x) > 0) on an open interval containing x0 . Let h(x) = f (x) − f (x0 ) − f ′ (x0 )(x − x0 ). Since h ′ (x) = f ′ (x) − f ′ (x0 ) = 0 at x = x0 , x = x0 is a CP of h. Now h ′′ (x) = f ′′ (x). Since h ′′ (x0 ) > 0, therefore h has a min value at x0 , so h(x) ≥ h(x0 ) = 0 for x near x0 . Since h(x) measures the distance y = f (x) lies above the tangent line y = f (x0 ) + f ′ (x0 )(x − x0 ) at x, therefore y = f (x) lies above that tangent line near x0 . Note: we must have h(x) > 0 for x near x0 , x 6= x0 , for otherwise there would exist x1 6= x0 , x1 near x0 , such that h(x1 ) = 0 = h(x0 ). If x1 > x0 , there would therefore exist x2 such that x0 < x2 < x1 and f ′ (x2 ) = f ′ (x0 ). Therefore there would exist x3 such that x0 < x3 < x2 and f ′ (x3 ) = 0, a contradiction. The same contradiction can be obtained if x1 < x0 .

38. Suppose that f has an inflection point at x0 . To be specific, suppose that f ′′ (x) < 0 on (a, x0 ) and f ′′ (x) > 0 on (x0 , b) for some numbers a and b satisfying a < x0 < b. If the graph of f has a non-vertical tangent line at x0 , then f ′ (x0 ) exists. Let F(x) = f (x) − f (x0 ) − f ′ (x0 )(x − x0 ).

2

2

we have

F(x) represents the signed vertical distance between the graph of f and its tangent line at x0 . To show that the graph of f crosses its tangent line at x0 , it is sufficient to show that F(x) has opposite signs on opposite sides of x0 . Observe that F(x0 ) = 0, and F ′ (x) = f ′ (x) − f ′ (x0 ), so that F ′ (x0 ) = 0 also. Since F ′′ (x) = f ′′ (x), the assumptions above show that F ′ has a local minimum value at x0 (by the First Derivative Test). Hence F(x) > 0 if a < x < x0 or x0 < x < b. It follows (by Theorem 6) that F(x) < 0 if a < x < x0 , and F(x) > 0 if x0 < x < b. This completes the proof for the case of a nonvertical tangent. If f has a vertical tangent at x0 , then its graph necessarily crosses the tangent (the line x = x0 ) at x0 , since the graph of a function must cross any vertical line through a point of its domain that is not an endpoint.

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INSTRUCTOR’S SOLUTIONS MANUAL

39.

f (x) = x n g(x) = −x n = − f (x), n−1

f n′ (x)

SECTION 4.5 (PAGE 244)

f (h) − f (0) = lim h −1 f (h) = 0 by h→0 h (a). Suppose that f (k) (0) = 0 for some k ≥ 1. Then

d) f ′ (0) = limh→0

n = 2, 3, 4, . . .

= nx = 0 at x = 0 If n is even, f n has a loc min, gn has a loc max at x = 0. If n is odd, f n has an inflection at x = 0, and so does gn .

f (k) (h) − f (k) (0) h→0 h = lim h −1 f (k) (h) h→0   1 = lim h −1 Pk f (h) = 0 h→0 h

f (k+1) (0) = lim

40. Let there be a function f such that f ′ (x0 ) = f ′′ (x0 ) = ... = f (k−1) (x0 ) = 0,

f (k) (x0 ) 6= 0

for some k ≥ 2.

If k is even, then f has a local min value at x = x0 when f (k) (x0 ) > 0, and f has a local max value at x = x0 when f (k) (x0 ) < 0. If k is odd, then f has an inflection point at x = x0 .

41.

f (x) = a)



e−1/x 0

2

if x 6= 0 if x = 0

lim x −n f (x) = lim

x→0+

x→0+

e−1/x xn

2

(put y = 1/x)

2

= lim y n e−y = 0 by Theorem 5 of Sec. 4.4

by (b). Thus f (n) (0) = 0 for n = 1, 2, . . . by induction.

e) Since f ′ (x) < 0 if x < 0 and f ′ (x) > 0 if x > 0, therefore f has a local min value at 0 and − f has a loc max value there. f) If g(x) = x f (x) then g ′ (x) = f (x) + x f ′ (x), g ′′ (x) = 2 f ′ (x) + x f ′′ (x). In general, g (n) (x) = n f (n−1) (x) + x f (n) (x) (by induction). Then g (n) (0) = 0 for all n (by (d)). Since g(x) < 0 if x < 0 and g(x) > 0 if x > 0, g cannot have a max or min value at 0. It must have an inflection point there.

42. We are given that

y→∞

x −n

Similarly, limx→0− f (x) = 0, and lim x→0 x −n f (x) = 0. P b) If P(x) = nj=0 a j x j then by (a)

f (x) =

  n X 1 f (x) = a j lim x − j f (x) = 0. lim P x→0 x→0 x j =0

c) If x 6= 0 and P1 (t) = 2t 3 , then f ′ (x) =

2 −1/x 2 e x3

  1 = P1 f (x). x

  1 f (x) for some Assume that f (k) (x) = Pk x k ≥ 1, where Pk is a polynomial. Then       1 ′ 1 1 1 f (x) + P P1 f (x) P k x x x2 k x   1 = Pk+1 f (x), x

f (k+1) (x) = −

where Pk+1 (t) = t 2 Pk′ (t) + P1 (t)Pk (t) is a polynomial.   1 By induction, f (n) = Pn f (x) for n 6= 0, where n Pn is a polynomial.

(

1 x 2 sin , x 0,

if x 6= 0; if x = 0.

If x 6= 0, then 1 1 − cos x x 2 1 1 1 1 f ′′ (x) = 2 sin − cos − 2 sin . x x x x x f ′ (x) = 2x sin

If x = 0, then ′

1 −0 h = 0. h

h 2 sin

f (x) = lim

h→0

Thus 0 is a critical point of f . There are points x arbitrarily close to 0 where f (x) > 0, for example 2 x = , and other such points where f (x) < 0, (4n + 1)π 2 for example x = . Therefore f does not have (4n + 3)π a local max or min at x = 0. Also, there are points arbitrarily close to 0 where f ′′ (x) > 0, for example 1 x = , and other such points where f ′′ (x) < 0, (2n + 1)π 1 for instance x = . Therefore f does not have con2nπ stant concavity on any interval (0, a) where a > 0, so 0 is not an inflection point of f either.

133 Copyright © 2014 Pearson Canada Inc.

SECTION 4.5 (PAGE 244)

ADAMS and ESSEX: CALCULUS 8

Section 4.6 Sketching the Graph of a Function (page 252)

The function graphed in Fig. 4.2(b): is even, is asymptotic to y = 0 at ±∞, is increasing on (−1.7, 0) and (1.7, ∞) (approximately), is decreasing on (−∞, −1.7) and (0, 1.7) (approximately), has CPs at x = 0 (max) and ±1.7 (min) (approximately), is concave up on (−2.5, −1) and (1, 2.5) (approximately), is concave down on (−∞, −2.5), (−1, 1), and (2.5, ∞) (approximately), has inflections at ±2.5 and ±1 (approximately).

1. Function (d) appears to be the derivative of function (c), and function (b) appears to be the derivative of function (d). Thus graph (c) is the graph of f , (d) is the graph of f ′ , (b) is the graph of f ′′ , and (a) must be the graph of the other function g. y y (a) (b) 4 3 2 1

−5 −4 −3 −2 −1 −1

4 3 2 1

1 2 3

4

x

−2 −3 −4 −5

(c)

1 2

3 4

x The function graphed in Fig. 4.2(c): is even, is asymptotic to y = 2 at ±∞, is increasing on (0, ∞), is decreasing on (−∞, 0), has a CP at x = 0 (min), is concave up on (−1, 1) (approximately), is concave down on (−∞, −1) and (1, ∞) (approximately), has inflections at x = ±1 (approximately).

−2 −3 −4 −5

y

(d)

4 3 2 1

−5 −4 −3 −2 −1 −1

−5 −4 −3 −2 −1 −1

1 2 3

−2 −3 −4 −5

4

x

y

4 3 2 1

−5 −4 −3 −2 −1 −1

Fig. 4.6.1

1 2

3 4

x

−2 −3 −4 −5

The function graphed in Fig. 4.2(d): is odd, is asymptotic to y = 0 at ±∞, is increasing on (−1, 1), is decreasing on (−∞, −1) and (1, ∞), has CPs at x = −1 (min) and 1 (max), is concave down on (−∞, −1.7) and (0, 1.7) (approximately), is concave up on (−1.7, 0) and (1.7, ∞) (approximately), has inflections at 0 and ±1.7 (approximately).

2. (a)

y

(b)

4 3 2 1

−5 −4 −3 −2 −1 −1

1 2 3

4

x

y

−2 −3 −4 −5

−5 −4 −3 −2 −1 −1

(d)

4 3 2 1

−5 −4 −3 −2 −1 −1

4 3 2 1 1 2

3 4

x

1 2

3 4

x

−2 −3 −4 −5

−2 −3 −4 −5

(c)

y

1 2 3

4

x

y

4 3 2 1

−5 −4 −3 −2 −1 −1

Fig. 4.6.2

−2 −3 −4 −5

The function graphed in Fig. 4.2(a): is odd, is asymptotic to y = 0 at ±∞, is increasing on (−∞, −1) and (1, ∞), is decreasing on (−1, 1), has CPs at x = −1 (max) and 1 (min), is concave up on (−∞, −2) and (0, 2) (approximately), is concave down on (−2, 0) and (2, ∞) (approximately), has inflections at x = ±2 (approximately).

3.

f (x) = x/(1 − x 2 ) has slope 1 at the origin, so its graph must be (c). g(x) = x 3 /(1 − x 4 ) has slope 0 at the origin, but has the same sign at all points as does f (x), so its graph must be (b). √ h(x) = (x 3 − x)/ 1 + x 6 has no vertical asymptotes, so its graph must p be (d). k(x) = x 3 / |x 4 − 1| is positive for all positive x 6= 1, so its graph must be (a).

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.6 (PAGE 252)

4.

5. (a)

y

3 2

3 2

1

1

−5 −4 −3 −2 −1 −1 −2 −3 −4

(c)

(b)

y

1 2 3

4

x

y

−5 −4 −3 −2 −1 −1 −2 −3 −4

(d)

−2 −3 −4

3 4

SP CP f′ + 0 − 1 + −−−−−−−−−−−|−−−−−−−−−−− |−−−−−−−→x loc loc f ր ց ր max min

x

y 3 2 1

3 2 1 −5 −4 −3 −2 −1 −1

1 2

f (0) = 1 f (±1) = 0 f (2) = 1 lim x→∞ f (x) = 2, lim x→−∞ f (x) = −1

1 2 3

4

x

−5 −4 −3 −2 −1 −1

Fig. 4.6.4

1 2

3 4

f ′′ + 0 + 2 − −−−−−−−−−−−|−−−−−−−−−−|−−−−−−−−→x f ⌣ ⌣ infl ⌢

x

−2 −3 −4

0 must be a SP because f ′′ > 0 on both sides and it is a loc max. 1 must be a CP because f ′′ is defined there so f ′ must be too. y y=2

The function graphed in Fig. 4.4(a): is odd, is asymptotic to x = ±1 and y = x, is increasing on (−∞, −1.5), (−1, 1), and (1.5, ∞) (approximately), is decreasing on (−1.5, −1) and (1, 1.5) (approximately), has CPs at x = −1.5, x = 0, and x = 1.5, is concave up on (0, 1) and (1, ∞), is concave down on (−∞, −1) and (−1, 0), has an inflection at x = 0. The function graphed in Fig. 4.4(b): is odd, is asymptotic to x = ±1 and y = 0, is increasing on (−∞, −1), (−1, 1), and (1, ∞), has a CP at x = 0, is concave up on (−∞, −1) and (0, 1), is concave down on (−1, 0) and (1, ∞), has an inflection at x = 0. The function graphed in Fig. 4.4(c): is odd, is asymptotic to x = ±1 and y = 0, is increasing on (−∞, −1), (−1, 1), and (1, ∞), has no CP, is concave up on (−∞, −1) and (0, 1), is concave down on (−1, 0) and (1, ∞), has an inflection at x = 0. The function graphed in Fig. 4.4(d): is odd, is asymptotic to y = ±2, is increasing on (−∞, −0.7) and (0.7, ∞) (approximately), is decreasing on (−0.7, 0.7) (approximately), has CPs at x = ±0.7 (approximately), is concave up on (−∞, −1) and (0, 1) (approximately), is concave down on (−1, 0) and (1, ∞) (approximately), has an inflection at x = 0 and x = ±1 (approximately).

y = f (x)

1 (2,1) −1

1

x

y=−1

Fig. 4.6.5

6. According to the given properties: Oblique asymptote: y = x − 1. Critical points: x = 0, 2. Singular point: x = −1. Local max 2 at x = 0; local min 0 at x = 2. SP CP CP f ′ + −1 + 0 − 2 + −−−−−−−−−|−−−−−−−−−− | −−−−−−−−−|−−−−−−→x loc loc f ր ր max ց min ր Inflection points: x = −1, 1, 3. f′ + −1 − 1 + 3 − −−−−−−−−−− |−−−−−−−−−|−−−−−−−−−|−−−−−−→x f ⌣ infl ⌢ infl ⌣ infl ⌢ Since

lim

x→±∞



 f (x) + 1 − x = 0, the line y = x − 1 is an

oblique asymptote.

135 Copyright © 2014 Pearson Canada Inc.

SECTION 4.6 (PAGE 252)

ADAMS and ESSEX: CALCULUS 8

y

y

y = (x 2 − 1)3

y = f (x)

2 (1,1)

(3,1) −1

√ −1/ 5

x

2

√ 1/ 5 x

1

−1 −1 y=x−1

Fig. 4.6.7

8. Fig. 4.6.6

y = x(x 2 − 1)2 , y ′ = (x 2 − 1)(5x 2 − 1), y ′′ = 4x(5x 2 − 3). From y: Intercepts: (0, 0), (1, 0). Symmetry: odd (i.e., about the origin). 1 From y ′ : Critical point: x = ±1, ± √ . 5 CP

CP CP CP 1 1 y ′ + −1 − − √ + √ − 1 + 5 5 −−−−−−−−|−−−−−−−−− |−−−−−−−−− | −−−−−−−− | −−−−→x loc ց loc loc loc y ր max min ր max ց min ր

7.

From y ′′ : q Inflection points at

y = (x 2 − 1)3

x = 0, ±

y ′ = 6x(x 2 − 1)2

= 6x(x − 1)2 (x + 1)2

y ′′ = 6[(x 2 − 1)2 + 4x 2 (x 2 − 1)]

= 6(x 2 − 1)(5x 2 − 1) √ √ = 6(x − 1)(x + 1)( 5x − 1)( 5x + 1) From y: Asymptotes: none. Symmetry: even. Intercepts: x = ±1. From y ′ : CP: x = 0, x = ±1. SP: none.

3 5.

q q 3 y ′′ − − 35 + 0 − + 5 −−−−−−−−−−− | −−−−−−−−−|−−−−−−−−−|−−−−−−→x y ⌢ infl ⌣ infl ⌢ infl ⌣ y

CP CP CP y ′ − −1 − 0 + 1 + −−−−−−−|−−−−−−−|−−−−−−− | −−−→x abs y ց ց min ր ր √1 5

1 From y ′′ : y ′′ = 0 at x = ±1, x = ± √ . 5 y = x(x 2 − 1)2 y ′′ + −1 − − √1 + √1 − 1 + 5 5 −−−−−−−− |−−−−−−−− | −−−−−−−|−−−−−−−|−−−−→x y ⌣ infl ⌢ infl ⌣ infl ⌢ infl ⌣

136 Copyright © 2014 Pearson Canada Inc.

Fig. 4.6.8

q

3 1 5

x

INSTRUCTOR’S SOLUTIONS MANUAL

9.

SECTION 4.6 (PAGE 252)

y

2−x 2 2 4 = − 1, y ′ = − 2 , y ′′ = 3 . x x x x From y: Asymptotes: x = 0, y = −1. Symmetry: none obvious. Intercept: (2, 0). Points: (−1, −3). From y ′ : CP: none. SP: none. y=

y=

x −1 x +1 (−2,3)

ASY y′ − 0 − −−−−−−−−−−−|−−−−−−−−→x y ց ց

y=1

1

From y ′′ : y ′′ = 0 nowhere.

x=−1

ASY y ′′ − 0 + −−−−−−−−−−−− | −−−−−−−→x y ⌢ ⌣ y=

y

x

−1

Fig. 4.6.10

2−x x

(2,0) x

11.

−1

(−1,−3)

Fig. 4.6.9

10.

2 2 −4 x −1 =1− , y′ = , y ′′ = . 2 x +1 x +1 (x + 1) (x + 1)3 From y: Intercepts: (0, −1), (1, 0). Asymptotes: y = 1 (horizontal), x = −1 (vertical). No obvious symmetry. Other points: (−2, 3). From y ′ : No critical point. y=

ASY y′ + −1 + −−−−−−−−−−−−−− |−−−−−−−−−−→x y ր ր

x3 1+x (1 + x)3x 2 − x 3 3x 2 + 2x 3 ′ y = = 2 (1 + x) (1 + x)2 2 2 (1 + x) (6x + 6x ) − (3x 2 + 2x 3 )2(1 + x) y ′′ = (1 + x)4 2 2 6x(1 + x) − 6x − 4x 3 6x + 6x 2 + 2x 3 = = (1 + x)3 (1 + x)3 2 2x(3 + 3x + x ) = (1 + x)3 From y: Asymptotes: x = −1. Symmetry: none. Intercepts (0, 0). Points (−3/2, 27/4). 3 From y ′ CP: x = 0, x = − . 2 y=

CP ASY CP y ′ − − 23 + −1 + 0 + −−−−−−−−− |−−−−−−−−− |−−−−−−−−|−−−−−→x loc ր y ց min ր ր From y ′′ : y ′′ = 0 only at x = 0.

From y ′′ : No inflection point. ASY y ′′ + −1 − −−−−−−−−−−−−−−|−−−−−−−−−−→x y ⌣ ⌢

ASY y ′′ + −1 − 0 + −−−−−−−−−−−−|−−−−−−−−−−−− | −−−−−−−→x y ⌣ ⌢ infl ⌣

137 Copyright © 2014 Pearson Canada Inc.

SECTION 4.6 (PAGE 252)

ADAMS and ESSEX: CALCULUS 8

y

y ′′ : y ′′ = 0 nowhere.

x3 y= 1+x 

3 27 −2, 4

ASY ASY √ √ y ′′ − − 2 + 2 − −−−−−−−−−− | −−−−−−−−− | −−−−−→x y ⌢ ⌣ ⌢



x=−1

y x

√ x=− 2

Fig. 4.6.11

12.

1 −2x 6x 2 − 8 , y′ = , y ′′ = . 2 2 2 4+x (4 + x ) (4 + x 2 )3 1 From y: Intercept: (0, 4 ). Asymptotes: y = 0 (horizontal). Symmetry: even (about y-axis). From y ′ : Critical point: x = 0.

√ x= 2 1/2

y=

x



CP + 0 − −−−−−−−−−−−−−− |−−−−−−−−−−→x abs y ր ց max 2 From y ′′ : y ′′ = 0 at x = ± √ . 3

1 −2,− 2







y=

1 2 − x2

1 2,− 2

y′

Fig. 4.6.13

2 2 √ −√ − + 3 3 −−−−−−−−−−−− | −−−−−−−−−−|−−−−−−−→x y ⌣ infl ⌢ infl ⌣ y ′′

+

y 1/4

y= −2 √ 3

1 4 + x2

√2 3

14. x

x x2 + 1 2x(x 2 + 3) , y′ = − 2 , y ′′ = . 2 −1 (x − 1) (x 2 − 1)3 From y: Intercept: (0, 0). Asymptotes: y = 0 (horizontal), x = ±1 (vertical). Symmetry: odd. Other points: (2, 23 ), (−2, − 32 ). From y ′ : No critical or singular points. y=

x2

Fig. 4.6.12

13.

2x 1 , y′ = 2 2−x (2 − x 2 )2 2 8x 2 4 + 6x 2 y ′′ = + = 2 2 2 3 (2 − x ) (2 − x ) (2 −√x 2 )3 From y: Asymptotes: y = 0, x = ± 2. Symmetry: even. Intercepts (0, 12 ). Points (±2, − 21 ). From y ′ : CP x = 0.

ASY ASY y′ − −1 − 1 − −−−−−−−−−−−− | −−−−−−−−−−−|−−−−−−−−→x y ց ց ց

y=

From y ′′ : y ′′ = 0 at x = 0.

ASY CP ASY √ √ y ′′ − − 2 − 0 + 2 + −−−−−−−−−|−−−−−−−−− | −−−−−−−−|−−−−−→x loc ր y ց ց min ր

ASY ASY y ′′ − −1 + 0 − 1 + −−−−−−−−−|−−−−−−−−− | −−−−−−−− |−−−−−→x y ⌢ ⌣ infl ⌢ ⌣

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.6 (PAGE 252)

y

16. y= x=−1



2 −2,− 3





x2 2 2, 3

x −1  x

x3 x 2 (x 2 − 3) ′′ 2x(x 2 + 3) , y′ = , y = . −1 (x 2 − 1)2 (x 2 − 1)3 From y: Intercept: (0, 0). Asymptotes: x = ±1 (vertical), y = √ x (oblique). Symmetry: odd. Other points:   √ 3 3 ± 3, ± . 2 √ From y ′ : Critical point: x = 0, ± 3. y=

x2

CP ASY CP ASY CP √ √ y ′ + − 3 − −1 − 0 − 1 − 3 + −−−−−−−− |−−−−−−−− | −−−−−−− | −−−−−−− | −−−−−−−|−−−→x loc ց loc ր y ր max ց ց ց min

x=1

From y ′′ : y ′′ = 0 at x = 0. ASY ASY y ′′ − −1 + 0 − 1 + −−−−−−−−−−−|−−−−−−−−−−|−−−−−−−−−−|−−−−−−−→x y ⌢ ⌣ infl ⌢ ⌣

Fig. 4.6.14

15.

x2 1 =1+ 2 x2 − 1 x −1 −2x y′ = 2 (x − 1)2 (x 2 − 1)2 − x2(x 2 − 1)2x 2(3x 2 + 1) = y ′′ = −2 (x 2 − 1)4 (x 2 − 1)3 From y: Asymptotes: y  = 1, x =  ±1. Symmetry: even. 4 Intercepts (0, 0). Points ±2, . 3 ′ From y : CP x = 0. y=

y

y=x

x=−1 √ − 3

ASY CP ASY y ′ + −1 + 0 − 1 − −−−−−−−−− | −−−−−−−−|−−−−−−−−−|−−−−−→x loc ց y ր ր max ց ′′ ′′ From y : y = 0 nowhere. ASY ASY + −1 − 1 + −−−−−−−−−− | −−−−−−−−− | −−−−−→x y ⌣ ⌢ ⌣

y=



4 −2, 3





4 2, 3

17.



y=1 x x=−1

x=1

Fig. 4.6.15

x3 −1

x2

Fig. 4.6.16

x2 x2 − 1

y=

x

x=1

y ′′

y

√ 3

x3 x3 + x − x x = =x− 2 x2 + 1 x2 + 1 x +1 (x 2 + 1)3x 2 − x 3 2x x 4 + 3x 2 x 2 (x 2 + 3) y′ = = = (x 2 + 1)2 (x 2 + 1)2 (x 2 + 1)2 2 2 3 4 2 2 (x + 1) (4x + 6x) − (x + 3x )2(x + 1)2x y ′′ = (x 2 + 1)4 5 3 4x + 10x + 6x − 4x 5 − 12x 3 = (x 2 + 1)3 2 2x(3 − x ) = (x 2 + 1)3 From y: Asymptotes: y = x (oblique). Symmetry: odd. Intercepts √ (0, 0). √ Points (± 3, ± 34 3). y=

139 Copyright © 2014 Pearson Canada Inc.

SECTION 4.6 (PAGE 252)

ADAMS and ESSEX: CALCULUS 8

y

From y ′ : CP: x = 0.

y=1

CP y′ + 0 + −−−−−−−−−−−|−−−−−−−−→x y ր ր

y= −1 √ 3

√ From y ′′ : y ′′ = 0 at x = 0, x = ± 3.

√1 3

x

Fig. 4.6.18

√ √ y ′′ + − 3 − 0 + 3 − −−−−−−−−−|−−−−−−−−|−−−−−−−− | −−−−→x y ⌣ infl ⌢ infl ⌣ infl ⌢

19. y

y= √

√  √ 3 3 − 3,− 4

x2 x2 + 1

√  3 3 3, 4



x3 +1

x2

3 x2 − 4 = x −1− x +1 x +1 3 (x + 1)2 + 3 y′ = 1 + = (x + 1)2 (x + 1)2 6 y ′′ = − (x + 1)3 From y: Asymptotes: y = x − 1 (oblique), x = −1. Symmetry: none. Intercepts (0, −4), (±2, 0). From y ′ : CP: none. y=

x

ASY y′ + −1 + −−−−−−−−−−−|−−−−−−−−→x y ր ր

y=x

From y ′′ : y ′′ = 0 nowhere. Fig. 4.6.17

18.

2x 2(1 − 3x 2 ) x2 , y′ = 2 , y ′′ = . 2 +1 (x + 1) (x 2 + 1)3 From y: Intercept: (0, 0). Asymptotes: y = 1 (horizontal). Symmetry: even. From y ′ : Critical point: x = 0. y=

ASY y ′′ + −1 − −−−−−−−−−−−− | −−−−−−−→x y ⌣ ⌢

x2

CP y′ − 0 + −−−−−−−−−−−−−− |−−−−−−−−−−→x abs y ց ր min

y

−2

−1

2

1 From y ′′ : y ′′ = 0 at x = ± √ . 3

y= y=x−1

1 1 − −√ + − √ 3 3 −−−−−−−−−−−− | −−−−−−−−−−|−−−−−−−→x y ⌢ infl ⌣ infl ⌢ y ′′

140 Copyright © 2014 Pearson Canada Inc.

Fig. 4.6.19

−4

x2 − 4 x +1

x

INSTRUCTOR’S SOLUTIONS MANUAL

20.

SECTION 4.6 (PAGE 252)

x2 − 2 ′ 2x −2(3x 2 + 1) , y = 2 , y ′′ = . 2 2 x −1 (x − 1) √ (x 2 − 1)3 From y: Intercept: (0, 2), (± 2, 0). Asymptotes: y = 1 (horizontal), x = ±1 (vertical). Symmetry: even. From y ′ : Critical point: x = 0.

From y ′ : CP: none.

y=

ASY ASY y′ + −1 + 1 + −−−−−−−−−|−−−−−−−−−− | −−−−−→x y ր ր ր

ASY CP ASY f ′ − −1 − 0 + 1 + −−−−−−−−−|−−−−−−−−−|−−−−−−−−− | −−−−−→x loc ր f ց ց min ր

From y ′′ : y ′′ = 0 at x = 0. ASY ASY y ′′ + −1 − 0 + 1 − −−−−−−−−− |−−−−−−−− | −−−−−−−|−−−−−→x y ⌣ ⌢ infl ⌣ ⌢

From y ′′ : y ′′ = 0 nowhere. ASY ASY y′ − −1 + 1 − −−−−−−−−−−−|−−−−−−−−−−− | −−−−−−−→x y ⌢ ⌣ ⌢

y

y

x=1 x=−1

x=1

x=−1 2

√ − 2

√ 2

x y=x

y=

x2 − 2 x2 − 1

x

2

−2

y=1

y=

x 3 − 4x x2 − 1

Fig. 4.6.21

Fig. 4.6.20

21.

x(x − 2)(x + 2) x 3 − 4x = 2 x −1 x2 − 1 2 2 (x − 1)(3x − 4) − (x 3 − 4x)2x y′ = (x 2 − 1)2 4 2 3x − 7x + 4 − 2x 4 + 8x 2 = (x 2 − 1)2 x4 + x2 + 4 = (x 2 − 1)2 (x 2 − 1)2 (4x 3 + 2x) − (x 4 + x 2 + 4)2(x 2 − 1)2x y ′′ = (x 2 − 1)4 5 3 4x − 2x − 2x − 4x 5 − 4x 3 − 16x = (x 2 − 1)3 3 −6x − 18x x2 + 3 = = −6x 2 2 3 (x − 1) (x − 1)3 From y: Asymptotes: y = x (oblique), x = ±1. Symmetry: odd. Intercepts (0, 0), (±2, 0). y=

22.

x2 − 1 1 2 6 = 1 − 2 , y ′ = 3 , y ′′ = − 4 . x2 x x x From y: Intercepts: (±1, 0). Asymptotes: y = 1 (horizontal), x = 0 (vertical). Symmetry: even. From y ′ : No critical points. y=

ASY y′ − 0 + −−−−−−−−−−−−−− |−−−−−−−−−−→x y ց ր From y ′′ : y ′′ is negative for all x.

141 Copyright © 2014 Pearson Canada Inc.

SECTION 4.6 (PAGE 252)

ADAMS and ESSEX: CALCULUS 8

y

y y=1

−1

1

x y=x

y=

x2 − 1 x2

√ − 5

Fig. 4.6.22

23.

−1

1

√ 5

y=

x5 (x 2 − 1)2

x

Fig. 4.6.23

x5 2x 3 − x = x + (x 2 − 1)2 (x 2 − 1)2 2 − 1)2 5x 4 − x 5 2(x 2 − 1)2x (x y′ = (x 2 − 1)4 6 4 5x − 5x − 4x 6 x 4 (x 2 − 5) = = 2 3 (x − 1) (x 2 − 1)3 (x 2 − 1)3 (6x 5 − 20x 3 ) − (x 6 − 5x 4 )3(x 2 − 1)2 2x y ′′ = (x 2 − 1)6 7 5 3 6x − 26x + 20x − 6x 7 + 30x 5 = (x 2 − 1)4 3 2 4x (x + 5) = (x 2 − 1)4 From y: Asymptotes: y  = x, x = ±1. Symmetry: odd.  √ 25 √ Intercepts (0, 0). Points ± 5, ± 5 . 16 √ ′ From y : CP x = 0, x = ± 5. y=

CP ASY CP ASY CP √ √ 1 − 5 + y ′ + − 5 − −1 + 0 + −−−−−−−− | −−−−−−−− | −−−−−−− | −−−−−−− | −−−−−−− | −−−−−−−−→x loc ց loc y ր max ր ր ց min ր From y ′′ : y ′′ = 0 if x = 0. ASY ASY y ′′ − −1 − 0 + 1 + −−−−−−−−−|−−−−−−−−− | −−−−−−−− |−−−−−→x y ⌢ ⌢ infl ⌣ ⌣

24.

(x − 2)(x − 6) (2 − x)2 ′ , y =− , x3 x4 √ √ 2(x 2 − 12x + 24) 2(x − 6 + 2 3)(x − 6 − 2 3) ′′ y = = . x5 x5 y=

From y: Intercept: (2, 0). Asymptotes: y = 0 (horizontal), x = 0 (vertical). Symmetry: none obvious. Other points: (−2, −2), (−10, −0.144). From y ′ : Critical points: x = 2, 6. ASY CP CP y′ − 0 − 2 + 6 − −−−−−−−−−− | −−−−−−−−− |−−−−−−−−−|−−−−−−→x loc ր loc y ց ց min max ց √ From y ′′ : y ′′ = 0 at x = 6 ± 2 3.

√ √ y ′′ − 0 + 6 + 2 3 − 6 − 2 3 + −−−−−−−− | −−−−−−−−|−−−−−−−−−−−− | −−−−−−→x y ⌢ ⌣ infl ⌢ infl ⌣

142 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.6 (PAGE 252)

y

y

y=

(2 − x)2 x3

y=

(6,2/27) √ 6+2 3

2 (−10,−0.144)

x3

1 − 4x

x=2 √2 3

−3

x

3

2 −√ 3 x=−2

√ 6−2 3

x

Fig. 4.6.25

26. Fig. 4.6.24

25.

1 1 = x 3 − 4x x(x − 2)(x + 2) 3x 2 − 4 3x 2 − 4 y′ = − 3 =− 2 2 2 (x − 4x) x (x − 4)2 3 2 (x − 4x) (6x) − (3x 2 − 4)2(x 3 − 4x)(3x 2 − 4) y ′′ = − (x 3 − 4x)4 4 2 4 6x − 24x − 18x + 48x 2 − 32 =− (x 3 − 4x)3 2 2 12(x − 1) + 20 = x 3 (x 2 − 4)3 From y: Asymptotes: y = 0, x = 0, −2, 2. Symmetry:  odd. No intercepts.    2 16 1 Points: ± √ , ± √ , ±3, ± 15 3 3 3 2 From y ′ : CP: x = ± √ . 3 y=

x x = , x2 + x − 2 (2 + x)(x − 1) 2 3 −(x + 2) 2(x + 6x + 2) y′ = , y ′′ = . (x + 2)2 (x − 1)2 (x + 2)3 (x − 1)3 From y: Intercepts: (0, 0). Asymptotes: y = 0 (horizontal), x = 1, x = −2 (vertical). Other points: (−3, − 34 ), (2, 12 ). From y ′ : No critical point. y=

ASY ASY y′ − −2 − 1 − −−−−−−−−−−−− | −−−−−−−−−−−|−−−−−−−−→x y ց ց ց

From y ′′ : y ′′ = 0 if f (x) = x 3 + 6x + 2 = 0. Since f ′ (x) = 3x 2 + 6 ≥ 6, f is increasing and can only have one root. Since f (0) = 2 and f (−1) = −5, that root must be between −1 and 0. Let the root be r. ASY ASY y ′′ − −2 + r − 1 + −−−−−−−−−|−−−−−−−−− | −−−−−−−− |−−−−−→x y ⌢ ⌣ infl ⌢ ⌣ y

y=

x=−2

ASY CP ASY CP ASY 2 2 ′ √ √ y − −2 − − + 0 + − 2 − 3 3 −−−−−−−−|−−−−−−−−|−−−−−−−−− | −−−−−−−− | −−−−−−−− | −−−−→x loc ր loc ց y ց ց min ր max ց From

y ′′ :

y ′′

x x2 + x − 2

(2,1/2) x

r (−3,−3/4)

= 0 nowhere.

ASY ASY ASY − −2 + 0 − 2 + −−−−−−−−|−−−−−−−− | −−−−−−−|−−−−→x y ⌢ ⌣ ⌢ ⌣

x=1

y ′′

Fig. 4.6.26

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SECTION 4.6 (PAGE 252)

27.

ADAMS and ESSEX: CALCULUS 8

x 3 − 3x 2 + 1 3 1 =1− + 3 x3 x x 3 3 3(x 2 − 1) y′ = 2 − 4 = x x x4 6 12 2 − x2 y ′′ = − 3 + 5 = 6 x x x5 From y : Asymptotes: y = 1, x = 0. Symmetry: none. Intercepts: since limx→0+ y = ∞, and lim x→0− y = −∞, there are intercepts between −1 and 0, between 0 and 1, and between 2 and 3. 1 Points: (−1, 3), (1, −1), (2, − 83 ), (3, ). 27 From y ′ : CP: x = ±1.

From y ′′ : y ′′ = 0 at x = kπ , where k is an integer.

y=

y ′′ + −2π − −π + 0 − π + 2π − −−−−−−−− | −−−−−−− |−−−−−−|−−−−−−− | −−−−−−|−−−−→x y ⌣ infl ⌢ infl ⌣ infl ⌢ infl ⌣ infl ⌢ y

2π

CP ASY CP y ′ + −1 − 0 − 1 + −−−−−−−−|−−−−−−−−− | −−−−−−−− | −−−−→x loc ց loc ր y ր max ց min √ From y ′′ : y ′′ = 0 at x = ± 2.

y = x + sin x

π

π

2π

x

ASY √ √ y ′′ + − 2 − 0 + 2 − −−−−−−−−−|−−−−−−−−|−−−−−−−− |−−−−→x y ⌣ infl ⌢ ⌣ infl ⌢ Fig. 4.6.28

y

(−1,3)

y=

x 3 − 3x 2 + 1 x3

29.

y=1 x (1,−1)

Fig. 4.6.27

28.

y′

y ′′

y = x + sin x, = 1 + cos x, = − sin x. From y: Intercept: (0, 0). Other points: (kπ, kπ ), where k is an integer. Symmetry: odd. From y ′ : Critical point: x = (2k + 1)π , where k is an integer. CP CP CP f′ + −π + π − 3π + −−−−−−−−−−− | −−−−−−−−− | −−−−−−−−− | −−−−−−→x f ր ր ր ր

y = x + 2 sin x, y = 0 if x = 0

y ′ = 1 + 2 cos x,

y ′′ = −2 sin x.

2π 1 ± 2nπ y ′ = 0 if x = − , i.e., x = ± 2 3 y ′′ = 0 if x = ±nπ From y:Asymptotes: (none). Symmetry: odd.    2π 2π √ 8π √ 8π Points: ± , ± + 3 , ± ,± + 3 , 3 3 3 3  4π 4π √ ± ,± − 3 . 3 3 2π ± 2nπ . From y ′ : CP: x = ± 3 CP CP CP CP CP 4π 2π 2π 4π y ′ − − 8π + − + − + − + 3 3 3 3 3 −−−−−−−−|−−−−−−−−|−−−−−−−−− | −−−−−−−− | −−−−−−−− | −−−−→x loc ր loc ց loc ր loc ց loc ր y ց min max min max min From y ′′ : y ′′ = 0 at x = ±nπ . y ′′ + −2π − −π + 0 − π + 2π − −−−−−−−− |−−−−−−−|−−−−−−− | −−−−−−|−−−−−−− | −−−→x y ⌣ infl ⌢ infl ⌣ infl ⌢ infl ⌣ infl ⌢

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.6 (PAGE 252)

y

From y ′′ : y ′′ = 0 at x = −2. y ′′ − −2 + −−−−−−−−−−−− | −−−−−−−→x y ⌢ infl ⌣ y 1π 3

y=x

π

2π

4π 3

x

y = x ex

y = x + 2 sin x

x



−2,−

2 e2



1 −1,− e

Fig. 4.6.31



Fig. 4.6.29

30.

2

2

2

y = e−x , y ′ = −2xe−x , y ′′ = (4x 2 − 2)e−x . From y: Intercept: (0, 1). Asymptotes: y = 0 (horizontal). Symmetry: even. From y ′ : Critical point: x = 0. CP y′ + 0 − −−−−−−−−−−−−−− |−−−−−−−−−−→x abs y ր ց max 1 From y ′′ : y ′′ = 0 at x = ± √ . 2

CP CP CP 5π 9π π − + − 0 + 4|−−−−−−−−− 4|−−−−−→x −−−− | −−−−−−−4|−−−−−−−−− abs ց abs ր loc y ր max min max ց From y ′′ : y ′′ = 0 at x = (k + 12 )π , where k is an integer.

+

y 1

y= −1 √ 2

x

Fig. 4.6.30

31.

π 3π 5π y ′′ 0 − + − + 2 2 −−−−− | −−−−−−−|−−−−−−−−− | −−−−−−−−2|−−−−−−→x y ⌢ infl ⌣ infl ⌢ infl ⌣

2 e−x

√1 2

y = e−x sin x (x ≥ 0), y ′ = e−x (cos x − sin x), y ′′ = −2e−x cos x. From y: Intercept: (kπ, 0), where k is an integer. Asymptotes: y = 0 as x → ∞. π From y ′ : Critical points: x = + kπ , where k is an 4 integer.

y′

1 1 −√ − √ + 2 2 −−−−−−−−−−−− | −−−−−−−−−−|−−−−−−−→x y ⌣ infl ⌢ infl ⌣ y ′′

32.

y

π −π/4 √
/ 2 4 ,e

y = xe x , y ′ = e x (1 + x), y ′′ = e x (2 + x). From y: Asymptotes: y = 0 (at x = −∞). Symmetry: (0,  0).   none. Intercept 1 2 Points: −1, − , −2, − 2 , e e From y ′ : CP: x = −1. CP y′ − −1 + −−−−−−−−−−−|−−−−−−−−→x abs y ց ր min

y = e−x sin x

π

5π 4

π 2

3π 2 x

Fig. 4.6.32

145 Copyright © 2014 Pearson Canada Inc.

SECTION 4.6 (PAGE 252)

33.

y = x 2 e−x

ADAMS and ESSEX: CALCULUS 8

y

2

2

y ′ = e−x (2x − 2x 3 ) = 2x(1 − x 2 )e−x

2

(−2,4e−2 )

y=

2

y ′′ = e−x (2 − 6x 2 − 2x(2x − 2x 3 ))

x 2 ex

2

= (2 − 10x 2 + 4x 4 )e−x From y: Asymptotes: y = 0. Intercept:  (0, 0).  Symmetry: even. 1 Points ±1, e From y ′ : CP x = 0, x = ±1.

√ −2− 2

x

Fig. 4.6.34

CP CP CP y ′ + −1 − 0 + 1 − −−−−−−−−|−−−−−−−−|−−−−−−−−|−−−−−→x abs ց abs ր abs ց y ր max min max From y ′′ : y ′′ = 0 if 2x 4 − 5x 2 + 1 = 0 √ 5 ± 25 − 8 2 x = √4 5 ± 17 . = 4 s s √ √ 5 + 17 5 − 17 so x = ±a = ± , x = ±b = ± . 4 4

35.

ln x , x

1 − ln x y′ = x2  1 x2 − − (1 − ln x)2x 2 ln x − 3 x ′′ y = = 4 x x3 From y: Asymptotes: x = 0, y = 0. Symmetry:   Intercept: (1,  0).  none. 3 1 3/2 , e , 3/2 . Points: e, e 2e From y ′ : CP: x = e. y=

ASY CP y′ 0 + e − −−−−−−|−−−−−−−−−−−|−−−−−−−→x abs ց y ր max

y ′′ + −a − −b + b − a + −−−−−−−−|−−−−−−−− | −−−−−−− | −−−−−−−|−−−−→x y ⌣ infl ⌢ infl ⌣ infl ⌢ infl ⌣

From y ′′ : y ′′ = 0 at x = e3/2 . ASY y ′′ 0 − e3/2 + −−−−−−− | −−−−−−−−−−|−−−−−−−→x y ⌢ infl ⌣

y

(−1,1/e)

√ −2+ 2

(1,1/e)

y = x 2 e−x

2

y (e,1/e)

−a

−b

b

a

x

34.

y = x 2 e x , y ′ = (2x + x 2 )e x = x(2√ + x)e x , √ ′′ 2 x y = (x + 4x + 2)e = (x + 2 − 2)(x + 2 + 2)e x . From y: Intercept: (0, 0). Asymptotes: y = 0 as x → −∞. From y ′ : Critical point: x = 0, x = −2.

y=

CP CP y′ + −2 − 0 + −−−−−−−−−−−− | −−−−−−−−−−−|−−−−−−−−→x loc abs y ր ց ր max min √ From y ′′ : y ′′ = 0 at x = −2 ± 2. y ′′

√

e3/2

1

Fig. 4.6.33

√

ln x x

Fig. 4.6.35

36.

+ −2 − 2 − −2 + 2 + −−−−−−−−−−−|−−−−−−−−−−−−− |−−−−−−−→x y ⌣ infl ⌢ infl ⌣

x

ln x (x > 0), x2 1 − 2 ln x ′′ 6 ln x − 5 y′ = , y = . x3 x4 From y: Intercepts: (1, 0). Asymptotes: y = 0, since y=

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.6 (PAGE 252)

y

ln x ln x = 0, and x = 0, since lim = −∞. 2 x→0+ x x2 ′ 1/2 From y : Critical point: x = e . lim

x→∞

y= √

CP √ y′ 0 + e − −−−−−|−−−−−−−−−−− |−−−−−−−−→x abs y ր ց max

1 4 − x2

From y ′′ : y ′′ = 0 at x = e5/6 .

1/2

38. √ ( e,(2e)−1 )

e5/6

y=

x

Fig. 4.6.37

y

1

2

−2

y ′′ 0 − e5/6 + −−−−−− | −−−−−−−−−−|−−−−−−−−→x y ⌢ infl ⌣

x

x , y ′ = (x 2 + 1)−3/2 , y ′′ = −3x(x 2 + 1)−5/2 . x2 + 1 From y: Intercept: (0, 0). Asymptotes: y = 1 as x → ∞, and y = −1 as x → −∞. Symmetry: odd. From y ′ : No critical point. y ′ > 0 and y is increasing for all x. From y ′′ : y ′′ = 0 at x = 0. y= √

y ′′ + 0 − −−−−−−−−−−− | −−−−−−→x y ⌣ infl ⌢

ln x x2

y y=1

Fig. 4.6.36

37.

y= √

1 x2

y=−1

= (4 − x 2 )−1/2

4− x 1 y = − (4 − x 2 )−3/2 (−2x) = 2 (4 − x 2 )3/2 3 (4 − x 2 )3/2 − x (4 − x 2 )1/2 (−2x) 2 y ′′ = (4 − x 2 )3 4 + 2x 2 = (4 − x 2 )5/2 From y: Asymptotes: x = ±2. Domain −2 < x < 2. Symmetry: even. Intercept: (0, 21 ). From y ′ : CP: x = 0.

From y ′′ : y ′′ = 0 nowhere, y ′′ > 0 on (−2, 2). Therefore, y is concave up.

x

x

x2 + 1

Fig. 4.6.38

′

ASY CP ASY y′ −2 − 0 + 2 −−−−−−|−−−−−−−−−−− |−−−−−−−−−−− | −−→x abs y ց ր min

y= √

39.

y = (x 2 − 1)1/3 2 y ′ = x(x 2 − 1)−2/3 3 2 2 ′′ y = [(x 2 − 1)−2/3 − x(x 2 − 1)−5/3 2x] 3 3   x2 2 2 −5/3 = − (x − 1) 1+ 3 3 From y: Asymptotes: none. Symmetry: even. Intercepts: (±1, 0), (0, −1). From y ′ : CP: x = 0. SP: x = ±1. SP CP SP y ′ − −1 − 0 + 1 + −−−−−−−−− | −−−−−−−−|−−−−−−−−|−−−−−→x abs ր y ց ց min ր

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SECTION 4.6 (PAGE 252)

ADAMS and ESSEX: CALCULUS 8

y

From y ′′ : y ′′ = 0 nowhere. y ′′ − −1 + 1 − −−−−−−−−−− | −−−−−−−−− | −−−−−→x y ⌢ infl ⌣ infl ⌢

−1 1 e , e


x

y 1 −1 e, e

y = (x 2 − 1)1/3




y = x ln |x|

−1

Fig. 4.6.40

1 x

41. −1

Fig. 4.6.39

sin x . 1 + x2 Curve crosses asymptote at infinitely many points: x = nπ (n = 0, ±1, ±2, . . .). y sin x y= 1 + x2 1 y = 0 is an asymptote of y =

y=

40. According to Theorem 5 of Section 4.4,

1+x 2

lim x ln x = 0.

x→0+

x

Thus, lim x ln |x| = lim x ln x = 0.

x→0

x→0+

y=−

If f (x) = x ln |x| for x 6= 0, we may define f (0) such that f (0) = lim x ln |x| = 0. Then f is continuous on

1 1+x 2

x→0

the whole real line and

Fig. 4.6.41 f ′ (x) = ln |x| + 1,

1 sgn (x). f ′′ (x) = |x|

From f : Intercept: (0, 0), (±1, 0). Asymptotes: none. Symmetry: odd. 1 From f ′ : CP: x = ± . SP: x = 0. e CP SP CP 1 1 ′ f + − − 0 − + e e| −−−−−−→x −−−−−−−−−−− | −−−−−−−−− | −−−−−−−−− loc loc f ր ց max ց min ր From f ′′ : f ′′ is undefined at x = 0. f ′′ − 0 + −−−−−−−−−−−−−−|−−−−−−−−−−→x f ⌢ infl ⌣

Section 4.7 (page 258)

Graphing with Computers

1. The longest (rightmost) of the exponential stripes in the given figure seems to begin at about (0.71, 35.7). Accordingly, the plot command > plot([exp(x)*ln(1+1/exp(x)), > 0.71*exp(x-35.7)], x=33..38, > y =0..2, style=point, sym-

bol=[circle, point], > color = [red, black], numpoints = 1500); will produce the curve as shown in the given figure (in red), and also the exponential curve (black) conforming to the rightmost stripe as shown in this figure:

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.8 (PAGE 264)

2 1.8 1.6 1.4 1.2 y 1 0.8 0.6 0.4 0.2 0 33

4. Since there are 64 − 52 − 2 = 10 bits left to represent the exponent the absolute value of the exponent, the smallest possible exponent is   −1111111111 = − 1 + 2 + 22 + · · · + 29 ,

that is, −1023 in base 10. The smallest positive mantissa is 0.000 · · · 001 = 2−52 , so the smallest positive binary floating-point number is 34

35

x

36

37

2−52 × 2−1023 = 2−52−1023

38

= 10−1075 log10 2 ≈ 10−324 .

Fig. 4.7.1

5. As in the previous exercise, there are 10 bits available

(The red curves appear gray here.) Other exponential stripes can be handled similarly

2. The square of any number will tend to require an increased number of digits to represent it—especially for squares of numbers with large numbers of digits to begin with. However on a computer the number of digits is fixed so the least significant digits are discarded. The resulting number is different from the square of the original number. Consequently the square root will not be quite the same and the expression cannot be expected to vanish exactly.

3. (a) We have   p g(x) = ln 2x − 22x − 1   1 = − ln √ 2x − 22x − 1 ! √ 2x + 22x − 1 √ √ = − ln (2x − 22x − 1)(2x + 22x − 1)   p = − ln 2x + 22x − 1 = f (x). (b) The problem is with g(x). When x grows large enough, the argument of the square root is evaluated as 22x , as the computer discards the 1. When this happens, the argument of the logarithm vanishes and the computer could be expected to return −∞. However we now arrive at the case of the previous exercise. The computer will return a variety of complex numbers, infinite values, and finite real values, produced by the computer evaluation of the logarithm of the expression in question 2. The computer only plots the finite real values, but all of them are completely spurious. 22x (1

2−2x ).

(c) The argument of the square root is − The computer will begin to encounter serious difficulties for values of x beginning where 2−2x ≈ ǫ, that is, −2x = −52 or x = 26. This is evident in the figure.

for the exponent, so the largest possible exponent is 1111111111 in base 2, or 1023 in base 10. The largest possible mantissa is 0.111 · · · 111 (52 digits) 1 1 1 = + 2 + · · · + 52 2 2 2 1 = 1 − 52 ≈ 1. 2 Thus the largest positive floating-point number is approximately 21023 = 101023 log10 2 ≈ 10308 .

Section 4.8 (page 264)

Extreme-Value Problems

1. Let the numbers be x and 7 − x. Then 0 ≤ x ≤ 7. The product is P(x) = x(7 − x) = 7x − x 2 . P(0) = P(7) = 0 and P(x) > 0 if 0 < x < 7. Thus maximum P occurs at a CP: 0=

dP 7 = 7 − 2x ⇒ x = . dx 2

The maximum product is P(7/2) = 49/4.

2. Let the numbers be x and

8 where x > 0. Their sum is x

8 . Since S → ∞ as x → ∞ or x → 0+, the x minimum sum must occur at a critical point: S=x+

0=

√ dS 8 = 1 − 2 ⇒ x = 2 2. dx x

√ √ 8 Thus, the smallest possible sum is 2 2 + √ = 4 2. 2 2

3. Let the numbers be x and 60 − x. Then 0 ≤ x ≤ 60.

Let P(x) = x 2 (60 − x) = 60x 2 − x 3 . Clearly, P(0) = P(60) = 0 amd P(x) > 0 if 0 < x < 60. Thus maximum P occurs at a CP: 0=

dP = 120x − 3x 2 = 3x(40 − x). dx

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SECTION 4.8 (PAGE 264)

ADAMS and ESSEX: CALCULUS 8

Therefore, x = 0 or 40. Max must correspond to x = 40. The numbers are 40 and 20.

Since A(x) → −∞ as x → ±∞ the maximum must occur at a critical point: 0=

4. Let the numbers be x and 16 − x. Let

P(x) = x 3 (16 − x)5 . Since P(x) → −∞ as x → ±∞, so the maximum must occur at a critical point: 0 = P ′ (x) = 3x 2 (16 − x)5 − 5x 3 (16 − x)4 = x 2 (16 − x)4 (48 − 8x).

The critical points are 0, 6 and 16. Clearly, P(0) = P(16) = 0, and P(6) = 216 × 105 . Thus, P(x) is maximum if the numbers are 6 and 10.

P Hence, the width and the length are and 2 P P (P − ) = . Since the width equals the length, it 2 2 is a square.

9. Let the dimensions of the isosceles triangle be as shown. Then 2x + 2y = P (given constant). The area is q

A = xh = x y 2 − x 2 = x

5. Let the numbers be x and 10 − x. We want to minimize S(x) = x 3 + (10 − x)2 ,

0 ≤ x ≤ 10.

S(0) = 100 and S(10) = 1, 000. For CP: 0 = S ′ (x) = 3x 2 − 2(10 − x) = 3x 2 + 2x − 20. √ The only positive CP is x = (−2 + 4 + 240)/6 ≈ 2.270. Since S(2.270) ≈ 71.450, the minimum value of S is about 71.45.

6. If the numbers are x and n − x, then 0 ≤ x ≤ n and the sum of their squares is

dA P = P − 2x ⇒ x = dx 2

s 

P −x 2

2

− x 2.

Evidently, y ≥ x so 0 ≤ x ≤ P/4. If x = 0 or x = P/4, then A = 0. Thus the maximum of A must occur at a CP. For max A: s dA P2 Px 0= = − Px − r , dx 4 P2 2 − Px 4 P2 P − 2Px − Px = 0, or x = . Thus y = P/3 and 2 6 the triangle is equilateral since all three sides are P/3.

i.e.,

S(x) = x 2 + (n − x)2 . y

Observe that S(0) = S(n) = n 2 . For critical points: 0 = S ′ (x) = 2x − 2(n − x) = 2(2x − n) ⇒ x = n/2.

x

Since S(n/2) = n 2 /2, this is the smallest value of the sum of squares.

7. Let the dimensions of a rectangle be x and y. Then the area is A = x y and the perimeter is P = 2x + 2y. Given A we can express P = P(x) = 2x +

2A , x

0=

dP 2A =2− 2 dx x

⇒

x 2 = A = x y ⇒ x = y.

Thus min P occurs for x = y, i.e., for a square.

8. Let the width and the length of a rectangle of given perimeter 2P be x and P − x. Then the area of the rectangle is A(x) = x(P − x) = Px − x 2 .

x

Fig. 4.8.9

10. Let the various dimensions be as shown in the figure. Since h = 10 sin θ and b = 20 cos θ , the area of the triangle is A(θ ) = 12 bh = 100 sin θ cos θ

(0 < x < ∞).

Evidently, minimum P occurs at a CP. For CP:

y

h

= 50 sin 2θ

for 0 < θ <

π . 2

π Since A(θ ) → 0 as θ → 0 and θ → , the maximum 2 must be at a critial point: 0 = A′ (θ ) = 100 cos 2θ ⇒ 2θ =

π π ⇒θ = . 2 4

Hence, the largest possible area is    π A(π/4) = 50 sin 2 = 50 m2 . 4

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.8 (PAGE 264)

(Remark: alternatively, we may simply observe that the largest value of sin 2θ is 1; therefore the largest possible area is 50(1) = 50 m2 .)

10

therefore P(x) is concave down on [0, R], so it must 2R have an absolute maximum value at x = √ . The largest 5 perimeter is therefore s     2R 4R 2 10R 2R =4 √ + R2 − = √ units. P √ 5 5 5 5

10

h

√

(x,

θ

θ b/2

R 2 −x 2 )

R

b/2 x

Fig. 4.8.10

11. Let the corners of the rectangle be as shown. √

The area of the rectangle is A = 2x y = 2x R 2 − x 2 (for 0 ≤ x ≤ R). If x = 0 or x = R then A = 0; otherwise A > 0. Thus maximum p A must occur at 2a critical  point: dA x 2 2 0= =2 R −x − √ ⇒ R 2 − 2x 2 = 0. dx R2 − x 2 R Thus x = √ and the maximum area is 2 s 2 R R = R 2 square units. 2√ R2 − 2 2 y (x,y)

R x

x

Fig. 4.8.11

Fig. 4.8.12

13. Let the upper right s corner be (x, y) as shown. Then x2 , so x ≤ a. a2 The area of the rectangle is s x2 A(x) = 4x y = 4bx 1 − 2 , a x ≥ 0 and y = b 1 −

Clearly, A = 0 if x = 0 or x = a, so maximum A must occur at a critical  point:  2 s 2x     x2 dA a2  = 4b  1 − 0= − s   2 dx a  x2  2 1− 2 a x2 x2 a b Thus 1 − 2 − 2 = 0 and x = √ . Thus y = √ . a a 2 2 a b The largest area is 4 √ √ = 2ab square units. 2 2

12. Let x be as shown in the figure. The perimeter of the

x2 y2 + =1 a 2 b2

rectangle is p P(x) = 4x + 2 R 2 − x 2

(0 ≤ x ≤ a).

y

(x,y)

(0 ≤ x ≤ R).

For critical points: x

dP −2x =4+ √ dx R2 − x 2 p 2R ⇒2 R 2 − x 2 = x ⇒ x = √ . 5 0=

Since d2 P −2R 2 = 8). w−8 w−8 Clearly, A → ∞ if w → ∞ or w → 8+. Thus minimum A occurs at a critical point: dA 100 100w =4+ − dw w − 8 (w − 8)2 100w = 4(w2 − 16w + 64) + 100w − 800

0=

that is, half the area of the triangle ABC. A C

w2 − 16w − 136 = 0 √ √ 16 ± 800 w= = 8 ± 10 2. 2

b y x C

1 (10)(10 sin θ ) = 50 sin θ cm2 , 2

a−x a

Fig. 4.8.14(a)

B

A

D

B

Fig. 4.8.14(b)

(b) This part has the same answer as part (a). To see this, let C D ⊥ AB, and solve separate problems for the largest rectangles in triangles AC D and BC D as shown. By part (a), both maximizing rectangles have the same height, namely half the length of C D. Thus, their union is a rectangle of area half of that of triangle ABC.

15. NEED FIGURE If the sides of the triangle are 10 cm, 10 cm, and√2x cm, then the area of the triangle is A(x) = x 100 − x 2 cm2 , where 0 ≤ x ≤ 10. Evidently A(0) = A(10) = 0 and A(x) > 0 for 0 < x < 10. Thus A will be maximum at a critical point. For a critical point   p 1 √ (−2x) 0 = A′ (x) = 100 − x 2 − x 2 100 − x 2 2 2 100 − x − x = √ . 100 − x 2 2 Thus the √ critical point is given by 2x = 100, so 50. The maximum area of the triangle is x √ = A( 50) = 50 cm2 .

√ Since w > 0 we must have w = 8 + 10 2. √ 100 Thus h = 4 + √ = 4 + 5 2. 10 2 √ √ The billboard should be 8 + 10 2 m wide and 4 + 5 2 m high.

w 2

4 h−4

h

w−8

4 2

Fig. 4.8.17

18. Let x be the side of the cut-out squares. Then the volume of the box is V (x) = x(70 − 2x)(150 − 2x)

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(0 ≤ x ≤ 35).

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.8 (PAGE 264)

Since V (0) = V (35) = 0, the maximum value will occur at a critical point:

21. Head for point C on road x km east of A. Travel time is

2

′

T =

0 = V (x) = 4(2625 − 220x + 3x ) = 4(3x − 175)(x − 15) 175 ⇒ x = 15 or . 3

122 + x 2 10 − x + . 15 39

12 10 + = 1.0564 hrs 15 39 √ 244 T (10) = = 1.0414 hrs 15 For critical points: T (0) =

We have

The only critical point in [0, 35] is x = 15. Thus, the largest possible volume for the box is V (15) = 15(70 − 30)(150 − 30) = 72, 000 cm3 .

0=

150 x x

√

1 x dT 1 = − √ dx 15 122 + x 2 39 p ⇒ 13x = 5 122 + x 2

⇒ (132 − 52 )x 2 = 52 × 122 ⇒ x = 5

150−2x 70−2x

70

5 13 + = 0.9949 < 15 39 (Or note that T (5) =

Fig. 4.8.18

d2T dt 2

19. Let the rebate be $x. Then number of cars sold per month is x  = 2000 + 4x. 50 The profit per car is 1000 − x, so the total monthly profit is 2000 + 200

P = (2000 + 4x)(1000 − x) = 4(500 + x)(1000 − x)



T (0) T (10).

√ x2 122 + x 2 − √ 1 122 + x 2 = 2 2 15 12 + x 122 = >0 15(122 + x 2 )3/2

so any critical point is a local minimum.) To minimize travel time, head for point 5 km east of A. A

C

x

= 4(500, 000 + 500x − x 2 ).

10−x

B

39 km/h

For maximum profit: 0=

dP = 4(500 − 2x) ⇒ x = 250. dx

12 15 km/h

d2 P = −8 < 0 any critical point gives a local dx2 max.) The manufacturer should offer a rebate of $250 to maximize profit.

√

122 +x 2

(Since

P

20. If the manager charges $(40+ x) per room, then (80−2x) rooms will be rented. The total income will be $(80 − 2x)(40 + x) and the total cost will be $(80 − 2x)(10) + (2x)(2). Therefore, the profit is P(x) = (80 − 2x)(40 + x) − [(80 − 2x)(10) + (2x)(2)] = 2400 + 16x − 2x 2

for x > 0.

Fig. 4.8.21

22. This problem is similar to the previous one except that the 10 in the numerator of the second fraction in the expression for T is replaced with a 4. This has no effect on the critical point of T , namely x = 5, which now lies outside the appropriate interval 0 ≤ x ≤ 4. Minimum T must occur at an endpoint. Note that

If P ′ (x) = 16 − 4x = 0, then x = 4. Since P ′′ (x) = −4 < 0, P must have a maximum value at x = 4. Therefore, the manager should charge $44 per room.

12 4 + = 0.9026 15 39 p 1 T (4) = 122 + 42 = 0.8433. 15 T (0) =

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SECTION 4.8 (PAGE 264)

ADAMS and ESSEX: CALCULUS 8

The minimum travel time corresponds to x = 4, that is, to driving in a straight line to B.

23. The time for the trip is 3000/(v − 100), so the total energy needed for the trip is

Since

v3 E = 3000k , v − 100

for A.

where k is a constant. Evidently we must have v > 100 or the trip would be impossible. The minimum value of E will occur at a critical point where 0=

1 1 , A(1) = > A(0). For CP: 16 4π   dA x 1−x 1 1 π 1 0= = − ⇒x + . = ⇒x= dx 2π 8 2π 8 8 4+π Now A(0) =

dE 2v 3 − 300v 2 = . dv (v − 100)2

d2 A 1 1 + > 0, the CP gives local minimum = dx2 2π 8

a) For max total area use none of wire for the square, i.e., x = 1. π 4 b) For minimum total area use 1 − = m 4+π 4+π for square. 1 metre 1−x

x

The minimum thus occurs at v = 150 knots, and the time for the flight at this speed would be 3000/50 = 60 hours, or 2.5 days. This is much slower than commercial airliners can travel; the time for the flight at that speed is not much shorter than a fast ship could cross the Atlantic ocean.

s r s x=C=2πr

24. (a) As shown in the previous exercise, the energy for a flight with airspeed v into the headwind is E h = kℓv 3 /(v − u). Similarly, the energy for a flight of the same distance with airspeed w and a tailwind of speed u is E t = kℓw3 /(w + u).

(b) Since E t′ = aℓw2 (2w + 3u)/(w + u)2 > 0 for all w > 0, E t is an increasing function of w. Thus its smallest value occurs for the smallest value of w that permits flight, namely w = s. The minimum energy for the tailwind trip is E t = kℓs 3 /(s + u). (c) The minimum energy for the tailwind part of the round trip in part (b) is independent of u. However, for the headwind part, the minimum E h = v = 3u/2 only applies as long as v ≥ s. Otherwise the plane cannot fly. If u > 2s/3, then v = 3u/2 and the least total energy for the trip is kℓ(s 3 /(s + u) + 27u 2 /4). If u < 2s/3, then the minimum value for E h at v = 3u/2 implies a speed too slow to stay airborne. As E h′ > 0 when v > 3u/2 the the least value for E h that is admissible happens for v = s. Thus the total energy becomes 2kℓs 4 /(s 2 − u 2 ).

25. Use x m for the circle and 1 − x m for square. The sum of areas is

2

π x2 1−x + 2 4π 4 x2 (1 − x)2 (0 ≤ x ≤ 1) = + 4π 42

A = πr 2 + s 2 =



1−x=P=4s

Fig. 4.8.25

26. Let the dimensions of the rectangle be as shown in the figure. Clearly, x = a sin θ + b cos θ, y = a cos θ + b sin θ. Therefore, the area is A(θ ) = x y = (a sin θ + b cos θ )(a cos θ + b sin θ ) = ab + (a 2 + b2 ) sin θ cos θ 1 = ab + (a 2 + b2 ) sin 2θ 2

for 0 ≤ θ ≤

π . 2

π . Since 4 π A′′ (θ ) = −2(a 2 + b2 ) sin 2θ < 0 when 0 ≤ θ ≤ , 2 π therefore A(θ ) must have a maximum value at θ = . 4 Hence, the area of the largest rectangle is If A′ (θ ) = (a 2 + b2 ) cos 2θ = 0, then θ =

A

π  4

π  1 = ab + (a 2 + b2 ) sin 2 2 1 1 2 2 = ab + (a + b ) = (a + b)2 2 2

sq. units.

a b (Note: x = y = √ + √ indicates that the rectangle 2 2 containing the given rectangle with sides a and b, has largest area when it is a square.)

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 4.8 (PAGE 264)

x θ a θ

x

a y b

θ

y

b

Fig. 4.8.26

27. Let the line have intercepts x, y as shown. Let θ be angle shown. The length of line is L=

√ 9 3 + cos θ sin θ

Fig. 4.8.28 (0 < θ <

If L ′ (θ ) = 0, then

π ). 2

π Clearly, L → ∞ if θ → 0+ or θ → −. 2 Thus the minimum length occurs at a critical point. For CP: √

dL 9 sin θ 3 cos θ = − ⇒ tan3 θ = dθ cos2 θ sin2 θ π ⇒θ = 6

0=



1 √ 3

a sin θ b cos θ − =0 cos2 θ sin2 θ a sin3 θ − b cos3 θ =0 cos2 θ sin2 θ a sin3 θ − b cos3 θ = 0 b tan3 θ = a b1/3 tan θ = 1/3 . a

⇔ ⇔ 3

⇔ ⇔

π Clearly, L(θ ) → ∞ as θ → 0+ or θ → −. Thus, the  1/3 2 b −1 minimum must occur at θ = tan . Using the a 1/3 1/3 b triangle above for tan θ = 1/3 , it follows that a

Shortest line segment has length √ √ 3 + = 8 3 units. L=√ 3/2 1/2 9

y

cos θ = √ Y

a 1/3 a 2/3 + b2/3

,

b1/3 sin θ = √ . a 2/3 + b2/3

Hence, the minimum is √ (9, 3)

θ 9

√ 3

L(θ ) = 

θ X

Fig. 4.8.27

of L = x + y, where x and y are as shown in the figure below: a b x= , y= . cos θ sin θ Thus, L = L(θ ) =

a b + cos θ sin θ



0 1 = ( p − 1)a p−1 ∞ if p < 1 Z a − p+1 a x x − p d x = lim c→0+ − p + 1

Z

(e−x − e−2x ) d x

 R 1 −e−x + e−2x R→∞ 2 0   1 1 1 = lim −e−R + e−2R + 1 − = sq. units. R→∞ 2 2 2 = lim



eu du

First assume that p 6= 1. Since a > 0 we have

Fig. 6.5.23 Z

R→∞ −1

Hence, the total area is I1 + I2 = 1 square unit.

y=ln x

Area of shaded region =

−1/R

eu du

1 = lim (e−1/R − e−1 ) = 1 − . R→∞ e

x

R

24.

−1

C→0+ −1/C

1 = lim (e−1 − e−1/C ) = . C→0+ e Z R Z −2 −1/x I2 = lim x e d x = lim

c→0+ units2

1

Z

∞

−p

0

y

c

a − p+1 c1− p = + lim c→0+ p − 1 1− p  1− p  a = 1 − p if p < 1  ∞ if p > 1.

1 y=e−x y=e−2x x

If p = 1 both integrals diverge as shown in Examples 2 and 6(a). Fig. 6.5.24

25.

∞

 4 2 Area = − dx 2x + 1 x +2 1   R = lim 2 ln(2x + 1) − ln(x + 2) R→∞ 1   2R + 1 = lim 2 ln − 0 = 2 ln 2 sq. units. R→∞ R+2 Z

Z

1

28.

2

29.

Z

26. The required area is Area = =

Z

∞

x −2 e−1/x d x

0

Z

0

1

x −2 e−1/x d x +

= I1 + I2 .

Z

∞ 1

x −2 e−1/x d x

Z 0 Z 1 x sgn x −x x dx = dx + dx −1 x + 2 −1 x + 2 0 x +2   Z 0 Z 1 2 2 = −1 + dx + 1− dx x + 2 x + 2 −1 0 0 1 16 = (−x + 2 ln |x + 2|) + (x − 2 ln |x + 2|) = ln . 9 −1 0

x 2 sgn (x − 1) d x Z 1 Z 2 = −x 2 d x + x2 dx 0

0

1

1 2 x3 x3 1 8 1 = − + = − + − = 2. 3 0 3 1 3 3 3 241

Copyright © 2014 Pearson Canada Inc.

SECTION 6.5 (PAGE 367)

30. Since

ADAMS and ESSEX: CALCULUS 8

x2 1 ≤ 3 for all x ≥ 0, therefore x +1

35.

x5

x2 dx +1 0 Z 1 Z ∞ x2 x2 = d x + dx 5 x5 + 1 0 x +1 1 Z 1 Z ∞ dx x2 ≤ dx + 5 x3 0 x +1 1 = I1 + I2 .

I =

∞

Z

x5

Since I1 is a proper integral (finite) and I2 is a convergent improper integral, (see Theorem 2), therefore I converges.

31.

32.

Z

∞ 2

√

x x dx ≥ x2 − 1

Z

2

∞

2x on [0, π/2], we have π Z ∞ Z π/2 | sin x| sin x d x ≥ dx x2 x2 0 0 Z 2 π/2 d x ≥ = ∞. π 0 x

37. Since sin x ≥

0

finite. Since x 3 ≥ x on [1, ∞), we have ∞

3

e−x d x ≤

1

Thus

Z

∞

∞ 1

e−x d x =

1 . e

y

y=sin x 2x y= π

π 2

38.

Fig. 6.5.37 √   √ 2 √ x x x ≤2 = , Since 0 ≤ 1 − cos x = 2 sin2 2 2 2 Z π2 Z π2 dx dx for x ≥ 0, therefore √ ≥2 , which x 1 − cos x 0 0 diverges to infinity.

Z

0

1 1 1 1 ≤ √ . On [1, ∞), √ ≤ 2. x x + x2 x x + x2 Z

Z

1

1

0 ∞

x

39. On (0, π/2], sin x < x, and so csc x ≥ 1/x. Thus

3

e−x d x converges.

34. On [0,1], √ Thus,

Z

The given integral diverges to infinity.

dx √ = I1 = ∞. x

Since I1 is a divergent improper integral, I diverges. Z 1 Z ∞  Z ∞ 3 3 e−x d x = e−x d x. + 0 0 1 Z 1 −x 3 e Now d x is a proper integral, and is therefore Z

The given integral diverges to infinity. sin x ≤ 1. Then Since sin x ≤ x for all x ≥ 0, thus x Z π Z π Z π sin x sin x I = d x = lim dx ≤ (1) d x = π. ǫ→0+ ǫ x x 0 0 Hence, I converges.

1 1 √ ≥ √ on [1, ∞). 1 + Zx 2 x Z ∞ ∞ dx dx Since √ diverges to infinity, so must √ . x 1 + x 1 1 Z ∞ dx Therefore √ also diverges to infinity. 1+ x 0 √ x x 1 Since 2 ≥ √ for all x > 1, therefore x −1 x I =

33.

36.

ex e−1 ≥ on [−1, 1]. Thus x +1 x +1 Z 1 Z ex 1 1 dx dx ≥ = ∞. e −1 x + 1 −1 x + 1

0

Therefore ∞ − ∞.)

Z 1 dx dx ≤ √ √ x + x2 x Z0 ∞ dx dx √ ≤ . x2 x + x2 1

π/2

Z

csc x d x >

Z

0

π/2

dx = ∞. x

π/2

csc x d x must diverge. (It is of the form

−π/2

40. Since ln x grows more slowly than any positive power of

Since both ofZ these integrals are convergent, therefore so ∞ dx . is their sum √ x + x2 0

x, therefore we have ln x ≤ kx 1/4 for some constant k 1 1 and every x ≥ 2. Thus, √ for x ≥ 2 ≥ 3/4 kx x ln x Z ∞ dx and √ diverges to infinity by comparison with x ln x Z ∞2 1 dx . k 2 x 3/4

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INSTRUCTOR’S SOLUTIONS MANUAL

41.

Z

∞

0

dx = xe x

1

Z

Z

+

0

1

Z

1

∞

dx . But xe x

1 dx ≥ xe x e

0

SECTION 6.5 (PAGE 367)

Z

1

0

44. We have

dx = ∞. x

Thus the given integral must diverge to infinity.

42. We are given that

Z

∞

1√ 2 π.

2

e−x d x =

0

∞

0

2

x 2 e−x d x = lim

Z

R

2

x 2 e−x d x

R→∞ 0 2 d V = xe−x

U=x dx 2 dU = d x V = − 12 e−x   Z R 1 R −x 2 1 2 = lim − xe−x + e dx R→∞ 2 2 0 Z ∞0 2 1 2 1 = − lim Re−R + e−x d x 2 R→∞ 2 0 1√ 1√ =0+ π= π. 4 4

b) Similarly, Z

∞ 0

2

x 4 e−x d x = lim

Z

R

R→∞ 0

2

x 4 e−x d x −x 2

U = x3 d V = xe dx 2 dU = 3x 2 d x V = − 12 e−x R   Z 3 R 2 −x 2 1 2 = lim − x 3 e−x + x e dx R→∞ 2 2 0 Z ∞0 1 3 2 2 = − lim R 3 e−R + x 2 e−x d x 2 R→∞ 2 0   3√ 3 1√ =0+ π = π. 2 4 8

43. Since f (x) = O(x p ) as x → 0+, we have | f (x) ≤ K x p for all x in the interval (0, a) for some constants K and a ≤ 1. If 0 < ǫ < a it follows that ǫ

0

≤K

so E(ǫ) =

O(ǫ p+1 )

Z

0

f (x) d x

ǫ

x p dx =

1/ǫ 

ǫ

Now we must show that no value of k larger than 1/2 will work. If k > 1/2, then Z ǫ Z ǫ √ dx dx I1 (ǫ) = √ > √ = ǫ. 2 x +x 0 0 2 x

I1 (ǫ) > ǫ (1/2)−k → ∞ as ǫ → 0+, and I1 (ǫ) is not ǫk O(ǫ k ).

Thus

45. Since f is continuous on [a, b], there exists a positive constant K such that | f (x)| ≤ K for a ≤ x ≤ b. If a < c < b, then Z b Z b f (x) d x − f (x) d x c a Z a f (x) d x ≤ K (c − a) → 0 as c → a + . = c

b

Z

c

f (x) d x =

c→a+ c

Z

b

f (x) d x.

a

Similarly Z c Z b f (x) d x f (x) d x − a a Z c f (x) d x ≤ K (b − c) → 0 as c → b − . = b

Thus lim

c→b− a Z ∞

f (x) d x =

Z

b

f (x) d x.

a

t x−1 e−t dt.

0

a) Since limt→∞ t x−1 e−t/2 = 0, there exists T > 0 such that t x−1 e−t/2 ≤ 1 if t ≥ T . Thus Z ∞ Z ∞ e−t dt = 2e−T /2 t x−1 e−t dt ≤ 0≤ T

and K ǫ p+1 , p+1

Z

Thus lim

46. Ŵ(x) =

Z |E(ǫ) =

Z

dx √ x + x2 Z ∞ Z dx dx + = I1 (ǫ) + I2 (ǫ) = √ √ 2 x + x x + x2 1/ǫ 0 Z ǫ √ dx |I1 (ǫ)| < √ =2 ǫ x Z0 ∞ dx |I2 (ǫ)| < = ǫ. 2 1/ǫ x 0 ǫ

−

Thus I1 (ǫ) = O(ǫ 1/2 ) as ǫ → 0+, and I2 (ǫ) = O(ǫ) as ǫ → 0+. Since ǫ < ǫ 1/2 (for 0 < ǫ < 1), both I1 (ǫ) and I2 (ǫ) are O(ǫ 1/2 ) as ǫ → 0+, and so, therefore, is E(ǫ).

a) First we calculate Z

∞

Z

E(ǫ) =

Z

∞

T

t x−1 e−t dt converges by the comparison

T

theorem. If x > 0, then 0≤

as ǫ → 0+.

Z

0

T

t x−1 e−t dt <

Z

T

t x−1 dt

0

243 Copyright © 2014 Pearson Canada Inc.

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ADAMS and ESSEX: CALCULUS 8

converges by Theorem 2(b). Thus the integral defining Ŵ(x) converges. Z

0

c→0+ R→∞

c

U = tx d V = e−t dt dU = xt x−1 d x V = −e−t ! R Z R = lim −t x e−t + x t x−1 e−t dt c→0+ c c R→∞ Z ∞ =0+x t x−1 e−t dt = xŴ(x). 0

c) Ŵ(1) =

Z

∞

0

      1 9 5 1 1 + 1+ + (1 + 1) + 1 + + = 4.75 2 2 4 4 2         1 1 9 25 49 M4 = 1+ + 1+ + 1+ + 1+ 2 16 16 16 16 = 4.625 1 T8 = (T4 + M4 ) = 4.6875 2       1 1 9 25 M8 = 1+ + 1+ + 1+ 4 64 64 64       49 81 121 + 1+ + 1+ + 1+ 64 64 64     169 225 + 1+ + 1+ = 4.65625 64 64 1 T16 = (T8 + M8 ) = 4.671875. 2 T4 =

∞

t x e−t dt Z R = lim t x e−t dt

b) Ŵ(x + 1) =

The approximations are

e−t dt = 1 = 0!.

By (b), Ŵ(2) = 1Ŵ(1) = 1 × 1 = 1 = 1!. In general, if Ŵ(k + 1) = k! for some positive integer k, then Ŵ(k + 2) = (k + 1)Ŵ(k + 1) = (k + 1)k! = (k + 1)!. Hence Ŵ(n + 1) = n! for all integers n ≥ 0, by induction.   Z ∞ 1 = t −1/2 e−t dt Let t = x 2 d) Ŵ 2 0 dt =Z2x d x Z ∞ ∞ √ 1 −x 2 2 = e 2x d x = 2 e−x d x = π x 0   0   1 1√ 1 3 = Ŵ = Ŵ π. 2 2 2 2

The exact errors are I − T4 = −0.0833333; I − T8 = −0.0208333; I − T16 = −0.0052083.

If f (x) = 1 + x 2 , then f ′′ (x) = 2 = K , and 1 K (2 − 0) = . Therefore, the error bounds are 12 3  2 1 ≈ 0.0833333; 2   1 1 2 − T8 | ≤ ≈ 0.0208333; 3 4   1 1 2 − T16 | ≤ ≈ 0.0052083. 3 8  2 1 1 − M4 | ≤ ≈ 0.0416667; 6 2  2 1 1 − M8 | ≤ ≈ 0.0104167. 6 4

Trapezoid : |I − T4 | ≤ |I |I

Section 6.6 The Trapezoid and Midpoint Rules (page 375)

I − M4 = 0.0416667; I − M8 = 0.0104167;

Midpoint : |I |I

1 3

Note that the actual errors are equal to these estimates since f is a quadratic function.

1. The exact value of I is

2. The exact value of I is

  2 x 3 (1 + x 2 ) d x = x + 3 0 0 8 = 2 + ≈ 4.6666667. 3

I =

Z

2

244 Copyright © 2014 Pearson Canada Inc.

I =

Z

1 0

1 e−x d x = −e−x

1 = 1 − ≈ 0.6321206. e

0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 6.6 (PAGE 375)

The approximations are

The approximations are

T4 = 14 ( 12 e0 + e−1/4 + e−1/2 + e−3/4 + 21 e−1 ) ≈ 0.6354094

M4 = 14 (e−1/8 + e−3/8 + e−5/8 + e−7/8 ) ≈ 0.6304774

T8 = 21 (T4 + M4 ) ≈ 0.6329434

M8 = 81 (e−1/16 + e−3/16 + e−5/16 + e−7/16 + e−9/16 + e−11/16 + e−13/16 + e−15/16 ) ≈ 0.6317092

T16 = 12 (T8 + M8 ) ≈ 0.6323263. The exact errors are I − T4 = −0.0032888; I − T8 = −0.0008228; I − T16 = −0.0002057.

I − M4 = 0.0016432; I − M8 = 0.0004114;

If f (x) = e−x , then f (2) (x) = e−x . On [0,1], | f (2) (x)| ≤ 1. Therefore, the error bounds are:   1 1 2 12 n   1 1 − T4 | ≤ ≈ 0.0052083; 12 16   1 1 ≈ 0.001302; − T8 | ≤ 12 64   1 1 − T16 | ≤ ≈ 0.0003255. 12 256   1 1 2 − Mn | ≤ 24 n   1 1 − M4 | ≤ ≈ 0.0026041; 24 16   1 1 − M8 | ≤ ≈ 0.000651. 24 64

Trapezoid : |I − Tn | ≤ |I |I |I Midpoint : |I |I |I

  π π 3π 1 π + 0 + sin + sin + sin ≈ 0.9871158 8 8 4 8 2   π 3π 5π 7π π sin + sin + sin + sin ≈ 1.0064545 M4 = 8 16 16 16 16 1 T8 = (T4 + M4 ) ≈ 0.9967852 2  π π 3π 5π sin M8 = + sin + sin 16 32 32 32 7π 9π 11π + sin + sin + sin 32 32 32  13π 15π + sin + sin ≈ 1.0016082 32 32 1 T16 = (T8 + M8 ) ≈ 0.9991967. 2 T4 =

The actual errors are I − T4 ≈ 0.0128842; I − T8 ≈ 0.0032148; I − T16 ≈ 0.0008033.

I − M4 ≈ −0.0064545; I − M8 ≈ −0.0016082;

If f (x) = sin x, then f ′′ (x) = − sin x, and | f ′′ (x)| ≤ 1 = K . Therefore, the error bounds are:   π 2 1 π −0 ≈ 0.020186; 12 2 8     1 π π 2 − T8 | ≤ −0 ≈ 0.005047; 12 2 16     1 π π 2 − T16 | ≤ −0 ≈ 0.001262. 12 2 32     2 1 π π − M4 | ≤ −0 ≈ 0.010093; 24 2 8     1 π π 2 − M8 | ≤ −0 ≈ 0.002523. 24 2 16

Trapezoid : |I − T4 | ≤ |I |I Midpoint : |I |I

Note that the actual errors satisfy these bounds.

Note that the actual errors satisfy these bounds.

4. The exact value of I is

3. The exact value of I is

I =

Z

π/2 0

sin x d x = 1.

I =

Z

0

1

dx = tan−1 1 + x2

1 π x = ≈ 0.7853982. 4 0 245

Copyright © 2014 Pearson Canada Inc.

SECTION 6.6 (PAGE 375)

ADAMS and ESSEX: CALCULUS 8

The approximations are    1 1 16 4 16 1 1 T4 = (1) + + + + 4 2 17 5 25 2 2 ≈ 0.7827941   1 64 64 64 64 M4 = + + + 4 65 73 89 113 ≈ 0.7867001 1 T8 = (T4 + M4 ) ≈ 0.7847471 2 1 256 256 256 256 M8 = + + + + 8 257 265 281 305  256 256 256 256 + + + 337 377 425 481 ≈ 0.7857237 1 T16 = (T8 + M8 ) ≈ 0.7852354. 2 The exact errors are I − T4 = 0.0026041; I − M4 = −0.0013019; I − T8 = 0.0006511; I − M8 = −0.0003255; I − T16 = 0.0001628. 1 −2x , then f ′ (x) = and Since f (x) = 2 1+x (1 + x 2 )2 2 6x − 2 f ′′ (x) = . On [0,1], | f ′′ (x)| ≤ 4. Therefore, (1 + x 2 )3 the error bounds are

Trapezoid : |I |I |I |I Midpoint : |I |I |I

  4 1 2 − Tn | ≤ 12 n   4 1 − T4 | ≤ ≈ 0.0208333; 12 16   4 1 − T8 | ≤ ≈ 0.0052083; 12 64   4 1 − T16 | ≤ ≈ 0.001302. 12 256   4 1 2 − Mn | ≤ 24 n   4 1 − M4 | ≤ ≈ 0.0104167; 24 16   4 1 − M8 | ≤ ≈ 0.0026042. 24 64

The exact errors are much smaller than these bounds. In part, this is due to very crude estimates made for | f ′′ (x)|.

5.

2 [3 + 2(5 + 8 + 7) + 3] = 46 2 1 T8 = [3 + 2(3.8 + 5 + 6.7 + 8 + 8 + 7 + 5.2) + 3] = 46.7 2

T4 =

6. 7.

8.

M4 = 2(3.8 + 6.7 + 8 + 5.2) = 47.4 2 T4 = 100 × [0 + 2(5.5 + 5 + 4.5) + 0] = 3, 000 km2 2 1 T8 = 100 × [0 + 2(4 + 5.5 + 5.5 + 5 + 5.5 + 4.5 + 4) + 0] 2 = 3, 400 km2 M4 = 100 × 2(4 + 5.5 + 5.5 + 4) = 3, 800 km2

9. We have  T4 = 0.4 12 (1.4142) + 1.3860 + 1.3026 + 1.1772  + 12 (0.9853) ≈ 2.02622

M4 = (0.4)(1.4071 + 1.3510 + 1.2411 + 1.0817) ≈ 2.03236 T8 = (T4 + M4 )/2 ≈ 2.02929 M8 = (0.2)(1.4124 + 1.3983 + 1.3702 + 1.3285 + 1.2734 + 1.2057 + 1.1258 + 1.0348) ≈ 2.02982 T16 = (T8 + M8 )/2 ≈ 2.029555.

10. The approximations for I =

Z

1

2

e−x d x are

0

 1 −1/256 e + e−9/256 + e−25/256 + e−49/256 + 8  e−81/256 + e−121/256 + e−169/256 + e−225/256

M8 =

T16

≈ 0.7473  1 1 = (1) + e−1/256 + e−1/64 + e−9/256 + e−1/16 + 16 2 e−25/256 + e−9/64 + e−49/256 + e−1/4 + e−81/256 +

e−25/64 + e−121/256 + e−9/16 + e−169/256 + e−49/64 +  1 e−225/256 + e−1 2 ≈ 0.74658. 2

2

Since f (x) = e−x , we have f ′ (x) = −2xe−x , 2 2 f ′′ (x) = 2(2x 2 − 1)e−x , and f ′′′ (x) = 4x(3 − 2x 2 )e−x . Since f ′′′ (x) 6= 0 on (0,1), therefore the maximum value of | f ′′ (x)| on [0, 1] must occur at an endpoint of that interval. We have f ′′ (0) = −2 and f ′′ (1) = 2/e, so | f ′′ (x)| ≤ 2 on [0, 1]. The error bounds are  2   1 2 1 ⇒ |I − M8 | ≤ n 24 64 ≈ 0.00130.  2   2 1 2 1 |I − Tn | ≤ ⇒ |I − T16 | ≤ 12 n 12 256 ≈ 0.000651.

|I − Mn | ≤

246 Copyright © 2014 Pearson Canada Inc.

2 24

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 6.6 (PAGE 375)

14. Let y = f (x). We are given that m 1 is the midpoint of

According to the error bounds, Z

11.

1

e

−x 2

0

[x0 , x1 ] where x1 − x0 = h. By tangent line approximate in the subinterval [x0 , x1 ],

d x = 0.747,

accurate to two decimal places, with error no greater than 1 in the third decimal place. Z π/2 sin x sin x I = d x. Note that lim = 1. x→0 x x 0 

π 1 16 π 8 π 16 3π 4 π + sin + sin + sin + sin 16 2 π 16 π 8 3π 16 π 4   16 5π 8 3π 16 7π 1 2 + sin + sin + sin + 5π 16 3π 8 7π 16 2 π ≈ 1.3694596  π 32 π 32 3π 32 5π 32 7π M8 = sin + sin + sin + sin 16 π 32 3π 32 5π 32 7π 32 9π 32 11π 32 13π 32 sin + sin + sin + 9π 32 11π 32 13π 32  32 15π + sin ≈ 1.3714136 15π 32 T16 = (T8 + M8 )/2 ≈ 1.3704366, I ≈ 1.370. T8 =

12. The exact value of I is I =

Z

0

The approximation is

1

1 1 x 3 = . x dx = 3 0 3 2

  1 1 2 1 2 T1 = (1) (0) + (1) = . 2 2 2

13.

The actual error is I − T1 = − 16 . However, since f (x) = x 2 , then f ′′ (x) = 2 on [0,1], so the error estimate here gives 1 2 (1)2 = . |I − T1 | ≤ 12 6 Since this is the actual size of the error in this case, the constant “12” in the error estimate cannot be improved (i.e., cannot be made larger).  2 Z 1 1 1 1 2 I = x d x = . M1 = (1) = . The actual 3 2 4 0 1 1 1 error is I − M1 = − = . 3 4 12 Since the second derivative of x 2 is 2, the error estimate is 2 1 |I − M1 | ≤ (1 − 0)2 (12 ) = . 24 12 Thus the constant in the error estimate for the Midpoint Rule cannot be improved; no smaller constant will work for f (x) = x 2 .

f (x) ≈ f (m 1 ) + f ′ (m 1 )(x − m 1 ). The error in this approximation is E(x) = f (x) − f (m 1 ) − f ′ (m 1 )(x − m 1 ). If f ′′ (t) exists for all t in [x0 , x1 ] and | f ′′ (t)| ≤ K for some constant K , then by Theorem 11 of Section 4.9,

|E(x)| ≤

K (x − m 1 )2 . 2

Hence, | f (x) − f (m 1 ) − f ′ (m 1 )(x − m 1 )| ≤

K (x − m 1 )2 . 2

We integrate both sides of this inequlity. Noting that x1 − m 1 = m 1 − x0 = 21 h, we obtain for the left side Z

x1

f (x) d x −

x0

Z

Z

x1

x1

f (m 1 ) d x x0

f (m 1 )(x − m 1 ) d x ′

− x0 Z x x1 (x − m 1 )2 1 ′ f (x) d x − f (m 1 )h − f (m 1 ) = x0 2 x0 Z x1 f (x) d x − f (m 1 )h . = x0

Integrating the right-hand side, we get

x K K (x − m 1 )3 1 (x − m 1 )2 d x = 2 3 x0 2 x0  3  K h h3 K 3 = + = h . 6 8 8 24

Z

x1

x1

f (x) d x − f (m 1 )h

Hence, Z

x0

Z =

x1

x0

[ f (x) − f (m 1 ) − f ′ (m 1 )(x − m 1 )] d x

K 3 ≤ h . 24

247 Copyright © 2014 Pearson Canada Inc.

SECTION 6.6 (PAGE 375)

ADAMS and ESSEX: CALCULUS 8

A similar estimate holds on each subinterval [x j −1 , x j ] for 1 ≤ j ≤ n. Therefore, Z

b

a

X Z n f (x) d x − Mn =

xj

x j −1 j =1 Z x n X j

≤



j =1 n X

≤

j =1

x j −1

 f (x) d x − f (m j )h

3.

0

f (x) d x − f (m j )h

  π π π 3π π 0 + 4 sin + 2 sin + 4 sin + sin 24 8 4 8 2 ≈ 1.0001346  π π 3π π π S8 = + 2 sin + 4 sin + 2 sin 0 + 4 sin 48 16 8 16 4  5π 3π 7π π + 4 sin + 2 sin + 4 sin + sin 16 8 16 2 ≈ 1.0000083.

S4 =

K 3 K 3 K (b − a) 2 h = nh = h 24 24 24

because nh = b − a.

Section 6.7 Simpson’s Rule 1.

I =

Z

2

0

(1 + x 2 ) d x =

(page 380)

Errors: I − S4 ≈ −0.0001346; I − S8 ≈ −0.0000083.

14 ≈ 4.6666667 3

4. The exact value of I is

     1 1 9 S4 = 1+4 1+ + 2(1 + 1) + 4 1 + 6 4 4  14 + (1 + 4) = 3        1 1 1 9 S8 = 1+4 1+ +2 1+ +4 1+ 12 16 4 16     9 25 +2 1+ + 2 (1 + 1) + 4 1 + 16 4    49 14 +4 1+ + (1 + 4) = 16 3

I =

1

e 0

=1−

0

1

1 dx π −1 = tan x = ≈ 0.7853982. 1 + x2 4 0

The actual errors are

2. The exact value of I is I =

Z

The approximations are         1 16 4 16 1 1+4 +2 +4 + S4 = 12 17 5 25 2 ≈ 0.7853922        1 64 16 64 S8 = 1+4 +2 +4 + 24 65 17 73          1 4 64 16 64 2 +4 +2 +4 + 5 89 25 113 2 ≈ 0.7853981.

The errors are zero because Simpson approximations are exact for polynomials of degree up to three.

Z

These errors are evidently much smaller than the corresponding errors for the corresponding Trapezoid Rule approximations. Z π/2 I = sin x d x = 1.

−x

d x = −e

I − S4 = 0.0000060; I − S8 = 0.0000001,

1 −x

accurate to 7 decimal places. These errors are evidently much smaller than the corresponding errors for the corresponding Trapezoid Rule approximation.



0

1 ≈ 0.6321206. e

5.

The approximations are 1 0 (e + 4e−1/4 + 2e−1/2 + 4e−3/4 + e−1 ) 12 ≈ 0.6321342 1 0 S8 = (e + 4e−1/8 + 2e−1/4 + 4e−3/8 + 24 2e−1/2 + 4e−5/8 + 2e−3/4 + 4e−7/8 + e−1 ) ≈ 0.6321214.

S4 =

The actual errors are I − S4 = −0.0000136; I − S8 = −0.0000008.

6.

1 [3 + 4(3.8 + 6.7 + 8 + 5.2) + 2(5 + 8 + 7) + 3] 3 ≈ 46.93

S8 =

1 S8 = 100 × [0 + 4(4 + 5.5 + 5.5 + 4) + 2(5.5 + 5 + 4.5) + 0] 3 ≈ 3, 533 km2

7. If f (x) = e−x , then f (4) (x) = e−x , and | f (4) (x)| ≤ 1 on [0, 1]. Thus

  1(1 − 0) 1 4 ≈ 0.000022 180 4   1(1 − 0) 1 4 |I − S8 | ≤ ≈ 0.0000014. 180 8

248 Copyright © 2014 Pearson Canada Inc.

|I − S4 | ≤

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 6.7 (PAGE 380)

If f (x) = sin x, then f (4) (x) = sin x, and | f (4) (x)| ≤ 1 on [0, π/2]. Thus

S2n =

1((π/2) − 0)  π 4 |I − S4 | ≤ ≈ 0.00021 180 8   4 1((π/2) − 0) π |I − S8 | ≤ ≈ 0.000013. 180 16

8. Let I =

Z

Hence, 4T2n − Tn Tn + 2Mn 2T2n + Mn = = . 3 3 3

9. We use the results of Exercise 9 of Section 7.6 and Exercise 8 of this section.

b

f (x) d x, and the interval [a, b] be subdia

vided into 2n subintervals of equal length h = (b − a)/2n. Let y j = f (x j ) and x j = a + j h for 0 ≤ j ≤ 2n, then S2n =

1 3

=

1 3



b−a 2n





b−a 2n



y0 + 4y1 + 2y2 + · · ·  + 2y2n−2 + 4y2n−1 + y2n y0 + 4

2n−1 X j =1

yj − 2

n−1 X j =1

y2 j + y2n

and T2n = Tn =

1 2



1 2



b−a 2n



b−a n



y0 + 2 y0 + 2

2n−1 X j =1

n−1 X j =1

y j + y2n



 y2 j + y2n .

Since T2n = 12 (Tn + Mn ) ⇒ Mn = 2T2n − Tn , then Tn + 2Mn Tn + 2(2T2n − Tn ) 4T2n − Tn = = 3 3 3 2T2n + 2T2n − Tn 4T2n − Tn 2T2n + Mn = = . 3 3 3 Hence,



I =

Z

1.6

f (x) d x 0

0.4 (1.4142 + 4(1.3860) + 2(1.3026) + 4(1.1772) 3 + 0.9853) ≈ 2.0343333 S8 = (T4 + 2M4 )/3 ≈ 2.0303133 S16 = (T8 + 2M8 )/3 ≈ 2.0296433. S4 =

10. The approximations for I =

Z

1

0

   1 1 S8 = 1 + 4 e−1/64 + e−9/64 + e−25/64 + 3 8     e−49/64 + 2 e−1/16 + e−1/4 + e−9/16 + e−1 S16

≈ 0.7468261    1 1 = 1 + 4 e−1/256 + e−9/256 + e−25/256 + 3 16

e−49/256 + e−81/256 + e−121/256 + e−169/256 +   e−225/256 + 2 e−1/64 + e−1/16 + e−9/64 + e−1/4 +   e−25/64 + e−9/16 + e−49/64 + e−1

≈ 0.7468243. 2

Tn + 2Mn 2T2n + Mn 4T2n − Tn = = . 3 3 3 Using the formulas of T2n and Tn obtained above, 4T2n − Tn 3     2n−1 X 1 4 b−a = y0 + 2 y j + y2n 3 2 2n j =1    n−1 X 1 b−a − y0 + 2 y2 j + y2n 2 n j =1    2n−1 n−1 X X 1 b−a = y0 + 4 yj − 2 y2 j + y2n 3 2n j =1 j =1 = S2n .

2

e−x d x are

2

If f (x) = e−x , then f (4) (x) = 4e−x (4x 4 − 12x 2 + 3). On [0,1], | f (4) (x)| ≤ 12, and the error bounds are   12(1) 1 4 180 n   12 1 4 |I − S8 | ≤ ≈ 0.0000163 180 8  4 12 1 |I − S16 | ≤ ≈ 0.0000010. 180 16 |I − Sn | ≤

Comparing the two approximations, I =

Z

0

1

2

e−x d x = 0.7468,

accurate to 4 decimal places.

249 Copyright © 2014 Pearson Canada Inc.

SECTION 6.7 (PAGE 380)

11.

I =

Z

0

1

ADAMS and ESSEX: CALCULUS 8

" #  4 1 1 4 1 5 4 x d x = . S2 = 0 +4 +1 = . 5 6 2 24 4

If f (x) = x 4 , then f (4) (x) = 24.   1 24(1 − 0) 1 4 = Error estimate: |I − S2 | ≤ . 180 2 120 1 5 1 Actual error: |I − S2 | = − = . 5 24 120 Thus the error estimate cannot be improved.

so I = 2

4.

12. The exact value of I is Z

I =

1

0

x3 dx =

The approximation is S2 =

1 3

1 1 x 4 = . 4 0 4

    3 1 1 1 + 13 = . 03 + 4 2 2 4

The actual error is zero. Hence, Simpson’s Rule is exact for the cubic function f (x) = x 3 . Since it is evidently exact for quadratic functions f (x) = Bx 2 + C x + D, it must also be exact for arbitrary cubics f (x) = Ax 3 + Bx 2 + C x + D.

Section 6.8 Other Aspects of Approximate Integration (page 386) 1.

2.

Z

dx Let x = u 3 1/3 (1 + x) x 0 Z 1 Z 1 u 2 du u du =3 = 3 . 3 3 0 u(1 + u ) 0 1+u 1

I =

1

ex

−1

√

dx

=

1 − x2

Z

π/2

Z

0 −1

ex d x √ + 1 − x2

Z

0

1

I1 = I2 =

Z

1

0

0

1

2 −1

√

+ e1−u

2 − u2

2

du.

1 t2 2 dt dx = − 3 t   Z 0 1 2 dt = − 3  2 r t 1 1 1 + +1 t2 t2 Z 1 t dt =2 . 4 3 0 t +t +1

Z

π/2 0

=2

dx √ x +1

dx √ sin x

Z

1

0 1

=2

Z Z

1

=4 =4

Z

0

0 1

0

∞

Let x =

Let sin x = u 2 √ 2u du = cos x d x = 1 − u 4 d x

u du √ u 1 − u4

du p (1 − u)(1 + u)(1 + y 2

Let 1 − u = v 2 −du = 2v dv

v dv p v (1 + 1 − v 2 )(1 + (1 − v 2 )2 ) dv p . (2 − v 2 )(2 − 2v 2 + v 4 )

dx = x4 + 1

0

esin θ dθ.

−π/2

ex d x √ = I1 + I2 . 1 − x2

Z 1 2 2eu −1 u du =2 √ u 2 − u2 0 Z 1 2 2e1−u u du =2 √ u 2 − u2 0

Let x =

2

eu −1 du √ 2 − u2 2 e1−u du √ 2 − u2

Z

1

0

dx + x4 + 1

Z

∞ 1

dx = I1 + I2 . x4 + 1

1 dt and d x = − 2 in I2 , then t t

I2 =

In I1 put 1 + x = u 2 ; in I2 put 1 − x = u 2 : Z

0

x2 +

1

Z

Another possibility: I =

eu

6. Let

3. One possibility: let x = sin θ and get Z

1

∞

1

ex d x Let t 2 = 1 − x √ 1−x 0 2t dt = −d x Z 0 1−t 2 Z 1 e 2 =− 2t dt = 2 e1−t dt. t 1 0

Z

5.

Z

Z

Z

1

0

  Z 1 t2 1 dt − 2 = dt.  4 4 t 1 0 1+t +1 t

Hence, Z

∞ 0

250 Copyright © 2014 Pearson Canada Inc.

dx = x4 + 1 =

Z Z

0

1 1

0

1 x2 + 4 x + 1 1 + x4 2 x +1 d x. x4 + 1



dx

INSTRUCTOR’S SOLUTIONS MANUAL

7.

I =

Z

0

1√

SECTION 6.8 (PAGE 386)

2 ≈ 0.666667. 3

x dx =

Hence,

! 1 1 + ≈ 0.603553 2 2 r r ! 1 1 3 2T2 + + ≈ 0.643283 T4 = 4 4 4 r r r r ! 1 1 3 5 7 T8 = 4T4 + + + + ≈ 0.658130 8 8 8 8 8 r r r r  1 1 3 5 7 T16 = 8T8 + + + + 16 16 16 16 16 r  r r r 9 11 13 15 + + + + ≈ 0.663581. 16 16 16 16 1 T2 = 0+ 2

r

Hence, I ≈ 0.14, accurate to 2 decimal places. These approximations do not converge very quickly, because the 2 fourth derivative of e−1/t has very large values for some values of t near 0. In fact, higher and higher derivatives behave more and more badly near 0, so higher order methods cannot be expected to work well either.

The errors are I − T2 I − T4 I − T8 I − T16

≈ 0.0631 ≈ 0.0234 ≈ 0.0085 ≈ 0.0031.

Observe that, although these errors are decreasing, they are not decreasing like 1/n 2 ; that is, |I − T2n | >>

1 |I − Tn |. 4

8. Let Z

∞

1 t 1 dt dx = − 2 t   Z 0 Z 1 −1/t 2 1 e 2 = e−(1/t) − 2 dt = dt. t t2 1 0 2

e−x d x

Let x =

Observe that 2

lim

t→0+

9. Referring to Example 5, we have ex = 1 + x +

√ This is because the second derivative of f (x) = x is f ′′ (x) = −1/(4x 3/2 ), which is not bounded on [0, 1].

I =

   1 1 0 + 4(4e−4 ) + e−1 3 2 ≈ 0.1101549   1 1 S4 = 0 + 4(16e−16 ) + 2(4e−4 ) 3 4    16 −16/9 e + e−1 +4 9 ≈ 0.1430237    1 1 64 64 S8 = 0 + 4 64e−64 + e−64/9 + e−64/25 + 3 8 9 25     64 −64/49 16 e + 2 16e−16 + 4e−4 + e−16/9 + e−1 49 9 ≈ 0.1393877. S2 =

e−1/t t −2 h ∞ i = lim t→0+ e 1/t 2 t2 ∞ −2t −3 = lim t→0+ e 1/t 2 (−2t −3 ) 1 = lim = 0. t→0+ e 1/t 2

where Rn ( f ; 0, x) = and x. Now

xn x2 + ··· + + Rn ( f ; 0, x), 2! n! e X x n+1 , for some X between 0 (n + 1)!

|Rn ( f ; 0, −x 2 )| ≤

x 2n+2 (n + 1)!

if 0 ≤ x ≤ 1 for any x, since −x 2 ≤ X ≤ 0. Therefore Z 1 Z 1 1 2 ≤ R ( f ; 0, −x ) d x x 2n+2 d x n (n + 1)! 0

0

1 . = (2n + 3)(n + 1)!

This error will be less than 10−4 if (2n + 3)(n + 1)! > 10, 000. Since 15 × 7! > 10, 000, n = 6 will do. Thus we use seven terms of the series (0 ≤ n ≤ 6): Z 1 2 e−x d x 0  Z 1 x4 x6 x8 x 10 x 12 ≈ 1 − x2 + − + − + dx 2! 3! 4! 5! 6! 0 1 1 1 1 1 1 =1− + − + − + 3 5 × 2! 7 × 3! 9 × 4! 11 × 5! 13 × 6! ≈ 0.74684 with error less than 10−4 .

251 Copyright © 2014 Pearson Canada Inc.

SECTION 6.8 (PAGE 386)

10. We are given that previous exercise

∞

Z

Z 01

∞

2

e−x d x =

1

2

e−x d x =

1√ 2 π

2 Dividing p the first two equations gives u = 3/5, so u = 3/5. Then 3 A/5 = 1/3, so A = 5/9, and finally, B = 8/9.

and from the

2

e−x d x = 0.74684. Therefore,

0

Z

ADAMS and ESSEX: CALCULUS 8

Z

∞

2

e−x d x −

0

Z

1

14. For any function f we use the approximation

2

e−x d x

Z

0

1√ π − 0.74684 2 = 0.139 (to 3 decimal places).

=

1

f (x) d x = 2

−1

Z

1

0

(bx 2 + d) d x = 2

A f (−u) + A f (u) = 2 A(bu 2 + d).



b +d 3

−1



−1

√

Z

√

f (x) d x ≈ f (−1/ 3) + f (1/ 3).

1

−1

    2 1 4 1 4 + √ = x4 dx ≈ − √ 9 3 3 −1 Z 1 2 2 2 Error = x 4 d x − = − ≈ 0.17778 9 5 9 −1     Z 1 1 1 cos x d x ≈ cos − √ + cos √ ≈ 1.67582 3 3 −1 Z 1 Error = cos x d x − 1.67582 ≈ 0.00712

≈ 1.68300 1

−1

cos x d x − 1.68300 ≈ 0.00006

Z

1

−1

e x d x − 2.35034 ≈ 0.00006.

1

Z

Z

1

−1 √ −1/ 3

x

−1

e dx ≈ e

Error =

Z

1

−1

√ 3

+ e1/

15.

I =

Z

1

≈ 2.34270

e x d x − 2.34270 ≈ 0.00771.

b d + + f 5 3 AF(−u) + B F(0) + AF(u) = 2 A(bu 4 + du 2 + f ) + B f. F(x) d x = 2

0

(bx 4 + d x 2 + f ) d x = 2

T21



T22 T32

These two expressions are identical provided 1 Au 4 = , 5

1 Au 2 = , 3

B A+ = 1. 2

2

e−x d x

0

4T10 − T00 ≈ 0.7471805 3 4T 0 − T10 = S4 = 2 ≈ 0.7468554 3 0 0 4T − T2 ≈ 0.7468261 = S8 = 3 3 1 1 16T2 − T1 = R2 = ≈ 0.7468337 15 1 1 16T3 − T2 = ≈ 0.7468242 15 64T32 − T22 = R3 = ≈ 0.7468241 63 ≈ 0.746824 to 6 decimal places.

T11 = S2 = R1 =

T31 

1

  1 0 1 −1 e + e ≈ 0.6839397 = T1 = R0 = (1) 2 2   1 1 0 1 T10 = T2 = e + e−1/4 + e−1 ≈ 0.7313703 2 2 2  1 0 −1/16 T2 = T4 = 2T2 + e + e−9/16 ≈ 0.7429841 4  1 0 T3 = T8 = 4T4 + e−1/64 + e−9/64 + e−25/64 + e−49/64 8 ≈ 0.7458656

symmetry, 1

Z

T00

13. If F(x) = ax 5 + bx 4 + cx 3 + d x 2 + ex + f , then, by

−1

Z

√ √ e x d x ≈ e− 3/5 + e 3/5 ≈ 2.35034

Error =

We have

Z

6

Error =

12. For any function f we use the approximation 1

i 8 p p 5h f (− 3/5) + f ( 3/5) + f (0). 9 9

 r ! r !6  6 5 3 3  + + 0 = 0.24000 x dx ≈ − 9 5 5 −1 Z 1 Error = x 6 d x − 0.24000 ≈ 0.04571 −1 " r ! r !# Z 1 5 8 3 3 cos x d x ≈ cos − + cos + 9 5 5 9 −1 1

Z

These two expressions √ are identical provided A = 1 and u 2 = 1/3, so u = 1/ 3.

Z

f (x) d x ≈

We have

11. If f (x) = ax 3 + bx 2 + cx + d, then, by symmetry, Z

1

T33 I

252 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

16. From Exercise 9 in Section 7.6, for I =

Z

SECTION 6.8 (PAGE 386)

1.6

The transformation is not because the derivative    suitable  1 1 1 of sin is − 2 cos , which has very large values t t t at some points close to 0. In order to approximate the integral I to an desired degree of accuracy, say with error less than ǫ in absolute value, we have to divide the integral into two parts: Z ∞ sin x I = dx 1 + x2 Zπt Z ∞ sin x sin x = d x + dx 2 1 + x 1 + x2 π t = I1 + I2 .

f (x) d x, 0

T00 = T1 = 1.9196

T10 = T2 = 2.00188 T20 = T4 = 2.02622

T30 = T8 = 2.02929.

Hence, R1 = T11 = T21 = R2 = T22 = T31 = T32 = R3 = T33 =

17.

4T10 − T00 = 2.0346684 3 4T20 − T10 = 2.0343333 = S4 3 16T21 − T11 = 2.0346684 15 4T30 − T20 = 2.0303133 = S8 3 16T31 − T21 = 2.0300453 15 2 2 64T3 − T2 = 2.0299719. 63

 2h  y0 + 4y2 + y4 3  h S4 = y0 + 4y1 + 2y2 + 4y3 + y4 3 16T21 − T11 2 T2 = 15 2h 16h (y + 4y 0 1 + 2y2 + 4y3 + y4 ) − 3 (y0 + 4y2 + y4 ) 3 15  h  14y0 + 64y1 + 24y2 + 64y3 + 14y4 45  2h  7y0 + 32y1 + 12y2 + 32y3 + 7y4 45

If t ≥ tan

Now let ZA be a numerical approximation to the proper t sin x integral d x, having error less than ǫ/2 in ab2 π 1+x solute value. Then

T11 = S2 = T21 = R2 = = = =

18. Let I =

Z

∞ π

π −ǫ , then 2 Z ∞ Z ∞ sin x dx d x < 2 1 + x 1 + x2 t t ∞ π ǫ = tan−1 (x) = − tan−1 (t) ≤ . 2 2 t

19.

|I − A| = |I1 + I2 − A| ≤ |I1 − A| + |I2 | ǫ ǫ ≤ + = ǫ. 2 2 Hence, A is an approximation to the integral I with the desired accuracy. sin x x cos c − sin x f (x) = , f ′ (x) = , x x2 x 2 (cos x − x sin x − cos x) − (x cos x − sin x)2x f ′′ (x) = x4 2 −x sin x − 2x cos x + 2 sin x = . x3 Now use l’Hˆopital’s Rule to get lim f ′′ (x)

x→0

−2x sin x − x 2 cos x − 2 cos x + 2x sin x + 2 cos x x→0 3x 2 cos x 1 = lim − =− . x→0 3 3 = lim

sin x dx 1 + x2

1 t dt dx = − 2 t Let x =

  1   Z 0 sin 1 t   − 2 dt = 1 t 1/π 1+ 2 t   1 Z 1/π sin t = dt. 1 + t2 0

20. If t is time and E is error, then for the trapezoid rule, t 2 E is approximately constant. Since E = 6 × 10−16 when t = 175.777 seconds, the time we would expect our computer to achieve an error of 10−32 is about s 10−16 175.7772 × 6 × −32 seconds, 10 which is about 1,365 years.

253 Copyright © 2014 Pearson Canada Inc.

SECTION 6.8 (PAGE 386)

ADAMS and ESSEX: CALCULUS 8

21. If t is time and E is error, then t 4 E is approximately constant for the Simpson’s Rule case. Since E = 3.15 × 10−30 when t = 175.777 seconds, we would expect our computer to achieve quadruple precision in time  1/4 10−30 4 175.777 × 3.15 × −32 10

5.

3 A B = + −1 2x − 1 2x + 1 2 Ax + A + 2Bx − B = 4x 2 − 1 n 3 2 A + 2B = 0 ⇒ A = −B = ⇒ A− B =3 2

4x 2

seconds, Z  Z 3 dx 3 dx dx = − 2 2x − 1 2x + 1 4x 2 − 1 3 2x − 1 = ln + C. 4 2x + 1

Z

or about 12 minutes if it were using Simpson’s Rule. Since our computer did the calculation more than 5,000 times faster than that, it must have been using an even higher-order method.

Review Exercises on Techniques of Integration (page 388) 1.

6.

4.

x d x Let u = x − 1 (x − 1)3 du  dx  Z Z = u+1 1 1 du = du = + u3 u2 u3 1 1 1 1 = − − 2 +C =− − + C. u 2u x − 1 2(x − 1)2 Z sin3 x cos3 x d x Z = sin3 x(1 − sin2 x) cos x d x Let u = sin x du = cos x d x Z u4 u6 3 5 = (u − u ) du = − +C 4 6 1 1 = sin4 x − sin6 x + C. 4 6 √ 1/3 Z √ (1 + x) √ d x Let u = 1 + x x dx du = √ 2 x Z = 2 u 1/3 du = 2( 34 )u 4/3 + C √ = 23 (1 + x)4/3 + C.

d V = cos 3x d x V = 13 sin 3x

= − 31 (x 2 + x − 2) cos 3x + 91 (2x + 1) sin 3x Z − 29 sin 3x d x

Z Z 1 2 x dx dx dx = − + 2x 2 + 5x + 2 3 2x + 1 3 x +2 2 1 = ln |x + 2| − ln |2x + 1| + C. 3 6

3.

U = x2 + x − 2 dU = (2x + 1) d x

U = 2x + 1 dU = 2 d x

Z

2.

(x 2 + x − 2) sin 3x d x

d V = sin 3x V = − 31 cos 3x Z = − 31 (x 2 + x − 2) cos 3x + 31 (2x + 1) cos 3x d x

x A B = + 2x + 1 x +2 2x 2 + 5x + 2 Ax + 2 A + 2Bx + B = 2x 2 + 5x + 2 n A + 2B = 1 ⇒ 2A + B = 0 Thus A = −1/3 and B = 2/3. We have

Z

Z

= − 31 (x 2 + x − 2) cos 3x + 91 (2x + 1) sin 3x 2 + cos 3x + C. 27

7.

Z √

1 − x2 dx x4

Let x = sin θ d x = cos θ dθ

cos2 θ dθ 4 Z sin θ = csc2 θ cot2 θ dθ

=

Z

Let v = cot θ dv = − csc2 θ dθ

v3 +C 3 !3 √ cot3 θ 1 1 − x2 =− +C = − + C. 3 3 x =−

Z

v 2 dv = −

254 Copyright © 2014 Pearson Canada Inc.

1 x

θ

√

1−x 2

Fig. RT.7

INSTRUCTOR’S SOLUTIONS MANUAL

8.

Z

x 3 cos(x 2 ) d x

=

1 2

Z

REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

Let w = x 2 dw = 2x d x

13.

9.

10.

11.

cos(x 2 ) + C.

√ 1+4x

x2 dx Let u = 5x 3 − 2 3 (5x − 2)2/3 du = 15x 2 d x Z 1 1 u −2/3 du = u 1/3 + C = 15 5 1 3 1/3 = (5x − 2) + C. 5

Z

1

Fig. RT.13

1 A B (A + B)x + (5 A − 3B) = + = + 2x − 15 x −3 x +5 x 2 + 2x − 15 n 1 1 A+B =0 ⇒ A= , B=− . ⇒ 5 A − 3B = 1 8 Z Z 8 Z 1 dx 1 dx dx = − x 2 + 2x − 15 8 x − 3 8 x +5 1 x − 3 = ln + C. 8 x + 5 Z

dx Let x = 2 tan θ (4 + x 2 )2 d x = 2 sec2 θ dθ Z Z 1 2 sec2 θ dθ = = cos2 θ dθ 16 sec4 θ 8 1 = (θ + sin θ cos θ ) + C 16   1 1 x −1 x = tan + + C. 16 2 8 4 + x2

14.

Z

15.

Z

sin3 x dx = 7 Zcos x Z

(u 3 + u 5 ) du =

u4 u6 + +C 4 6

Let u = tan x du = sec2 x d x

1 1 tan4 x + tan6 x + C. 4 6

Z

2

Fig. RT.11

(1 + sin 2x) d x 1 2

tan3 x sec4 x d x

tan3 x(1 + tan2 x) sec2 x d x

=

=x−

Z

16. We have

θ

Z

=

=

4+x 2

(sin x + cos x)2 d x =

cos x d x Let u = sin x 1 + sin2 x du = cos x d x Z du = = tan−1 u + C 1 + u2 = tan−1 (sin x) + C.

=

x

Z

2x

θ

x2

√

12.

Let 2x = tan θ 2x ln 2 d x = sec2 θ dθ

=

U =w d V = cos w dw dU = dw Z V = sin w = 12 w sin w − 12 sin w dw 1 2

√ 2x 1 + 4x d x

Z 1 sec3 θ dθ ln 2  1  = sec θ tan θ + ln | sec θ + tan θ | + C 2 ln 2  √ 1  x√ 2 1 + 4x + ln(2x + 1 + 4x ) + C. = 2 ln 2

w cos w dw

= 21 x 2 sin(x 2 ) +

Z

cos 2x + C.

x2 dx (3 + 5x 2 )3/2

q Let x = 53 tan u q d x = 35 sec2 u du q Z ( 3 tan2 u)( 3 sec2 u) du 5 5 (3)3/2 sec3 u

Z

1 = √ (sec u − cos u) du 5 5 1 = √ (ln | sec u + tan u| − sin u) + C 5 5 √ √  √ 2  1 5x 5x 5x + 3 = √ ln √ + √ − √ +C 3 3 5 5 5x 2 + 3 p √  1 x = √ ln 5x 2 + 3 + 5x − √ + C0 , 5 5 5 5x 2 + 3 √ 1 where C0 = C − √ ln 3. 5 5

255 Copyright © 2014 Pearson Canada Inc.

REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

√

5x 2 +3

u

20.

√ 5x

√ 3

Fig. RT.16

17.

I =

Z

e−x sin 2x d x U = e−x dU = −e−x d x

1 1 = − e−x cos 2x − 2 2

d V = sin 2x d x 1 V = − cos 2x 2 Z

21.

e−x cos 2x d x

18.

19.

Z 2x 2 + 10x − 6x − 3 2x 2 + 4x − 3 d x = dx 2 x + 5x x 2 + 5x  Z  6x + 3 2− = dx x(x + 5) 6x + 3 A B (A + B)x + 5 A = + = x(x + 5) x x +5 x(x + 5) n 3 27 A+ B =6 ⇒ ⇒A= , B= . 5ZA = 3 5 Z Z 5 3 27 dx dx I = 2 dx − − 5 x 5 x +5 3 27 = 2x − ln |x| − ln |x + 5| + C. 5 5 Z I = cos(3 ln x) d x I =

= x cos(3 ln x) + 3

dV = dx V =x

Z

x ln(1 + x 2 ) dx 1 + x2

Let u = ln(1 + x 2 ) 2x d x du = 1 + x2

=

22.

Z

23.

Z

Z

U = cos(3 ln x) 3 sin(3 ln x) d x dU = − Zx

1 A Bx + C = + 2 +x x 4x + 1 A(4x 2 + 1) + Bx 2 + C x = 4x 3 + x  4A + B = 0 ⇒ B = −4. ⇒ C = 0, A = 1 Z Z Z 1 dx x dx d x = − 4 3 4x + x x 4x 2 + 1 1 = ln |x| − ln(4x 2 + 1) + C. 2

4x 3

Z u2 1 u du = +C 2 4 2 1 = ln(1 + x 2 ) + C. 4

U = e−x dU = −e−x d x

d V = cos 2x d x 1 V = sin 2x 2   1 −x 1 1 −x 1 = − e cos 2x − e sin 2x + I 2 2 2 2 1 −x 1 1 −x = − e cos 2x − e sin 2x − I 2  4  4 1 −x 2 I = −e cos 2x + sin 2x + C. 5 5

ADAMS and ESSEX: CALCULUS 8

sin2 x cos4 x d x Z 1 1 2 = 2 (1 − cos 2x)[ 2 (1 + cos 2x)] d x Z 1 = (1 + cos 2x − cos2 2x − cos3 2x) d x 8 Z 1 1 1 = x+ sin 2x − (1 + cos 4x) d x 8 Z 16 16 1 − (1 − sin2 2x) cos 2x d x 8 1 x 1 1 x sin 2x − − sin 4x − sin 2x = + 8 16 16 64 16 1 + sin3 2x + C 48 x sin 4x sin3 2x = − + + C. 16 64 48 √ x2 dx √ Let x = 2 sin θ √ 2 − x2 d x = 2 cos θ dθ Z 2 = 2 sin θ dθ = θ − sin θ cos θ + C √ x x 2 − x2 = sin−1 √ − + C. 2 2

sin(3 ln x) d x √

U = sin(3 ln x) dV = dx 3 cos(3 ln x) d x V =x dU = x  = x cos(3 ln x) + 3 x sin(3 ln x) − 3I

2 x

θ

1 3 I = x cos(3 ln x) + x sin(3 ln x) + C. 10 10

256 Copyright © 2014 Pearson Canada Inc.

√

2−x 2

Fig. RT.23

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

24. We have I =

Z

= tan4 x sec x d x tan3

U= x dU = 3 tan2 x Zsec2 x d x

= tan3 x sec x − 3 = tan3 x sec x − 3

d V = tan x sec x d x V = sec x

27.

tan2 x sec3 x d x

Z

tan2 x(tan2 x + 1) sec x d x

tan2 x sec x d x

U = tan x d V = tan x sec x d x dU = sec2 xZd x V = sec x 3 = tan x sec x − sec x d x Z = tan x sec x − (tan2 x + 1) sec x d x

28.

= tan x sec x − J − ln | sec x + tan x| + C J = 21 tan x sec x − 12 ln | sec x + tan x| + C. I =

25.

1 4

+

tan3 x sec x − 3 8

3 8

tan x sec x

ln | sec x + tan x| + C.

x2 dx Let u = 4x + 1 (4x + 1)10 du = 4 d x  Z  u−1 2 1 1 du = 4 4 u 10 Z 1 = (u −8 − 2u −9 + u −10 ) du 64 1 −8 1 −9 1 −7 u + u − u +C =− 448 256 576   1 1 1 1 = − + C. + − 64 7(4x + 1)7 4(4x + 1)8 9(4x + 1)9

Z

26. We have Z

x sin−1

x  2

U = sin−1 dU = √

=

dx x 

x2 2

x2 = 2 x2 = 2 x2 = 2

dx

2

dV = x dx x2 V = 2

4 − x2  x  1 Z x2 dx sin−1 − √ Let x = 2 sin u 2 2 4 − x2 d x = 2 cos u du Z   −1 x 2 sin − 2 sin u du 2 x  Z sin−1 − (1 − cos 2u) du 2 x  sin−1 − u + sin u cos u + C 2

 x  1 p x2 − 1 sin−1 + x 4 − x 2 + C. 2 2 4

Z

sin5 (4x) d x Z = (1 − cos2 4x)2 sin 4x d x

Let u = cos 4x du = −4 sin 4x d x

Z 1 (1 − 2u 2 + u 4 ) du 4   1 2 3 1 5 =− u− u + u +C 4 3 5 1 1 1 cos5 4x + C. = − cos 4x + cos3 4x − 4 6 20 We have Z Z dx x dx I = = Let u = x 2 x 6 − 2x 4 + x 2 x 5 − 2x 3 + x du = 2x d x Z Z du 1 du 1 = = 2 u 3 − 2u 2 + u 2 u(u − 1)2 A B C 1 = + + u(u − 1)2 u u − 1 (u − 1)2 A(u 2 − 2u + 1) + B(u 2 − u) + Cu = u 3 − 2u 2 + u ( A+B =0 ⇒ −2 A − B + C = 0 ⇒ A = 1, B = −1, C = 1. Z A=1 Z Z 1 1 1 du du du = − 2 u 3 − 2u 2 + u 2 u 2 u −1 Z 1 du + 2 (u − 1)2 1 1 1 1 +K = ln |u| − ln |u − 1| − 2 2 2u−1 2 x 1 1 − + K. = ln 2 2 |x − 1| 2(x 2 − 1) =−

3 = tan Z x sec x − 3I − 3J where

J=



29.

Z

dx 2 + ex Z e−x d x = Let u = 2e−x + 1 2e−x + 1 du = −2e−x d x Z 1 1 du =− = − ln(2e−x + 1) + C. 2 u 2

30. Let

In =

Z

x n 3x d x U = xn dU = nx n−1 d x

d V = 3x d x 3x V = ln 3

x n 3x n − In−1 . ln 3 ln 3 Z 3x I0 = 3x d x = + C. ln 3 =

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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

34. We have

Hence, I3 =

Z

Z

x 3 3x d x

sin2 x cos x d x Let u = sin x 2 − sin x du = cos x d x Z 2 u du Let 2 − u = v = 2−u du = −dv  Z Z  4 − 4v + v 2 4 =− dv = − + 4 − v dv v v 2 v = −4 ln |v| + 4v − +C 2 1 = −4 ln |2 − u| + 4(2 − u) − (2 − u)2 + C 2 1 = −4 ln(2 − sin x) − 2 sin x − sin2 x + C1 . 2

Z

35.

32. We have Z = = = =

33.

 Z  2x + 1 x2 + 1 1 − d x = dx x 2 + 2x + 2 x 2 + 2x + 2 Z 2x + 1 d x Let u = x + 1 x− (x + 1)2 + 1 du = d x Z 2u − 1 x− du u2 + 1 x − ln |u 2 + 1| + tan−1 u + C x − ln(x 2 + 2x + 2) + tan−1 (x + 1) + C.

Z

dx √ Let x = sin θ x2 1 − x2 dx = Z Z cos θ dθ cos θ dθ = = csc2 θ dθ sin2 θ cos θ √ 1 − x2 + C. = − cot θ + C = − x

√

1−x 2

Fig. RT.33

√

2x

1−4x 2

Fig. RT.35 Z

e1/x dx x2

=−

x

x3 dx

Let 2x = sin θ 1 − 4x 2 2 d x = cos θ dθ Z Z sin3 θ cos θ dθ 1 1 = (1 − cos2 θ ) sin θ dθ = 16 cos θ 16   1 1 = − cos θ + cos3 θ + C 16 3 1p 1 (1 − 4x 2 )3/2 − = 1 − 4x 2 + C. 48 16 √

θ

37.

θ

Z

1

36.

1

x 3 (ln x)2 d x

U = (ln x)2 dV = x3 dx 1 2 V = x4 dU = ln x d x 4 x Z 1 1 4 2 3 x ln x d x = x (ln x) − 4 2 U = ln x dV = x3 dx 1 1 dU = d x V = x4 x 4 Z 1 4 1 4 1 2 = x (ln x) − x ln x + x3 dx 4 8 8   1 1 x4 (ln x)2 − ln x + + C. = 4 2 8

   x 3 3x 3 x 2 3x 2 x3x 1 − − − I0 + C 1 ln 3 ln 3 ln 3 ln 3 ln 3 ln 3  3  2 x 3x 6x 6 = 3x − + − + C1 . ln 3 (ln 3)2 (ln 3)3 (ln 3)4 =

31.

ADAMS and ESSEX: CALCULUS 8

Z

1 x 1 du = − 2 d x x

Let u =

eu du = −eu + C = −e1/x + C.

Z

x +1 √ dx x2 + 1 Z p dx = x2 + 1 + √ Let x = tan θ x2 + 1 d x = sec2 θ dθ Z p = x 2 + 1 + sec θ dθ p = x 2 + 1 + ln | sec θ + tan θ | + C p p = x 2 + 1 + ln(x + x 2 + 1) + C.

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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

42. Assume that x ≥ 1 and let x = sec u and √

d x = sec u tan u du. Then

1+x 2 x

x2 dx √ x2 − 1 Z Z sec3 u tan u du = sec3 u du = tan u 1 1 = sec u tan u + ln | sec u + tan u| + C 2 2 p 1 p 2 1 = x x − 1 + ln |x + x 2 − 1| + C. 2 2

Z

θ 1

Fig. RT.37

38.

Z

e(x

=3

1/3 )

Z

Let x = u 3 d x = 3u 2 du

u 2 eu du = 3I2

Differentiation shows that this solution is valid for x ≤ −1 also.

See solution to #16 of Section 6.6 for Z n u In = u e d x = u n eu − n In−1 .

= 3[u 2 eu − 2(ueu − eu )] + C 1/3

39.

= e(x ) (3x 2/3 − 6x 1/3 + 6) + C.  Z  Z 9x − 3 x3 − 3 d x = 1 + d x. I = x 3 − 9x x 3 − 9x

43.

(x + 1 − 1) d x Let u = x + 1 (x + 1)2 − 2 du = d x Z Z u−1 1 du 2 = du = ln |u − 2| − . u2 − 2 2 u2 − 2

I =

A 9x − 3 B C = + + 3 x x −3 x +3 x − 9x Ax 2 − 9 A + Bx 2 + 3Bx + C x 2 − 3C x = x 3 − 9x ( ( A = 1/3 A+ B+C = 0 ⇒ 3B − 3C = 9 ⇒ B = 4/3 C = −5/3. −9 A = −3

40.

41.

Z

10 x+2 d x √ x +2

=−

u − √2 1 1 2 I = ln |u − 2| − √ ln √ +K 2 2 2 u + 2 x + 1 − √2 1 1 2 = ln |x + 2x − 1| − √ ln √ + K. 2 2 2 x + 1 + 2

(1 − 2u 2 + u 4 )u 9 du

u 10

+

u 12

−

u 14

+C

10 6 14 cos12 x cos10 x cos14 x − − + C. = 6 10 14

Z

Thus we have

√

x +2 dx du = √ 2 x +2 Z √ 2 2 10u + C = 10 x+2 + C. = 2 10u du = ln 10 ln 10 Z sin5 x cos9 x d x Z = (1 − cos2 x)2 cos9 x sin x d x Let u = cos x du = − sin x d x Z =−

Let u =

x dx = x 2 + 2x − 1

A 1 B = √ + √ u2 − 2 u− 2 u+ 2 √ √ Au + 2 A + Bu − 2B = u2 − 2  A+ B =0 ⇒ √ 2(A − B) = 1 1 ⇒ A = −B = √ . 2 2

Thus we have Z Z Z 4 5 dx dx dx 1 + − I =x+ 3 x 3 x −3 3 x +3 1 4 5 = x + ln |x| + ln |x − 3| − ln |x + 3| + K . 3 3 3 √

Z

44.

Z

2x − 3

Let u = 4 − 3x + x 2 du = (−3 + 2x) d x Z p √ du = √ = 2 u + C = 2 4 − 3x + x 2 + C. u √

4 − 3x + x 2

dx

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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

45.

Z

= = = = =

46. Let

x 2 sin−1 2x d x U = sin−1 2x dV = x2 dx 2 dx x3 dU = √ V = 2 1 − 4x 3 Z 2 x3 dx x3 −1 sin 2x − √ Let v = 1 − 4x 2 3 3 1 − 4x 2 dv = −8x d x   Z 2 1 − v x3 1 sin−1 2x − dv − 3 3 4v 1/2 8 Z   3 x 1 −1 −1/2 sin 2x + v − v 1/2 dv 3 48 1√ 1 x3 sin−1 2x + v − v 3/2 + C 3 24 72 x3 1p 1 sin−1 2x + 1 − 4x 2 − (1 − 4x 2 )3/2 + C. 3 24 72 √

3x = sec u and

√

3 d x = sec u tan u du. Then

Z √ 2 3x − 1 dx x 1 Z tan u √ sec u tan u du 3 = 1 √ sec u 3 Z Z = tan2 u du = (sec2 u − 1) du p √ = tan u − u + C = 3x 2 − 1 − sec−1 ( 3x) + C   p 1 = 3x 2 − 1 + sin−1 √ + C1 . 3x

47.

48.

Z

Z 1 cos4 x sin4 x d x = sin4 2x d x 16 Z 1 = (1 − cos 4x)2 d x 64  Z  1 1 + cos 8x = 1 − 2 cos 4x + dx 64 2   1 3x sin 4x sin 8x = − + +C 64 2 2 16   1 sin 8x = 3x − sin 4x + + C. 128 8

Z p =

x− Z q 1 4

Z

1 4

cos2 u du = 18 u +

u

√

Let x − dx = 1 8

1 2

1 2

=

1 2

1 x− 2

x−x 2

Fig. RT.48

49.

50.

51.

Z

dx √ Let x = u 2 (4 + x) x d x =Z 2u du Z du 2u du =2 = (4 + u 2 )u 4 + u2 √ 2 x u = tan−1 + C = tan−1 + C. 2 2 2 Z x  x tan−1 dx 3 x  U = tan−1 dV = x dx 3 x2 3 dx V = dU = 2 9 + x2 Z   2 2 x x x 3 = tan−1 dx − 2 3 2 9 + x2  x  3 Z  x2 9 = tan−1 − 1− dx 2 3 2 9 + x2  x  3x x  9 x2 tan−1 − + tan−1 + C. = 2 3 2 2 3 Z x4 − 1 I = dx x 3 + 2x 2 Z 4 x + 2x 3 − 2x 3 − 4x 2 + 4x 2 − 1 = dx x 3 + 2x 2   Z 4x 2 − 1 x −2+ 3 d x. = x + 2x 2 4x 2 − 1 B A C = + 2 + x 3 + 2x 2 x x x +2 Ax 2 + 2 Ax + Bx + 2B + C x 2 = x 3 + 2x 2 ( ( A = 1/4 A+C =4 ⇒ 2 A + B = 0 ⇒ B = −1/2 C = 15/4. 2B = −1 Z Z 1 1 x2 dx − 2x + − 2 4 x 2 x2 1 1 = − 2x + ln |x| + + 2 4 2x

I =

x2 dx − (x − 12 )2 d x

1 2

Thus

sin u

Z dx 15 dx + 2 x 4 x +2 15 ln |x + 2| + K . 4

cos u du

sin u cos u + C p = 81 sin−1 (2x − 1) + 41 (2x − 1) x − x 2 + C.

=

ADAMS and ESSEX: CALCULUS 8

52. Let u = x 2 and du = 2x d x; then we have I =

Z

dx = x(x 2 + 4)2

260 Copyright © 2014 Pearson Canada Inc.

Z

x dx 1 = x 2 (x 2 + 4)2 2

Z

du . u(u + 4)2

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

56. We have

Since 1 A B C = + + u(u + 4)2 u u + 4 (u + 4)2 A(u 2 + 8u + 16) + B(u 2 + 4u) + Cu = u(u + 4)2 ( A+B =0 1 1 1 ⇒ 8 A + 4B + C = 0 ⇒ A = , B=− , C =− , 16 16 4 16 A = 1

I =

Z Z Z 1 du 1 du 1 du − − 32 u 32 u+4 8 (u + 4)2 1 u 1 1 ln + +C = 32 u + 4 8 u + 4 1 x 2 1 = ln 2 + + C. 2 32 x +4 8(x + 4)

54.

therefore I =− Z

I =

x2 2 x2 = 2

−1

e2 tan x dx 1 + x2

i 1 h sin(ln x) + cos(ln x) + C. 2x

Let u = 2 tan−1 x 2 dx du = 1 + x2 Z 1 u 1 1 −1 u = e du = e + C = e2 tan x + C. 2 2 2

58.

 8x − 2 dx x+ 2 x −7   Z Z 1 1 dx dx + 4+ √ √ + 4− √ √ 7 x+ 7 7 x− 7     √ √ 1 1 + 4 + √ ln |x + 7| + 4 − √ ln |x − 7| + C. 7 7

Z 

=

57.

dx U = sin(ln x) dV = 2 x cos(ln x) 1 dU = dx V =− x x Z sin(ln x) cos(ln x) =− + dx x x2 dx U = cos(ln x) dV = 2 x sin(ln x) −1 dU = − dx V = x x sin(ln x) cos(ln x) =− − − I, x x

55.

therefore

sin(2 ln x) dx x

Let u = 2 ln x 2 du = d x x Z 1 1 = sin u du = − cos u + C 2 2 1 = − cos(2 ln x) + C. 2 Since Z sin(ln x) I = dx x2

x 3 − 7x + 8x − 2 dx x2 − 7  Z  8x − 2 d x. = x+ 2 x −7 Z

√ 8x − 2 A B (A + B)x + (B − A) 7 = √ + √ = x2 − 7 x2 − 7 x+ 7 x− 7 ( A+B =8 1 1 ⇒ B − A = − √2 ⇒ A = 4 + √ , B = 4 − √ , 7 7 7

I =

53.

x3 + x − 2 dx = x2 − 7

Since

therefore

Z

Z

Z

ln(3 + x 2 ) x dx 3 + x2

Let u = ln(3 + x 2 ) 2x d x du = 3 + x2 Z 2 1 1 u2 = +C = ln(3 + x 2 ) + C. u du = 2 4 4 Z Z cos7 x d x = (1 − sin2 x)3 cos x d x Let u = sin x du = cos x d x Z Z =

(1 − u 2 )3 du =

(1 − 3u 2 + 3u 4 − u 6 ) du

= u − u 3 + 35 u 5 − 71 u 7 + C

59.

= sin x − sin3 x + 53 sin5 x − 17 sin7 x + C. Z sin−1 (x/2) d x Let u = sin−1 (x/2) (4 − x 2 )1/2 dx dx du = p = √ 4 − x2 2 1 − (x 2 /4) Z 2 u2 1  −1 +C = sin (x/2) + C. = u du = 2 2

60. We have Z

Z

4

tan (π x) d x = tan2 (π x)[sec2 (π x) − 1] d x Z Z = tan2 (π x) sec2 (π x) d x − [sec2 (π x) − 1] d x =

1 1 tan3 (π x) − tan(π x) + x + C. 3π π

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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

61.

(x + 1) d x √ 2 Z x + 6x + 10 (x + 3 − 2) d x p = Let u = x + 3 (x + 3)2 + 1 du = d x Z (u − 2) du √ = u2 + 1 Z p du Let u = tan θ = u2 + 1 − 2 √ u2 + 1 du = sec2 θ dθ Z p = x 2 + 6x + 10 − 2 sec θ dθ p = x 2 + 6x + 10 − 2 ln | sec θ + tan θ | + C   p p = x 2 + 6x + 10 − 2 ln x + 3 + x 2 + 6x + 10 + C.

Z

√

x 2 +6x+10

63.

ADAMS and ESSEX: CALCULUS 8

√ x3 dx Let x = 2 tan θ 7/2 √ + 2) d x = 2 sec2 θ dθ √ Z √ 2 2 tan3 θ 2 sec2 θ dθ = √ 7 Z 8 2 sec θ 1 3 2 = √ sin θ cos θ dθ 2 2Z 1 = √ (1 − cos2 θ ) cos2 θ sin θ dθ Let u = cos θ 2 2 du ! = − sin θ dθ Z 3 5 u 1 1 u − +C = √ (u 4 − u 2 ) du = √ 3 2 2 2 2 5  !5 !3  √ √ 1 1 2 1 2 +C = √ √ − √ 3 2 2 5 2 + x2 2 + x2

Z

(x 2

=

2 1 + C. − 3(2 + x 2 )3/2 5(2 + x 2 )5/2

x+3

√

θ

2+x 2

1

x

Fig. RT.61

θ

√ 2

Fig. RT.63

62.

Z

Z

65.

Z

e x (1 − e2x )5/2 d x

Let e x = sin u e x d x = cos u du  3 Z Z 1 (1 + cos 2u)3 du = cos6 u du = 2 Z 1 = (1 + 3 cos 2u + 3 cos2 2u + cos3 2u) du 8 Z 3 3 u sin 2u + = + (1 + cos 4u) du+ 8 Z 16 16 1 (1 − sin2 2u) cos 2u du 8 5u 3 3 sin 2u + sin 2u + sin 4u + = 16 16 64 16 1 3 sin 2u + C − 48 5 1 = sin−1 (e x ) + sin[2 sin−1 (e x )]+ 16 4 3 1 −1 x sin[4 sin (e )] − sin3 [2 sin−1 (e x )] + C 64 48 5 1 p = sin−1 (e x ) + e x 1 − e2x 16 2   3 xp + e 1 − e2x 1 − 2e2x 16 3/2 1 3x  − e 1 − e2x + C. 6

 Z  1 3 x2 d x = 1 + dx 2x 2 − 3 2 2x 2 − 3 √ Z   x 3 1 1 = + √ √ −√ √ dx 2 4 2x + 3 √2x − √3 √ x 3 2x − 3 = + √ ln √ √ + C. 2 4 2 2x + 3

64.

x 1/2 d x Let x = u 6 1 + x 1/3 d x = 6u 5 du Z u8 =6 du u2 + 1 Z 8 u + u6 − u6 − u4 + u4 + u2 − u2 − 1 + 1 =6 du u2 + 1  Z  1 u6 − u4 + u2 − 1 + 2 =6 du u +1 ! u7 u5 u3 =6 − + − u + tan−1 u + C 7 5 3 √ 6 6 = x 7/6 − x 5/6 + 2 x − 6x 1/6 + 6 tan−1 x 1/6 + C. 7 5

66. We have Z

dx x(x 2 + x + 1)1/2

262 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

=

Z

dx x[(x +

1 2 2)

√

+

3 1/2 4]

REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

√ 1 3 Let x + = tan θ 2 2 √ 3 dx = sec2 θ dθ 2

3 sec2 θ dθ 2 = √   √  3 1 3 tan θ − sec θ 2 2 2 Z Z 2 sec θ dθ dθ = √ =2 √ 3 tan θ − 1 3 sin θ − cos θ Z √ 3 sin θ + cos θ dθ =2 3 sin2 θ − cos2 θ Z √ Z sin θ dθ cos θ dθ =2 3 + 2 2 θ − cos2 θ 2 θ − cos2 θ 3 sin 3 sin Z √ Z sin θ dθ cos θ dθ =2 3 + 2 3 − 4 cos2 θ 4 sin2 θ − 1 Let u = cos θ , du = − sin θ dθ in the first integral; let v = sin θ , dv = cos θ dθ in the second integral. Z √ Z dv du = −2 3 + 2 3 − 4u 2 4v 2 − 1 √ Z Z 3 du 1 du =− − 3 1 2 2 2 − v2 4 −u √ 4 √    cos θ + 3 3 1 2 2 =− √ ln √ 2 2 3 3 cos θ − 2 1   sin θ + 1 1 2 + C − (2) ln 1 2 2 sin θ − 2  √   cos θ − 3 sin θ − 1 1 2 2 = ln  √   + C. 2 cos θ + 3 sin θ + 1 2 2

67.

Z

√ 2x + 1 3 Since sin θ = √ and cos θ = √ , 2 2 2 x +x +1 2 x +x +1 therefore

Z

dx x(x 2 + x + 1)1/2

√ 1 (x + 2) − 2 x 2 + x + 1 = ln √ +C. 2 (x + 2) + 2 x 2 + x + 1

68.

69.

Z

1+x √ d x Let x = u 2 1+ x d x = 2u du Z u(1 + u 2 ) =2 du 1+u Z 3 2 u + u − u 2 − u + 2u + 2 − 2 du =2 1+u  Z  2 =2 u2 − u + 2 − du 1+u  3  u u2 =2 − + 2u − 2 ln |1 + u| + C 3 2 √ √ 2 3/2 = x − x + 4 x − 4 ln(1 + x) + C. 3 Z x dx Let u = x 2 4x 4 + 4x 2 + 5 du = 2x d x Z 1 du = 2 Z 4u 2 + 4u + 5 1 du = Let w = 2u + 1 2 (2u + 1)2 + 4 dw = 2du Z   1 dw 1 −1 w = = tan +C 4 8 2 w2 + 4   1 1 + C. = tan−1 x 2 + 8 2 Z x dx Let u = x 2 − 4 (x 2 − 4)2 du = 2x d x Z 1 du 1 = +C =− 2 2 u 2u 1 1 =− +C =− 2 + C. 2 2(x − 4) 2x − 8

70. Use the partial fraction decomposition

1 A Bx + C = + 2 x3 + x2 + x x x +x +1 A(x 2 + x + 1) + Bx 2 + C x = x3 + x2 + x ( A+ B =0 ⇒ A + C = 0 ⇒ A = 1, B = −1, C = −1. A=1 Therefore, Z dx 3 + x2 + x x Z Z dx x +1 = − dx x x2 + x + 1 Z u + 21 = ln |x| − du u 2 + 43  1  = ln |x| − ln x 2 + x + 1 − 2

Let u = x + du = d x

1 √ tan−1 3



1 2

2x + 1 √ 3



+ C.

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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 388)

71.

72.

Z

x 2 tan−1 x d x

U = tan−1 x d V = x 2 d x dx x3 dU = V = 1 + x2 3 Z x3 1 x3 dx −1 = tan x − 3 3 1 + x2 Z 3 3 1 x +x−x x tan−1 x − dx = 3 3 x2 + 1 x3 1 1 = tan−1 x − x 2 + ln(1 + x 2 ) + C. 3 6 6 Z e x sec(e x ) d x Let u = e x du = e x d x Z

A B 1 = + (3z − 1)(z + 3) 3z − 1 z+3 Az + 3 A + 3Bz − B = (3z − 1)(z + 3)  n A = 3/10 A + 3B = 0 ⇒ ⇒ B = −1/10. 3A − B = 1

Z

dx −1

Let x = (u + 1)3 d x = 3(u + 1)2 du  Z Z  (u + 1)2 1 =3 du = 3 u +2+ du u u  2  u =3 + 2u + ln |u| + C 2 3 = (x 1/3 − 1)2 + 6(x 1/3 − 1) + 3 ln |x 1/3 − 1| + C. 2 x 1/3

76.

77.

Z

1 dz dz 3 − 5 3z − 1 5 z+3 1 1 = ln |3z − 1| − ln |z + 3| + C 5 5 1 3 tan−1 (x/2) − 1 + C. = ln 5 tan−1 (x/2) + 3

I =

74.

Z

dx Ztan x + sin x cos x d x = sin x(1 + cos x)

Let z = tan(x/2), 1 − z2 , 1 + z2

2 dz 1 + z2 2z sin x = 1 + z2 dx =

1 − z 2 2 dz 1 + z2 1 + z2 =   2z 1 − z2 1 + 1 + z2 1 + z2 Z Z 2 1 1 − z2 (1 − z ) dz = dz = z(1 + z 2 + 1 − z 2 ) 2 z 1 z2 = ln |z| − +C 2 4 x 1  x 2 1 + C. tan = ln tan − 2 2 4 2 Remark: Since x sin2 2 x 2 = 1 − cos x , tan = x 2 1 + cos x cos2 2 the answer can also be written as 1 1 − cos x 1 1 − cos x ln − · + C. 4 1 + cos x 4 1 + cos x Z

= ln | sec(e x ) + tan(e x )| + C. Z x 2 dz dx Let z = tan , d x = I = 4 sin x − 3 cos x 2 1 + z2 1 − z2 2z cos x = , sin x = 1 + z2 1 + z2 2 dz Z 1 + z2 = 8z 3 − 3z 2 − 2 1 + z2 Z Z1+z dz dz =2 =2 . 2 3z + 8z − 3 (3z − 1)(z + 3)

Thus

Z

cos x =

sec u du = ln | sec u + tan u| + C

=

73.

75.

ADAMS and ESSEX: CALCULUS 8

78.

Z

x dx

Z

x dx p Let u = 2x + 1 3 − 4x − 4x 2 4 − (2x + 1)2 du = 2 d x Z 1 u −1 = du √ 4 4 − u2 u  p 1 1 4 − u 2 − sin−1 +C =− 4 4 2   p 1 1 −1 1 2 =− 3 − 4x − 4x − sin x+ + C. 4 4 2 √ Z x d x Let x = u 2 1+x d x = 2u du  Z Z  u 2 du 1 =2 = 2 1 − du 1 + u2 1 + u2   √ √ = 2 u − tan−1 u + C = 2 x − 2 tan−1 x + C. Z √ 1 + e x d x Let u 2 = 1 + e x 2u du = e x d x  Z Z  2 2 2u du 2+ 2 = = du u2 − 1 u −1  Z  1 1 2+ = − du u−1 u+1 u − 1 +C = 2u + ln u + 1 √ 1 + ex − 1 √ = 2 1 + e x + ln √ + C. 1 + ex + 1 √

264 Copyright © 2014 Pearson Canada Inc.

=

INSTRUCTOR’S SOLUTIONS MANUAL

79.

I =

Z

x4 dx = x3 − 8

Z 

x+

8x x3 − 8



OTHER REVIEW EXERCISES 6 (PAGE 389)

If a + c = 1, b + a + d = 0, c − a = 0, and d + c − b = 0, then a = c = −d = 1/2 and b = 0. Thus

d x.

8x A Bx + C = + 2 3 x −8 x −2 x + 2x + 4 Ax 2 + 2 Ax + 4 A + Bx 2 − 2Bx + C x − 2C = x3 − 8 ( ( A+B =0 B = −A ⇒ 2 A − 2B + C = 8 ⇒ C = 2 A 4 A − 2C = 0 6A = 8 Thus A = 4/3, B = −4/3, C = 8/3. We have Z Z x2 4 dx 4 x −2 I = + − dx 2 3 x −2 3 x 2 + 2x + 4 Z x2 4 4 x +1−3 = + ln |x − 2| − dx 2 3 3 (x + 1)2 + 3 x2 4 2 = + ln |x − 2| − ln(x 2 + 2x + 4) 2 3 3 4 −1 x + 1 + √ tan √ + K. 3 3

Z

x

e cos x d x =

1 x 2 e (sin x

J =

2.

xe x sin x d x =

x r e−x d x Z R = lim x r e−x d x 0

c

c→0+

3.

0

because lim R→∞ R r e−R = 0 for any r . In order to ensure that limc→0+ cr e−c = 0 we must have limc→0+ cr = 0, so we need r > 0. π/2 Z π/2 csc x d x = lim − ln | csc x + cot x| c→0+

c→0+

d V = e x cos x d x V = 21 e x (sin x + cos x) Z = 21 xe x (sin + cos x) − 21 e x (sin x + cos x) d x

=

c

= lim ln | csc c + cot c| = ∞ (diverges)

xe x cos x d x U=x dU = d x

= 21 xe x (sin + cos x)

i ex h x sin x − (x − 1) cos x + C. 2

∞

0

Z

i ex h x cos x + (x − 1) sin x + C. 2

U = xr d V = e−x d x r−1 dU = r x dr V = −e−x R Z ∞ x r−1 e−x d x = lim −x r e−x + r c→0+ 0 c R→∞ Z ∞ = lim cr e−c + r x r−1 e−x d x

Now

− 14 e x (sin x − cos x + sin x + cos x) x 1 1 x 2 xe (sin x + cos x) − 2 e sin x + C.

4.

i d xh e (ax + b) cos x + (cx + d) sin x dx h = e x (ax + b) cos x + (cx + d) sin x + a cos x + c sin x i − (ax + b) sin x + (cx + d) cos x h  = e x (a + c)x + b + a + d cos x   i + (c − a)x + d + c − b sin x

Z

∞

dx = lim R→∞ x + x3

1

5.

Z

1√

x ln x d x

0

=

Z

=4

Z

R 1



x 1 − x 1 + x2



dx

 R 1 ln |x| − ln(1 + x 2 ) R→∞ 2   1 1 R2 ln 2 = lim ln + ln 2 = R→∞ 2 1 + R2 2

= lim

+C

Other Review Exercises 6 (page 389) 1.

Z

Z

c→0+ R→∞

+ cos x) + C,

e x sin x d x = 21 e x (sin x − cos x) + C.

xe x cos x d x =

If a + c = 0, b + a + d = 0, c − a = 1, and d + c − b = 0, then b = c = −a = 1/2 and d = 0. Thus

80. By the procedure used in Example 4 of Section 7.1, Z

Z

I =

1



Let x = u 2 d x = 2u du

u(2 ln u)2u du

0

Z

1

u 2 ln u du

0

d V = u 2 du u3 V = 3 ! 1 Z 1 u3 1 2 ln u − u du 3 3 c c

U = ln u du dU = u = 4 lim

c→0+

=−

4 4 4 lim c3 ln c − (1 − c3 ) = − 3 c→0+ 9 9

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OTHER REVIEW EXERCISES 6 (PAGE 389)

6.

7.

8.

1

Z 1 dx dx = ∞ (diverges) √ > 0 x 1 − x2 0 x Z 1 dx Therefore diverges. √ −1 x 1 − x 2 Z ∞ Z 1 Z ∞ dx I = + = I1 + I2 √ x = xe 0 0 1 Z 1 Z 1 dx dx √ x < √ =2 I1 = xe x Z0 ∞ Z0 ∞ 1 dx e−x d x = I2 = √ x < e xe 1 1 Thus I converges, and I < 2 + (1/e). R 60 Volume = 0 A(x) d x. The approximation is Z

T6 =

10 h 10, 200 + 2(9, 200 + 8, 000 + 7, 100 2 i + 4, 500 + 2, 400) + 100

≈ 364, 000 m3 .

9.

S6 =

10 h 10, 200 + 4(9, 200 + 7, 100 + 2, 400) 3 i + 2(8, 000 + 4, 500) + 100

≈ 367, 000 m3 Z 1p I = 2 + sin(π x) d x

10.

1 h√

p p T4 = 2 + 2( 2 + sin(π/4) + 2 + sin(π/2) 8 p √ i + 2 + sin(3π/4) + 2

≈ 1.609230 p 1 hp 2 + sin(π/8) + 2 + sin(3π/8) M4 = 4 i p p 2 + sin(5π/8) + 2 + sin(7π/8) ≈ 1.626765 I ≈ 1.6

T8 S8 I

12.

I

T4 T8 S8 I

13.

 0.730 2.198 + 1.001 + 1.332 + 1.729 + 2 2 = 5.526   1 S4 = 0.730 + 2.198 + 4(1.001 + 1.729) + 2(1.332) 3 = 5.504.

a) T4 = 1



b) If T8 = 5.5095, then S8 =

4T8 − T4 = 5.504. 3

c) Yes, S4 = S8 suggests that Sn may be independent of n, which is consistent with a polynomial of degree not exceeding 3.

Challenging Problems 6 (page 389)

1.

a) Long division of x 2 + 1 into x 4 (1 − x)4 = x 8 − 4x 7 + 6x 6 − 4x 5 + x 4 yields x 4 (1 − x)4 4 = x 6 − 4x 5 + 5x 4 − 4x 2 + 4 − 2 . 2 x +1 x +1 Integrating both sides over [0, 1] leads at once to

0

11.

ADAMS and ESSEX: CALCULUS 8

1 = (T4 + M4 ) ≈ 1.617996 2 1 = (T4 + 2M4 ) ≈ 1.62092 3 ≈ 1.62 Z ∞ x2 = d x Let x = 1/t 5 3 1/2 x + x + 1 d x = −(1/t 2 ) dt Z 2 Z 2 4 t dt (1/t ) dt = = 5 3 5 2 0 t +t +1 0 (1/t ) + (1/t ) + 1 ≈ 0.4444 M4 ≈ 0.4799 ≈ 0.4622 M8 ≈ 0.4708 ≈ 0.4681 S16 ≈ 0.4680 ≈ 0.468 to 3 decimal places

Z

1

0

x 4 (1 − x)4 22 22 dx = − 4tan−1 1 = − π. 2 x +1 7 7

x 4 (1 − x)4 22 > 0 on (0, 1), − π > 0, and so 7 x2 + 1 22 π< . 7 Z 1 b) If I = x 4 (1 − x)4 d x, then since 1 < x 2 + 1 < 2 Since

0

on (0, 1), we have I >

Z

0

1

x 4 (1 − x)4 I dx > . 2 x2 + 1

Thus I > (22/7) − π > I /2, or 22 22 I −I 100 100 1 1 = (1 + · · ·) > 300 300 1 1 −K /200 (e + · · ·) < . = 100 100

a) Let f (x) = Ax 5 + Bx 4 + C x 3 + Dx 2 + E x + F. Then Z

1

0

In each case the · · · represent terms much less than the first term (shown) in the sum. Evidently M100 is smallest if k is much greater than 100, and is therefore the best approximation. T100 appears to be the worst.

2a + 2b + c = 1.

b) If m = h = 1/2, we obtain

For very large K , the value of I is very small (I < 1/K ). However, T100

2b 1 = , 4 3

7 16 f (m − h) + f (m − 21 h) 90 45  2 16 7 + f (m) + f (m + 21 h) + f (m + h) . 15 45 90 f (x) d x ≈ 2h

m−h

= 0.

2a +

Solving these equations, we get a = 7/90, b = 16/45, and c = 2/15. The approximation for the integral of any function f on [m − h, m + h] is

d) Since limn→∞ In = 0, we must have n 1 1X 1− e j =0 j !

2b 1 = , 16 5

2a +

1

0

8.

 2 1 7 0 16 −1/8 e + e + e−1/4 2 90 45 15 16 −3/8 7 −1/2 + e + e 45 45  16 −5/8 2 16 7 + e + e−3/4 + e−7/8 + e−1 45 15 45 90 ≈ 0.63212055883.

e−x d x ≈

a) f ′ (x) < 0 on [1, ∞), and lim x→∞ f (x) = 0. Therefore Z ∞ Z ∞ | f ′ (x)| d x = − f ′ (x) d x 1 1 Z R = − lim f ′ (x) d x R→∞ 1

= lim ( f (1) − f (R)) = f (1). R→∞

Also h i 2h a f (−h) + b f (−h/2) + c f (0) + b f (h/2) + a f (h) h   = 2 a 2Bh 5 + 2Dh 3 + 2F ! # 2Bh 5 2Dh 3 +b + + 2F + cFh . 16 4

Thus Z ∞ Z ′ ≤ f (x) cos x d x R

Thus lim

268 Copyright © 2014 Pearson Canada Inc.

Z

R→∞ 1

R

∞ R

| f ′ (x)| d x → 0 as R → ∞.

f ′ (x) cos x d x exists.

INSTRUCTOR’S SOLUTIONS MANUAL

b)

Z

∞

CHALLENGING PROBLEMS 6 (PAGE 389)

But since | sin x| ≥ sin2 x = 21 (1 − cos(2x)), we have

f (x) sin x d x

1

U = f (x) d V = sin x d x dU = f ′ (x) d x V = − cos x R Z ∞ f ′ (x) cos x d x = lim f (x) cos x + R→∞ 1 Z1 ∞ = − f (1) cos(1) + f ′ (x) cos x d x; 1

the integral converges.

c) f (x) = 1/x satisfies the conditions of part (a), so Z ∞ sin x d x converges x 1 by part (b). Similarly, it can be shown that Z ∞ cos(2x) d x converges. x 1

Z

∞ 1

| sin x| dx ≥ x

Z

∞ 1

1 − cos(2x) . 2x

R∞ The latter integral diverges because 1 (1/x) d x R∞ diverges to infinity while 1 (cos(2x))/(2x) d x converges. Therefore Z

1

∞

| sin x| dx x

diverges to infinity.

269 Copyright © 2014 Pearson Canada Inc.

SECTION 7.1 (PAGE 399)

CHAPTER 7. GRATION

ADAMS and ESSEX: CALCULUS 8

APPLICATIONS OF INTE-

3. By slicing: Z

V =π

Section 7.1 Volumes by Slicing—Solids of Revolution (page 399)

1

(x − x 4 ) d x ! 1 3π x 5 x2 = − cu. units. 2 5 0 10

0

=π

1. By slicing: V =π

1

Z

0

x4 dx =

π cu. units. 5

By shells:

By shells: V = 2π

Z

= 2π

1

y(1 −

√

y) d y ! 1 π 2y 5/2 y2 = − cu. units. 2 5 5 0

0

y

1

Z

√ y( y − y 2 ) d y 0 ! 1 3π 2y 5/2 y 4 = = 2π − cu. units. 5 4 0 10

V = 2π

y

√ y= x

(1,1)

y=x 2

y=x 2

x

x

x

x

Fig. 7.1.3

4. Slicing:

Fig. 7.1.1

2. Slicing: V =π

Z

0

V =π

1

=π y− Shells: V = 2π = 2π

Z

 1 π 1 2 y = cu. units. 2 2 0

1

3

x dx x4 4

 1 = π cu. units. 2 0

0

(y − y 4 ) d y



 1 3π 1 2 1 5 y − y = cu. units. 2 5 10 0

Z

1

V = 2π

Shells:

0



1

=π

(1 − y) d y



Z

= 2π

y

0



x(x 1/2 − x 2 ) d x

 1 3π 2 5/2 1 4 x − x = cu. units. 5 4 10 0 y

(1,1) y=x 2

√ y= x 1 x

Fig. 7.1.2

x

Fig. 7.1.4

270 Copyright © 2014 Pearson Canada Inc.

y=x 2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.1 (PAGE 399)

y

5. a) About the x-axis: V =π

Z

Z

2 0

(1,1)

x 2 (2 − x)2 d x

y=x y=x 2

2

(4x 2 − 4x 3 + x 4 ) d x 0 ! 2 4x 3 16π x 5 4 =π = −x + cu. units. 3 5 0 15 =π

x

Fig. 7.1.6

b) About the y-axis: V = 2π = 2π

7. a) About the x-axis:

2

Z

x 2 (2 − x) d y

0



2x 3 3

−

V = 2π

 2 x4 4

= 8π cu. units. 3 0

= 2π

Z

3

0



y(4y − y 2 − y) d y

y3 −

y

(a)

b) About the y-axis:

y=2x−x 2

 3 27π y 4 = cu. units. 4 0 2

Z 3h i (4y − y 2 )2 − y 2 d y 0 Z 3 =π (15y 2 − 8y 3 + y 4 ) d y 0 ! 3 y 5 108π 3 4 = π 5y − 2y + = cu. units. 5 0 5

V =π 2 x

y

y

y=2x−x 2

(b)

(3,3) x=y

2 x

x=4y−y 2

Fig. 7.1.5

x

6. Rotate about Fig. 7.1.7

a) the x-axis V =π =π

Z

1

0



b) the y-axis V = 2π = 2π

(x 2 − x 4 ) d x

8. Rotate about

 1 2π 1 3 1 5 x − x = cu. units. 3 5 15 0

Z

1

0



a) the x-axis

x(x − x 2 ) d x

 1 1 3 1 4 π x − x = cu. units. 3 4 6 0

V =π

Z

Z0

π

[(1 + sin x)2 − 1] d x

π

=π (2 sin x + sin2 x) d x 0  π  π π = −2π cos x + x − sin 2x 2 4 0 1 2 = 4π + π cu. units. 2

271 Copyright © 2014 Pearson Canada Inc.

SECTION 7.1 (PAGE 399)

ADAMS and ESSEX: CALCULUS 8

y

b) the y-axis

(1/3,3)

V = 2π

Z

π

x sin x d x

3x+3y=10

0

U=x d V = sin x d x dU = d x V = − cos x π Z π   cos x d x = 2π −x cos x +

1 y= x

x

0

0

Fig. 7.1.10

= 2π 2 cu. units.

9. a) About the x-axis: V =π

Z

0

1

= 4π − π = 4π − π

4− Z

11. 1 (1 + x 2 )2

π/4

0

Z

π/4



(3,1/3)

dx

sec2 θ dθ sec4 θ

Let x = tan θ d x = sec2 θ dθ

Z

1

V = 2 × 2π (2 − x)(1 − x) d x 0 Z 1 = 4π (2 − 3x + x 2 ) d x 0



= 4π 2x −

cos2 θ dθ

y

y

0

π/4 π = 4π − (θ + sin θ cos θ ) 2 0 π 15π π2 π2 − = − cu. units. = 4π − 8 4 4 8 b) About the y-axis:  Z 1  1 x 2− V = 2π dx 1 + x2 0   1 1 = 2π x 2 − ln(1 + x 2 ) 2 0   1 = 2π 1 − ln 2 = 2π − π ln 2 cu. units. 2 y

y=2

 1 10π x 3 3x 2 = + cu. units. 2 3 0 3 x+y=1

x=2

x

x

Fig. 7.1.11

12.

V =π

Z

1 −1

[(1)2 − (x 2 )2 ] d x

  1 1 = π x − x 5 5 −1 8π = cu. units. 5 y

y=

1 1+x 2

y=1 x2 y=1−x 2

x

1

dx

x

x

Fig. 7.1.9 Fig. 7.1.12

10. By symmetry, rotation about the x-axis gives the same volume as rotation about the y-axis, namely  Z 3  10 1 V = 2π x −x − dx 3 x 1/3   3 5 2 1 3 = 2π x − x − x 3 3 1/3 512π = cu. units. 81

x

13. The volume remaining is V = 2 × 2π = 2π

272 Copyright © 2014 Pearson Canada Inc.

Z

Z

2 1

3√ 0

p x 4 − x2 dx

u du =

3 4π 3/2 u 3 0

Let u = 4 − x 2 du = −2x d x √ = 4π 3 cu. units.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.1 (PAGE 399)

4 3 32π π2 = cu. units., 3 3 √ 32π therefore the volume removed is − 4π 3 cu. units. 3 The percentage removed is

Since the volume of the ball is

15. The volume remaining is

√ 32π √ ! − 4π 3 3 3 3 × 100 = 100 1 − ≈ 35. 32π 8 3 About 35% of the volume is removed. y

x

b

y

√

4−x 2

y=

1

Z

 x xh 1 − dx b a  b  2 x 3 x − = 2π h 2 3b a   2 a3 = π h(b2 − a 2 ) − π h b2 − 3 b   3 1 2a = π h b2 − 3a 2 + cu. units. 3 b

V = 2π

h

x y b + h =1

dx 2 x

x=a

dx b x

Fig. 7.1.13

x

Fig. 7.1.15

16. Let a circular disk with radius a have centre at point 14. The radius of the hole is the remaining volume is

q

R 2 − 14 L 2 . Thus, by slicing,

   L2 R2 − x 2 − R2 − dx 4 −L/2  2  L/2 L 1 = 2π x − x 3 4 3 0 π 3 = L cu. units (independent of R). 6

V =π

Z

L/2 

y

√

y=

(a, 0). Then the disk is rotated about the y-axis which is one of its tangent lines. The volume is: Z 2a p V = 2 × 2π x a 2 − (x − a)2 d x Let u = x − a 0 du = d x Z a p 2 2 = 4π (u + a) a − u du Z−a Z ap a p = 4π u a 2 − u 2 du + 4π a a 2 − u 2 du −a −a   1 = 0 + 4π a π a 2 = 2π 2 a 3 cu. units. 2 (Note that the first integral is zero because the integrand is odd and the interval is symmetric about zero; the second integral is the area of a semicircle.)

R 2 −x 2

y R

q L 2

L2 R2 − 4

(x−a)2 +y 2 =a 2 x

2a a

x

L

Fig. 7.1.14

Fig. 7.1.16

273 Copyright © 2014 Pearson Canada Inc.

SECTION 7.1 (PAGE 399)

ADAMS and ESSEX: CALCULUS 8

17. Volume of the smaller piece: Z

19. The volume of the ellipsoid is

a

  x2 b2 1 − 2 d x a 0  3  a 4 x = 2π b2 x − 2 = π ab2 cu. units. 3a 3 0

(a 2 − x 2 ) d x b   a x 3 = π a2 x − 3 b   a 3 − b3 = π a 2 (a − b) − 3 π 2 2 = (a − b)[3a − (a + ab + b2 )] 3 π = (a − b)2 (2a + b) cu. units. 3

V =π

y

Z

V = 2π

a

y

y=b b

√

y=

x

a

1−

x2 a2

dx

a 2 −x 2

dx b

q

a x

x

x

Fig. 7.1.19

20. The cross-section at height p y is an annulus (ring)

Fig. 7.1.17

18. Let the centre of the bowl be at (0, 30). Then the vol-

2 2 having p inner radius b − a − y and outer radius 2 2 b + a − y . Thus the volume of the torus is

Z

a

q q h i (b + a 2 − y 2 )2 − (b − a 2 − y 2 )2 d y −a Z q a 4b a 2 − y 2 d y = 2π

ume of the water in the bowl is Z 20 h i 302 − (y − 30)2 d y V =π 0 Z 20 =π 60y − y 2 d y

V =π

0

π a2 = 8π b = 2π 2 a 2 b cu. units.. 4

0

  20 1 = π 30y 2 − y 3 3 0

We used the area of a quarter-circle of radius a to evaluate the last integral.

3

≈ 29322 cm . y

21. a) Volume of revolution about the x-axis is V =π

30

Z

∞

e−2x d x

0

R π e−2x = cu. units. R→∞ −2 0 2

= π lim

20

b) Volume of revolution about the y-axis is x 2 +(y−30)2 =302

x

V = 2π Fig. 7.1.18

Z

∞

xe−x d x

0

R = 2π lim (−xe−x − e−x ) = 2π cu. units. R→∞

274 Copyright © 2014 Pearson Canada Inc.

0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.1 (PAGE 399)

y

25. Since all isosceles right-angled triangles having leg length a cm are congruent, S does satisfy the condition for being a prism given in early editions. It does not satisfy the condition in this edition because one of the line segments joining vertices of the triangular cross-sections, namely the x-axis, is not parallel to the line joining the vertices of the other end of the hypotenuses of the two bases.

1 y=e−x

dx x

x

The volume os S is still the constant cross-sectional area a 2 /2 times the height b, that is, V = a 2 b/2 cm3 .

Fig. 7.1.21

26. Using heights f (x) estimated from the given graph, we obtain

22. The volume is V =π

Z

1

R ∞ x 1−2k x −2k d x = π lim R→∞ 1 − 2k 1

R 1−2k π + . R→∞ 1 − 2k 2k − 1

= π lim

In order for the solid to have finite volume we need 1 − 2k < 0,

k>

that is,

V =π

1

2 f (x) d x

πh 2 ≈ 3 + 4(3.8)2 + 2(5)2 + 4(6.7)2 + 2(8)2 3 i + 4(8)2 + 2(7)2 + 4(5.2)2 + 32 ≈ 938 cu. units.

27. Using heights f (x) estimated from the given graph, we obtain

1 . 2

V = 2π ≈

R∞

x 1−k d x. This improper integral converges if 1 − k < −1, i.e., if k > 2. The solid has finite volume only if k > 2.

23. The volume is V = 2π

Z 9

1

y

Z

9

x f (x) d x 1

2π h 1(3) + 4(2)(3.8) + 2(3)(5) + 4(4)(6.7) + 2(5)(8) 3 i + 4(6)(8) + 2(7)(7) + 4(8)(5.2) + 9(3) ≈ 1537 cu. units.

28. Using heights f (x) estimated from the given graph, we obtain V = 2π

y=x −k

dx 1

x

x

Z

9 1

(x + 1) f (x) d x

2π h ≈ 2(3) + 4(3)(3.8) + 2(4)(5) + 4(5)(6.7) + 2(6)(8) 3 i + 4(7)(8) + 2(8)(7) + 4(9)(5.2) + 10(3) ≈ 1832 cu. units.

29. The region is symmetric about x = y so has the same

Fig. 7.1.23

24. A solid consisting of points on parallel line segments between parallel planes will certainly have congruent cross-sections in planes parallel to and lying between the two base planes, any solid satisfying the new definition will certainly satisfy the old one. But not vice versa; congruent cross-sections does not imply a family of parallel line segments giving all the points in a solid. For a counterexample, see the next exercise. Thus the earlier, incorrect definition defines a larger class of solids than does the current definition. However, the formula V = Ah for the volume of such a solid is still valid, as all congruent cross-sections still have the same area, A, as the base region.

volume of revolution about the two coordinate axes. The volume of revolution about the y-axis is Z 8 V = 2π x(4 − x 2/3 )3/2 d x Let x = 8 sin3 u 0 d x = 24 sin2 u cos u du Z π/2 = 3072π sin5 u cos4 u du 0 Z π/2 = 3072π (1 − cos2 u)2 cos4 u sin u du Let v = cos u 0 dv = − sin u du Z 1 = 3072π (1 − v 2 )2 v 4 dv 0 Z 1 = 3072π (v 4 − 2v 6 + v 8 ) dv 0   1 2 1 8192π = 3072π − + = cu. units. 5 7 9 105

275 Copyright © 2014 Pearson Canada Inc.

SECTION 7.1 (PAGE 399)

ADAMS and ESSEX: CALCULUS 8

R = sin α, so R = (x + h) sin α. x +h Using the result of Exercise #17, the volume of liquid displaced by the ball is Note that

4 π R 3 . Expressing this volume 3 as the “sum” (i.e., integral) of volume elements that are concentric spherical shells having thickness dr and varying radius r , and therefore having surface area kr 2 and volume kr 2 dr , we obtain

30. The volume of the ball is

4 π R3 = 3

R

Z

0

kr 2 dr =

k 3 R . 3

Thus k = 4π .

R

V =

We would like to consider V as a function of x for −2R ≤ x ≤ R since V = 0 at each end of this interval, and V > 0 inside the interval. However, the actual interval of values of x for which the above formulation makes physical sense is smaller: x must satisfy −R ≤ x ≤ h tan2 α. (The left inequality signifies nonsubmersion of the ball; the right inequality signifies that the ball is tangent to the glass somewhere below the rim.) We look for a critical point of V , considered as a function of x. (As noted above, R is a function of x.) We have

dr r

0=

Fig. 7.1.30

31. Let the ball have radius R, and suppose its centre is x units above the top of the conical glass, as shown in the figure. (Clearly the ball which maximizes liquid overflow from the glass must be tangent to the cone along some circle below the top of the cone — larger balls will have reduced displacement within the cone. Also, the ball will not be completely submerged.)

π (R − x)2 (2R + x). 3

   dV π dR = − 1 (2R + x) 2(R − x) dx 3 dx   d R + (R − x)2 2 +1 dx

dR (4R + 2x + 2R − 2x) = 4R + 2x − (R − x). dx Thus   R 6R sin α = 3(R + x) = 3 R + −h sin α 2R sin2 α = R sin α + R − h sin α h sin α h sin α R= = . 1 − 2 sin2 α + sin α cos 2α + sin α This value of R yields a positive value of V , and corresponds to x = R(2 sin α − 1). Since sin α ≥ sin2 α, −R ≤ x =

h sin2 α h sin α(2 sin α − 1) ≤ = h tan2 α. 1 + sin α − 2 sin2 α cos2 α

Therefore it gives the maximum volume of liquid displaced. x R h h sec α (h+x) cos α α

32. Let P be the point (t, 25 − t). The line through P perpen-

dicular to AB has equation y = x + 25 − 2t, and meets the curve x y = 1 at point Q with x-coordinate s equal to the positive root of s 2 + ( 52 − 2t)s = 1. Thus,  1 5 s= 2t − + 2 2

Fig. 7.1.31

276 Copyright © 2014 Pearson Canada Inc.

s



5 − 2t 2

2

 +4 .

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.2 (PAGE 403)

y A(1/2,2)

Section 7.2 (page 403)

√ 2 dt

1 y= x

More Volumes by Slicing

P 5 x+y= 2

Q

1.

V =

Z

2 3 2 x = 6 m3 2 0

2

3x d x =

0

2. A horizontal slice of thickness dz at height a has volume

B(2,1/2)

d V = z(h − z) dz. Thus the volume of the solid is s

t

x

Z

V =

h

(z(h − z) dz =

0



hz 2 z3 − 2 3

Fig. 7.1.32 The volume element at P has radius √ P Q = 2(t − s)   s  2 √ 5 1 5 = 2 − − 2t + 4 4 2 2 and thickness

0

3. A horizontal √ slice of thickness dz at height a has volume d V = π z 1 − z 2 dz. Thus the volume of the solid is 1

Z

V =

0

π = 2

√

2 dt. Hence, the volume of the solid is s  2 2 Z 2 √  √ 5 1 5 2 − − 2t + 4 2 dt V =π 4 2 2 1/2  s   2 √ Z 2 25 5 5  −  = 2 2π − 2t + 4 + 4 2 1/2 16  2 # 1 5 − 2t + 4 dt Let u = 2t − 25 4 2 du = 2 dt   √ Z 3/2 41 5 p u2 = 2π − du u2 + 4 + 4 4 −3/2 16   3/2 √ 1 41 − = 2π u + u 3 16 12 −3/2 √ Z 5 2π 3/2 p 2 u + 4 du Let u = 2 tan v 4 −3/2 du = 2 sec2 v dv √ Z −1 (3/4) tan √ 33 2π = − 5 2π sec3 v dv 4 tan−1 (−3/4) √ √ Z tan−1 (3/4) 33 2π − 10 2π = sec3 v dv 4 0 √ √  33 2π = − 5 2π sec v tan v+ 4  tan−1 (3/4) ln | sec v + tan v| 0    √ 33 15 = 2π −5 + ln 2 − 0 − ln 1 4 16   √ 57 = 2π − 5 ln 2 cu. units. 16

 h h3 units3 . = 6

4. 5.

p z 1 − z 2 dz

let u − 1 − z 2

1 π π 2 3/2 u = units3 . u du = 2 3 3

1√

Z

0

0

3 x 3 26 = cu. units 3 1 3 1 Z 6 Z 6 (16 + 6z − z 2 ) dz (2 + z)(8 − z) dz = V =

V =

3

Z

x2 dx =

0

0

 6 z 3 2 = 16z + 3z − = 132 ft3 3 0 

6. The area of an √ equilateral triangle of edge √ √ √  1 2

3 2

A(x) = x the solid is V =

Z

1

4

x =

3 4 x

√

x is sq. units. The volume of

√ √ √ 4 15 3 3 3 2 x dx = x = cu. units. 4 8 8 1

7. The area of cross-section at height y is A(y) =

 y 2π(1 − (y/ h)) (π a 2 ) = π a 2 1 − sq. units. 2π h

The volume of the solid is Z h  y π a2 h V = π a2 1 − dy = cu. units. h 2 0

8. Since V = 4, we have 4=

Z

0

2

kx 3 d x = k

2 x 4 = 4k. 4 0 277

Copyright © 2014 Pearson Canada Inc.

SECTION 7.2 (PAGE 403)

ADAMS and ESSEX: CALCULUS 8

13. The cross-section at distance y from the vertex of the

Thus k = 1.

9. The volume between height 0 and height z is z3 =

10.

Z

z3.

Thus

partial cone is a semicircle of radius y/2 cm, and hence area π y 2/8 cm2 . The volume of the solid is

z

A(t) dt,

0

V =

where A(t) is the cross-sectional area at height t. Differentiating the above equation with respect to z, we get 3z 2 = A(z). The cross-sectional area at height z is 3z 2 sq. units. Z z This is similar to Exercise 7. We have 4z = A(t) dt,

Z

12

0

π 2 π 123 y dy = = 72π cm3 . 8 24 z

0

11.

so A(z) = 4. Thus the square cross-section at height z has side 2 units. Z r q 2 V =2 2 r 2 − y2 d y 0   r Z r y 3 16r 3 =8 (r 2 − y 2 ) d y = 8 r 2 y − = cu. units. 3 0 3 0 z

√

2

r 2 −y 2

x

x=

r 2 −y 2

(12, 12, 0) Fig. 7.2.13

14. The volume of a solid of given height h and given crosssectional area A(z) at height z above the base is given by Z h V = A(z) dz.

15. Let the x-axis be along the diameter shown in the fig-

12. The area equilateral triangle of base 2y is √ of an √ 1 2 3y) =

x

If two solids have the same height h and the same area function A(z), then they must necessarily have the same volume.

Fig. 7.2.11 2 (2y)(

12 y

0

y

√

y

3y . Hence, the solid has volume Z r√ V =2 3(r 2 − x 2 ) d x 0   r √ 1 = 2 3 r 2 x − x 3 3 0 4 3 = √ r cu. units. 3

ure, with the origin at the centre of the base. The crosssection perpendicular to the x-axis at x is a rectangle √ a+b a−b + x. having base 2 r 2 − x 2 and height h = 2 2 Thus the volume of the truncated cylinder is r

  p a−b a+b + x dx (2 r 2 − x 2 ) 2 2r −r Z r 2 (a + b) p πr = (a + b) r 2 − x 2 d x = cu. units. 2 −r

V =

Z

y √ 3y x 2y

h

r

x 2 +y 2 =r 2

x y Fig. 7.2.12

r √ y = r2 − x2

Fig. 7.2.15

278 Copyright © 2014 Pearson Canada Inc.

x

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.2 (PAGE 403)

16. The plane z = k meets the ellipsoid in the ellipse  2 k a b c x2 y2   2  +   2  = 1 k k 2 2 a 1− b 1− c c  x 2

that is,

z

+

 y 2

=1−

45◦ 20 x

which has area

√

y=

y 400−x 2

x

 2  k A(k) = π ab 1 − . c 

Fig. 7.2.17

The volume of the ellipsoid is found by summing volume elements of thickness dk:

18. The solution is similar to that of Exercise 15 except that

  2  k π ab 1 − dk c −c   c 1 = π ab k − 2 k 3 3c

V =

Z

c

4 = π abc cu. units. 3

the legs of the right-triangular cross-sections are y − 10 √ √ instead of y, and x goes from −10 3 to 10 3 instead of −20 to 20. The volume of the notch is

−c

Z

√ 10 3

1 p ( 400 − x 2 − 10)2 d x 2 0 Z 10√3   p = 500 − x 2 − 20 400 − x 2 d x

V =2

z

0

c

√ 4, 000π = 3, 000 3 − ≈ 1, 007 cm3 . 3

2 2 x2 + by2 + cz 2 =1 a2 (one-eighth of the solid is shown)

k A(k)

19. The hole has the shape of two copies of the trunb

y

a

x

Fig. 7.2.16

cated cylinder of Exercise 13, placed base to base, √ with a + b = 3 2 in and r = 2 in. Thus the volume of wood (the volume of the hole) is √ removed √ V = 2(π 22 )(3 2/2) = 12 2π in3 .

20. One eighth of the region lying inside both cylinders is 17. Cross-sections of the wedge removed perpendicular to the x-axis are isosceles, right triangles. The volume of the wedge removed from the log is Z

20

1 p ( 400 − x 2 )2 d x 2 0   20 x 3 16, 000 = 400x − = cm3 . 3 3

V =2

shown in the figure. If the region is sliced by a horizontal plane at height z, then the intersection is a rectangle with area p p A(z) = b2 − z 2 a 2 − z 2 . The volume of the whole region is

0

V =8

Z

0

b

p

p b2 − z 2 a 2 − z 2 dz.

279 Copyright © 2014 Pearson Canada Inc.

SECTION 7.2 (PAGE 403)

ADAMS and ESSEX: CALCULUS 8

z

5.

y = x 2/3 , r

b

√

A(z)√

b2 −z 2

a 2 −z 2

a

y

6.

x

Fig. 7.2.20

21. By the result given in Exercise 18 with a = 4 cm and b = 2 cm, the volume of wood removed is V =8

Z

0

2p

4 − z2

p 16 − z 2 dz ≈ 97.28 cm3 .

(We used the numerical integration routine in Maple to evaluate the integral.)

7.

Section 7.3 Arc Length and Surface Area (page 410)

1.

y = 2x − 1, y ′ = 2, ds = Z 3√ √ L= 5 d x = 2 5 units.

p 1 + 22 d x

8.

1

2.

y = ax + b, A ≤ x ≤ B, y ′ = a. The length is L=

3.

4.

Z

B A

2 −1/3 x , 3 √

4 −2/3 9x 2/3 + 4 x dx = dx 9 3|x|1/3 √ Z 1 9x 2/3 + 4 d x Let u = 9x 2/3 + 4 L=2 3x 1/3 0 du = 6x −1/3 d x Z 13 3/2 √ 1 2(13 ) − 16 = u du = units. 9 4 27

ds =

z

y′ =

p p 1 + a 2 d x = 1 + a 2 (B − A) units.

√ √ y = 23 x 3/2 , y ′ = x, ds = 1 + x d x 8 Z 8 √ 52 2 L= 1 + x d x = (1 + x)3/2 = units. 3 3 0 0

3√ y 2 = (x − 1)3 , y = (x − 1)3/2 , y ′ = x −1 2 Z 2r Z 2√ 9 1 1 + (x − 1) d x = 9x − 5 d x L= 4 2 1 1 2 1 133/2 − 8 = (9x − 5)3/2 = units. 27 27

9.

1+

q 2(x + 1)3 = 3(y − 1)2 , y = 1 + 23 (x + 1)3/2 q y ′ = 32 (x + 1)1/2 , r r 3x + 3 3x + 5 ds = 1 + dx = dx 2 2 √ 0 Z 0√ 1 2 3/2 L= √ 3x + 5 d x = (3x + 5) 9 2 −1 −1 √   2 3/2 = 5 − 23/2 units. 9 1 x2 1 x3 + , y′ = − 2 12 x 4 x s  2 2  2 1 x x ds = 1 + − 2 dx = 4 x 4   3 Z 4 2 x 1 x L= + 2 dx = − 4 12 x 1 y=

 1 + 2 dx x  4 1 = 6 units. x 1

x3 1 1 + , y′ = x 2 − 2 3 4x 4x s     1 1 2 ds = 1 + x 2 − 2 d x = x 2 + 2 d x 4x 4x   3  2 Z 2 1 59 1 x 2 x + 2 dx = L= − = 24 units. 4x 3 4x 1 1 y=

ln x x2 1 x − , y′ = − 2 4 2x 2 s  2   x x 1 1 − dx = + dx ds = 1 + 2x 2 2x 2    e Z e 1 x x 2 ln x L= + dx = + 2x 2 2 4 1 1 2 2 1 e −1 e +1 = + = units. 2 4 4 y=

10. If y = x 2 −

1

280 Copyright © 2014 Pearson Canada Inc.

ln x 1 then y ′ = 2x − and 8 8x   1 2 1 + (y ′ )2 = 2x + . 8x

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.3 (PAGE 410)

Thus the arc length is given by

14.

s

  1 2 dx s= 1 + 2x − 8x 1  Z 2 1 = 2x + dx 8x 1   2 1 1 = x 2 + ln x = 3 + ln 2 units. 8 8 1 Z

11.

12.

Z

a

s=

s=

Z

=

Z

π/4 p

2

L=

π/4

π/6

15. 1 + tan2 x d x

π/4 sec x d x = ln | sec x + tan x| π/6

  √ 1 2 = ln( 2 + 1) − ln √ + √ 3 3 √ 2+1 units. = ln √ 3

Z

4

s

1+

2

Z

4

x 2/3 + y 2/3 = x 2/3 . By symmetry, the curve has congruent arcs in the four quadrants. For the first quadrant arc we have 3/2  y = a 2/3 − x 2/3  1/2  2 3  2/3 y′ = − x −1/3 . a − x 2/3 2 3 Thus the length of the whole curve is

L=4

13.

Z

0

2p

1 + 4x 2 d x

Z

= 4a

y = x 2 , 0 ≤ x ≤ 2, y ′ = 2x. length =

4e2x dx − 1)2

(e2x

e2x + 1 dx 2x 2 e −1 Z 4 x 4 x e + e−x e − e−x = d x = ln x −x 2 e −e     2 1 1 = ln e4 − 4 − ln e2 − 2 e e  8  e − 1 e2 e4 + 1 = ln = ln units. 4 4 e e −1 e2 =

Z a p 1 + sinh2 x d x = cosh x d x 0 0 a ea − e−a = sinh x = sinh a = units. 2 0

π/6

ex − 1 , 2≤x ≤4 ex + 1 x x e + 1 (e + 1)e x − (e x − 1)e x y′ = x e −1 (e x + 1)2 2e x . = 2x e −1 The length of the curve is y = ln

a

0

1/3

= 4a 1/3 Let 2x = tan θ 2 d x = sec2 θ dθ

Z 1 x=2 = sec3 θ 2 x=0  x=2 1 sec θ tan θ + ln | sec θ + tan θ | = 4 x=0  2 p 1 p 2 = 2x 1 + 4x + ln(2x + 1 + 4x 2 ) 4 0 √  1 √ = 4 17 + ln(4 + 17) 4 √ √ 1 = 17 + ln(4 + 17) units. 4

s Z

1+

a 2/3 − x 2/3 dx x 2/3

a

x −1/3 d x 3 2/3 a x = 6a units. 2 0 0

16. The required length is L=

Z

0

1p

1 + (4x 3 )2 d x =

Z

0

1p

1 + 16x 6 d x.

Using a calculator we calculate some Simpson’s Rule approximations as described in Section 7.2: S2 ≈ 1.59921 S8 ≈ 1.60025

S4 ≈ 1.60110 S16 ≈ 1.60023.

To four decimal places the length is 1.6002 units.

281 Copyright © 2014 Pearson Canada Inc.

SECTION 7.3 (PAGE 410)

17.

y = x 1/3 , 1 ≤ x ≤ 2, y ′ = Length = have

R2

ADAMS and ESSEX: CALCULUS 8

1 −2/3 x . 3 r

f (x) d x, where f (x) =

1

T4 = 1.03406 T8 = 1.03385 T16 = 1.03378

22.

1+

1 . We 9x 4/3

q y = x 3/2 , 0 ≤ x ≤ 1. ds = 1 + 49 x d x. The area of the surface of rotation about the x-axis is

S = 2π

M4 = 1.03363 M8 = 1.03374 M16 = 1.00376.

=

Thus the length is approximately 1.0338 units. y ′ = −3x/y. Thus

9x 2

1+

ds =

3 − 3x 2

dx =

s

3 + 6x 2 3 − 3x 2

4

1

0

s

=

d x.

3 + 6x 2 d x ≈ 8.73775 units 3 − 3x 2

(with a little help from Maple’s numerical integration routine.)

Z

x2 dx = 4 − 2x 2

1+

Z

20.

21.

1 0

s

4 − x2 dx 4 − 2x 2

x2

4− d x ≈ 1.05810 units 4 − 2x 2

(with a little help from Maple’s numerical integration routine). Z 2 p S = 2π |x| 1 + 4x 2 d x Let u = 1 + 4x 2 0 du = 8x d x  17 Z π 17 √ π  2 3/2 = u du = u 4 1 4 3 1 √ π = (17 17 − 1) sq. units. 6 √ y = x 3 , 0 ≤ x ≤ 1. ds = 1 + 9x 4 d x. The area of the surface of rotation about the x-axis is Z

1

S = 2π π 18

Z

10 √

=

0

1

p x 3 1 + 9x 4 d x u du =

128π 243

Let u = 1 + 9x 4 du = 36x 3 d x

π (103/2 − 1) sq. units. 27

a

Z

0

Z

3/2 0

1+

9x dx 4

p u 4 1 + u 2 du

Z tan−1 (3/2)

Let 9x = 4u 2 9 d x = 8u du Let u = tan v du = sec2 v dv

tan4 v sec3 v dv

0

Z tan−1 (3/2) 0

(sec7 v − 2 sec5 v + sec3 v) dv.

n−2 1 secn−2 v tan v + n−1 n−1

Z

secn−2 v dv

sec3 v dv =

1 (sec a tan a + ln | sec a + tan a|. 2

We have

√ The length of the short arc from (0, 1) to (1, 1/ 2) is s

128π 243

r

(see Exercise 36 of Section 7.1) to reduce the powers of secant down to 3, and then use

y ′ = −x/(2y). Thus ds =

0

secn v dv =

19. For the ellipse x 2 + 2y 2 = 2, we have 2x + 4yy ′ = 0, so s

x

3/2

At this stage it is convenient to use the reduction formula

The circumference of the ellipse is Z

1

128π = 243

18. For the ellipse 3x 2 + y 2 = 3, we have 6x + 2yy ′ = 0, so s

Z

I = = =

= =

Z

a

(sec7 v − 2 sec5 v + sec3 v) dv a  Z a Z a sec5 v tan v 5 5 + − 2 sec v dv + sec3 v dv 6 6 0 0 0 # " a Z 3 a 7 sec3 v tan v sec5 a tan a 3 − sec v dv +4 6 6 4 0 0 Z a + sec3 v dv 0 Z sec5 a tan a 7 sec3 a tan a 1 a − + sec3 v dv 6 24 8 0 sec5 a tan a 7 sec3 a tan a sec a tan a + ln | sec a + tan a| − + . 6 24 16 0

Substituting a = ar ctan(3/2) now gives the following value for the surface area: √ √ ! 28 13π 8π 3 + 13 S= + ln sq. units. 81 243 2

282 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.3 (PAGE 410)

23. If y = x 3/2 , 0 ≤ x ≤ 1, is rotated about the y-axis, the surface area generated is

26. S = 2π

r

1

Z

x 1+

0

9x dx 4

Let u = 1 + du =

=

Z

32π 81

13/4

1



32π = 81 64π = 81

9x 4

9 dx 4

√ (u − 1) u du

 13/4 2 5/2 2 3/2 u − u 5 3 1

(13/4)5/2 − 1 (13/4)3/2 − 1 − 5 3

!

sq. units.

24. We have Z

1

p e x 1 + e2x d x

Let e x = tan θ e x d x = sec2 θ dθ Z x=1 p Z x=1 2 2 = 2π 1 + tan θ sec θ dθ = 2π sec3 θ dθ

S = 2π

0

x=0

x=0

  x=1 = π sec θ tan θ + ln | sec θ + tan θ | . x=0

Since

p 1 + e2 , √ x = 0 ⇒ tan θ = 1, sec θ = 2,

x = 1 ⇒ tan θ = e, sec θ =

therefore  p  p √ √ S = π e 1 + e2 + ln | 1 + e2 + e| − 2 − ln | 2 + 1| √   p √ 1 + e2 + e 2 sq. units. = π e 1 + e − 2 + ln √ 2+1

25. If y = sin x, 0 ≤ x ≤ π , is rotated about the x-axis, the surface area generated is S = 2π

Z

= 2π

Z

= 2π

Z

π 0 1 −1

p sin x 1 + cos2 d x

Let u = cos x du = − sin x d x

p 1 + u 2 du

π/4

−π/4

Let u = tan θ du = sec2 θ dθ Z π/4 sec3 θ dθ = 4π sec3 θ dθ 0

  π/4 = 2π sec θ tan θ + ln | sec θ + tan θ | 0 √ √  = 2π 2 + ln(1 + 2) sq. units.

 2 2  2 2 x 1 1 x 1 + (y ′ )2 = 1 + − 2 + 2 = 4 x 4 x  2  Z 4 3 x 1 x 1 S = 2π + + 2 dx 12 x 4 x 1 ! Z 4 x 1 x5 + + 3 dx = 2π 48 3 x 1  4  6 x2 1 x + − 2 = 2π 288 6 2x 1 275 = π sq. units. 8 x3 1 + , 1 ≤ x ≤ 4, we have x  212 1 x + 2 d x. ds = 4 x The surface generated by rotating the curve about the y-axis has area  Z 4  2 1 x + 2 dx S = 2π x 4 x 1  4  4 x + ln |x| = 2π 16   1 255 = 2π + ln 4 sq. units. 16

27. For y =

28. The area of the cone obtained by rotating the line y = (h/r )x, 0 ≤ x ≤ r , about the y-axis is √ r Z r q r 2 + h 2 x 2 x 1 + (h/r )2 d x = 2π S = 2π r 2 0 0 p = πr r 2 + h 2 sq. units.

29. For the circle (x − b)2 + y 2 = a 2 we have 2(x − b) + 2y Thus s ds =

1+

dy =0 dx

⇒

dy x −b =− . dx y

(x − b)2 a a dx = dx = p dx 2 y2 y a − (x − b)2

(if y > 0). The surface area of the torus obtained by rotating the circle about the line x = 0 is Z b+a a xp S = 2 × 2π d x Let u = x − b 2 b−a a − (x − b)2 du = d x Z a u+b = 4π a √ du a2 − u 2 −a Z a du = 8π ab √ by symmetry a2 − u 2 0 a u = 8π ab sin−1 = 4π 2 ab sq. units. a 0 283

Copyright © 2014 Pearson Canada Inc.

SECTION 7.3 (PAGE 410)

ADAMS and ESSEX: CALCULUS 8

30. The top half of x 2 + 4y 2 = 4 is y = dy −x = √ , and dx 2 4 − x2 S = 2 × 2π =π

Z

=π

Z

0

Z

2

√

0

2p

π/3

4 − x2 2

s



1p 4 − x 2 , so 2

x

2

dx √ 2 4 − x2 r 16 Let x = sin θ 3 r 16 cos θ dθ dx = 3 1+

16 − 3x 2 d x

4 (4 cos θ ) √ cos θ dθ 3

0

Z 16π π/3 cos2 θ dθ = √ 3 0   π/3 8π = √ θ + sin θ cos θ 3 0 √ 2π(4π + 3 3) = sq. units. √ 3 3

31. For the ellipse x 2 + 4y 2 = 4 we have dx + 8y = 0 2x dy

⇒

dx y = −4 . dy x

The arc length element on the ellipse is given by

ds = =

s

1+ s 1+



dx dy

2

32. As in Example 4, the arc length element for the ellipse is

ds =

v u u 2 a 2 − b2 2  2 ua − x t dy a2 dx = 1+ d x. 2 2 dx a −x

s

To get the area of the ellipsoid, we must rotate both the upper and lower semi-ellipses (see the figure for Exercise 20 of Section 8.1):

q 16y 2 1 4 + 12y 2 d y. d y = x2 x

If the ellipse is rotated about the y-axis, the resulting surface has area Z 1 q 1 S = 2 × 2π x 4 + 12y 2 d y x 0 Z 1q √ = 8π 1 + 3y 2 d y Let 3y = tan θ √ 0 3d y = sec2 θ dθ Z π/3 8π sec3 θ dθ = √ 3 0  π/3 8π  = √ sec θ tan θ + ln | sec θ + tan θ | 2 3 0 √  8π  √ = √ 2 3 + ln(2 + 3) 2 3 √ ! ln(2 + 3) = 8π 1 + sq. units. √ 2 3

r

 x 2 

c−b 1− + a 0 r    x 2 c+b 1− ds a v u 2 2 u Z a u a2 − a − b x 2 t 2 a = 8π c dx a2 − x 2 0   1 = 8π c of the circumference of the ellipse 4 = 8π ca E(ε)

S = 2 × 2π

√

R π/2 √ a 2 − b2 and E(ε) = 0 1 − ε 2 sin t dt a as defined in Example 4. where ε =

33. From Example 3, the length is π/2

s

π2 cos2 t dt 1+ 4 0 s Z 10 π/2 π2 π2 = 1+ − sin2 t dt π 0 4 4 s Z π/2 5p π2 = 4 + π2 1− sin2 t dt π 4 + π2 0   5p π 2 4+π E √ . = π 4 + π2

10 s= π

dy

Z a 

Z

34. Let the equation of the sphere be x 2 + y 2 = R 2 . Then the surface area between planes x = a and x = b (−R ≤ a < b ≤ R) is S = 2π = 2π

Z

Z

= 2π R

284 Copyright © 2014 Pearson Canada Inc.

b a b a

Z

p

p

a

b

s

R2 − x 2 1 + R2 − x 2 √



dy dx

R R2

− x2

2

dx

dx

d x = 2π R(b − a) sq. units.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.4 (PAGE 417)

Thus, the surface area depends only on the radius R of the sphere, and the distance (b − a) between the parellel planes.

b) The surface area is S = 2π

y

> 2π

Z

Z1 ∞ 1

a

b

x

Fig. 7.3.34

c) Covering a surface with paint requires applying a layer of paint of constant thickness to the surface. Far to the right, the horn is thinner than any prescribed constant, so it can contain less paint than would be required to cover its surface.

1. The mass of the wire is

35. If the curve y = x k , 0 < x ≤ 1, is rotated about the

m=

y-axis, it generates a surface of area S = 2π = 2π

1

0

Z

0

p x 1 + k 2 x 2(k−1) d x

1p

x 2 + k 2 x 2k d x.

If k ≤ −1, we have S ≥ 2π k

Z

1

x k d x, which is infinite.

0

xk

If k ≥ 0, the surface area S is finite, since is bounded on (0, 1] in that case. Hence we need only consider the case −1 < k < 0. In this case 2 < 2 − 2k < 4, and S = 2π

Z

Z

1 0

p x 1 + k 2 x 2(k−1) d x

1p

x 2−2k + k 2 x k d x Z 1 p 2 < 2π 1 + k x k d x < ∞. = 2π

0

0

36.

Thus the area is finite if and only if k > −1. Z 1 r 1 S = 2π |x| 1 + 2 d x x 0 Z 1p x 2 + 1 d x Let x = tan θ = 2π 0 d x = sec2 θ dθ Z π/4 = 2π sec3 θ dθ 0

  π/4 = π sec θ tan θ + ln | sec θ + tan θ | 0 √ √ = π [ 2 + ln( 2 + 1)] sq. units.

37.

a) Volume V = π

R ∞ dx 1 x 2 = π cu. units.

r 1 1 1 + 4 dx x x dx = ∞. x

Section 7.4 Mass, Moments, and Centre of Mass (page 417)

x 2 +y 2 =R 2

Z

∞

Z

0

=−

L

δ(s) ds =

Z

L

sin

0

πs ds L

L L π s 2L cos = . π L 0 π

Since δ(s) is symmetric about s = L/2 (that is, δ((L/2) − s) = δ((L/2) + s)), the centre of mass is at the midpoint of the wire: s¯ = L/2.

2. A slice of the wire of width d x at x has volume d V = π(a + bx)2 d x. Therefore the mass of the whole wire is Z L m= δ0 π(a + bx)2 d x 0 Z L (a 2 + 2abx + b2 x 2 ) d x = δ0 π 0   1 = δ0 π a 2 L + abL 2 + b2 L 3 . 3 Its moment about x = 0 is Z L M x=0 = xδ0 π(a + bx)2 d x 0 Z L = δ0 π (a 2 x + 2abx 2 + b2 x 3 ) d x 0   1 1 2 2 2 = δ0 π a L + abL 3 + b2 L 4 . 2 3 4 Thus, the centre of mass is   1 2 2 2 1 δ0 π a L + abL 3 + b2 L 4 2 3 4   x¯ = 1 δ0 π a 2 L + abL 2 + b2 L 3 3   1 2 2 1 2 2 a + abL + b L L 2 3 4 = . 1 a 2 + abL + b2 L 2 3

285 Copyright © 2014 Pearson Canada Inc.

SECTION 7.4 (PAGE 417)

ADAMS and ESSEX: CALCULUS 8

3. The mass of the plate is m = σ0 × area = The moment about x = 0 is Z a p M x=0 = xσ0 a 2 − x 2 d x σ0 = 2

a2

π σ0 a 2 . 4

5. The mass of the plate is a2

Let u = − du = −2x d x

0

Z

3 3 π a and y¯ = a. Hence, the centre of mass 16 8 3 3 is located at ( π a, a). 16 8 Thus, x¯ =

√ u du

x2 m=2

0

σ0 a 3 4 4a M x=0 = = . By symmetry, Thus x¯ = m 3 π σ0 a 2 3π ¯ = x. y ¯  Thus the centre of mass of the plate is 4a 4a , . 3π 3π y

y=

4.

a

x

Fig. 7.4.3 √ A vertical strip has area d A = a 2 − x 2 d x. Therefore, the mass of the quarter-circular plate is Z a p (σ0 x) a 2 − x 2 d x Let u = a 2 − x 2 m= 0 du = −2x d x   a 2 Z a2 √ 1 1 1 2 = σ0 u du = σ0 u 3/2 = σ0 a 3 . 2 2 3 3 0 0 The moment about x = 0 is Z a p M x=0 = σ0 x 2 a 2 − x 2 d x 0

= σ0 a

4

a4

Z

π/2

2

Let x = a sin θ d x = a cos θ dθ

sin θ cos θ dθ π/2

σ0 sin2 2θ dθ 4 0 Z σ0 a 4 π/2 π σ0 a 4 . = (1 − cos 4θ ) dθ = 8 16 0 =

4

Let u = 4 − y du = −d y

(4 − u)u 1/2 du

 4 8 3/2 2 5/2 256k u − u = 15 . 3 5 0

M y=0 = 2

Z

4 0

Z

4

0



p ky 2 4 − y d y

Let u = 4 − y du = −d y

(16u 1/2 − 8u 3/2 + u 5/2 ) du

 4 4096k 32 3/2 16 5/2 2 7/2 u − u + u = 105 . 3 5 7 0

15 16 4096k · = . The centre of mass of the 105 256k 7 plate is (0, 16/7).

Thus y¯ =

y 4

√

x=

4−y x

density ky

2

−2

Fig. 7.4.5

6. A vertical strip at h has area d A = (2 − 23 h) dh. Thus, the mass of the plate is

2

0

Z



p ky 4 − y d y

By symmetry, M x=0 = 0, so x¯ = 0.

= 2k

dx x

0

0

= 2k

= 2k

a 2 −x 2

4

Z

= 2k

a 2 σ0 a 3 σ0 2 3/2 = u = . 2 3 3 0

√

Z

The moment about y = 0 is Z a 1 M y=0 = σ0 x(a 2 − x 2 ) d x 2 0  2 2  a 1 a x x 4 1 = σ0 − = a 4 σ0 . 2 2 4 0 8

   Z 3 h2 2 h− (5h) 2 − h dh = 10 dh 3 3 0 0  3  2 h 3 h = 10 = 15 kg. − 2 9 0

m=

Z

3

The moment about x = 0 is

 Z 3 h3 h2 − dh 3 0  3  3 h h 4 45 = 10 − = kg-m. 3 12 0 2

M x=0 = 10

286 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.4 (PAGE 417)

The moment about y = 0 is   Z 3  1 2 1 M y=0 = 10 2 − h h − h 2 dh 3 3 0 2  Z 3 2 1 h − h 2 + h 3 dh = 10 3 9 0  2  3 h 2h 3 h 4 15 = 10 − + = kg-m. 2 9 36 0 2     45 15 3 1 2 2 Thus, x¯ = = and y¯ = = . The centre 15 2 15 2 of mass is located at ( 32 , 12 ).

Since the mass is symmetric about the y-axis, and the plate is symmetric about both the x- and y-axis, therefore the centre of mass must be located at the centre of the square. y

√a 2

r

a y= √ −x 2 √a dr 2 x

Fig. 7.4.8

y

2 2 y=2− 3 x

9.

dh h

3

x

M x=0 =

Fig. 7.4.6

M y=0

7. The mass of the plate is m=

Z

m=

a

kx a d x =

0

By symmetry, y¯ = a/2. Z M x=0 =

ka 3 . 2

b

Z

a

Z

b

a

1 = 2

Z

  σ (x) g(x) − f (x) d x

  xσ (x) g(x) − f (x) d x b

  xσ (x) (g(x))2 − ( f (x))2 d x a   M x=0 M y=0 , . Centre of mass: m m y

a 0

y=g(x)

ka 4 . 3

kx 2 a d x =

ka 4

2 2a Thus x¯ = · = . The centre of mass of the 3  3 ka 3 2a a plate is , . 3 2

density ρ(x)

y= f (x) a

y

a

x

Fig. 7.4.9

density

kx

10. The slice of the brick shown in the figure has volume a

8.

b

d V = 50 d x. Thus, the mass of the brick is

x

Fig. 7.4.7   a A vertical strip has area d A = 2 √ − r dr . Thus, the 2 mass is √  Z a/ 2   a m=2 kr 2 √ − r dr 2 0  Z a/√2  k a = 4k √ r − r 2 dr = √ a 3 g. 2 3 2 0

m=

Z

0

20

20 kx50 d x = 25kx 2 = 10000k g. 0

The moment about x = 0, i.e., the yz-plane, is M x=0 = 50k

Z

0

20

x2 dx =

50 = (8000)k g-cm. 3

50 3 20 kx 0 3

287 Copyright © 2014 Pearson Canada Inc.

SECTION 7.4 (PAGE 417)

ADAMS and ESSEX: CALCULUS 8

y

50 (8000)k 40 Thus, x¯ = 3 = . Since the density is inde10000k 3 5 pendent of y and z, y¯ = and z¯ = 5. Hence, the centre 2 of mass is located on the 20 cm long central axis of the brick, two-thirds of the way from the least dense 10 × 5 face to the most dense such face.

y

x

y+2R

y

5

−R dx

x

x

20 −2R

10

z

Fig. 7.4.11

Fig. 7.4.10

12. A slice at height z has volume d V = π y 2 dz and density kz g/cm3 . Thus, the mass of the cone is Z

m=

b

kzπ y 2 dz

0

  z 2 dz z 1− b 0  2  b z 2z 3 z4 = π ka 2 − + 2 2 3b 4b 0 1 = π ka 2 b2 g. 12

= π ka 2

11. Choose axes through the centre of the ball as shown in the following figure. The mass of the ball is

m=

Z

R

−R

(y + 2R)π(R 2 − y 2 ) d y 

= 4π R R 2 y −

 R y3 3

= 8 π R 4 kg. 3 0

By symmetry, the centre of mass lies along the y-axis; we need only calculate y¯ .

M y=0 =

Z

R

−R

= 2π = 2π

2

Z

b

The moment about z = 0 is Mz=0 = π ka

2

Z

b

z

0

2



z 1− b

2

dz =

1 π ka 2 b3 g-cm. 30

2b . Hence, the centre of mass is on the axis 5 of the cone at height 2b/5 cm above the base. Thus, z¯ =

z b

2

y(y + 2R)π(R − y ) d y

Z

R

y 2 (R 2 − y 2 ) d y ! R 3 4 y 5 2y R = − π R5 . 3 5 0 15

dz

0

4π R 5 3 R · = . The centre of mass is 4 15 8π R 10 on the line through the centre of the ball perpendicular to the plane mentioned in the problem, at a distance R/10 from the centre of the ball on the side opposite to the plane.

z z
y=a 1− b y a

Thus y¯ =

Fig. 7.4.12

13. By symmetry, y¯ = 0.

288 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.4 (PAGE 417)

8 16a 2ρ0 a 5 · = . 15 πρ0 a 4  15  16a 8a The centre of mass is , 0, . 15 15

z

so x¯ =

a

z

14. Assume the cone has its base in the x y-plane and its √

vertex at height b on the z-axis. By symmetry, the centre of mass lies on the z-axis. A cylindrical shell of thickness d x and radius x about the z-axis has height z = b(1 − (x/a)). Since it’s density is constant kx, its mass is  x dm = 2π bkx 2 1 − d x. a Also its centre of mass is at half its height,

a 2 −z 2

−a a

y

a x

Fig. 7.4.13 y¯ shell =

A horizontal slice of the solid √ at height z with thickness dz is a half-disk of radius a 2 − z 2 with centre of mass √ 4 a2 − z 2 at x¯ = , by Exercise 3 above. Its mass is 3π dm = ρ0 z dz

Thus its moment about z = 0 is

π 2 (a − z 2 ), 2

Hence

and its moment about x = 0 is d M x=0

 x 2 d x. d Mz=0 = y¯shell dm = π bkx 2 1 − a

Mz=0

Mz=0 =

πρ0 2

πρ0 = 2 and z¯ = Finally,

15. y x 2 +y 2 =a 2 ds

θ dθ −a

s

xa

a

(a 2 z 2 − z 4 ) dz ! a z 5 πρ0 a 5 a2 z 3 − = , 3 5 0 15

Fig. 7.4.15

0

8 πρ0 a 5 8a · . = 4 15 πρ0 a 15

Z 2ρ0 a z(a 2 − z 2 )3/2 dz 3 0 Z 2 ρ0 a 3/2 = u du 3 0   a 2 ρ0 2 5/2 2ρ0 a 5 = u = , 3 5 15 0

M x=0 =

Z

and z¯ = Mz=0 /m = b/5. The centre of mass is on the axis of the cone at height b/5 cm above the base.

Also,

Z

a

 x π kba 3 2π bkx 2 1 − dx = a 6 0 Z a  2 2 a3 x π kb = π bkx 2 1 − dx = a 30 0

m=

√ 2 2 πρ0 2 2 4 a −z = dm x¯ = z(a − z ) 2 3π 2ρ0 2 2 3/2 z(a − z ) . = 3

Thus the mass of the solid is Z πρ0 a 2 m= (a z − z 3 ) dz 2 0   a πρ0 a 4 z 4 πρ0 a 2 z 2 = − . = 2 2 4 0 8

b x 1− . 2 a

Consider the area element which is the thin half-ring shown in the figure. We have dm = ks π s ds = kπ s 2 ds. kπ 3 a . 3 Regard this area element as itself composed of smaller elements at positions given by the angle θ as shown. Then Z π  d M y=0 = (s sin θ )s dθ ks ds

Thus, m = Let u = a 2 − z 2 du = −2z dz

0

M y=0

= 2ks 3 ds, Z a ka 4 = 2k s 3 ds = . 2 0

289 Copyright © 2014 Pearson Canada Inc.

SECTION 7.4 (PAGE 417)

ADAMS and ESSEX: CALCULUS 8

ka 4 3 3a · = . By symmetry, x¯ = 0. 2 kπ a 3 2π   3a Thus, the centre of mass of the plate is 0, . 2π Therefore, y¯ =

17.

Z

∞

2

Ce−kr (4πr 2 ) dr 0 Z ∞ √ 2 = 4π C r 2 e−kr dr Let u = k r √ 0 du = k dr Z ∞ 4π C 2 = 3/2 u 2 e−u du k 0

m=

2

d V = ue−u du 2 V = − 12 e−u ! Z 2 R 4π C −ue−u 1 R −u 2 = 3/2 lim + e du R→∞ k 2 0 2 0   Z 4π C 1 ∞ −u 2 = 3/2 0 + e du 2 0 k √  π 3/2 5.57C 4π C π =C ≈ 3/2 . = 3/2 4 k k k U =u dU = du

16. y

ds

L π

s

θ x

18.

Fig. 7.4.16 L . Let s measure the π distance along the wire from the point where it leaves the positive x-axis. Thus, the density at position s is πs  δ(s) = sin g/cm. The mass of the wire is L

The radius of the semicircle is

m=

Z

L

0

0

L

L πs sin2 ds π L

Let θ = π s/L dθ = π ds/L

 2 Z π L sin2 θ dθ π 0  π 2 L2  = L g-cm. = θ − sin θ cos θ 2π 2 2π 0 =

Since the wire and the density function are both symmetric about the y-axis, we have M x=0 = 0.  L Hence, the centre of mass is located at 0, . 4

∞

2

0

L 2L L π s πs = ds = − cos g. sin L π L 0 π

Z

Z

0

 2  2 =√ 0 + lim (e0 − e−R = √ . R→∞ πk πk

Since an arc element ds at position s is at height L πs L , the moment of the wire about y = sin θ = sin π π L y = 0 is M y=0 =

1 m

r Ce−kr (4πr 2 ) dr 0 Z ∞ 4π C 2 = r 3 e−kr dr Let u = kr 2 3/2 −3/2 Cπ k 0 du = 2kr dr Z ∞ 3/2 4k 1 = √ ue−u du π 2k 2 0 U =u d V = e−u du dU = du V = −e−u ! R Z R 2 −u −u =√ e du lim −ue + π k R→∞

r¯ =

Section 7.5

1.

A=

πr 2 Z4 r

Centroids

p x r2 − x2 dx

(page 422)

Let u = r 2 − x 2 du = −2x d x r 2 Z r2 3/2 1 u r3 = u 1/2 du = = 2 0 3 0 3 r3 4 4r x¯ = · = = y¯ by symmetry. 3 πr2 3π  4r 4r The centroid is , . 3π 3π M x=0 =

290 Copyright © 2014 Pearson Canada Inc.

0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.5 (PAGE 422)

√ 2−1 π Thus x¯ = √ , and y¯ = √ . The ln(1√+ 2) 8 ln(1 ! + 2) 2−1 π centroid is √ , √ . ln(1 + 2) 8 ln(1 + 2)

y

√

y=

r

r 2 −x 2

y

y= √

dx x

1 1+x 2

x

r

Fig. 7.5.1

2. By symmetry, x¯ = 0. A horizontal strip at y √ has mass √

dm = 2 9 − y d y and moment d M y=0 = 2y 9 − y d y about y = 0. Thus, m=2

Z

0

9p

and M y=0 = 2

Z

=4

Z

9 0 3 0

9   2 3/2 9 − y d y = −2 (9 − y) = 36 3 0 p y 9 − y dy

Fig. 7.5.3

Let u 2 = 9 − y 2u du = −d y

3 648 . (9u 2 − u 4 ) du = 4(3u 3 − 15 u 5 ) = 5 0

18 648 = . Hence, the centroid is at Thus, y¯ = 5 × 36 5   18 0, . 5

4. The area of the sector is A = 18 πr 2 . Its moment about x = 0 is

y

9

M x=0 =

y=9−x 2 dy

3

x

Fig. 7.5.2

1

π/4

=

Z

0

√

dx 1 + x2

M y=0

0

x2 dx +

Z

r

√ r/ 2

p x r2 − x2 dx

r3 8 8r √ × = √ . By symmetry, the πr 2 3 2 3 2π   √ π centroid must lie on the line y = x tan = x( 2 − 1). 8 √ 8r ( 2 − 1) Thus, y¯ = √ . 3 2π

Let x = tan θ d x = sec2 θ dθ

y

sec θ dθ

0

π/4 √ = ln | sec θ + tan θ | = ln(1 + 2) 0

M x=0

√ r/ 2

Thus, x¯ =

3. The area and moments of the region are Z

Z

r r3 1 r3 = √ − (r 2 − x 2 )3/2 √ = √ . 6 2 3 3 2 r/ 2

y

−3

A=

x

1

1 1 p √ x dx = √ = 1 + x 2 = 2 − 1 2 1+x 0 0 1 Z 1 1 dx 1 π −1 = = tan x = . 2 2 0 1+x 2 8 0 Z

√

y=

y=x

√r

r

r 2 −x 2

x

2

Fig. 7.5.4

291 Copyright © 2014 Pearson Canada Inc.

SECTION 7.5 (PAGE 422)

ADAMS and ESSEX: CALCULUS 8

y

5. By symmetry, x¯ = 0. We have A=2

Z

0

√ 3 p

Let x = 2 sin θ d x = 2 cos θ dθ

π/3

 √ cos2 θ dθ − 3 0 π/3 √ −2 3 = 4(θ + sin θ cos θ ) 0 √ ! √ π 3 4π √ =4 + −2 3= − 3 3 4 3 Z √3 p 2 1 4 − x2 − 1 dx =2× 2 0 Z √3   p = 5 − x2 − 2 4 − x2 dx

 Z =2 4

M y=0

y=b

 4 − x2 − 1 dx

0

dx

as shown in the figure. The area and centroid of T1 are given by 4×1 = 2, 2 0+3+4 7 x¯1 = = , 3 3

A1 =

√

4−x 2 −1

4×2 = 4, 2 0+2+4 x¯2 = = 2, 3

Z

0−2+0 2 =− . 3 3

It follows that 7 ×2= 3 1 = ×2= 3

M1,x=0 =

√ x 3

14 3 2 3

M2,x=0 = 2 × 4 = 8 8 2 M2,y=0 = − × 4 = − . 3 3

A = 2 + 4 = 6, 14 38 M x=0 = +8= , 3 3

M y=0 =

2 8 − = −2. 3 3

38 19 −2 1 = , and y¯ = = − . The centroid 3×6 9 6 3  19 1 ,− . of the quadrilateral is 9 3 Thus x¯ =

1 2 π ab.

The

y

(3,1)

  2  Z a x x2 b2 1 − d x = b2 1 − 2 dx a a −a 0  3  a x 2 = b2 x − 2 = ab2 . 3a 3 0 1 2

y¯2 =

Since areas and moments are additive, we have for the whole quadrilateral

Fig. 7.5.5

M y=0 =

0+1+0 1 = . 3 3

A2 =

M1,y=0

moment about y = 0 is

y¯1 =

The area and centroid of T2 are given by

y

6. By symmetry, x¯ = 0. The area is A =

ax

7. The quadrilateral consists of two triangles, T1 and T2 ,

√ √ 9 3 − 4π 3 9 3 − 4π · Thus y¯ = √ = √ . The 3 4π!− 3 3 4π − 3 3 √ 9 3 − 4π √ . centroid is 0, 4π − 3 3

√ − 3

x2 a2

Fig. 7.5.6

Z √ √ = 5 3− 3−2 4 − x2 dx 0 ! √ √ √ π 3 4π = 4 3−4 + =3 3− . 3 4 3

y=

1−

x

−a

√ 3p

1

q

a

2ab2 2 4b Thus, y¯ = × = . 3 π ab 3π

T1 T2

(2,−2)

Fig. 7.5.7

292 Copyright © 2014 Pearson Canada Inc.

4 x

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.5 (PAGE 422)

8. The region is the union of a half-disk and atriangle.  The

4 centroid of the half-disk is known to be at 1, and 3π   2 2 that of the triangle is at , − . The area of the semi3 3 π circle is and the triangle is 2. Hence, 2   π  3π + 8 2 M x=0 = = ; (1) + (2) 2 3 6     π  4 2 2 M y=0 = + (2) − =− . 2 3π 3 3 π Since the area of the whole region is + 2, then 2 3π + 8 4 x¯ = and y¯ = − . 3(π + 4) 3(π + 4)

10. By symmetry, x¯ = y¯ = 0. The volume is V = 32 πr 3 . A thin slice of the solid at height z will have volume d V = π y 2 dz = π(r 2 − z 2 ) dz. Thus, the moment about z = 0 is Z

r

zπ(r 2 − z 2 ) dz  2 2  r r z πr 4 z 4 =π = − . 2 4 0 4

Mz=0 =

0

πr 4 3 3r × = . Hence, the centroid is 4 2πr 3 8 on the axis of the hemisphere at distance 3r/8 from the base.

Thus, z¯ =

y

z

√

1−(x−1)2

y=

√

y= 1

2

x

y=x−2

r 2 −z 2 dz y

−2

r

Fig. 7.5.8 x

9. A circular strip of the surface between heights y and y + d y has area

Fig. 7.5.10

r dy = 2π x d y = 2πr d y. cos θ x The total surface area is Z r d y = 2πr 2 . S = 2πr d S = 2π x

1 2 3 πr h. (See the following figure.) The  disk-shaped slice with vertical width dz has z , and therefore has volume radius y = r 1 − h

11. The cone has volume V =

0

The moment about y = 0 is r Z r 2 M y=0 = 2πr y d y = πr (y ) = πr 3 . 0

0

πr 3 r = . By symmetry, the centroid of the 2πr 2 2 hemispherical surface is on the axis of symmetry of the hemisphere. It is halfway between the centre of the base circle and the vertex. Thus y¯ =

y

 z 2 r2 d V = πr 2 1 − dz = π 2 (h − z)2 dz. h h We have πr 2 h2

Z

h

Mz=0 =

πr 2 h2

Z

h

=

θ (x,y)

dS r

πr 2 = 2 h

y

θ x

Fig. 7.5.9

x

0

0



z(h − z)2 dz

Let u = h − z du = −dz

(h − u)u 2 du

 h hu 3 u 4 πr 2 h 2 − = . 3 4 0 12

πr 2 h 2 3 h · = . The centroid of the 12 πr 2 h 4 solid cone is on the axis of the cone, at a distance above the base equal to one quarter of the height of the cone. Therefore z¯ =

293 Copyright © 2014 Pearson Canada Inc.

SECTION 7.5 (PAGE 422)

ADAMS and ESSEX: CALCULUS 8

z h

Thus y¯ =

π π  π , and the centroid is , . 8 2 8 y

dz y=sin x

z z
y=r 1− h

π

π/2

y

x

r

Fig. 7.5.13 Fig. 7.5.11

14. The area of the region is

12. A band  at height  z with vertical width dz has radius z y =r 1− , and has actual (slant) width h

ds =

s



1+

dy dz

2

dz =

s

1+

r2 dz. h2

A=

s

 r2 z 1 + 2 dz. d A = 2πr 1 − h h Thus the area of the conical surface is s Z p r2 h  z 1− dz = πr r 2 + h 2 . A = 2πr 1 + 2 h 0 h

Thus, x¯ =

0

π/2 = 1. cos x d x = sin x 0

Z

π/2

x cos x d x

0

U=x d V = cos x d x dU = d x V = sin x π/2 Z π/2 π = x sin x − sin x d x = − 1. 2 0 0

π − 1. The moment about y = 0 is 2 1 2

M y=0 =

The moment about z = 0 is

Thus, y¯ =

π/2

Z

cos2 x d x

0

  π/2 π 1 1 = . x + sin 2x 4 2 8 0

=

s

  Z z r2 h Mz=0 = 2πr 1 + 2 z 1− dz h 0 h s   h p 1 r 2 z2 z 3 = 2πr 1 + 2 − = πr h r 2 + h2. 2 3h 0 3 h

π/2

The moment about x = 0 is M x=0 =

Its area is

Z

π . The centroid is 8 y

√ πr h r 2 + h 2 1 h × = . By √ 2 2 3 3 πr r + h symmetry, x¯ = y¯ = 0. Hence, the centroid is on the axis of the conical surface, at distance h/3 from the base. π By symmetry, x¯ = . The area and y-moment of the 2 region are given by

1



 π π − 1, . 2 8

y=cos x

Thus, z¯ =

13.

A=

Z

x

π 2

x

Fig. 7.5.14

π

0Z

sin x d x = 2 π

1 sin2 x d x 2 0 π 1 π = (x − sin x cos x) = . 4 4 0

M y=0 =

dx

πr . By symmetry, x¯ = y¯ . An 2 element of the arc between x and x + d x has length

15. The arc has length L =

294 Copyright © 2014 Pearson Canada Inc.

ds =

dx r dx r dx . = = √ sin θ y r2 − x2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.5 (PAGE 422)

Thus M x=0 = Hence x¯ =

r2

Z

0

r

xr d x √ r2 − x2

r p = −r r 2 − x 2 = r 2 . 0

2r 2 = , and the centroid is · πr π y



 2r 2r , . π π

x 2 +y 2 =r 2

r

ds

Therefore, x¯ = y¯ = 2/15, and the centroid is (2/15, 2/15).

19. The region in figure (c) is the union of a half-disk of area π/2 and centroid (0, 4/(3π )) (by Example 1) and a triangle of area 1 and centroid (0, −1/3). Therefore its area is (π/2) + 1 and its centroid is (x, ¯ y¯ ), where x¯ = 0 and     π +2 π 4 −1 1 y¯ = +1 = . 2 2 3π 3 3 Therefore, the centroid is (0, 2/[3(π + 2)]).

20. The region in figure (d) is the union of three half-disks,

r

one with area π/2 and centroid (0, 4/(3π )), and two with areas π/8 and centroids (−1/2, −2/(3π )) and (1/2, −2/(3π )). Therefore its area is 3π/4 and its centroid is (x¯ , y¯ ), where

θ x x+d x

r

x

Fig. 7.5.15

16. The solid S in question consists of a solid cone C with vertex at the origin, height 1, and top a circular disk of radius 2, and a solid cylinder D of radius 2 and height 1 sitting on top of the cone. These solids have volumes VC = 4π/3, V D = 4π , and VS = VC + V D = 16π/3. By symmetry, the centroid of the solid lies on its vertical axis of symmetry; let us continue to call this the yaxis. We need only determine y¯ S . Since D lies between y = 1 and y = 2, its centroid satisfies y¯ D = 3/2. Also, by Exercise 11, the centroid of the solid cone satisfies y¯C = 3/4. Thus C and D have moments about y = 0:      4π 3 3 MC,y=0 = = π, M D,y=0 = (4π ) = 6π. 3 4 2

3π (x) ¯ = 4 3π ( y¯ ) = 4

    π π −1 π 1 (0) + + =0 2 8 2 8 2       π 4 π −2 π −2 1 + + = . 2 3π 8 3π 8 3π 2

Therefore, the centroid is (0, 2/(3π )).

21. By symmetry the centroid is (1, −2). y

(1,1)

y=2x−x 2

x

(1,−2)

y=−2

Thus M S,y=0 = π + 6π = 7π , and z¯ S = 7π/(16π/3) = 21/16. The centroid of the solid S is on its vertical axis of symmetry at height 21/16 above the vertex of the conical part.

17. The region in figure (a) is the union of a rectangle of area 2 and centroid (1, 3/2) and a triangle of area 1 and centroid (2/3, 2/3). Therefore its area is 3 and its centroid is (x, ¯ y¯ ), where   8 2 = 3x¯ = 2(1) + 1 3 3     11 3 2 3 y¯ = 2 +1 = . 2 3 3 Therefore, the centroid is (8/9, 11/9).

18. The √ region in figure (b) is the union of a square of area

( 2)2 = 2 and centroid (0, 0) and a triangle of area 1/2 and centroid (2/3, 2/3). Therefore its area is 5/2 and its centroid is (x, ¯ y¯ ), where   5 1 2 1 x¯ = 2(0) + = . 2 2 3 3

Fig. 7.5.21

22. The line segment from (1, 0) to (0, 1) has centroid ( 12 , 12 ) √ and length 2. By Pappus’s Theorem, the surface area of revolution about x = 2 is   √ 1 √ A = 2π 2 − 2 = 3π 2 sq. units. 2 y 1 r¯ 1 2

1

2

3

x

Fig. 7.5.22

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SECTION 7.5 (PAGE 422)

ADAMS and ESSEX: CALCULUS 8

1 1 3, 3

and area 12 . By Pappus’s Theorem the volume of revolution about x = 2 is   1 1 5π V = × 2π 2 − = cu. units. 2 3 3

23. The triangle T has centroid




26. The region bounded by y = 0 and y = ln(sin x) between x = 0 and x = π/2 lies below the x-axis, so A=−

y

π/2

ln(sin x) d x ≈ 1.088793

0

−1 π/2 x ln(sin x) d x ≈ 0.30239 A 0 Z 2 −1 π/2  y¯ = ln(sin x) d x ≈ −0.93986. 2A 0

x¯ =

1

T

Z

x

1

27. The area and moments of the region are

x=2

24.

Z

Fig. 7.5.23 √ s 3 The altitude h of the triangle is . Its centroid is at 2 h s height = √ above the base side. Thus, by Pappus’s 3 2 3 Theorem, the volume of revolution is √ !   s 3s π s3 s V = 2π × = cu. units. √ 2 2 4 2 3 √ h s 3 The centroid of one side is = above the base. 2 4 Thus, the surface area of revolution is √ ! √ 3s S = 2 × 2π (s) = s 2 π 3 sq. units. 4

R −1 1 dx = lim = A= 3 2 R→∞ 2 (1 + x) 2(1 + x) 0 Z0 ∞ x dx M x=0 = Let u = x + 1 (1 + x)3 0 du = d x Z ∞ u−1 du = u3 1   R 1 1 1 1 = lim − + 2 = 1 − = R→∞ u 2u 2 2 1 R Z ∞ 1 dx −1 = 1. M y=0 = = lim 6 5 R→∞ 10(1 + x) 0 2 0 (1 + x) 10 Z

∞


The centroid is 1, 15 . y

1

y=

1 (x + 1)3

s h x s

Fig. 7.5.27 Fig. 7.5.24

25. For the purpose of evaluating the integrals in this problem and the next, the definite integral routine in the TI-85 calculator√was used. For the region bounded by y = 0 and y = x cos x between x = 0 and x = π/2, we have Z π/2 √ A= x cos x d x ≈ 0.704038 0 Z 1 π/2 3/2 x¯ = x cos x d x ≈ 0.71377 A 0 Z π/2 1 y¯ = x cos2 x d x ≈ 0.26053. 2A 0

28. The surface q given by Z ∞ area is S = 2π

e−x

−∞

2

2

1 + 4x 2 e−2x d x. Since

2

lim 1 + 4x 2 e−2x = 1, this expression must be bounded

x→±∞

2

for all x, that is, 1 ≤ 1 + Z4x 2 e−2x ≤ K 2 for some con∞ √ 2 stant K . Thus, S ≤ 2π K e−x d x = 2K π π . The −∞

integral converges and the surface area is finite. Since the 2 whole curve y = e−x lies above the x-axis, its centroid would have to satisfy y¯ > 0. However, Pappus’s Theorem would then imply that the surface of revolution would have infinite area: S = 2π y¯ × (length of curve) = ∞. The curve cannot, therefore, have any centroid.

296 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.6 (PAGE 429)

29. By analogy with the formulas for the region a ≤ x ≤ b,

Triangle L M N has area 1 + tan t, and the x-coordinate of its centroid is

f (x) ≤ y ≤ g(y), the region c ≤ y ≤ d, f (y) ≤ x ≤ g(y) will have centroid (M x=0 / A, M y=0 / A), where A= M x=0 M y=0

Z d c

1 = 2 Z =

Z d h c

d

c

x¯ L M N − sec t − sec t + (1 + tan t) sin t + sec t + (1 − tan t) sin t = 3 2 sin t − sec t = . 3

 g(y) − f (y) d y

2 i 2  dy g(y) − f (y)

  y g(y) − f (y) d y.

Triangle L N P has area 1 − tan t, and the x-coordinate of its centroid is − sec t + sec t + sec t + (1 − tan t) sin t 3 sec t + (1 − tan t) sin t = . 3

30. Let us take L to be the y-axis and suppose that a plane

x¯ L N P =

curve C lies between x = a and x = b where 0 < a < b. Thus, r¯ = x, ¯ the x-coordinate of the centroid of C. Let ds denote an arc length element of C at position x. This arc length element generates, on rotation about L, a circular band of surface area d S = 2π x ds, so the surface area of the surface of revolution is S = 2π

Z

Therefore, x¯ L M N P =

x=b x=a

31.

x ds = 2π M x=0 = 2π rs. ¯

=

y

= = =

L

1 t (π/4) − t

t 1 √ 2

√

which is positive provided 0 < t < π/4. Thus the beam will rotate counterclockwise until an edge is on top.

P

x

Section 7.6 (page 429)

2

(π/4) − t

1h (2 sin t − sec t)(1 + tan t) 6 i + (sec t + sin t − sin t tan t)(1 − tan t) i 1h 3 sin t − 2 sec t tan t + sin t tan2 t 6   2 sin2 t sin t 3− + 6 cos2 t cos2 t i sin t h 2 2 3 cos t + sin t − 2 6 cos2 t h i i sin t sin t h 2 cos2 t − 1 = cos(2t) 2 2 6 cos t 6 cos t

N

1. M Fig. 7.5.31 We need to find the x-coordinate x¯ L M N P of the centre of buoyancy, that is, of the centroid of quadrilateral L M N P. From various triangles in the figure we can determine the x-coordinates of the four points:

Other Physical Applications

a) The pressure at the bottom is p = 9, 800 × 6 N/m2 . The force on the bottom is 4 × p = 235, 200 N. b) The pressure at depth h metres is 9, 800h N/m2 . The force on a strip between depths h and h + dh on one wall of the tank is d F = 9, 800h × 2 dh = 19, 600 h dh N. Thus, the total force on one wall is

x L = − sec t, x P = sec t, x M = − sec t + (1 + tan t) sin t x N = sec t + (1 − tan t) sin t

F = 19, 600

Z

0

6

h dh = 19, 600 × 18 = 352, 800 N.

297 Copyright © 2014 Pearson Canada Inc.

SECTION 7.6 (PAGE 429)

ADAMS and ESSEX: CALCULUS 8

h

dh

6 m θ h h+dh

26

24

2 m 2 m

Fig. 7.6.3 √

Fig. 7.6.1

2. A vertical slice of water at position y with thickness d y is in contact with √ the botttom over an area 8 sec θ d y = 45 101 d y m2 , which is at depth 1 x = 10 y + 1 m. The force exerted on this area is then √ 1 y + 1) 45 101 d y. Hence, the total force d F = ρg( 10 exerted on the bottom is

4. The height of each triangular √ face is 2 3 m and the

height of the pyramid is 2 2 m. Let the angle between r 2 the triangular face and the base be θ , then sin θ = 3 1 and cos θ = √ . 3

 Z 20  4√ 1 101 ρg y + 1 dy 5 10 0  2  20 √ y 4 101 (1000)(9.8) + y = 5 20 0

√ 2 2

F=

2 θ

4

≈ 3.1516 × 106 N.

√ 2 3

4

Fig. 7.6.4 20 y

y

1

front view of

dy

y

√ 10−2 2

dy 3

θ

x

θ

Fig. 7.6.2 x

√ √ x= 2y+10−2 2

and h + dh has area 26 200 dh dA = = 200 × dh. cos θ 24 The force on this strip is d F = 9, 800 h d A ≈ 2.12 × 106 h dh N. Thus the total force on the dam is 24 0

h dh ≈ 6.12 × 108 N.

60◦

2 side view of one face

3. A strip along the slant wall of the dam between depths h

Z

√ d y sec θ = 3d y

10

y x= 10 +1

F = 2.12 × 106

one face

4

Fig. 7.6.4 A vertical slice of water with thickness d y at a distance y from the vertex of the pyramid exerts a force on the shaded strip shown in the front view,√which has area √ √ 2 3y d y m2 and which is at depth 2y + 10 − 2 2 m. Hence, the force exerted on the triangular face is Z 2 √ √ √ F = ρg ( 2y + 10 − 2 2)2 3y d y 0 √  √ √ 2 2 2 3 = 2 3(9800) y + (5 − 2)y 3 0

298 Copyright © 2014 Pearson Canada Inc.

≈ 6.1495 × 105 N.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.6 (PAGE 429)

5. The unbalanced force is F = 9, 800 × 5 = 9, 800 × 5

Z

20

h dh 6



 20 h 2 ≈ 8.92 × 106 N. 2 6

The work done to empty the pool is Z 3 W = ρg h A(h) dh 0  Z 1 Z 3 = ρg 160h dh + 240h − 80h 2 dh 0

1

1   3   80 = 9800 80h 2 + 120h 2 − h 3 3 1 0 = 3.3973 × 106 N·m.

5 m

20 8 h 20 m 3

1 A(h)

6 m

Fig. 7.6.5

6. The spring force is F(x) = kx, where x is the amount

of compression. The work done to compress the spring 3 cm is 3 3 9 1 kx d x = kx 2 = k. 100 N·cm = W = 2 2 0 0 Z

200 Hence, k = N/cm. The work necessary to compress 9 the spring a further 1 cm is Z

W =

3

4

kx d x =



200 9



4 700 1 2 = x N·cm. 2 3 9

7. A layer of water in the tank between depths h and h + dh has weight d F = ρg d V = 4ρg dh. The work done to raise the water in this layer to the top of the tank is d W = h d F = 4ρgh dh. Thus the total work done to pump all the water out over the top of the tank is W = 4ρg

Z

0

6

Fig. 7.6.8

9. A layer of water between depths y and y + d y

has volume d V = π(a 2 − y 2 ) d y and weight d F = 9, 800π(a 2 − y 2 ) d y N. The work done to raise this water to height h m above the top of the bowl is d W = (h + y) d F = 9, 800π(h + y)(a 2 − y 2 ) d y N·m. Thus the total work done to pump all the water in the bowl to that height is Z a W = 9, 800π (ha 2 + a 2 y − hy 2 − y 3 ) d y 0   a a2 y 2 hy 3 y 4 2 = 9, 800π ha y + − − 2 3 4 0  3  4 2a h a = 9, 800π + 3 4   8h 3a + 8h 3 3 = 2450π a a + N·m. = 9, 800π a 12 3

h dh = 4 × 9, 800 × 18 ≈ 7.056 × 105 N·m.

y

a

dy

8. The horizontal cross-sectional area of the pool at depth h is A(h) =



160, if 0 ≤ h ≤ 1; 240 − 80h, if 1 < h ≤ 3.

Fig. 7.6.9

299 Copyright © 2014 Pearson Canada Inc.

SECTION 7.6 (PAGE 429)

ADAMS and ESSEX: CALCULUS 8

10. When the water surface is y m above the centre of the piston face (−R ≤ y ≤ R), a horizontal strip on the piston face at height z m above the√centre of the piston 2 2 face, having height √ dz has width 2 (R − z ) and so its area is d A = 2 R 2 − z 2 dz. The strip is y − z m below the surface of the water, and the pressure at that depth is ρg(y − z) = 9,800(y − z) N/m2 . Thus, the√force of the water on that strip is d F = 19,600(y − z) R 2 − z 2 dz Newtons. The total force of the water on the piston is F=

Z

y

−R

Section 7.7 Applications in Business, Finance, and Ecology (page 433) 1.

Cost = $4, 000 + = $11, 000.

Z

0

6x 2 2x 6− 3 + 6 10 10

1,000 



dx

2. The number of chips sold in the first year was p 19,600(y − z) R 2 − z 2 dz

= 19,600

Z

sin−1 (y/R)

(let z = R sin θ ) 1, 000 2

 sin−1 (y/R) R2 y R3 (θ + sin θ cos θ ) + cos3 θ = 19,600 2 3 −π/2  2 3/2  2 R πyR 1 2 −1 y = 19,600 y sin + + R − y2 N. 2 R 4 3 

W = 2ρg X

Z

= 9,800X

R 0

Z

0

q y R2 − y 2 d y

R2

let u = R 2 − y 2

√ 19,600 u du = X R 3 N·m. 3

Since no work is lost to friction, the work to push the piston in to distance X/2 is equal to this work needed to raise the water.

12. Let the time required to raise the bucket to height h m be t minutes. Given that the velocity is 2 m/min, then h t = . The weight of the bucket at time t is 2 h 16 kg − (1 kg/min)(t min) = 16 − kg. Therefore, 2 the work done required to move the bucket to a height of 10 m is  Z 10  h W =g 16 − dh 2 0   10 h 2 = 9.8 16h − = 1323 N·m. 4 0

te−t/10 dt = 100, 000 − 620, 000e−26/5

that is, about 96,580.

3. The monthly charge is x

Z

0

4 √ dt 1+ t

√

let t = u 2

 Z √x  u 1 =8 du = 8 du 1− 1+u 1+u 0 0√  √ =$8 x − ln(1 + x) . Z

11. Initially the water occupies the bottom half of a cylinder of length X. Half of this water (say, a bottom halfcylinder of length X/2, must be moved to fill the top half of a cylinder of length X/2. By symmetry, we can accomplish this by moving a thin horizontal slice of water of thickness d y at distance y below the central axis of the cylinder to height y above that axis, that p X is, we move a volume d V = 2 R 2 − y 2 d y of wa2 ter up a distance p 2y. The work required to do this is d W = 2ρg X y R 2 − y 2 d y N·m. Thus the work to raise the half-cylinder of water is

52

0

2

(y − R sin θ ) R cos θ dθ

−π/2

Z

x

4. The price per kg at time t (years) is $10 + 5t. Thus the revenue per year at time t is 400(10 + 5t)/(1 + 0.1t) $/year. The total revenue over the year is Z

0

1

400(10 + 5t) dt ≈ $4, 750.37. 1 + 0.1t

5. The present value of continuous payments of $1,000 per year for 10 years at a discount rate of 2% is V =

Z

10 0

1,000e−0.02t dt =

10 1,000 −0.02t e = $9,063.46. −0.02 0

6. The present value of continuous payments of $1,000 per year for 10 years at a discount rate of 5% is V =

Z

10 0

1,000e−0.05t dt =

10 1,000 −0.05t e = $7,869.39. −0.05 0

7. The present value of continuous payments of $1,000 per year for 10 years beginning 2 years from now at a discount rate of 8% is V =

Z

12 2

1,000e−0.08t dt =

300 Copyright © 2014 Pearson Canada Inc.

12 1,000 −0.08t e = $5,865.64. −0.08 2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.7 (PAGE 433)

8. The present value of continuous payments of $1,000 per year for 25 years beginning 10 years from now at a discount rate of 5% is V =

Z

35

1,000e−0.05t dt =

10

35 1,000 −0.05t e = $8,655.13. −0.05 10

9. The present value of continuous payments of $1,000 per year for all future time at a discount rate of 2% is V =

Z

∞ 0

1,000e−0.02t dt =

1, 000 = $50, 000. −0.02

10. The present value of continuous payments of $1,000 per year beginning 10 years from now and continuing for all future time at a discount rate of 5% is Z ∞ 1,000 −0.5 V = 1,000e−0.05t dt = e = $12,130.61. −0.05 10

11. After t years, money is flowing at $(1,000 + 100t) per year. The present value of 10 years of payments discounted at 5% is V = 100

10

Z

0

(10 + t)e−0.05t dt

d V = e−0.05t dt e−0.05t V = −0.05 10 Z 10 −0.05t e + 100 = 100(10 + t) e−0.05t dt −0.05 0 0.05 0 10 100 −0.05t = −4261.23 + e = $11, 477.54. −(0.05)2 0 U = 10 + t dU = dt

12. After t years, money is flowing at $1,000(1.1)t per year. The present value of 10 years of payments discounted at 5% is V = 1,000

Z

10

et ln(1.1) e−0.05t dt

0

13. The amount after 10 years is A = 5, 000

0

10

reach $1,000,000. The $5, 000(1.1)t dt deposited in the time interval [t, t + dt] grows for T − t years, so the balance after T years is Z

T

5, 000(1.1)t (1.06)T −t dt = 1, 000, 000  Z T 1, 000, 000 1.1 t T dt = (1.06) = 200 1.06 5, 000 0 # "  (1.06)T 1.1 T − 1 = 200 ln(1.1/1.06) 1.06 0

(1.1)T − (1.06)T = 200 ln

1.1 . 1.06

This equation can be solved by Newton’s method or using a calculator “solve” routine. The solution is T ≈ 26.05 years.

15. Let P(τ ) be the value at time τ < t that will grow to $P = P(t) at time t. If the discount rate at time τ is δ(τ ), then d P(τ ) = δ(τ )P(τ ), dτ or, equivalently, d P(τ ) = δ(τ ) dτ. P(τ ) Integrating this from 0 to t, we get Z t ln P(t) − ln P(0) = δ(τ ) dτ = λ(t), 0

and, taking exponentials of both sides and solving for P(0), we get P(0) = P(t)e−λ(t) = Pe−λ(t) . The present value of a stream of payments due at a rate P(t) at time t from t = 0 to t = T is Z T Z t −λ(t) P(t)e dt, where λ(t) = δ(τ ) dτ. 0

0

16. The analysis carried out in the text for the logistic growth

0

10 1,000 t (ln(1.1)−0.05 = e = $12, 650.23. ln(1.1) − 0.05

Z

14. Let T be the time required for the account balance to

10 5,000 0.05t e0.05t dt = e = $64,872.13. 0.05 0

model showed that the total present value of future harvests could be maximized by holding the population size x at a value that maximizes the quadratic expression  x − δx. Q(x) = kx 1 − L

If the logistic model d x/dt = kx(1 − (x/L)) is replaced with a more general growth model d x/dt = F(x), exactly the same analysis leads us to maximize Q(x) = F(x) − δx.

301 Copyright © 2014 Pearson Canada Inc.

SECTION 7.7 (PAGE 433)

ADAMS and ESSEX: CALCULUS 8

For realistic growth functions, the maximum will occur where Q ′ (x) = 0, that is, where F ′ (x) = δ.

A reasonable cost function C(x) will increase as x decreases (the whales are harder to find), and will exceed p if x ≤ x0 , for some positive population level x0 . The value of x that maximizes Q(x) must exceed x0 , so the model no longer predicts extinction, even for large discount rates δ. However, the optimizing population x may be so low that other factors not accounted for in the simple logistic growth model may still bring about extinction whether it is economically indicated or not.

17. We are given L = 80, 000, k = 0.12, and δ = 0.05.

According to the analysis in the text, the present value of future harvests will be maximized if the population level is maintained at x = (k − δ)

L 0.07 = (80, 000) = 23, 333.33 2k 0.24

The annual revenue from harvesting to keep the population at this level (given a price of $6 per fish) is   23, 333.33 = $11, 900. 6(0.12)(23, 333.33) 1 − 80, 000

Section 7.8

(page 445)

1. The expected winnings on a toss of the coin are $1 × 0.49 + $2 × 0.49 + $50 × 0.02 = $2.47.

18. We are given that k = 0.02, L = 150, 000, p = $10, 000. The growth rate at population level x is

2.   dx x = 0.02x 1 − . dt 150, 000

Probability

If you pay this much to play one game, in the long term you can expect to break even. P (a) We need 6n=1 K n = 1. Thus 21K = 1, and K = 1/21. (b) Pr(X ≤ 3) = (1/21)(1 + 2 + 3) = 2/7.

3. From the second previous Exercise, the mean winings is a) The maximum sustainable annual harvest is d x = 0.02(75, 000)(0.5) = 750 whales. dt x=L/2

b) The resulting annual revenue is $750 p = $7, 500, 000.

c) If the whole population of 75,000 is harvested and the proceeds invested at 2%, the annual interest will be 75, 000($10, 000)(0.02) = $15, 000, 000. d) At 5%, the interest would be (5/2)($15, 000) = $37, 500, 000. e) The total present value of all future harvesting revenue if the population level is maintained at 75,000 and δ = 0.05 is Z ∞ 7, 500, 000 e−0.05t 7, 500, 000 dt = = $150, 000, 000. 0.05 0

19. If we assume that the cost of harvesting 1 unit of population is $C(x) when the population size is x, then the effective income from 1 unit harvested is $( p − C(x)). Using this expression in place of the constant p in the analysis given in the text, we are led to choose x to maximize  h  i x Q(x) = p − C(x) kx 1 − − δx . L

µ = $2.47. Now σ 2 = 1 × 0.49 + 4 × 0.49 + 2,500 × 0.02 − µ2 ≈ 52.45 − 6.10 = 46.35. The standard deviation is thus σ ≈ $6.81.

4. Since Pr(X = n) = n/21, we have µ= σ2 =

6 X n=1

6 X n=1

nPr(X = n) =

1 × 1 + 2 × 2 +··· + 6 × 6 13 = ≈ 4.33 21 3

n 2 Pr(X = n) − µ2 =

12 + 23 + · · · + 63 − µ2 21

169 20 = 21 − = ≈ 2.22 9 9 √ 20 σ = ≈ 1.49. 3

5. The mean of X is µ=1×

9 1 + (2 + 3 + 4 + 5) × + 6 × 1160 ≈ 3.5833. 60 6

The expectation of X 2 is E(X 2 ) = 12 ×

9 1 +(22 +32 +42 +52 )× +62 ×1160 ≈ 15.7500. 60 6

Hence the standard deviation of X is √ 15.75 − 3.58332 ≈ 1.7059. 9 2 29 Also Pr(X ≤ 3) = + = ≈ 0.4833. 60 6 60

302 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.8 (PAGE 445)

6. (a) Calculating as we did to construct the probability function in Example 2, but using the different values for the probabilities of “1” and “6”, we obtain 9 9 × ≈ 0.0225 60 60 9 × 16 = 0.0500 f (3) = 2 × 60 9 1 f (4) = 2 × × 16 + = 0.0778 60 36 9 2 f (5) = 2 × × 16 + = 0.1056 60 36 9 3 f (6) = 2 × × 16 + = 0.1333 60 36 4 9 × 1160 + = 0.1661 f (7) = 2 × 60 36 3 11 × 16 + = 0.1444 f (8) = 2 × 60 36 11 2 f (9) = 2 × × 16 + = 0.1167 60 36 11 1 f (10) = 2 × × 16 + = 0.0889 60 36 11 f (11) = 2 × × 16 = 0.0611 60 11 × 1160 = 0.0336. f (12) = 60 f (2) =

(b) Multiplying each value f (n) by n and summing, we get 12 X µ= n f (n) ≈ 7.1665. n=2

Similarly, 2

E(X ) =

12 X n=2

2

n f (n) ≈ 57.1783,

so the standard deviation of X is σ=

q

E(X 2 ) − µ2 ≈ 2.4124.

The mean is somewhat larger than the value (7) obtained for the unweighted dice, because the weighting favours more 6s than 1s showing if the roll is repeated many times. The standard deviation is just a tiny bit smaller than that found for the unweighted dice (2.4152); the distribution of probability is just slightly more concentrated around the mean here.

7.

(a) The sample space consists of the eight triples (H, H, H ), (H, H, T ), (H, T, H ), (T, H, H ), (H, T, T ), (T, H, T ), (T, T, H ), and (T, T, T ).

(b) We have Pr(H, H, H ) = (0.55)3 = 0.166375

Pr(H, H, T ) = Pr(H, T, H ) = Pr(T, H, H ) = (0.55)2 (0.45) = 0.136125 Pr(H, T, T ) = Pr(T, H, T ) = Pr(T, T, H ) = (0.55)(0.45)2 = 0.111375 Pr(T, T, T ) = (0.45)3 = 0.091125.

(c) The probability function f for X is given by f (0) = (0.45)3 = 0.911125

f (1) = 3 × (0.55)(0.45)2 = 0.334125

f (2) = 3 × (0.55)2 (0.45) = 0.408375 f (3) = (0.55)3 = 0.166375.

(d) Pr(X ≥ 1) = 1 − Pr(X = 0) = 0.908875. (e) E(X) = 0× f (0)+1× f (1)+2× f (2)+3× f (3) = 1.6500.

8. The number of red balls in the sack must be 0.6 × 20 = 12. Thus there are 8 blue balls. (a) The probability of pulling out one blue ball is 8/20. If you got a blue ball, then there would be only 7 blue balls left among the 19 balls remaining in the sack, so the probability of pulling out a second blue ball is 7/19. Thus the probability of pulling out two 8 7 14 blue balls is × = . 20 19 95 (b) The sample space for the three ball selection consists of all eight triples of the form (x, y, z), where each of x, y, z is either R(ed) or B(lue). Let X be the number of red balls among the three balls pulled out. Arguing in the same way as in (a), we calculate Pr(X = 0) = Pr(B, B, B) =

7 6 14 8 × × = 20 19 18 285

≈ 0.0491 Pr(X = 1) = Pr(R, B, B) + Pr(B, R, B) + Pr(B, B, R) 12 8 7 28 =3× × × = ≈ 0.2947 20 19 18 95 Pr(X = 2) = Pr(R, R, B) + Pr(R, B, R) + Pr(B, R, R) 11 8 44 12 =3× × × = ≈ 0.4632 20 19 18 95 12 11 10 11 Pr(X = 3) = Pr(R, R, R) = × × = 20 19 18 57 ≈ 0.1930 Thus the expected value of X is E(X) = 0 × =

14 28 44 11 +1× +2× +3× 285 95 95 57

9 = 1.8. 5

9. We have f (x) = C x on [0, 3].

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SECTION 7.8 (PAGE 445)

ADAMS and ESSEX: CALCULUS 8

a) C is given by

c) We have

1=

Z

3 C 9 C x d x = x 2 = C. 2 2 0

3 0

2 Hence, C = . 9 b) The mean is

Since

variance is

3

0

3

Z

2 = 9

Z

2 9

0

x2 dx =

3 2 3 x = 2. 27 0

3 9 2 4 x dx = x = , the 36 0 2 3

σ 2 = E(X 2 ) − µ2 =

9 1 −4 = , 2 2

and the standard deviation is σ = 1/ 2. c) We have Pr(µ − σ ≤ X ≤ µ + σ ) =

2 9

Z

µ+σ

x dx

µ−σ

(µ + σ )2 − (µ − σ )2 4µσ = ≈ 0.6285. 9 9

10. We have f (x) = C x on [1, 2]. a) To find C, we have Z

2

Cx dx =

1

2 Hence, C = . 3 b) The mean is Z

2

2 = 3

Z

µ = E(X) =

Since

E(X 2 )

variance is

µ+σ

x dx

µ−σ

a) C is given by

1=

1

Z

0

Hence, C = 3.

1 C C 3 Cx dx = x = . 3 3 0 2

b) The mean, variance, and standard deviation are Z

1

3 x3 dx = 4 0 Z 1 9 3 9 3 σ 2 = E(X 2 ) − µ2 = 3 x4 dx − = − = 16 5 16 80 0 p σ = 3/80. c) We have Pr(µ − σ ≤ X ≤ µ + σ ) = 3

1=

Z

(µ + σ )2 − (µ − σ )2 4µσ = ≈ 0.5875. 3 3

µ = E(X) = 3

√

=

=

2 3

11. We have f (x) = C x 2 on [0, 1].

µ = E(X) = E(X 2 )

Pr(µ − σ ≤ X ≤ µ + σ ) =

2 3

1

2 3 C 2 x = C. 2 2 1

x2 dx = 2

1

2 14 2 3 x = ≈ 1.556. 9 1 9

2 1 4 5 x d x = x = , the 6 1 2 3

5 196 13 σ 2 = E(X 2 ) − µ2 = − = 2 81 162

r

µ+σ

x2 dx

µ−σ

= (µ + σ )3 − (µ − σ )3 r !3 r !3 3 3 3 3 = + − − ≈ 0.668. 4 80 4 80

12. We have f (x) = C sin x on [0, π ]. a) To find C, we calculate 1=

Z

Hence, C =

π 0

1 . 2

π C sin x d x = −C cos x = 2C. 0

b) The mean is µ = E(X) =

and the standard deviation is σ =

Z

13 ≈ 0.283. 162

304 Copyright © 2014 Pearson Canada Inc.

1 2

Z

π

x sin x d x

0

U =x d V = sin x d x dU = d x V = − cos x π Z π   1 = −x cos x + cos x d x 2 0 0 π = = 1.571. 2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.8 (PAGE 445)

Since

c) We have

E(X 2 ) =

1 2

Z

π

Z

x 2 sin x d x

0

U = x2 d V = sin x d x dU = 2x d x V = − cos x π   Z π 1 2 x cos x d x −x cos x + 2 = 2 0 0 U=x d V = cos x d x dU = d x V = sin x π Z π    1 2 sin x d x π + 2 x sin x − = 2

Let u = x − 12 du = d x    Z σ 1 σ σ3 = 12 − u 2 du = 12 − 4 4 3 0   12 1 1 = √ − ≈ 0.626. 20 4 60

0

0

1 = (π 2 − 4). 2

14. It was shown in Section 6.1 (p. 349) that

Hence, the variance is

Z

π2 − 4 π2 π2 − 8 σ 2 = E(X 2 )− µ2 = − = ≈ 0.467 2 4 4 and the standard deviation is

σ =

s

If In =

π2 − 8 ≈ 0.684. 4

n −x

x e

Z

∞

=−

1 2

Z

d x = −x e

x n e−x d x, then

Z

∞ 0

Let u = kx; then

µ+σ

sin x d x

µ−σ

i cos(µ + σ ) − cos(µ − σ )

Z

∞

0

x n e−kx d x =

1=

0

C(x − x 2 ) d x = C

Hence, C = 6.

k n+1

1

u n e−u du =

0

1 n! In = n+1 . k n+1 k

a) To find C, observe that 1=C

x2 x3 − 2 3

 1 = C. 6 0

1 µ = E(X) = 6 (x 2 − x 3 ) d x = 2 0 Z 1 1 σ 2 = E(X 2 ) − µ2 = 6 (x 3 − x 4 ) d x − 4 0 3 1 1 = − = 10 4 20 p σ = 1/20.

Z

∞

0

xe−kx d x =

C . k2

Hence, C = k 2 . b) The mean is µ = E(X) = k 2

Z

∞

E(X 2 )

Z

∞

b) The mean, variance, and standard deviation are Z

∞

Z

1

Now let f (x) = C xe−kx on [0, ∞).

a) C is given by 

if n ≥ 1.

e−x d x = 1, therefore In = n! for n ≥ 1.

13. We have f (x) = C(x − x 2 ) on [0, 1].

1

x n−1 e−x d x.

R→∞

2 = sin µ sin σ = sin σ ≈ 0.632.

Z

+n

Z

0

Since I0 =

Pr(µ − σ ≤ X ≤ µ + σ ) =

n −x

In = lim −R n e−R + n In−1 = n In−1

c) Then

1h

(1/2)+σ

Pr(µ − σ ≤ X ≤ µ + σ ) = 6 (x − x 2 ) d x (1/2)−σ   # Z (1/2)+σ " 1 1 2 =6 − x− dx 4 2 (1/2)−σ

Since

=

k2

0

then the variance is

x 2 e−kx d x = k 2 3 −kx

x e

0

σ 2 = E(X 2 ) − µ2 =

dx = k

2



2 k3



6 k4



=

2 . k



=

6 , k2

6 4 2 − 2 = 2 2 k k k

and the standard deviation is σ =

√

2 . k

305 Copyright © 2014 Pearson Canada Inc.

SECTION 7.8 (PAGE 445)

ADAMS and ESSEX: CALCULUS 8

c) Finally, Pr(µ − σ ≤ X ≤ µ + σ ) Z µ+σ = k2 xe−kx d x Let u = kx µ−σ du = k d x Z k(µ+σ ) −u = ue du

16. No. The identity constant C.

k(µ−σ )

k(µ+σ ) Z k(µ+σ ) + e−u du = −ue−u k(µ−σ ) k(µ−σ ) √ √ √ √ = −(2 + 2)e−(2+ 2) + (2 − 2)e−(2− 2) √ −(2+ 2)

−e ≈ 0.738.

15.

+e

17.

Z

0

∞

2

e−x d x =

C 2

√

Z

∞

2

−∞

e−x d x =

Thus C = 2/ π.

U=x dU = d x =−

2 1 +√ π π

−∞

C d x = 1 is not satisfied for any

1 2 2 √ e−(x−µ) /2σ σ 2π Z ∞ 1 2 2 xe−(x−µ) /2σ d x mean = √ σ 2π −∞

f µ,σ (x) =

Let z =

x −µ σ

1 dx σ

Z ∞ 1 2 = √ (µ + σ z)e−z /2 dz 2π −∞ Z ∞ µ 2 = √ e−z /2 dz = µ 2π −∞   variance = E (x − µ)2 Z ∞ 1 2 2 (x − µ)2 e−(x−µ) /2σ d x = √ σ 2π −∞ Z ∞ 1 2 = √ σ 2 z 2 e−z /2 dz = σ Var(Z) = σ σ 2π −∞

√ C π . 2

b) The mean, variance, and standard deviation are Z ∞ 2 ∞ 1 e−x 2 −x 2 xe dx = − √ = √ µ= √ π 0 π 0 π Z ∞ 1 2 2 σ2 = − + √ x 2 e−x d x π π 0

∞

dz =

√ −(2− 2)

a) We have 1=C

Z

18. Since f (x) = 2 π

2

d V = xe−x d x 2 V = − 21 e−x ! ∞ Z 1 ∞ −x 2 x −x 2 e dx − e + 2 2 0 0 √   1 1 1 π 0+ · = − 2 2 2 π

Z

∞

0

2 > 0 on [0, ∞) and π(1 + x 2 )

2 π  dx 2 −1 tan (R) = = 1, = lim R→∞ π 1 + x2 π 2

therefore f (x) is a probability density function on [0, ∞). The expectation of X is

1 2 =− +√ π π r 1 1 σ= − ≈ 0.426. 2 π

µ = E(X) =

2 π

Z

0

∞

x dx 1 + x2

1 = lim ln(1 + R 2 ) = ∞. R→∞ π

c) We have Z µ+σ 2 2 Pr(µ − σ ≤ X ≤ µ + σ ) = √ e−x d x π µ−σ √ Let x = z/ 2 √ d x = dz/ 2 r Z √ 2(µ+σ ) 2 2 = e−z /2 dz. √ π 2(µ−σ ) √ √ But 2(µ − σ ) ≈ 0.195 and 2(µ + σ ) ≈ 1.40. Thus, if Z is a standard normal random variable, we obtain by interpolation in the table on page 386 in the text, Pr(µ − σ ≤ X ≤ µ + σ ) = 2Pr(0.195 ≤ Z ≤ 1.400) ≈ 2(0.919 − 0.577) ≈ 0.68.

No matter what the cost per game, you should be willing to play (if you have an adequate bankroll). Your expected winnings per game in the long term is infinite.

19.

a) The density function for the uniform distribution on [a, b] is given by f (x) = 1/(b − a), for a ≤ x ≤ b. By Example 5, the mean and standard deviation are given by µ=

b+a , 2

σ =

b−a √ . 2 3

b+a b−a + √ > b, and similarly, 2 3 µ − 2σ < a, therefore Pr(|X − µ| ≥ 2σ ) = 0. Since µ + 2σ =

306 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 7.9 (PAGE 453)

b) For f (x) = ke−kx on [0, ∞), we know that 1 (Example 6). Thus µ − 2σ < 0 and µ = σ = k 3 µ + 2σ = . We have k   3 Pr(|X − µ| ≥ 2σ ) = Pr X ≥ k Z ∞ =k e−kx d x 3/k ∞ = e−3 ≈ 0.050. = −e−kx

from the table in this section.

22. If X is the random variable giving the spinner’s value, then Pr(X = 1/4) = 1/2 and the density function for the other values of X is f (x) = 1/2. Thus the mean of X is µ = E(X) = Also,  Z 1  1 1 1 1 19 + Pr X = x 2 f (x) d x = + = 16 4 32 6 96 0 9 11 19 − = . σ 2 = E(X 2 ) − µ2 = 96 64 192

E(X 2 ) =

3/k

1 2 2 √ e(x−µ) /2σ , which has mean µ σ 2π and standard deviation σ , we have

c) For f µ,σ (x) =

Pr(|X − µ| ≥ 2σ ) = 2Pr(X ≤ µ − 2σ ) Z µ−2σ 1 2 2 =2 √ e−(x−µ) /2σ d x σ 2π −∞ x −µ Let z = σ 1 dz = d x σ Z −2 2 −z 2 = √ e dz 2π −∞ = 2Pr(Z ≤ −2) ≈ 2 × 0.023 = 0.046

Thus σ =

1.

1 1 = (see Example 6). Then µ 20 Z ∞ Z 12 1 1 Pr(T ≥ 12) = e−t/20 dt = 1 − e−t/20 dt 20 12 20 0 12 −12/20 −t/20 ≈ 0.549. = 1+e =e

where k =

0

21. If X is distributed normally, with mean µ = 5, 000, and

⇒

3.

standard deviation σ = 200, then

Pr(X ≥ 5500) Z ∞ 1 2 2 = e−(x−5000) /(2×200 ) d x √ 200 2π 5500 x − 5000 Let z = 200 dx dz = 200 Z ∞ 1 −z 2 /2 = √ e dz 2π 5/2 = Pr(Z ≥ 5/2) = Pr(Z ≤ −5/2) ≈ 0.006

11/192.

y = 0 is a constant solution. Otherwise y dy = dx 2x dy dx 2 = y x 2 ln y = ln x + C1 ⇒ y2 = C x

20. The density function for T is f (t) = ke−kt on [0, ∞),

The probability that the system will last at least 12 hours is about 0.549.

√

Section 7.9 First-Order Differential Equations (page 453)

2.

from the table in this section.

  Z 1 1 1 3 1 1 Pr X = + x f (x) d x = + = . 4 4 8 4 8 0

4.

y = 1/3 is a constant solution. Otherwise 3y − 1 dy = dZ x x Z dy dx = 3y − 1 x 1 1 ln |3y − 1| = ln |x| + ln C 3 3 3y − 1 =C x3 1 y = (1 + C x 3 ). 3

dy x2 = 2 ⇒ dx y x3 y3 = + C1 , 3 3

y2 d y = x 2 d x or

x 3 − y3 = C

y = 0 is a constant solution. Otherwise, dy = x 2 y2 Z dx Z dy = x2 dx y2 1 1 1 − = x3 + C y 3 3 3 ⇒ y=− 3 . x +C

307 Copyright © 2014 Pearson Canada Inc.

SECTION 7.9 (PAGE 453)

5.

6.

ADAMS and ESSEX: CALCULUS 8

Y = 0 is a constant solution. Otherwise dY dY = tY ⇒ = t dt dt Y 2 t 2 ln Y = + C1 , or Y = Cet /2 2 dx = e x sin t dt Z Z e−x d x =

Hence,  Z  1 1 1 dy = + + dy y 2 (1 − y) y y2 1−y 1 = ln |y| − − ln |1 − y|. y

Z

Therefore,

sin t dt

−e−x = − cos t − C ⇒ x = − ln(cos t + C).

7.

11.

dy d x Z dy 1 + y2 tan−1 y ⇒ y

12. We have

1+y = Ce2x 1−y

8.

9.

or

y=

Ce2x − 1 Ce2x + 1

(linear) Z  2 1 µ = exp − dx = 2 x x 1 dy 2 − 3 y=1 x2 dx x d y =1 dx x2 y = x + C, so y = x 3 + C x 2 x2

d 2 dy (x y) = x 2 + 2x y dx  d x   2y 1 dy + = x2 =1 = x2 dx x x2 Z ⇒ x2y = dx = x + C

=x +C = tan(x + C).

dy dy = 2 + ey ⇒ = dt dt 2 + ey Z Z e−y d y = dt 2e−y + 1 1 − ln(2e−y + 1) = t + C1 2 or

y 1 − = x + K. 1− y y

dy 2y 1 + = . Let 2 d x x x Z 2 d x = 2 ln x = ln x 2 , then eµ = x 2 , and µ= x

= 1 + y2 Z = dx

2e−y + 1 = C2 e−2t ,

10.

dy 2 − y = x2 dx x

y = 1 and y = −1 are constant solutions, Otherwise dy dy = 1 − y2 ⇒ = dx dx 1 − y2   1 1 1 + dy = dx 2 1+ y 1− y 1 1 + y ln = x + C1 2 1 − y

ln

⇒

  1 y = − ln Ce−2t − 2

y = 0 and y = 1 are constant solutions. For the other solutions we have dy = y 2 (1 − y) d x Z Z dy = dx = x + K. y 2 (1 − y) Expand the left side in partial fractions:

13.

1 C + 2. x x

 Z dy 2 d x = e2x + 2y = 3 µ = exp dx d 2x (e y) = e2x (y ′ + 2y) = 3e2x dx 3 3 e2x y = e2x + C ⇒ y = + Ce−2x 2 2

14. We have and

1 A B C = + 2+ 2 y (1 − y) y y 1− y A(y − y 2 ) + B(1 − y) + C y 2 = y 2 (1 − y) ( − A + C = 0; ⇒ A − B = 0; ⇒ A = B = C = 1. B = 1.

y=

R dy + y = e x . Let µ = d x = x, then eµ = e x , dx

  d x dy dy (e y) = e x + ex y = ex + y = e2x dx dx dx Z 1 2x x 2x ⇒ e y = e d x = e + C. 2 Hence, y =

308 Copyright © 2014 Pearson Canada Inc.

1 x e + Ce−x . 2

INSTRUCTOR’S SOLUTIONS MANUAL

15.

SECTION 7.9 (PAGE 453)

Z  dy +y=x µ = exp 1 d x = ex dx d x (e y) = e x (y ′ + y) = xe x dx Z

ex y =

19.

xe x d x = xe x − e x + C

y = x − 1 + Ce−x

16. We have

R dy + 2e x y = e x . Let µ = 2e x d x = 2e x , then dx

d  2ex  x x dy + 2e x e2e y e y = e2e dx d x   dy x x = e2e + 2e x y = e2e e x . dx

y(1) = 3e ⇒ 3 = 1 + C ⇒ C = 2

y = (x + 2)e1/x .

20.

e

Z

2e x

e

esin x y =

ex d x

Let u = 2e x du = 2e x d x Z 1 1 x = eu du = e2e + C. 2 2

y=

18.

dy + 10y = 1, y dt Z µ = 10 dt = 10t

1 10




=

21.

y(x) = 2 +

Z

x

0

t dt y(t)

H⇒

y(0) = 2

x dy = , i.e. y d y = x d x dx y y2 = x 2 + C

22 = 02 + C H⇒ p y = 4 + x 2.

dy d 10t (e y) = e10t + 10e10t y = e10t dt dt 1 10t e10t y(t) = e +C 10
2e e e 2 1 y 10 = 10 ⇒ = +C ⇒ C = 10 10 10 1 1 1−10t y= + e . 10 10

22.

dy + 3x 2 y = x 2 , dx Z µ = 3x 2 d x = x 3

23.

d x3 3 dy 3 3 (e y) = e x + 3x 2 e x y = x 2 e x dx d x Z 1 3 3 3 ex y = x 2ex d x = ex + C 3 1 2 y(0) = 1 ⇒ 1 = + C ⇒ C = 3 3 1 2 −x 3 y= + e . 3 3

2x d x = x 2 + C

y = (x 2 − π 2 )e− sin x .

2 10

y(0) = 1

y(π ) = 0

y(π ) = 0 ⇒ 0 = π 2 + C ⇒ C = −π 2

1 x Hence, y = + Ce−2e . 2

17.

y ′ + (cos x)y = 2xe− sin x , Z µ = cos x d x = sin x

d sin x (e y) = esin x (y ′ + (cos x)y) = 2x dx Z

Therefore, 2e x

x 2 y ′ + y = x 2 e1/x , y(1) = 3e 1 y ′ + 2 y = e1/x Zx 1 1 µ= dx = − x x2   d  −1/x  1 −1/x ′ e y =e y + 2y =1 dx x Z e−1/x y = 1 d x = x + C

y(x) = 1 +

Z

x

0

C=4

(y(t))2 dt 1 + t2

H⇒

y(0) = 1

y2 dy = , i.e. d y/y 2 = d x/(1 + x 2 ) dx 1 + x2 1 − = tan−1 x + C y − 1 = 0 + C H⇒ C = −1

y = 1/(1 − tan−1 x). y(x) = 1 +

Z

1

x

y(t) dt t (t + 1)

H⇒

y(1) = 1

dy y = , for x > 0 dx x(x + 1) dy dx dx dx = = − y x(x + 1) x x +1 x ln y = ln + ln C x +1 Cx y= , H⇒ 1 = C/2 x +1 2x y= . x +1

309 Copyright © 2014 Pearson Canada Inc.

SECTION 7.9 (PAGE 453)

24.

y(x) = 3 + dy = e−y , dx ey = x + C

Z

x

e−y dt

ADAMS and ESSEX: CALCULUS 8

0

dv = mg − kv has constant solution dt v = mg/k. However this solution does not satisfy the initial condition v(0) = 0. For other solutions we separate variables:

28. The equationt m

y(0) = 3

H⇒

i.e. e y d y = d x H⇒

3 = y(0) = ln C y = ln(x + e3 ).

y = ln(x + C) C = e3

H⇒

Z

Z dv = dt k g− v m k m − ln g − v = t + C. k m

25. Since a > b > 0 and k > 0,

lim x(t) = lim

  ab e(b−a)kt − 1

Since v(0) = 0, therefore C = −

be(b−a)kt − a ab(0 − 1) = = b. 0−a

t→∞

t→∞

remains positive for all t > 0, so m ln k

26. Since b > a > 0 and k > 0, ab lim x(t) = lim

t→∞

t→∞

= lim



e(b−a)kt

(b−a)kt − a be   ab 1 − e(a−b)kt

mg , the constant solution of the k differential equation noted earlier. This limiting velocity can be obtained directly from the differential equation by dv = 0. setting dt √ y = mg/k is a constant solution of the equation. For other solutions we proceed by separation of variables: t→∞

dv = mg − kv 2 dt k dv = g − v2 dt m dv = dt k g − v2 Z m Z dv k kt = dt = + C. mg 2 m m −v k

m

is indeterminate (0/0) if a = b. If a = b the original differential equation becomes dx = k(a − x)2 , dt which is separable and yields the solution 1 = a−x

Z

dx =k (a − x)2

Z

dt = kt + C.

1 1 1 = kt + . Since x(0) = 0, we have C = , so a a−x a Solving for x, we obtain x=

a 2 kt . 1 + akt

This solution also results from evaluating the limit of solution obtained for the case a 6= b as b approaches a (using l’Hˆopital’s Rule, say).

=t

Note that lim v(t) =

29.


ab e(b−a)kt − 1 , be(b−a)kt − a

k v m

k v m = e−kt/m g  mg  ⇒ v = v(t) = 1 − e−kt/m . k

27. The solution given, namely x=

g−

g−

 −1

b − ae(a−b)kt ab(1 − 0) = = a. b−0 t→∞

g

m k ln g. Also, g − v k m

Let a 2 = mg/k, where a > 0. Thus, we have Z

dv kt +C = a 2 − v 2 m a + v kt 1 = ln +C 2a a − v m r a + v 2akt kg ln = + C1 = 2 t + C1 a−v m m √ a+v = C2 e2t kg/m . a−v

310 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

30.

SECTION 7.9 (PAGE 453)

Assuming v(0) = 0, we get C2 = 1. Thus √ a + v = e2t kg/m (a − v)  √   √  v 1 + e2t kg/m = a e2t kg/m − 1 r  mg  2t √kg/m = −1 e k √ r 2t kg/m mg e −1 v= √ k e2t kg/m + 1 r mg as t → ∞. This also follows from Clearly v → k dv setting = 0 in the given differential equation. dt The balance in the account after t years is y(t) and y(0) = 1000. The balance must satisfy dy y2 = 0.1y − dt 1, 000, 000 105 y − y 2 dy = 106 Z Zdt dt dy = 106 105 y − y 2  Z  1 1 1 t C + dy = 6 − 5 y 10 105 105 − y 10 t ln |y| − ln |105 − y| = −C 10 105 − y = eC−(t/10) y 105 y = C−(t/10) . e +1 Since y(0) = 1000, we have 105 1000 = y(0) = C e +1 and y=

105 99e−t/10

The balance after 1 year is y=

105 99e−1/10

+1

C = ln 99,

⇒

+1

.

≈ $1, 104.01.

As t → ∞, the balance can grow to lim y(t) = lim

t→∞

t→∞

105 e(4.60−0.1t) + 1

=

⇒ ⇒

y+x

dy = 0, dx

or

dy y =− . dx x

Curves that intersect these hyperbolas at right angles dy x must therefore satisfy = , or x d x = y d y, a sepdx y arated equation with solutions x 2 − y 2 = C, which is also a family of rectangular hyperbolas. (Both families are degenerate at the origin for C = 0.)

32. Let x(t) be the number of kg of salt in the solution in the tank after t minutes. Thus, x(0) = 50. Salt is coming into the tank at a rate of 10 g/L × 12 L/min = 0.12 kg/min. Since the contents flow out at a rate of 10 L/min, the volume of the solution is increasing at 2 L/min and thus, at any time t, the volume of the solution is 1000 + 2t L. Therefore the conx(t) centration of salt is L. Hence, salt is being 1000 + 2t removed at a rate x(t) 5x(t) kg/L × 10 L/min = kg/min. 1000 + 2t 500 + t Therefore, dx 5x = 0.12 − dt 500 + t dx 5 + x = 0.12. dt 500 + t Z

5 dt = 5 ln |500 + t| = ln(500 + t)5 for 500 + t t > 0. Then eµ = (500 + t)5 , and

Let µ =

i dh dx (500 + t)5 x = (500 + t)5 + 5(500 + t)4 x dt dy   5x dx = (500 + t)5 + dy 500 + t = 0.12(500 + t)5 .

Hence, 105 0+1

= $100, 000.

For the account to grow to $50,000, t must satisfy 100, 000 99e−t/10 + 1 99e−t/10 + 1 = 2 t = 10 ln 99 ≈ 46 years.

50, 000 = y(t) =

31. The hyperbolas x y = C satisfy the differential equation

(500 + t)5 x = 0.12

Z

(500 + t)5 dt = 0.02(500 + t)6 + C

⇒ x = 0.02(500 + t) + C(500 + t)−5 . Since x(0) = 50, we have C = 1.25 × 1015 and x = 0.02(500 + t) + (1.25 × 1015 )(500 + t)−5 .

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SECTION 7.9 (PAGE 453)

ADAMS and ESSEX: CALCULUS 8

3. The barrel is generated by revolving x = a − by 2 ,

After 40 min, there will be x = 0.02(540) + (1.25 × 1015 )(540)−5 = 38.023 kg

(−2 ≤ y ≤ 2), about the y-axis. Since the top and bottom disks have radius 1 ft, we have a − 4b = 1. The volume of the barrel is

of salt in the tank. V =2

Review Exercises 7 (page 454)

Z

2

0

π(a − by 2 )2 d y

! 2 2aby 3 b2 y 5 = 2π a y − + 3 5 0   16 32 == 2π 2a 2 − ab + b2 . 3 5 2

1. 3 cm

3 cm 1 cm

5 cm

3 cm

1 cm 5 cm

1 cm Fig. R-7.1 The volume of thread that can be wound on the left spool is π(32 − 12 )(5) = 40π cm3 . The height of the winding region of the right spool at distance r from the central axis of the spool is of the form h = A + Br . Since h = 3 if r = 1, and h = 5 if r = 3, we have A = 2 and B = 1, so h = 2 + r . The volume of thread that can be wound on the right spool is   3 3 r 3 100π r (2 + r ) dr = 2π r 2 + 2π = cm3 . 3 1 3 1 Z

The right spool will hold thread.

100 (1, 000) = 833.33 m of 3 × 40

2. Let A(y) be the cross-sectional area of the bowl at height y above the bottom. When the depth of water in the bowl is Y , then the volume of water in the bowl is V (Y ) =

Z

Y

Since V = 16 and a = 1 + 4b, we have 

16 32 2π 2(1 + 4b) − b(1 + 4b) + b2 3 5 60 128b2 + 80b + 15 − = 0. π 2



= 16

Solving this quadratic gives two solutions, b ≈ 0.0476 and b ≈ −0.6426. Since the second of these leads to an unacceptable negative value for a, we must have b ≈ 0.0476, and so a = 1 + 4b ≈ 1.1904.

4. A vertical slice parallel to the top ridge of the solid at distance x to the right of the centre is a rectangle of base √ √ 2 100 − x 2 cmï‰and height 3(10 − x) cm. Thus the solid has volume Z

10 √

p 3(10 − x)2 100 − x 2 d x 0 √ Z 10 p √ Z = 40 3 100 − x 2 d x − 4 3

V =2

0

Let u = 100 − x 2 du = −2x d x √ 100π √ Z 100 √ = 40 3 −2 3 u du 4   0 √ 4 = 1, 000 3 π − cm3 . 3

A(y) d y.

0

The water evaporates at a rate proportional to exposed surface area. Thus dV = k A(Y ) dt d V dY = k A(Y ) dY dt dY A(Y ) = k A(Y ). dt

60◦

10 cm

Hence dY/dt = k; the depth decreases at a constant rate.

312 Copyright © 2014 Pearson Canada Inc.

Fig. R-7.4

x

10 0

p x 100 − x 2 d x

INSTRUCTOR’S SOLUTIONS MANUAL

5. The arc length of y = s=

Z

0

REVIEW EXERCISES 7 (PAGE 454)

y

1 cosh(ax) from x = 0 to x = 1 is a

1q

1 + sinh2 (ax) d x =

1 1 1 = sinh(ax) = sinh a. a a 0

Z

1

cosh(ax) d x

1

0

x

1 sinh a = 2, that is, sinh a = 2a. Solving this a by Newton’s Method or a calculator solve function, we get a ≈ 2.1773.

We want

6. The area of revolution of y = x-axis is

S = 2π

Z

6

s



dy dx

√

Fig. R-7.8 Let the disk have centre (and therefore centroid) at (0, 0). Its area is 9π . Let the hole have centre (and therefore centroid) at (1, 0). Its area is π . The remaining part has area 8π and centroid at (x, ¯ 0), where

x, (0 ≤ x ≤ 6), about the

(9π )(0) = (8π )x¯ + (π )(1). Thus x¯ = −1/8. The centroid of the remaining part is 1/8 ft from the centre of the disk on the side opposite the hole.

2

y 1+ dx r 6√ 1 x 1+ dx = 2π 4x 0 Z 6r 1 = 2π x + dx 4 0  3/2 6   4π 1 = 4π 125 − 1 = 62π sq. units. = x+ 3 4 3 8 8 3 0 0

3

Z

9. Let the area of cross-section of the cylinder be A. When the piston is y cm above the base, the volume of gas in the cylinder is V = Ay, and its pressure P(y) satisfies P(y)V = k (constant). The force exerted by the piston is F(y) = P(y)A =

kA k = . Ay y

We are told that F = 1, 000 N when y = 20 cm. Thus k = 20, 000 N·cm. The work done by the piston as it descends to 5 cm is Z 20 20 20, 000 d y = 20, 000 ln ≈ 27, 726 N·cm. W = y 5 5

7. The region is a quarter-elliptic disk with semi-axes a = 2 and b = 1. The area of the region is A = π ab/4 = π/2. The moments about the coordinate axes are

10. We are told that for any a > 0, M x=0 =

Z

0

Z

s

x 1−

x2 4

dx

x2

8/(3π ), 4/(3π ) .

Z a h 0

Z 2 i 2  f (x) − g(x) d x = 2π

a 0

h i x f (x)−g(x) d x.

Differentiating both sides of this equation with respect to a, we get  2  2 h i f (a) − g(a) = 2a f (a) − g(a) ,

1√

or, equivalently, f (a) + g(a) = 2a. Thus f and g must satisfy f (x) + g(x) = 2x

Thus x¯ = M x=0 / A = 8/(3π ) and y¯ = M y=0 / A  = 4/(3π ). The centroid is

8.

π

Let u = 1 − 4 x du = − d x 2

4 u du = 3 0  Z  1 2 x2 = 1− dx 2 0 4   2 2 1 x 3 = . = x− 2 12 0 3 =2

M y=0

2

11.

for every x > 0.

Z dy 3y dy dx = ⇒ =3 dx x −1 y x −1 ⇒ ln |y| = ln |x − 1|3 + ln |C|

⇒ y = C(x − 1)3 . Since y = 4 when x = 2, we have 4 = C(2 − 1)3 = C, so the equation of the curve is y = 4(x − 1)3 .

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REVIEW EXERCISES 7 (PAGE 454)

ADAMS and ESSEX: CALCULUS 8

12. The ellipses 3x 2 + 4y 2 = C all satisfy the differential

c) Using the result of Example 4 in Section 7.1, we calculate the volume of the first bead: Z π π −2kx V1 = e (1 − cos(2x)) d x 2 0 π π π e−2kx π e−2kx (2 sin(2x) − 2k cos(2x)) = − −4k 0 2 4(1 + k 2 ) 0 π π −2kπ −2kπ = (1 − e )− (k − ke ) 4k 4(1 + k 2 ) π (1 − e−2kπ ). = 4k(1 + k 2 )

equation

6x + 8y

dy = 0, dx

or

dy 3x =− . dx 4y

A family of curves that intersect these ellipses at right dy 4y angles must therefore have slopes given by = . dx 3x Thus Z Z dy dx 3 =4 y x 3 ln |y| = 4 ln |x| + ln |C|.

By part (a) and Theorem 1(d) of Section 6.1, the sum of the volumes of the first n beads is π Sn = (1 − e−2kπ ) 4k(1 + k 2 )  n−1 i h  2 × 1 + e−2kπ + e−2kπ + · · · + e−2kπ

The family is given by y 3 = C x 4 .

13. The original $8,000 grows to $8, 000e0.08 in two years. Between t and t + dt, an amount $10, 000 sin(2π t) dt comes in, and this grows to $10, 000 sin(2π t)e0.04(2−t) dt by the end of two years. Thus the amount in the account after 2 years is 8, 000e0.08 +10, 000

Z

0

2

−2knπ π −2kπ 1 − e (1 − e ) 4k(1 + k 2 ) 1 − e−2kπ π −2knπ (1 − e ). = 4k(1 + k 2 )

=

Thus the total volume of all the beads is π V = lim Sn = cu. units.. n→∞ 4k(1 + k 2 )

sin(2π t)e0.04(2−t) dt ≈ $8, 798.85.

(We omit the details of evaluation of the integral, which is done by the method of Example 4 of Section 7.1.)

2. 10 m

Challenging Problems 7 (page 455)

1.

1m

a) The nth bead extends from x = (n − 1)π to x = nπ , and has volume Z nπ Vn = π e−2kx sin2 x d x (n−1)π Z π nπ = e−2kx (1 − cos(2x)) d x 2 (n−1)π

Fig. C-7.2

Let x = u + (n − 1)π Zd x = du i π π −2ku −2k(n−1)π h = e e 1 − cos(2u + 2(n − 1)π ) du 2 0 Z π −2k(n−1)π π −2ku e (1 − cos(2u)) du = e 2 0 = e−2k(n−1)π V1 .

Vn+1 e−2knπ V1 = −2k(n−1)π = e−2kπ , which deVn e V1 pends on k but not n. Thus

b) Vn+1 /Vn = 1/2 if −2kπ = ln(1/2) = − ln 2, that is, if k = (ln 2)/(2π ).

h(r ) = a(r 2 − 100)(r 2 − k 2 ), 2

2

where 0 < k < 10

h (r ) = 2ar (r − k ) + 2ar (r − 100) = 2ar (2r 2 − 100 − k 2 ). ′

2

The deepest point occurs where 2r 2 = 100 + k 2 , i.e., r 2 = 50 + (k 2 /2). Since this depth must be 1 m, we require  2   k k2 a − 50 50 − = −1, 2 2

or, equivalently, a(100 − k 2 )2 = 4. The volume of the pool is Z 10 V P = 2π a r (100 − r 2 )(r 2 − k 2 ) dr k   250, 000 1 = 2π a − 2, 500k 2 + 25k 4 − k 6 . 3 12

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INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 7 (PAGE 455)

The volume of the hill is   Z k 1 V H = 2π a r (r 2 −100)(r 2 −k 2 ) dr = 2π a 25k 4 − k 6 . 12 0 These two volumes must be equal, so k 2 = 100/3 and k ≈ 5.77 m. Thus a = 4/(100 − k 2 )2 = 0.0009. The volume of earth to be moved is V H with these values of a and k, namely "     # 1 100 4 100 2 − 2π(0.0009) 25 ≈ 140 m3 . 3 12 3

3.

y = ax + bx 2 + cx 3

y

x

f (x) = ax must satisfy f (h) = r , f ′ (h) = 0, ′ and f (x) > 0 for 0 < x < h. The first two conditions require that ah + bh 2 + ch 3 = r a + 2bh + 3ch 2 = 0,

from which we obtain by solving for b and c, c=

a=1 a+b+c = 2 b + 2c = m

0

+ cx 3

3r − 2ah , h2

 a + bx + cx 2 a) If f (x) = 2  p + qx + r x b + 2cx for 0 < f ′ (x) = q + 2r x for 1 <

for 0 ≤ x ≤ 1 , then for 1 ≤ x ≤ 3 x 0. Its two roots are x1 = h and x2 = h 2 a/(3ah − 6r ). a must be restricted so that x2 is not in the interval (0, h). If a < 2r/ h, then x2 < 0. If 2r/ h < a < 3r/ h, then x2 > h. If a > 3r/ h, then 0 < x2 < h. Hence the interval of acceptable values of a is 0 ≤ a ≤ 3r/ h. We have   13πr 2 h 3r 9πr 2 h V (0) = , V = . 35 h 14

1

3q

1 + (q + 2r x)2 d x

with the values of b, c, q, and r determined above. A plot of the graph of L(m) reveals a minimum value in the neighbourhood of m = −0.3. The derivative of L(m) is a horrible expression, but Mathematica determined its zero to be about m = −0.281326, and the corresponding minimum value of L is about√4.41748. The polygonal line ABC has length 3 2 ≈ 4.24264, which is only slightly shorter.

5. Let b = ka so that the cross-sectional curve is given by y = f (x) = ax(1 − x)(x + k). The requirement that f (x) ≥ 0 for 0 ≤ x ≤ 1 is satisfied provided either a > 0 and k ≥ 0 or a < 0 and k ≤ −1. The volume of the wall is Z 1 πa V (a, k) = 2π(15 + x) f (x) d x = (78 + 155k). 30 0 To minimize this expression for a > 0 we should take k = 0. This gives f (x) = ax 2 (1 − x). To minimize V (a, k) for a < 0 we should take k = −1. This gives f (x) = −ax(1 − x)2 . Since we want the maximum value of f to be 2 in either case, we calculate the critical points of these two possible functions. For a > 0 the CP is x = 2/3 and f (2/3) = 2 gives a = 27/2. The volume in this case is V (27/2, 0) = (27π/60)(78 − 0). For a < 0 the CP is x = 1/3 and f (1/3) = 2 gives a = −27/2. The volume in this case is V (−27/2, −1) = −(27π/60)(78 − 155) = (27π/60)(77). Thus the minimum volume occurs for f (x) = (27/2)x(1 − x)2 , i.e. b = −a = 27/2.

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CHALLENGING PROBLEMS 7 (PAGE 455)

ADAMS and ESSEX: CALCULUS 8

6. Starting with V1 (r ) = 2r , and using repeatedly the formula

Vn (r ) =

Z

r

8.

y

p Vn−1 ( r 2 − x 2 ) d x,

−r

Q

Maple gave the following results:

y = f (x)

V2 (r ) = πr 2 1 V4 (r ) = π 2r 4 2 1 V6 (r ) = π 3r 6 6 1 4 8 V8 (r ) = π r 24 1 5 10 V10 (r ) = π r 120

V1 (r ) = 2r 4 V3 (r ) = πr 3 3 8 2 5 V5 (r ) = π r 15 16 3 7 π r V7 (r ) = 105 32 4 9 V9 (r ) = π r 945

L P(x, y)

It appears that

x (L , 0)

1 n 2n π r , and n! n 2 V2n−1 (r ) = π n−1r 2n−1 1 · 3 · 5 · · · (2n − 1) 22n−1 (n − 1)! n−1 2n−1 = π r . (2n − 1)! V2n (r ) =

Fig. C-7.8

If Q = (0, Y ), then the slope of P Q is

These formulas predict that V11 (r ) =

211 5! 5 11 π r 11!

and

V12 (r ) =

1 6 12 π r , 6!

both of which Maple is happy to confirm.

7. With y and θ as defined in the statement of the problem, we have The needle crosses a line if y < 5 sin θ . The probability of this happening is the ratio of the area under the curve to the area of the rectangle in the figure, that is, 1 10π

2 Since |P Q| = L, we have (y − Y )2 = L 2 − √x . Since the slope d y/d x is negative at P, d y/d x = − L 2 − x 2 /x. Thus

and 0 ≤ θ < π.

0 ≤ y ≤ 10

Pr =

y−Y dy = f ′ (x) = . x −0 dx

Z

0

π

5 sin θ dθ =

1 . π

y=−

Z √

L2 − x2 d x = L ln x

L+

√

! p L2 − x2 − L 2 − x 2 +C. x

Since y = 0 when x = L, we have C = 0 and the equation of the tractrix is

y y = 10

L+

y = L ln

√

L2 − x2 x

!

−

p

L 2 − x 2.

Note that the first term can be written in an alternate way:

y = 5 sin x

π

y = L ln θ

Fig. C-7.7

316 Copyright © 2014 Pearson Canada Inc.



L−

√

x L2 − x2



−

p

L 2 − x 2.

INSTRUCTOR’S SOLUTIONS MANUAL

9.

CHALLENGING PROBLEMS 7 (PAGE 455)

a) S(a, a, c) is the area of the surface obtained by rotating the ellipse (x 2 /a 2 ) + (y 2 /c2 ) = 1 (where a >√ c) about the y-axis. Since y ′ = −cx/(a a 2 − x 2 ), we have s Z a c2 x 2 S(a, a, c) = 2 × 2π x 1+ 2 2 dx a (a − x 2 ) 0 p Z 4π a a 4 − (a 2 − c2 )x 2 = x dx √ a 0 a2 − x 2 Let x = a sin u d x = a cos u du Z p 4π π/2 a sin u a 4 − (a 2 − c2 )a 2 sin2 u du = a 0 Z π/2 p = 4π a sin u a 2 − (a 2 − c2 )(1 − cos2 u) du 0

Let v = cos u dv = − sin u du Z 1p c2 + (a 2 − c2 )v 2 dv. = 4π a

c) Since b =



S(a, b, c) ≈

   b−c a−b a+ c, we use a−c a−c



b−c a−c





a−b S(a, a, c)+ a−c



S(a, c, c).

d) We cannot evaluate S(3, 2, 1) even numerically at this stage. The double integral necessary to calculate it is not treated until a later chapter. (The value is approximately 48.882 sq. units.) However, using the formulas obtained above,

0

This integral can now be handled using tables or computer algebra. It evaluates to ! √ 2π ac2 a + a 2 − c2 2 S(a, a, c) = 2π a + √ ln . c a 2 − c2 b) S(a, c, c) is the area of the surface obtained by rotating the ellipse √ of part (a) about the y-axis. Since y ′ = −cx/(a a 2 − x 2 ), we have s Z a c2 x 2 y 1+ 2 2 dx S(a, c, c) = 2 × 2π a (a − x 2 ) 0 p Z 4π c a p 2 a 4 − (a 2 − c2 )x 2 = 2 a − x2 √ dx a a2 − x 2 Z0 a p 4π c a 4 − (a 2 − c2 )x 2 d x = 2 a 0 s Z a a 2 − c2 2 = 4π c 1− x dx a4 0 2π a 2 c c = 2π c2 + √ cos−1 . a a 2 − c2

S(3, 3, 1) + S(3, 1, 1) S(3, 2, 1) ≈ 2   √ 1 6π 18π = 18π + √ ln(3 + 8) + 2π + √ cos−1 (1/3) 2 8 8 ≈ 49.595 sq. units.

317 Copyright © 2014 Pearson Canada Inc.

SECTION 8.1 (PAGE 468)

ADAMS and ESSEX: CALCULUS 8

CHAPTER 8. CONICS, PARAMETRIC CURVES, AND POLAR CURVES Section 8.1 Conics

8. If x 2 + 4y 2 − 4y = 0, then   1 = 1, x 2 + 4 y2 − y + 4

(page 468)

1. The ellipse with foci (0, ±2) has major axis along the

y-axis and c = 2. If a = 3, then b2 = 9 − 4 = 5. The ellipse has equation x2 y2 + = 1. 5 9

0,

 1 , 2

y

a = −1 and principal axis x = 2. Its equation is (x − 2)2 = −4(y − 4) = 16 − 4y.

4. A parabola with focus at (0, −1) and principal axis

along y = −1 will have vertex at a point of the form (v, −1). Its equation will then be of the form (y + 1)2 = ±4v(x − v). The origin lies on this curve if 1 = ±4(−v 2 ). Only the − sign is possible, and in this case v = ±1/2. The possible equations for the parabola are (y + 1)2 = 1 ± 2x.

5. The hyperbola with semi-transverse axis a = 1 and foci

x 2 +4y 2 −4y=0

1 1 2

1

3. A parabola with focus (2, 3) and vertex (2, 4) has

9. If 4x 2 + y 2 − 4y = 0, then 4x 2 + y 2 − 4y + 4 = 4 4x 2 + (y − 2)2 = 4 x2 +

(y − 2)2 =1 4

This is an ellipse with semi-axes 1 and 2, centred at (0, 2). y 4

x2 = 1. 3

(−1,2)

4x 2 +y 2 −4y=0

(1,2)

2

6. The hyperbola with foci at (±5, 1) and asymptotes x = ±(y − 1) is rectangular, has centre at (0, 1) and has transverse axis along the line y = 1. Since c = 5 and a = b (because the asymptotes are perpendicular to each other) we have a 2 = b2 = 25/2. The equation of the hyperbola is 25 . 2

x

Fig. 8.1.9

10. If 4x 2 − y 2 − 4y = 0, then 4x 2 − (y 2 + 4y + 4) = −4,

7. If x 2 + y 2 + 2x = −1, then (x + 1)2 + y 2 = 0. This

x

Fig. 8.1.8

(0, ±2) has transverse axis along the y-axis, c = 2, and b2 = c2 − a 2 = 3. The equation is

represents the single point (−1, 0).



1 semi-major axis 1, semi-minor axis , and foci at 2  √  3 1 , ± . 2 2

(2, 1), and major axis along y = 1. If ǫ = 1/2, then a = c/ǫ = 4 and b2 = 16 − 4 = 12. The ellipse has equation (x − 2)2 (y − 1)2 + = 1. 16 12

x 2 − (y − 1)2 =

(y − 21 )2 x2 + = 1. 1 1 4

This represents an ellipse with centre at

2. The ellipse with foci (0, 1) and (4, 1) has c = 2, centre

y2 −

or

or

x2 (y + 2)2 − = −1. 1 4

This represents a hyperbola with centre at (0, −2), semitransverse√axis 2, semi-conjugate axis 1, and foci at (0, −2 ± 5). The asymptotes are y = ±2x − 2.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 8.1 (PAGE 468)

y


This is a hyperbola with centre − 23 , 1 , and asymptotes √ the straight lines 2x + 3 = ±2 2(y − 1). y

x −2 4x 2 −y 2 −4y=0

3 (− 2 ,1) (−3,1)

1

x 2 −2y 2 +3x+4y=2 x

Fig. 8.1.10

11. If x 2 + 2x − y = 3, then (x + 1)2 − y = 4.

Thus y = (x + 1)2 − 4. This is a parabola with vertex (−1, −4), opening upward. y

Fig. 8.1.13

14. If 9x 2 + 4y 2 − 18x + 8y = −13, then 9(x 2 − 2x + 1) + 4(y 2 + 2y + 1) = 0

x

⇔9(x − 1)2 + 4(y + 1)2 = 0.

This represents the single point (1, −1). x 2 +2x−y=3

15. If 9x 2 + 4y 2 − 18x + 8y = 23, then

(−1,−4)

9(x 2 − 2x + 1) + 4(y 2 + 2y + 1) = 23 + 9 + 4 = 36

Fig. 8.1.11

12. If x + 2y

+ 2y 2

9(x − 1)2 + 4(y + 1)2 = 36

= 1, then

(x − 1)2 (y + 1)2 + = 1. 4 9

  3 1 2 = −x 2 y +y+ 4 2  2 3 1 ⇔ x = −2 y+ . 2 2

This is an ellipse with centre (1, −1), and semi-axes 2 and 3. y

This represents a parabola with vertex at ( 32 , − 12 ), focus 1 13 at ( 11 8 , − 2 ) and directrix x = 8 . y

(−1,−1)

x 

x (3,−1)

13. If x 2 − 2y 2 + 3x + 4y = 2, then  3 2 9 − 2(y − 1)2 = 2 4
2 x + 32 (y − 1)2 − =1 9 9 x+

8

3 1 2 ,− 2



(1,−4)

Fig. 8.1.15

16. The equation (x − y)2 − (x + y)2 = 1 simplifies to

Fig. 8.1.12

4

(1,−1)

9x 2 +4y 2 −18x+8y=23

x+2y+2y 2 =1



(1,2)

4x y = −1 and hence represents a rectangular hyperbola with centre at the origin, asymptotes along the coordinate axes, transverse axis along
y = −x, conjugate axis along
y = x, vertices at 12 , − 21 and − 21 , 12 , semi-transverse √ and semi-conjugate q axes equal to 1/ 2, semi-focal sepa-

ration equal to 12 + 12 = 1, and hence foci at the points     √ √1 , − √1 and − √1 , √1 . The eccentricity is 2. 2

2

2

2

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SECTION 8.1 (PAGE 468)

ADAMS and ESSEX: CALCULUS 8

√ This is a rectangular√hyperbola with centre √ (0, − 2), semi-axes a = b = 2, and eccentricity 2. √ The semifocal separation is 2; the foci are at (±2, − 2). The √ asymptotes are u = ±(v + 2). In terms of the √ original coordinates, the centre is (1, −1), √ the foci are (± 2 + 1, ± 2 − 1), and the asymptotes are x = 1 and y = −1.

y

x



1 1 2,2



y

(x−y)2 −(x+y)2 =1

xy+x−y=2

Fig. 8.1.16 (1,−1)

x

17. The parabola has focus at (3, 4) and principal axis along y = 4. The vertex must be at a point of the form (v, 4), in which case a = ±(3 − v) and the equation of the parabola must be of the form (y − 4)2 = ±4(3 − v)(x − v). This curve passes through the origin if 16 = ±4(v 2 − 3v). We have two possible equations for v: v 2 − 3v − 4 = 0 and v 2 − 3v + 4 = 0. The first of these has solutions v = −1 or v = 4. The second has no real solutions. The two possible equations for the parabola are (y − 4)2 = 4(4)(x + 1)

(y − 4)2 = 4(−1)(x − 4)

or or

y 2 − 8y = 16x

y 2 − 8y = −4x

18. The foci of the ellipse are (0, 0) and (3, 0), so the centre is (3/2, 0) and c = 3/2. The semi-axes a and b must satisfy a 2 − b2 = 9/4. Thus the possible equations of the ellipse are (x − (3/2))2 y2 + 2 = 1. 2 (9/4) + b b

19. For x y + x − y = 2 we have A = C = 0, B = 1. We

therefore rotate the coordinate axes (see text pages 407– 408) through angle θ = π/4. (Thus cot 2θ = 0 = (A − C)/B.) The transformation is 1 x = √ (u − v), 2

Fig. 8.1.19

20. We have x 2 + 2x y + y 2 = 4x − 4y + 4 and

A = 1, B = 2, C = 1, D = −4, E = 4 and F = −4. We rotate the axes through angle θ satisfyπ ing tan 2θ = B/(A − C) = ∞ ⇒ θ = . Then A′ = 2, 4 √ B ′ = 0, C ′ = 0, D ′ = 0, E ′ = 4 2 and the transformed equation is √ 2u 2 + 4 2v − 4 = 0

⇒

  √ 1 u 2 = −2 2 v − √ 2

which represents a parabola with vertex at  (u, v) = 0, √1 and principal axis along u = 0. 2 The distance a from the √ √ focus to the vertex is given by 4a = 2 2, so a√= 1/ 2 and the focus is at (0, 0). The directrix is v = 2. 1 1 Since x = √ (u − v) and y = √ (u + v), the vertex 2 2 of the parabola in terms of x y-coordinates is (− 12 , 12 ), and the focus is (0, 0). The directrix is x − y = 2. The principal axis is y = −x. y

y=−x

1 y = √ (u + v). 2

x 2 +2xy+y 2 =4x−4y+4 (−1/2,1/2)

The given equation becomes x

1 2 1 1 (u − v 2 ) + √ (u − v) − √ (u + v) = 2 2 2 2 √ u 2 − v 2 − 2 2v = 4  √ 2 u2 − v + 2 = 2 √ (v + 2)2 u2 − = 1. 2 2

320 Copyright © 2014 Pearson Canada Inc.

Fig. 8.1.20

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 8.1 (PAGE 468)

21. For 8x 2 + 12x y + 17y 2 = 20, we have A = 8, B = 12, C = 17, F = −20. Rotate the axes through angle θ where B 12 4 tan 2θ = =− =− . A−C 9 3 Thus cos 2θ = 3/5, sin 2θ = −4/5, and 2 cos2 θ − 1 = cos 2θ =

3 5

⇒

cos2 θ =

4 . 5

2 1 We may therefore take cos θ = √ , and sin θ = − √ . 5 5 The transformation is therefore 2 1 2 1 x = √ u+ √ v u=√ x−√ y 5 5 5 5 1 2 1 2 y = −√ u + √ v v=√ x+√ y 5 5 5 5 The coefficients of the transformed equation are       4 2 1 A′ = 8 + 12 − + 17 =5 5 5 5 B′ = 0       2 4 1 − 12 − + 17 = 20. C′ = 8 5 5 5

Then A′ = 0, B ′ = 0, C ′ = 5, D ′ = transformed equation is 5v 2 +

√

5u = 0

⇒

√ 5, E ′ = 0 and the

1 v2 = − √ u 5

which represents a parabola with vertex at (u, v) = (0, 0),   1 1 focus at − √ , 0 . The directrix is u = √ and the 4 5 4 5 2 1 principal axis is v = 0. Since x = √ u − √ v and 5 5 2 1 y = √ u + √ v, in terms of the x y-coordinates, the ver5 5   1 1 tex is at (0, 0), the focus at − , − . The directrix 10 20 is 2x + y = 41 and the principal axis is 2y − x = 0. y

x x 2 −4xy+4y 2 +2x+y=0

The transformed equation is

x=2y

u2 + v 2 = 1. 5u 2 + 20v 2 = 20, or 4 This is an ellipse with centre √ (0, 0), semi-axes a = 2 and b = 1, and foci at u = ± 3, v = 0. In terms of the original coordinates, the centre is (0, 0), √ √ ! 2 3 3 the foci are ± √ , − √ . 5 5

Fig. 8.1.22

y

8x 2 +12xy+17y 2 =20

p

x 2 + y 2. The distance from P to D is x + p. Thus

23. The distance from P to F is

p

x 2 + y2 =ǫ x+p x 2 + y 2 = ǫ 2 (x 2 + 2 px + p2 )

x

(1 − ǫ 2 )x 2 + y 2 − 2 pǫ 2 x = ǫ 2 p2 . Fig. 8.1.21

22. We have

y

x 2 −4x y +4y 2 +2x + y

= 0 and A = 1, B = −4, C = 4, D = 2, E = 1 and F = 0. We rotate the axes through angle θ satisfying tan 2θ = B/(A−C) = 43 . Then p 5 3 sec 2θ = 1 + tan2 2θ = ⇒ cos 2θ = 3 5 r r  1 + cos 2θ 4 2   = = √ ;  cos θ = 2 5 r r5 ⇒  1 − cos 2θ 1 1   sin θ = = =√ . 2 5 5

P=(x,y) x=− p D F

x

Fig. 8.1.23

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SECTION 8.1 (PAGE 468)

ADAMS and ESSEX: CALCULUS 8

y

24. Let the equation of the parabola be y 2 = 4ax. The fo-

cus F is at (a, 0) and vertex at (0, 0). Then the distance from √ the vertex to the focus is a. At x = a, y = 4a(a) = ±2a. Hence, ℓ = 2a, which is twice the distance from the vertex to the focus.

ℓ a c

y

x x2 y2 − =1 a 2 b2

y 2 =4ax ℓ

Fig. 8.1.26 x

(a,0)

27. S2

Fig. 8.1.24

C2

25.

c2 ℓ2 We have 2 + 2 = 1. Thus a b

B F2

  c2 ℓ2 = b 2 1 − 2 but c2 = a 2 − b2 a   a 2 − b2 b2 = b2 1 − = b2 2 . 2 a a

P F1

S1

C1

Therefore ℓ = b2 /a.

y b

A

x2 y2 + =1 a 2 b2

ℓ c

a x

V

Fig. 8.1.27

Fig. 8.1.25

x2 y2 − = 1. The a2 b2 vertices are at (±a, 0) and the foci are at (±c, 0) where √ √ c = a 2 + b2 . At x = a 2 + b2 ,

26. Suppose the hyperbola has equation

a 2 + b2 y2 − =1 a2 b2 2 2 2 (a + b )b − a 2 y 2 = a 2 b2 y=± Hence, ℓ =

b2 . a

b2 . a

Let the spheres S1 and S2 intersect the cone in the circles C1 and C2 , and be tangent to the plane of the ellipse at the points F1 and F2 , as shown in the figure. Let P be any point on the ellipse, and let the straight line through P and the vertex of the cone meet C1 and C2 at A and B respectively. Then P F1 = P A, since both segments are tangents to the sphere S1 from P. Similarly, P F2 = P B. Thus P F1 + P F2 = P A + P B = AB = constant (distance from C1 to C2 along all generators of the cone is the same.) Thus F1 and F2 are the foci of the ellipse.

28. Let F1 and F2 be the points where the plane is tangent to the spheres. Let P be an arbitrary point P on the hyperbola in which the plane intersects the cone. The spheres are tangent to the cone along two circles as shown in the figure. Let P AV B be a generator of the cone (a straight line lying on the cone) intersecting these two circles at A and B as shown. (V is the vertex of the cone.) We have P F1 = P A because two tangents to a sphere from

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 8.2 (PAGE 474)

a point outside the sphere have equal lengths. Similarly, P F2 = P B. Therefore P F2 − P F1 = P B − P A = AB = constant, F

since the distance between the two circles in which the spheres intersect the cone, measured along the generators of the cone, is the same for all generators. Hence, F1 and F2 are the foci of the hyperbola.

P

Y

C V

X

Q

P F1 A

A

Fig. 8.1.29

V B

Section 8.2

Parametric Curves

(page 474)

1. If x = t, y = 1 − t, (0 ≤ t ≤ 1) then

F2

x + y = 1. This is a straight line segment. y

Fig. 8.1.28 1

x=t y=1−t (0≤t≤1) x

1

Fig. 8.2.1

29. Let the plane in which the sphere is tangent to the cone meet AV at X. Let the plane through F perpendicular to the axis of the cone meet AV at Y . Then V F = V X, and, if C is the centre of the sphere, FC = XC. Therefore V C is perpendicular to the axis of the cone. Hence Y F is parallel to V C, and we have Y V = V X = V F. If P is on the parabola, F P ⊥ V F, and the line from P to the vertex A of the cone meets the circle of tangency of the sphere and the cone at Q, then

2. If x = 2 − t and y = t + 1 for 0 ≤ t < ∞, then

y = 2 − x + 1 = 3 − x for −∞ < x ≤ 2, which is a half line. y

x=2−t y=t+1

(2,1)

F P = P Q = Y X = 2V X = 2V F.

x

Fig. 8.2.2 Since F P = 2V F, F P is the semi-latus rectum of the parabola. (See Exercise 18.) Therefore F is the focus of the parabola.

3. If x = 1/t, y = t − 1, (0 < t < 4), then y = is part of a hyperbola.

1 − 1. This x

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SECTION 8.2 (PAGE 474)

ADAMS and ESSEX: CALCULUS 8

y

y





1 4 ,3 t=4

bx=ay

1 y= x − 1

t=0 a

x

t=1 x y=−1

bx=−ay

Fig. 8.2.3

4. If x =

Fig. 8.2.6

1 t and y = for −∞ < t < ∞, then 1 + t2 1 + t2 1 + t2 1 x 2 + y2 = = =x (1 + t 2 )2 1 + t2   1 1 2 + y2 = . ⇔ x− 2 4

7. If x = 3 sin π t, y = 4 cos π t, (−1 ≤ t ≤ 1), then x2 y2 + = 1. This is an ellipse. 9 16 y

t=0

x2 y2 9 + 16 =1

This curve consists of all points of the circle with centre at ( 21 , 0) and radius 21 except the origin (0, 0). y

x

t=∞ t=−∞

t=−1 t=1

t=0 x

Fig. 8.2.7

x=1/(1+t 2 ) y=t/(1+t 2 )

8. If x = cos sin s and y = sin sin s for −∞ < s < ∞, then

Fig. 8.2.4

5. If x = 3 sin 2t, y = 3 cos 2t, (0 ≤ t ≤ π/3), then x 2 + y 2 = 9. This is part of a circle. y t=0

x 2 + y 2 = 1. The curve consists of the arc of this circle extending from (a, −b) through (1, 0) to (a, b) where a = cos(1) and b = sin(1), traversed infinitely often back and forth. y x=cos sin s

x 2 +y 2 =9

y=sin sin s x

1 rad

π t= 3

x

Fig. 8.2.5

6. If x = a sec t and y = b tan t for −

π π < t < , then 2 2

x2 y2 − 2 = sec2 t − tan2 t = 1. 2 a b The curve is one arch of this hyperbola.

Fig. 8.2.8

9. If x = cos3 t, y = sin3 t, (0 ≤ t ≤ 2π ), then x 2/3 + y 2/3 = 1. This is an astroid.

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INSTRUCTOR’S SOLUTIONS MANUAL

y

SECTION 8.2 (PAGE 474)

and

t=π/2

1 = (cos2 t+sin2 t)2 = cos4 t+sin4 t+2 cos2 t sin2 t.

x 2/3 +y 2/3 =1

Hence,

t=0

t=π

1 + (x − y)2 = 2(cos4 t + sin4 t) = 2(x + y).

t=2π

(b) If x = sec4 t and y = tan4 t, then t=3π/2

(x − y)2 = (sec4 t − tan4 t)2

= (sec2 t + tan2 t)2

Fig. 8.2.9 √

10. If x = 1 − 4 − t 2 and y = 2 + t for −2 ≤ t ≤ 2 then (x − 1)2 = 4 − t 2 = 4 − (y − 2)2 . The parametric curve is the left half of the circle of radius 4 centred at (1, 2), and is traced in the direction of increasing y. √ y 4−t 2

x=1− y=2+t −2≤t≤2

= sec4 t + tan4 t + 2 sec2 t tan2 t

and 1 = (sec2 t−tan2 t)2 = sec4 t+tan4 t−2 sec2 t tan2 t. Hence, 1 + (x − y)2 = 2(sec4 t + tan4 t) = 2(x + y). (c) Similarly, if x = tan4 t and y = sec4 t, then 1 + (x − y)2 = 1 + (y − x)2

= (sec2 t − tan2 t)2 + (sec4 t − tan4 t)2

(1,2)

= 2(tan4 t + sec4 t) = 2(x + y).

These three parametric curves above correspond to different parts of the parabola 1+(x − y)2 = 2(x + y), as shown in the following diagram.

x

Fig. 8.2.10

y

11.

x = cosh t, y = sinh t represents the right half (branch) of the rectangular hyperbola x 2 − y 2 = 1.

12.

x = 2 − 3 cosh t, y = −1 + 2 sinh t represents the left half (branch) of the hyperbola

x=tan 4 t y=sec4 t The parabola 2(x+y)=1+(x−y)2

1 x=cos4 t y=sin4 t

(y + 1)2 (x − 2)2 − = 1. 9 4

x=sec4 t y=tan4 t 1

x

Fig. 8.2.14

13.

x = t cos t, y = t sin t, (0 ≤ t ≤ 4π ) represents two revolutions of a spiral curve winding outwards from the origin in a counterclockwise direction. The point on the curve corresponding to parameter value t is t units distant from the origin in a direction making angle t with the positive x-axis.

14. (a) If x = cos4 t and y = sin4 t, then

15. The slope of y =

at x is m = 2x. Hence the parabola can be parametrized x = m/2, y = m 2 /4, (−∞ < m < ∞).

16. If (x, y) is any point on the circle x 2 + y 2 = R 2 other

than (R, 0), then the line from (x, y) to (R, 0) has slope y m= . Thus y = m(x − R), and x−R x 2 + m 2 (x − R)2 = R 2

(x − y)2 = (cos4 t − sin4 t)2 h i2 = (cos2 t + sin2 t)(cos2 t − sin2 t) = (cos2 t − sin2 t)2 4

4

2

2

= cos t + sin t − 2 cos t sin t

x2

(m 2 + 1)x 2 − 2x Rm 2 + (m 2 − 1)R 2 = 0 h i (m 2 + 1)x − (m 2 − 1)R (x − R) = 0 ⇒

x=

(m 2 − 1)R or x = R. m2 + 1

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SECTION 8.2 (PAGE 474)

ADAMS and ESSEX: CALCULUS 8

y

The parametrization of the circle in terms of m is given by (m 2 − 1)R x= m2 + 1   2 2Rm (m − 1)R − R =− 2 y=m 2 m +1 m +1

Y b

P = (x, y)

T

a t

X x

where −∞ < m < ∞. This parametrization gives every point on the circle except (R, 0). y (x,y) slope m x (R,0)

Fig. 8.2.18 x 2 +y 2 =R 2

19. If x =

Fig. 8.2.16

3t 3t 2 , y= , (t 6= −1), then 3 1+t 1 + t3

x 3 + y3 =

17. y T a t

P = (x, y)

27t 3 27t 3 3 (1 + t ) = = 3x y. (1 + t 3 )3 (1 + t 3 )2

As t → −1, we see that |x| → ∞ and |y| → ∞, but x+y=

X x

3t (1 + t) 3t = → −1. 3 1+t 1 − t + t2

Thus x + y = −1 is an asymptote of the curve. y

t=1

Fig. 8.2.17 t→∞

Using triangles in the figure, we see that the coordinates of P satisfy x = a sec t,

y = a sin t.

t=0

x

folium of Descartes

The Cartesian equation of the curve is t→−1−

y2 a2 + = 1. a2 x2 The curve has two branches extending to infinity to the left and right of the circle as shown in the figure.

20. Let C0 and P0 be the original positions of the centre of

18. The coordinates of P satisfy x = a sec t, The Cartesian equation is

Fig. 8.2.19

y = b sin t.

the wheel and a point at the bottom of the flange whose path is to be traced. The wheel is also shown in a subsequent position in which it makes contact with the rail at R. Since the wheel has been rotated by an angle θ ,

y2 a2 + = 1. b2 x2

O R = arc S R = aθ.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 8.2 (PAGE 474)

y

Thus, the new position of the centre is C = (aθ, a). Let P = (x, y) be the new position of the point; then

Tt

x = O R − P Q = aθ − b sin(π − θ ) = aθ − b sin θ, y = RC + C Q = a + b cos(π − θ ) = a − b cos θ. These are the parametric equations of the prolate cycloid. y

P

Q

Pt =(x,y)

b θ

θt

b

S

C0

t

Ct

t

A a

C

x

a O R

x

P0

Fig. 8.2.21 Fig. 8.2.20

If a = 2 and b = 1, then x = 2 cos t, y = 0. This is a straight line segment. If a = 4 and b = 1, then

y

x=aθ −b sin θ y=a−b cos θ x 2π a

x = 3 cos t + cos 3t = 3 cos t + (cos 2t cos t − sin 2t sin t)   = 3 cos t + (2 cos2 t − 1) cos t − 2 sin2 t cos t

= 2 cos t + 2 cos3 t − 2 cos t (1 − sin2 t) = 4 cos3 t y = 3 sin t + sin 3t = 3 sin t − sin 2t cos t − (cos 2t sin t)   = 3 sin t − 2 sin t cos2 t − (1 − 2 sin2 t) sin t

Fig. 8.2.20

= 2 sin t − 2 sin t + 2 sin3 t + 2 sin3 t = 4 sin3 t This is an astroid, similar to that of Exercise 11.

22. a) From triangles in the figure, 21. Let t and θt be the angles shown in the figure below. Then arc ATt = arc Tt Pt , that is, at = bθt . The centre C t of the rolling circle is Ct = (a − b) cos t, (a − b) sin t . Thus

x = |T X| = |OT | tan t = tan t
y = |OY | = si n π2 − t = |OY | cos t

x − (a − b) cos t = b cos(θt − t) y − (a − b) sin t = −b sin(θt − t). Since θt − t =

= |OT | cos t cos t = cos2 t. y y=1 T

a a−b t −t = t, therefore b b

Y

1 2

  (a − b)t x = (a − b) cos t + b cos b   (a − b)t y = (a − b) sin t − b sin . b

X

P = (x, y)

t

O

x

Fig. 8.2.22

327 Copyright © 2014 Pearson Canada Inc.

SECTION 8.2 (PAGE 474)

b)

23.

ADAMS and ESSEX: CALCULUS 8

1 1 = sec2 t = 1 + tan2 t = 1 + x 2 . Thus y = . y 1 + x2

x = sin t,

y = sin(2t)

27.

y

x

  1 1 x = 1+ cos t − cos(nt) n n   1 1 y = 1+ sin t − sin(nt) n n represents a cycloid-like curve that is wound around the circle x 2 + y 2 = 1 instead of extending along the xaxis. If n ≥ 2 is an integer, the curve closes after one revolution and has n − 1 cusps. The left figure below shows the curve for n = 7. If n is a rational number, the curve will wind around the circle more than once before it closes. y

Fig. 8.2.23 x

24.

x = sin t,

y = sin(3t)

y Fig. 8.2.27 x

28.

Fig. 8.2.24

25.

x = sin(2t),

y = sin(3t)





1 1 x = 1+ cos t + cos((n − 1)t) n n   1 1 sin t − sin((n − 1)t) y = 1+ n n represents a cycloid-like curve that is wound around the  2 inside circle x 2 + y 2 = 1 + (2/n) and is externally

tangent to x 2 + y 2 = 1. If n ≥ 2 is an integer, the curve closes after one revolution and has n cusps. The figure shows the curve for n = 7. If n is a rational number but not an integer, the curve will wind around the circle more than once before it closes. y

y

x

x Fig. 8.2.25

26.

x = sin(2t),

y = sin(5t)

Fig. 8.2.28 y

Section 8.3 Smooth Parametric Curves and Their Slopes (page 479) x

Fig. 8.2.26

1.

x = t2 + 1 y = 2t − 4 dx dy = 2t =2 dt dt No horizontal tangents. Vertical tangent at t = 0, i.e., at (1, −4).

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 8.3 (PAGE 479)

2.

x = t 2 − 2t y = t 2 + 2t dx dy = 2t − 2 = 2t + 2 dt dt Horizontal tangent at t = −1, i.e., at (3, −1). Vertical tangent at t = 1, i.e., at (−1, 3).

3.

x = t 2 − 2t y = t 3 − 12t dx dy = 2(t − 1) = 3(t 2 − 4) dt dt Horizontal tangent at t = ±2, i.e., at (0, −16) and (8, 16). Vertical tangent at t = 1, i.e., at (−1, −11).

4.

x = t 3 − 3t y = 2t 3 + 3t 2 dx dy = 3(t 2 − 1) = 6t (t + 1) dt dt Horizontal tangent at t = 0, i.e., at (0, 0). Vertical tangent at t = 1, i.e., at (−2, 5). At t = −1 (i.e., at (2, 1)) both d x/dt and d y/dt change sign, so the curve is not smooth there. (It has a cusp.) 2

x = te−t /2 y = e−t dx dy 2 2 = (1 − t 2 )e−t /2 = −2te−t dt dt Horizontal tangent at t = 0, i.e., at (0, 1). Vertical tangent at t = ±1, i.e. at (±e−1/2 , e−1 ).

6.

x = sin t y = sin t − t cos t dx dy = cos t = t sin t dt dt Horizontal tangent at t = nπ , i.e., at (0, −(−1)n nπ ) (for integers n). Vertical tangent at t = (n + 12 )π , i.e. at (1, 1) and (−1, −1).

8.

9.

x = t4 − t2 y = t 3 + 2t dy dx = 4t 3 − 2t = 3t 2 + 2 dt dt 3(−1)2 + 2 5 dy = =− . At t = −1; dx 4(−1)3 − 2(−1) 2

11.

x = cos(2t) y = sin t dx dy = −2 sin(2t) = cos t dt dt π dy cos(π/6) 1 At t = ; = =− . 6 dx −2 sin(π/3) 2

12.

y = te2t dy = e2t (1 + 2t) dt dy e−4 (1 − 4) 3 At t = −2; = =− . dx 2 2e−4

13.

x = t 3 − 2t = −1 y = t + t 3 = 2 at t = 1 dx dy = 3t 2 − 2 = 1 = 1 + 3t 2 = 4 at t = 1 dt dt Tangent line: x = −1 + t, y = 2 + 4t. This line is at (−1, 2) at t = 0. If you want to be at that point at t = 1 instead, use

2

5.

7.

10.

x = −1 + (t − 1) = t − 2,

14.

x = sin(2t) y = sin t dx dy = 2 cos(2t) = cos t dt dt Horizontal tangent at t = (n + 12 )π , i.e., at (0, ±1). √ Vertical tangent at t = 12 (n + 21 )π , i.e., at (±1, 1/ 2) and √ (±1, −1/ 2). 3t 3t 2 y= 3 1+t 1 + t3 3 dx 3(1 − 2t ) dy 3t (2 − t 3 ) = = dt dt (1 + t 3 )2 (1 + t 3 )2 Horizontal tangent at t = 0 and t = 21/3 , i.e., at (0, 0) and (21/3 , 22/3 ). Vertical tangent at t = 2−1/3 , i.e., at (22/3 , 21/3 ). The curve also approaches (0, 0) vertically as t → ±∞. x=

x = t3 + t dx = 3t 2 + 1 dt dy At t = 1; = dx

y = 1 − t3 dy = −3t 2 dt −3(1)2 3 =− . 3(1)2 + 1 4

x = e2t dx = 2e2t dt

15.

y = 2 + 4(t − 1) = 4t − 2.

π 1 −√ 4 2 1 dx = 1 + sin t = 1 + √ dt 2 1 π y = 1 − sin t = 1 − √ at t = 4 2 1 π dy = − cos t = − √ at t = dt 2 4  π 1 1 Tangent line: x = − √ + 1 + √ t, 4 2 2 1 t y =1− √ − √ . 2 2 x = t − cos t =

x = t 3 − t, y = t 2 is at (0, 1) at t = −1 and t = 1. Since dy 2t ±2 = 2 = = ±1, dx 3t − 1 2 the tangents at (0, 1) at t = ±1 have slopes ±1.

16.

x = sin t, y = sin(2t) is at (0, 0) at t = 0 and t = π . Since  dy 2 cos(2t) 2 if t = 0 = = −2 if t = π , dx cos t the tangents at (0, 0) at t = 0 and t = π have slopes 2 and −2, respectively.

329 Copyright © 2014 Pearson Canada Inc.

SECTION 8.3 (PAGE 479)

17.

18.

ADAMS and ESSEX: CALCULUS 8

x = t3 y = t2 dx dy = 3t 2 = 2t both vanish at t = 0. dt dt dy 2 dx 3t = has no limit as t → 0. = → 0 as dx 3t dy 2 t → 0, but d y/dt changes sign at t = 0. Thus the curve is not smooth at t = 0. (In this solution, and in the next five, we are using the Remark following Example 2 in the text.) x dx dt y dy dt

The tangent is horizontal at t = 2, (i.e., (0, −4)), and is vertical at t = 1 (i.e., at (−1, −3). Observe that d 2 y/d x 2 > 0, and the curve is concave up, if t > 1. Similarly, d 2 y/d x 2 < 0 and the curve is concave down if t < 1. y x=t 2 −2t y=t 2 −4t

4

= (t − 1)

= 4(t − 1)3

x

3

= (t − 1)

= 3(t − 1)2

t=1

both vanish at t = 1.

t=2

dx 4(t − 1) = → 0 as t → 1, and d y/dt does not dy 3 change sign at t = 1, the curve is smooth at t = 1 and therefore everywhere.

Since

19.

20.

x = t sin t y = t3 dx dy = sin t + t cos t = 3t 2 both vanish at t = 0. dt dt 3t 2 6t dy = lim = lim = 0, lim t→0 sin t + t cos t t→0 2 cos t − t sin t t→0 d x but d x/dt changes sign at t = 0. d x/d y has no limit at t = 0. Thus the curve is not smooth at t = 0.

x = t3 y = t − sin t dx dy 2 = 3t = 1 − cos t both vanish at t = 0. dt dt dx 3t 2 6t lim = lim = lim = 6 and d y/dt does t→0 d y t→0 1 − cos t t→0 sin t not change sign at t = 0. Thus the curve is smooth at t = 0, and hence everywhere.

21. If x = t 2 − 2t and y = t 2 − 4t, then

dx dy = 2(t − 1), = 2(t − 2) dt dt d2x d2 y = 2 =2 2 dt dt d2 y 1 d dy = dx2 d x/dt dt d x 1 d t −2 1 = = . 2(t − 1) dt t − 1 2(t − 1)3

Fig. 8.3.21

22. If x = f (t) = t 3 and y = g(t) = 3t 2 − 1, then f ′ (t) = 3t 2 , f ′′ (t) = 6t; g ′ (t) = 6t, g ′′ (t) = 6. Both f ′ (t) and g ′ (t) vanish at t = 0. Observe that 6t 2 dy = 2 = . dx t 3t Thus, lim

t→0+

dy = ∞, dx

lim

t→0−

dy = −∞ dx

and the curve has a cusp at t = 0, i.e., at (0, −1). Since d2 y (3t 2 )(6) − (6t)(6t) 2 = =− 4 1 the curve does not approach the origin, and the petals become less distinct as C increases.

29. If r = 1/θ for θ > 0, then lim y = lim

θ →0+

θ →0+

sin θ = 1. θ

Thus y = 1 is a horizontal asymptote. y

y=1

37. r = C + cos θ sin(3θ )

r=1/θ

For C < 1 there appear to be 6 petals of 3 different sizes. For C ≥ 1 there are only 4 of 2 sizes, and these coalesce as C increases.

x

38.

Fig. 8.5.29

y

30. The graph of r = cos nθ has 2n leaves if n is an even

integer and n leaves if n is an odd integer. The situation for r 2 = cos nθ is reversed. The graph has 2n leaves if n is an odd integer (provided negative values of r are allowed), and it has n leaves if n is even.

r = ln(θ ) x

31. If r = f (θ ), then x = r cos θ = f (θ ) cos θ y = r sin θ = f (θ ) sin θ.

32. r = cos θ cos(mθ )

Fig. 8.5.38

For odd m this flower has 2m petals, 2 large ones and 4 each of (m − 1)/2 smaller sizes. For even m the flower has m + 1 petals, one large and 2 each of m/2 smaller sizes.

33. r = 1 + cos θ cos(mθ )

These are similar to the ones in Exercise 32, but the curve does not approach the origin except for θ = π in the case of even m. The petals are joined, and less distinct. The smaller ones cannot be distinguished.

We will have [ln θ1 , θ1 ] = [ln θ2 , θ2 ] if θ2 = θ1 + π

and

ln θ1 = − ln θ2 ,

that is, if ln θ1 + ln(θ1 + π ) = 0. This equation has solution θ1 ≈ 0.29129956. The corresponding intersection point has Cartesian coordinates (ln θ1 cos θ1 , ln θ1 sin θ1 ) ≈ (−1.181442, −0.354230).

339 Copyright © 2014 Pearson Canada Inc.

SECTION 8.5 (PAGE 489)

ADAMS and ESSEX: CALCULUS 8

y

39. y

x

r = 1/θ

A

x r=θ

Fig. 8.6.2 r = ln(θ )

3.

Fig. 8.5.39 The two intersections of r = ln θ and r = 1/θ for 0 < θ ≤ 2π correspond to solutions θ1 and θ2 of 1 ln θ1 = , θ1

1 Area = 4 × 2 = 2a 2

π/4

Z

π/4 sin 2θ = a 2 sq. units. 2 0 y

1 . ln θ2 = − θ2 + π

r 2 =a 2 cos 2θ x

The first equation has solution θ1 ≈ 1.7632228, giving the point (−0.108461, 0.556676), and the second equation has solution θ2 ≈ 0.7746477, giving the point (−0.182488, −0.178606).

Section 8.6 Slopes, Areas, and Arc Lengths for Polar Curves (page 493)

a 2 cos 2θ dθ

0

Fig. 8.6.3

4.

Area = =

1 2

Z

0

π/3

sin2 3θ dθ =

1 4

Z

π/3 0

(1 − cos 6θ ) dθ

  π/3 π 1 1 = θ − sin 6θ sq. units. 4 6 12 0 y

1. Area =

1 2

Z

2π

0

θ dθ =

(2π )2 4

π θ= 3

= π 2.

y

A x

√ r= θ r=sin 3θ θ =0

θ =2π x

Fig. 8.6.4

Fig. 8.6.1

5.

Z 1 π/8 Total area = 16 × cos2 4θ dθ 2 0 Z π/8 =4 (1 + cos 8θ ) dθ 0

2.

1 Area = 2

Z

2π 0

2π θ 3 4 θ 2 dθ = = π 3 sq. units. 6 0 3

340 Copyright © 2014 Pearson Canada Inc.



=4 θ+

 π/8 sin 8θ π = sq. units. 8 2 0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 8.6 (PAGE 493)

y

8.

r=cos 4θ π/8

x

Z 1 2 1 π/2 2 πa + 2 × a (1 − sin θ )2 dθ 2 2 0  Z π/2  π a2 1 − cos 2θ = + a2 1 − 2 sin θ + dθ 2 2 0   π/2 π a2 3 1 = + a 2 θ + 2 cos θ − sin 2θ 2 2 4 0   5π 2 = − 2 a sq. units. 4

Area =

y

Fig. 8.6.5 r=a

6. The circles r = a and r = 2a cos θ intersect

4×



x

A

at θ = ±π/3. By symmetry, the common area is 4×(area of sector−area of right triangle) (see the figure), i.e.,

r=a(1−sin θ )

√  √   1 2 1 a 3a 4π − 3 3 2 πa − = a sq. units. 6 22 2 6

Fig. 8.6.8

y r=2a cos θ

r=a

9. For intersections: 1 + cos θ = 3 cos θ . Thus 2 cos θ = 1 and θ = ±π/3. The shaded area is given by

A

x

Fig. 8.6.6 π

π 2  Z π 1 + cos 2θ π 1 − 2 cos θ + = dθ − 2 2 π/2   π   π 3 π sin 2θ = π− − 2 sin θ − − 2 2 4 2 π/2 π = + 2 sq. units. 4

7. Area = 2 ×

1 2 

Z

π/2

(1 − cos θ )2 dθ −

2×

1 2

 Z π/2 cos2 θ dθ (1 + cos θ )2 dθ − 9 π/3 π/3  Z π  1 + cos 2θ 1 + 2 cos θ + = dθ 2 π/3 Z 9 π/2 (1 + cos 2θ ) dθ − 2 π/3     π 3 2π sin 2θ = + 2 sin θ + 2 3 4

Z

π

π/3

 π/2 9 sin 2θ − θ+ 2 2 π/3 √ √ ! π √ 3 9 3 π = − 3− + sq. units. = 4 8 4 2 4 

y

y

π/3 r=3 cos θ

r=1−cos θ r=1+cos θ

r=1

x

x

Fig. 8.6.7

Fig. 8.6.9

341 Copyright © 2014 Pearson Canada Inc.

SECTION 8.6 (PAGE 493)

ADAMS and ESSEX: CALCULUS 8

π 5π and ± , 6 6 the area inside the lemniscate and outside the circle is Z i 1 π/6 h 4× 2 cos 2θ − 12 dθ 2 0 π/6 √ π π − = 3− = 2 sin 2θ sq. units. 3 3

10. Since r 2 = 2 cos 2θ meets r = 1 at θ = ±

dr = aeaθ . dθ √ + a 2 e2aθ dθ = 1 + a 2 eaθ dθ . The length of

13. r = eaθ , (−π ≤ θ ≤ π ). √ ds = e2aθ the curve is Z

π

−π

p 1 + a 2 eaθ dθ =

√

1 + a 2 aπ (e − e−aπ ) units. a

0

y

14.

r=1

A

A x r 2 =2 cos 2θ

Z

π 2π/3

=

Z

(1 + 2 cos θ )2 dθ π

2π/3



 1 + 4 cos θ + 2(1 + cos 2θ ) dθ

π π =3 + sin 2θ + 4 sin θ 3 2π/3 2π/3 √ √ √ 3 3 3 =π −2 3+ =π− sq. units. 2 2 π 

y 2π/3

2π

0

=a

Z

=a

Z

p a 2 + a 2 θ 2 dθ

2π 0

p

1 + θ 2 dθ

θ =2π

Let θ = tan u dθ = sec2 u dθ

sec3 u du

θ =0

 θ =2π a sec u tan u + ln | sec u + tan u| 2 θ =0 i θ =2π p ah p = θ 1 + θ 2 + ln | 1 + θ 2 + θ | 2 θ =0 i p ah p = 2π 1 + 4π 2 + ln(2π + 1 + 4π 2 ) units. 2

11. r = 0 at θ = ±2π/3. The shaded area is 1 2

Z

=

Fig. 8.6.10

2×

s=

15.

r 2 = cos 2θ dr = −2 sin 2θ 2r dθ s

⇒

dr sin 2θ =− dθ r

√ sin2 2θ dθ = sec 2θ dθ cos 2θ Z π/4 √ Length = 4 sec 2θ dθ.

ds =

cos 2θ + 0

r=1+2 cos θ

y

r 2 =cos 2θ

3 x

1

x

−2π/3

Fig. 8.6.15

Fig. 8.6.11

12.

s=

Z

π

0

Z

π

s 

dr dθ

2

+ r 2 dθ

p θ 4 + θ 2 dθ

=

Z

0

π

p 4θ 2 + θ 4 dθ

Let u = 4 + θ 2 du = 2θ dθ 4+π 2 Z 4+π 2 √ 1 1 = u du = u 3/2 2 4 3 4 i 1h 2 3/2 (4 + π ) − 8 units. = 3 =

0

16. If r 2 = cos 2θ , then 2r

dr sin 2θ dr = −2 sin 2θ ⇒ = −√ dθ dθ cos 2θ

and ds =

342 Copyright © 2014 Pearson Canada Inc.

s

cos 2θ +

sin2 2θ dθ dθ = √ . cos 2θ cos 2θ

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 8.6 (PAGE 493)

a) Area of the surface generated by rotation about the x- axis is Sx = 2π

Z

π/4

r sin θ ds

0

Z

π/4 √

dθ cos 2θ sin θ √ cos 2θ 0 π/4 √ = (2 − 2)π sq. units. = −2π cos θ

= 2π

0

b) Area of the surface generated by rotation about the y- axis is

i.e., at P1 =



√ π 2, 4



and P2 = (0, 0).

dr For r 2 = 2 sin 2θ we have 2r = 4 cos 2θ . At P1 we dθ √ have r = 2 and dr/dθ = 0. Thus the angle ψ between the curve and the radial line θ = π/4 is ψ = π/2. For r = 2 cos θ we have dr/dθ = −2 sin θ , so the angle between this curve and the radial line θ = π/4 satisfies r tan ψ = = −1, and ψ = 3π/4. The two dr/dθ θ =π/4

3π π π curves intersect at P1 at angle − = . 4 2 4 The Figure shows that at the origin, P2 , the circle meets the lemniscate twice, at angles 0 and π/2. y

S y = 2π = 4π

π/4

Z

r cos θ ds

−π/4 π/4 √

Z

0

cos 2θ cos θ √

dθ cos 2θ

π/4 √ = 4π sin θ = 2 2π sq. units.

x

0

r=2 cos θ

r 2 =2 sin 2θ

Fig. 8.6.18

17. For r = 1 + sin θ , 1 + sin θ r = . tan ψ = dr/dθ cos θ √ If θ = π/4, then tan ψ = 2 +√1 and ψ = 3π/8. If θ = 5π/4, then tan ψ = 1 − 2 and ψ = −π/8. The line y = x meets the cardioid r = 1 + sin θ at the origin at an angle of 45◦ , and also at first and third quadrant points at angles of 67.5◦ and −22.5◦ as shown in the figure. y

19. The curves r = 1 − cos θ and r = 1 − sin θ intersect on

the rays θ = π/4 and θ = 5π/4, as well as at the origin. At the origin their cusps clearly intersect at right angles. For r = 1 − cos θ , tan ψ √1 = (1 − cos θ )/ sin θ . At θ = π/4, tan ψ1 = 2 − √ 1, so ψ1 = π/8. At θ = 5π/4, tan ψ1 = −( 2 + 1), so ψ1 = −3π/8. For r = 1 − sin θ , tan ψ2 =√(1 − sin θ )/(− cos θ ). At θ = π/4, tan ψ2 = 1√ − 2, so ψ2 = −π/8. At θ = 5π/4, tan ψ2 = 2 + 1, so ψ2 = 3π/8. At π/4 the curves intersect at angle π/8−(−π/8) = π/4. At 5π/4 the curves intersect at angle 3π/8 − (−3π/8) = 3π/4 (or π/4 if you use the supplementary angle). y

r=1+sin θ

r=1−cos θ

θ = π/4

r=1−sin θ

ψ

x x

Fig. 8.6.17

18. The two curves r 2 = 2 sin 2θ and r = 2 cos θ intersect where

2 sin 2θ = 4 cos2 θ

Fig. 8.6.19

20. We have r = cos θ + sin θ . For horizontal tangents:

4 sin θ cos θ = 4 cos θ (sin θ − cos θ ) cos θ = 0 ⇔ sin θ = cos θ or cos θ = 0,

 dy d  = cos θ sin θ + sin2 θ dθ dθ = cos2 θ − sin2 θ + 2 sin θ cos θ cos 2θ = − sin 2θ ⇔ tan 2θ = −1.

0=

2

⇔

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SECTION 8.6 (PAGE 493)

ADAMS and ESSEX: CALCULUS 8

π 3π Thus θ = − or . The tangents are horizontal at 8  8     π π π cos − sin ,− and 8 8 8       3π 3π 3π + sin , . cos 8 8 8 For vertical tangent:  dx d  2 0= = cos θ + cos θ sin θ dθ dθ = −2 cos θ sin θ + cos2 θ − sin2 θ ⇔ sin 2θ = cos 2θ ⇔ tan 2θ = 1. Thus are vertical tangents at  There   θ = π/8 of 5π/8. π π π + sin , and cos 8 8 8       5π 5π 5π cos + sin , . 8 8 8 y

22. We have r 2 = cos 2θ , and 2r zontal tangents:

dr = −2 sin 2θ . For horidθ

⇔

  d sin 2θ r sin θ = r cos θ + sin θ − dθ r cos 2θ cos θ = sin 2θ sin θ

⇔

cos θ = 0 or

0=

⇔ (cos2 θ − sin2 θ ) cos θ = 2 sin2 θ cos θ cos2 θ = 3 sin2 θ.

There are no points on the curve where cos θ = 0. Therefore, horizontal tangents occur only where tan2 θ = 1/3. There are horizontal tangents at    1 π 1 5π √ ,± and √ , ± . 6 6 2 2 For vertical tangents:   sin 2θ d r cos θ = −r sin θ + cos θ − 0= dθ r ⇔ cos 2θ sin θ = − sin 2θ cos θ ⇔ (cos2 θ − sin2 θ ) sin θ = −2 sin θ cos2 θ ⇔

sin θ = 0

or

3 cos2 θ = sin2 θ.

There are no points on the curve where tan2 θ = 3, so the only vertical tangents occur where sin θ = 0, that is, at the points with polar coordinates [1, 0] and [1, π ].

r=cos θ +sin θ

y x r 2 =cos 2θ

Fig. 8.6.20

x

Fig. 8.6.22

21.

r = − cot θ . r = 2 cos θ . tan ψ = dr/dθ For horizontal tangents we want tan ψ = − tan θ . Thus we want − tan θ = − cot θ , and √ so θ = ±π/4 or ±3π/4. The tangents are horizontal at [ 2, ±π/4]. For vertical tangents we want tan ψ = cot θ . Thus we want − cot θ = cot θ , and so θ = 0, ±π/2, or π . There are vertical tangents at the origin and at [2, 0]. y θ =π/4 r=2 cos θ 2 x

θ =−π/4

Fig. 8.6.21

sin 2θ = 2 cos 2θ For horizontal tangents:

23. r = sin 2θ . tan ψ =

1 2

tan 2θ .

tan 2θ = −2 tan θ 2 tan θ = −2 tan θ 2 1 − tan  θ 

tan θ 1 + (1 − tan2 θ ) = 0

tan θ (2 − tan2 θ ) = 0. √ √ Thus θ = 0, π , ± tan−1 2, π ± tan−1 2. There are horizontal tangents at the origin and the points # " √ # " √ √ √ 2 2 2 2 , ± tan−1 2 and , π ± tan−1 2 . 3 3 Since the rosette r = sin 2θ is symmetric about x = y, there must be vertical tangents at the origin and at the points " √ # " √ # 2 2 2 2 −1 1 −1 1 , ± tan √ and , π ± tan √ . 3 3 2 2

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INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 8 (PAGE 494)

y

y r=2(1−sin θ ) x x r=sin 2θ

Fig. 8.6.23

24. We have r = eθ and

Fig. 8.6.25

dr = eθ . For horizontal tangents: dθ

d r sin θ = eθ cos θ + eθ sin θ dθ π tan θ = −1 ⇔ θ = − + kπ, 4

0= ⇔

where k = 0, ±1, ±2, . . .. At the points [ekπ −π/4 , kπ − π/4] the tangents are horizontal. For vertical tangents: d r cos θ = eθ cos θ − eθ sin θ dθ π tan θ = 1 ↔ θ = + kπ. 4

0= ⇔

At the points [ekπ +π/4 , kπ + π/4] the tangents are vertical. 1 − sin θ . cos θ For horizontal tangents tan ψ = − cot θ , so

26.

dy dx = f ′ (θ ) cos θ − f (θ ) sin θ, = f ′ (θ ) sin θ + f (θ ) cos θ dθ r dθ 2  2  ds = f ′ (θ ) cos θ − f (θ ) sin θ + f ′ (θ ) sin θ + f (θ ) cos θ dθ   2 2 = f ′ (θ ) cos2 θ − 2 f ′ (θ ) f (θ ) cos θ sin θ + f (θ ) sin2 θ 1/2  2  2 2 2 ′ ′ dθ + f (θ ) sin θ + 2 f (θ ) f (θ ) sin θ cos θ + f (θ ) cos θ r 2  2  = f ′ (θ ) + f (θ ) dθ.

Review Exercises 8 (page 494) 1.

25. r = 2(1 − sin θ ), tan ψ = −

1 − sin θ sin θ =− cos θ cos θ cos θ = 0, or 2 sin θ = 1.

cos θ 1 − sin θ = cos θ sin θ sin2 θ − sin θ = cos2 θ = 1 − sin2 θ

⇔

x2 y2 − =1 4 9 Hyperbola, transverse axis along the x-axis. Semi-transverse axis a = 2, semi-conjugate axis b = 3. √ c2 = a 2 + b2 = 13. Foci: (± 13, 0). Asymptotes: 3x ± 2y = 0.

3.

⇔

x + y 2 = 2y + 3 ⇔ (y − 1)2 = 4 − x Parabola, vertex (4, 1), opening to the left, principal axis y = 1. a = −1/4. Focus: (15/4, 1).

4. 2x 2 + 8y 2 = 4x − 48y

2(x 2 − 2x + 1) + 8(y 2 + 6y + 9) = 74

−

(y + 3)2 (x − 1)2 + = 1. 37 37/4

2 sin2 θ − sin θ − 1 = 0 (sin θ − 1)(2 sin θ + 1) = 0

The solutions here are θ = π/2 (the origin again), θ = −π/6 and θ = −5π/6. There are vertical tangents at [3, −π/6] and [3, −5π/6].

x2 + y2 = 1 2 √ Ellipse, semi-major axis a = 2, along the x-axis. Semiminor axis b = 1. c2 = a 2 − b2 = 1. Foci: (±1, 0). x 2 + 2y 2 = 2

2. 9x 2 − 4y 2 = 36

−

The solutions are θ = ±π/2, ±π/6, and ±5π/6. θ = π/2 corresponds to the origin where the cardioid has a cusp, and therefore no tangent. There are horizontal tangents at [4, −π/2], [1, π/6], and [1, 5π/6]. For vertical tangents tan ψ = cot θ , so

x = r cos θ = f (θ ) cos θ , y = r sin θ = f (θ ) sin θ .

Ellipse, √ centre (1, √ −3), major axis along y = −3. a = 37, b√= 37/2, c2 = a 2 − b2 = 111/4. Foci: (1 ± 111/2, −3).

5.

x = t, y = 2 − t, (0 ≤ t ≤ 2). Straight line segment from (0, 2) to (2, 0).

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REVIEW EXERCISES 8 (PAGE 494)

ADAMS and ESSEX: CALCULUS 8

6.

x = 2 sin(3t), y = 2 cos(3t), (0 ≤ t ≤ 2) Part of a circle of radius 2 centred at the origin from the point (0, 2) clockwise to (2 sin 6, 2 cos 6).

7.

x = cosh t, y = sinh2 t. Parabola x 2 − y = 1, or y = x 2 − 1, traversed left to right.

8.

x = et , y = e−2t , (−1 ≤ t ≤ 1). Part of the curve x 2 y = 1 from (1/e, e2 ) to (e, 1/e2 ).

9.

x = cos(t/2), y = 4 sin(t/2), (0 ≤ t ≤ π ). The first quadrant part of the ellipse 16x 2 + y 2 = 16, traversed counterclockwise.

10.

11.

13.

x = t 3 − 3t y = t3 dx dy = 3(t 2 − 1) = 3t 2 dt dt Horizontal tangent at t = 0, i.e., at (0, 0). Vertical tangent at t = ±1, i.e., at (2, −1) and (−2, 1).  dy t2 >0 = 2 1 if |t| < 1

t=1

x

t=−1

x=

t=−1

√ t=± 3

t=0

Fig. R-8.13

14.

x = t 3 − 3t y dx dy 2 = 3(t − 1) dt dt Horizontal tangent at t (−2, 16). Vertical tangent at t =

= t 3 − 12t = 3(t 2 − 4)

= ±2, i.e., at (2, −16) and

±1, i.e., at (2, 11) and (−2, −11).

x  t2 − 4 dy > 0 if |t| > 2 or |t| < 1 = 2 < 0 if 1 < |t| < 2 dx t −1 Slope → 1 as t → ±∞. y (−2,16)

Slope t=1

Fig. R-8.11

12.

x = t 3 − 3t y = t 3 + 3t dx dy = 3(t 2 − 1) = 3(t 2 + 1) dt dt Horizontal tangent: none. Vertical tangent at t = ±1, i.e., at (2, −4) and (−2, 4).

(2,11)

(−2,−11)

 dy t2 + 1 > 0 if |t| > 1 Slope = 2 < 0 if |t| < 1 dx t −1 Slope → 1 as t → ±∞. y (−2,4)

x

(2,−16)

Fig. R-8.14

15. The curve x = t 3 − t, y = |t 3 | is symmetric about x = 0

x=t 3 −3t y=t 3 +3t

x (2,−4)

Fig. R-8.12

x=t 3 −3t y=t 3 −12t

since x is an odd function and y is an even function. Its self-intersection occurs at a nonzero value of t that makes x = 0, namely, t = ±1. The area of the loop is A=2

Z

t=1

t=0

(−x) d y = −2

Z

0

1

(t 3 − t)3t 2 dt

  1 3 1 = −t 6 + t 4 = sq. units. 2 2 0

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INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 8 (PAGE 494)

y

y

r =θ x

x = t3 − t y = |t 3 | t=±1

Fig. R-8.19

x

t=0

Fig. R-8.15

20. r = |θ |,

(−2π ≤ θ ≤ 2π ) y r=|θ |

16. The volume of revolution about the y-axis is V =π =π

t=1

Z

x2 dy

t=0 Z 1 0

= 3π = 3π

Z

x

(t 6 − 2t 4 + t 2 )3t 2 dt 1

(t 8 − 2t 6 + t 4 ) dt  1 2 1 8π − + = cu. units. 9 7 5 105

0

Fig. R-8.20

21. r = 1 + cos(2θ ) 17.

y

x = et − t, y = 4et/2 , (0 ≤ t ≤ 2). Length is Z

2p

(et − 1)2 + 4et dt Z 2p Z 2 = (et + 1)2 dt = (et + 1) dt 0 0 2 = (et + t) = e2 + 1 units.

L=

0

r=1+cos 2θ

x

Fig. R-8.21

0

22. r = 2 + cos(2θ )

y

18. Area of revolution about the x-axis is r=2+cos(2θ )

S = 2π = 8π =

Z



4et/2 (et + 1) dt  2 2 3t/2 e + 2et/2 3 0

x

16π 3 (e + 3e − 4) sq. units. 3 Fig. R-8.22

19. r = θ,

−3π 2

≤θ ≤

3π 2




23. r = 1 + 2 cos(2θ )

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REVIEW EXERCISES 8 (PAGE 494)

ADAMS and ESSEX: CALCULUS 8

y

27. r = 1 +

√

2 sin θ approaches the origin in the directions √ for which sin θ = −1/ 2, that is, θ = −3π/4 and θ = −π/4. The smaller loop corresponds to values of θ between these two values. By symmetry, the area of the loop is

r=1+2 cos 2θ

x

A=2× =

Fig. R-8.23

Z

1 2

Z

−π/4 −π/2

−π/4 −π/2

√ (1 + 2 2 sin θ + 2 sin2 θ ) dθ

√ (2 + 2 2 sin θ − cos(2θ )) dθ

 −π/4 √ 1 = 2θ − 2 2 cos θ − sin(2θ ) 2 −π/2 π 1 π −3 = −2+ = sq. units. 2 2 2 

24. r = 1 − sin(3θ )

y

r=1−sin(3θ )

y

√ r=1+ 2 sin θ

π/6

x

−3π/4

Fig. R-8.24

28. r cos θ = x = 1/4 and r = 1 + cos θ intersect where

1 π/3 (1 + 2 cos(2θ ))2 dθ A=2× 2 0 Z π/3 [1 + 4 cos(2θ ) + 2(1 + cos(4θ ))] dθ =

1 4 cos θ 4 cos2 θ + 4 cos θ − 1 = 0 √ √ −4 ± 16 + 16 ± 2−1 cos θ = = . 8 2

1 + cos θ =

Z

0

 π/3 1 = 3θ + 2 sin(2θ ) + sin(4θ ) 2 0 √ 3 3 =π+ sq. units. 4 

26. Area of a small loop: A=2× =

Z

1 2

π/2 π/3

π/2 π/3

x

Fig. R-8.27

25. Area of a large loop:

Z

−π/4

√ Only ( 2 − 1)/2 is between −1 √ and 1, so is a possible 2−1 value of cos θ . Let θ0 = cos−1 . Then 2 v p !2 u √ √ u 2−1 1+2 2 t = . sin θ0 = 1 − 2 2 By symmetry, the area inside r = 1 + cos θ to the left of the line x = 1/4 is

(1 + 2 cos(2θ ))2 dθ

[1 + 4 cos(2θ ) + 2(1 + cos(4θ ))] dθ

  π/2 1 = 3θ + 2 sin(2θ ) + sin(4θ ) 2 π/3 √ π 3 3 = − sq. units. 2 4

π

 1 + cos(2θ ) dθ + cos θ0 sin θ0 2 θ0   π 3 1 = (π − θ0 ) + 2 sin θ + sin(2θ ) 2 4 θ0 p √ √ ( 2 − 1) 1 + 2 2 + 4 ! q ! √ √ √ 2 − 1 2 − 9 3 = π − cos−1 + 1+2 2 sq. units. 2 2 8

A=2×

1 2

348 Copyright © 2014 Pearson Canada Inc.

Z



1 + 2 cos θ +

INSTRUCTOR’S SOLUTIONS MANUAL

x=1/4

y

CHALLENGING PROBLEMS 8 (PAGE 494)

r=1+cos θ

θ0

C1

x

A1 S1

Fig. R-8.28 C

Challenging Problems 8 (page 494)

P

F1

F2 C2 A2

1. The surface of the water is elliptical (see Problem 2 below) whose semi-minor axis is 4 cm, the radius of the cylinder, and whose semi-major axis is 4 sec θ cm because of the tilt of the glass. The surface area is that of the ellipse x = 4 sec θ cos t, y = 4 sin t,

(0 ≤ t ≤ 2π ).

This area is A=4

Z

Fig. C-8.2 Let P be any point on C. Let A1 A2 be the line through P that lies on the cylinder, with A1 on C1 and A2 on C2 . Then P F1 = P A1 because both lengths are of tangents drawn to the sphere S1 from the same exterior point P. Similarly, P F2 = P A2 . Hence P F1 + P F2 = P A1 + P A2 = A1 A2 ,

t=π/2

x dy

t=0 π/2

Z

(4 sec θ cos t)(4 cos t) dt Z π/2 = 32 sec θ (1 + cos(2t)) dt = 16π sec θ cm2 .

=4

S2

0

0

θ

4 cm

which is constant, the distance between the centres of the two spheres. Thus C must be an ellipse, with foci at F1 and F2 .

3. Given the foci F1 and F2 , and the point P on the ellipse, construct N1 P N2 , the bisector of the angle F1 P F2 . Then construct T1 PT2 perpendicular to N1 N2 at P. By the reflection property of the ellipse, N1 N2 is normal to the ellipse at P. Therefore T1 T2 is tangent there. N2 T1 P

4 sec θ cm

T2

θ θ N1 F1

F2

Fig. C-8.1

Fig. C-8.3

2. Let S1 and S2 be two spheres inscribed in the cylinder, one on each side of the plane that intersects the cylinder in the curve C that we are trying to show is an ellipse. Let the spheres be tangent to the cylinder around the circles C1 and C2 , and suppose they are also tangent to the plane at the points F1 and F2 , respectively, as shown in the figure.

4. Without loss of generality, choose the axes and axis scales so that the parabola has equation y = x 2 . If P is the point (x0 , x02 ) on it, then the tangent to the parabola at P has equation y = x02 + 2x0 (x − x0 ),

349 Copyright © 2014 Pearson Canada Inc.

CHALLENGING PROBLEMS 8 (PAGE 494)

ADAMS and ESSEX: CALCULUS 8

which intersects the principal axis x = 0 at (0, −x02 ). Thus R = (0, −x02 ) and Q = (0, x02 ). Evidently the vertex V = (0, 0) bisects R Q. y

6.

y P = [r, θ ] [a, θ0 ]

Q

P=

(x0 , x02 )

r a θ θ0

x

V

x

Fig. C-8.6

R

Fig. C-8.4 To construct the tangent at a given point P on a parabola with given vertex V and principal axis L, drop a perpendicular from P to L, meeting L at Q. Then find R on L on the side of V opposite Q and such that QV = V R. Then P R is the desired tangent.

5.

L

a) Let L be a line not passing through the origin, let [a, θ0 ] be the polar coordinates of the point L that is closest to the origin. If P = [r, θ ] is point on the line, then, from the triangle in the ure, a = cos(θ − θ0 ), r

or r =

and on any fig-

a . cos(θ − θ0 )

b) As shown in part (a), any line not passing through the origin has equation of the form

y b

r = g(θ ) =

c

a = a sec(θ − θ0 ), cos(θ − θ0 )

for some constants a and θ0 . We have

2 ft 2 ft

g ′ (θ ) = a sec(θ − θ0 ) tan(θ − θ0 )

a x

g ′′ (θ ) = a sec(θ − θ0 ) tan2 (θ − θ0 )  2  2 + a sec3 (θ − θ0 ) g(θ ) + 2 g ′ (θ ) − g(θ )g ′′ (θ ) = a 2 sec2 (θ − θ0 ) + 2a 2 sec2 (θ − θ0 ) tan2 (θ − θ0 )

− a 2 sec2 (θ − θ0 ) tan2 (θ − θ0 ) − a 2 sec4 (θ − θ0 ) h   i = a 2 sec2 (θ − θ0 ) 1 + tan2 (θ − θ0 ) − sec4 (θ − θ0 ) = 0.

Fig. C-8.5 y2 x2 Let the ellipse be 2 + 2 = 1, with a = 2 and foci at a b (0, ±2) so that c = 2 and b2 = a 2 + c2 = 8. The volume of the barrel is V =2

Z

= 8π

2 0



π x 2 d y = 2π

Z

0

2

  y2 4 1− dy 8

c) If r = g(θ ) is the polar equation of the tangent to r = f (θ ) at θ = α, then g(α) = f (α) and g ′ (α) = f ′ (α). Suppose that 

2  2 f (α) + 2 f ′ (α) − f (α) f ′′ (α) > 0.

By part (b) we have

 2 y 3 40π 3 y− = ft . 24 0 3



350 Copyright © 2014 Pearson Canada Inc.

2  2 g(α) + 2 g ′ (α) − g(α)g ′′ (α) = 0.

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 8 (PAGE 494)

Subtracting, and using g(α) = f (α) and g ′ (α) = f ′ (α), we get f ′′ (α) < g ′′ (α). It follows that f (θ ) < g(θ ) for values of θ near α; that is, the graph of r = f (θ ) is curving to the origin side of its tangent at α. Similarly, if 

8. Take the origin at station O as shown in the figure. Both of the lines L 1 and L 2 pass at distance 100 cos ǫ from the origin. Therefore, by Problem 6(a), their equations are 100 cos ǫ 100 cos ǫ 
 = sin(θ + ǫ) cos θ − π2 − ǫ 100 cos ǫ 100 cos ǫ 
 = r= . π sin(θ − ǫ) cos θ − 2 + ǫ

L1 :

 2 2 f (α) + 2 f ′ (α) − f (α) f ′′ (α) < 0,

r=

L2 :

then the graph is curving to the opposite side of the tangent, away from the origin.

The search area A(ǫ) is, therefore,

7. x(t)

0

B

A x

θ R

r

π 4 +ǫ

 1002 cos2 ǫ 1002 cos2 ǫ − dθ π sin2 (θ − ǫ) sin2 (θ + ǫ) 4 −ǫ Z π +ǫ   4 2 = 5, 000 cos ǫ π csc2 (θ − ǫ) − csc2 (θ + ǫ) dθ

1 A(ǫ) = 2

Z



−ǫ

4

 = 5, 000 cos2 ǫ cot π4 + 2ǫ − 2 cot π4 + cot π4 − 2ǫ # "

sin π4 + 2ǫ cos π4 + 2ǫ 2
+
−2 = 5, 000 cos ǫ sin π4 + 2ǫ cos π4 + 2ǫ 
 = 10, 000 cos2 ǫ csc π2 + 4ǫ − 1 = 10, 000 cos2 ǫ(sec(4ǫ) − 1) mi2 .

For ǫ = 3◦ = π/60, we have A(ǫ) ≈ 222.8 square miles. Also A′ (ǫ) = −20, 000 cos ǫ sin ǫ(sec(4ǫ) − 1)

+ 40, 000 cos2 ǫ sec(4ǫ) tan(4ǫ) A (π/60) ≈ 8645. ′

Fig. C-8.7 When the vehicle is at position x, as shown in the figure, the component of the gravitational force on it in the direction of the tunnel is ma(r ) cos θ = −

When ǫ = 3◦ , the search area increases at about 8645(π/180) ≈ 151 square miles per degree increase in ǫ. y

mg mgr cos θ = − x. R R

ǫ L2 L1

By Newton’s Law of Motion, this force produces an acceleration d 2 x/dt 2 along the tunnel given by m

Area A(ǫ)

100 mi

d2x mg x, =− dt 2 R

ǫ π/4

that is

x

O d2x + ω2 x = 0, dt 2

where

g ω2 = . R

This is the equation of simple harmonic motion, with √ period T = 2π/ω = 2π R/g. For R ≈ 3960 mi ≈ 2.09 × 107 ft, and g ≈ 32 ft/s2 , we have T ≈ 5079 s ≈ 84.6 minutes. This is a rather short time for a round trip between Atlanta and Baghdad, or any other two points on the surface of the earth.

Fig. C-8.8

9. The easiest way to determine which curve is which is to calculate both their areas; the outer curve bounds the larger area. The curve C1 with parametric equations x = sin t,

y=

1 sin(2t), 2

(0 ≤ t ≤ 2π )

351 Copyright © 2014 Pearson Canada Inc.

CHALLENGING PROBLEMS 8 (PAGE 494)

C1 is the outer curve, and the area between the curves is 1/3 sq. units.

has area A1 = 4 =4 =4 Let u = cos t du = − sin t dt =4

Z

ADAMS and ESSEX: CALCULUS 8

t=π/2

y dx

t=0 Z π/2 0

Z

π/2

Z

1

1 sin(2t) cos t dt 2

y

sin t cos2 t dt

0

0

x u 2 du =

4 sq. units. 3

The curve C2 with polar equation r 2 = cos(2θ ) has area π/4 Z 4 π/4 = 1 sq. units. A2 = cos(2θ ) dθ = sin(2θ ) 2 0 0

352 Copyright © 2014 Pearson Canada Inc.

Fig. C-8.9

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 9.1 (PAGE 503)

CHAPTER 9. SEQUENCES, SERIES, AND POWER SERIES Section 9.1 Sequences and Convergence (page 503) 1.

2.

3.

4.

5.

6.

7.

8.

  2n 2 2 8 9 = 1, , , . . . is bounded, n2 + 1 n2 + 1 5 5 positive, increasing, and converges to 2.     2n 4 3 8 = 1, , , , . . . is bounded, positive, n2 + 1 5 5 17 decreasing, and converges to 0.     (−1)n 7 13 4− = 5, , , . . . is bounded, positive, and n 2 3 converges to 4.         1 1 1 sin = sin 1, sin , sin , . . . is bounded, n 2 3 positive, decreasing, and converges to 0.   2     n −1 1 3 8 15 = n− = 0, , , , . . . is bounded n n 2 3 4 below, positive, increasing, and diverges to infinity.       n  e e 2 e e 3 = , , , . . . is bounded, positive, πn π π π decreasing, and converges to 0, since e < π .  n     √ e e n = . Since e/ π > 1, the sequence √ π n/2 π is bounded below, positive, increasing, and diverges to infinity.     −1 2 −3 (−1)n n = , , , . . . is bounded, alternaten e e2 e3 ing, and converges to 0. 



 = 2−

14.

5 −2 5 − 2n 2 lim = lim n =− . 7 3n − 7 3 3− n

15.

4 n− n2 − 4 n lim = lim = ∞. 5 n+5 1+ n

16.

n2 lim 3 = lim n +1



9. {2n /n n } is bounded, positive, decreasing, and converges

17. lim(−1)n

18.

(n!)2 1 2 3 = ··· (2n)! n+1 n+2 n+3 (n + 1)2 an+1 = Also, an (2n + 2)(2n + 1)   (n!)2 is positive, decreasing, (2n)! to 0.

 n n 1 ≤ . 2n 2 1 < . Thus the sequence 2 bounded, and convergent

11. {n cos(nπ/2)} = {0, −2, 0, 4, 0, −6, . . .} is divergent. 12.

  sin 2 sin 3 = sin 1, , , . . . is bounded and conn 2 3 verges to 0.

n sin n o

13. {1, 1, −2, 3, 3, −4, 5, 5, −6, . . .} is divergent.

1+

1 n3

= 0.

n = 0. n3 + 1

2 1 √ 1− √ + 2 n2 − 2 n + 1 1 n n n lim = lim =− . 1 1 1 − n − 3n 2 3 − −3 n2 n

19. lim

en − e−n 1 − e−2n = lim = 1. n −n e +e 1 + e−2n

1 sin x cos x = lim = lim = 1. x→0+ x x→0+ 1 n     −3 n n−3 n lim = lim 1 + = e−3 by l’Hˆopital’s n n Rule. n x lim = lim x→∞ ln(n + 1) ln(x + 1) 1  = lim x + 1 = ∞. = lim  x→∞ x→∞ 1 x +1

20. lim n sin 21.

22.

√

√

23. lim( n + 1 − n) = lim √ 24.

to 0.

10.

1 n

25.

n+1−n √ = 0. n+1+ n

  p n 2 − (n 2 − 4n) √ lim n − n 2 − 4n = lim n + n 2 − 4n 4n 4 r = lim √ = lim = 2. 2 4 n + n − 4n 1+ 1− n p p lim( n 2 + n − n 2 − 1) n 2 + n − (n 2 − 1) = lim √ √ n2 + n + n2 − 1 n+1 ! = lim r r 1 1 n 1+ + 1− 2 n n = lim r

1+ 1+

1 + n

1 n r

1−

1 n2

=

1 . 2

353 Copyright © 2014 Pearson Canada Inc.

SECTION 9.1 (PAGE 503)

26. If an =



n−1 n+1

n

ADAMS and ESSEX: CALCULUS 8

Thus, {an } is increasing by induction. Observe that a1 < 5 and a2 < 5. If ak < 5 then

, then

  n n−1 n n lim an = lim n n+1      1 n 1 n lim 1 + = lim 1 − n n −1 e = = e−2 (by Theorem 6 of Section 3.4). e 

27.

(n!)2 (1 · 2 · 3 · · · n)(1 · 2 · 3 · · · n) = (2n)! 1 · 2 · 3 · · · n · (n + 1) · (n + 2) · · · 2n  n 1 2 3 n 1 = · · ··· ≤ . n+1 n+2 n+3 n+n 2 Thus lim an = 0. an =

28. We have lim and lim

lim

4n n!

n2 = 0 since 2n grows much faster than n 2 2n

ak+1 =

Therefore, an < 5 for all n, by induction. Since {an } is increasing and bounded above, it converges. Let lim an = a. Then a=

32.

√ 15 + 2a ⇒ a 2 − 2a − 15 = 0 ⇒ a = −3, or a = 5.

Since a > a1 , we must have lim an = 5.     1 1 n so ln an = n ln 1 + Let an = 1 + . n n   1 = x ln(x + 1) − x ln x, then a) If f (x) = x ln 1 + x x f ′ (x) = ln(x + 1) + − ln x − 1 x + 1   1 x +1 − = ln x x +1 Z x+1 dt 1 = − t x +1 x Z x+1 1 1 > dt − x +1 x x +1 1 1 = − = 0. x +1 x +1

= 0 by Theorem 3(b). Hence,

   n 2 22n n2 n 2 2n 4n = lim n · = lim n = 0. lim n! 2 n! 2 n!

πn ⇒ 0 < an < (π/4)n . Since π/4 < 1, 1 + 22n therefore (π/4)n → 0 as n → ∞. Thus lim an = 0. √ Let a1 = 1 and an+1√= 1 + 2an for n = 1, 2, 3, . . .. Then we have a2 = 3 > a1 . If ak+1 > ak for some k, then

29. an = 30.

ak+2 =

p p 1 + 2ak+1 > 1 + 2ak = ak+1 .

Thus, {an } is increasing by induction. Observe that a1 < 3 and a2 < 3. If ak < 3 then ak+1 =

p p √ √ 1 + 2ak < 1 + 2(3) = 7 < 9 = 3.

Therefore, an < 3 for all n, by induction. Since {an } is increasing and bounded above, it converges. Let lim an = a. Then a=

31.

√ √ 1 + 2a ⇒ a 2 − 2a − 1 = 0 ⇒ a = 1 ± 2.

√ Since a = 1 − 2 < 0, √ it is not appropriate. Hence, we must have lim an = 1 + 2. √ Let a1 = 3 and an+1√= 15 + 2an for n = 1, 2, 3, . . .. Then we have a2 = 21 > 3 = a1 . If ak+1 > ak for some k, then ak+2 =

p p 15 + 2ak+1 > 15 + 2ak = ak+1 .

p p √ 15 + 2ak < 15 + 2(5) = 25 = 5.

Since f ′ (x) > 0, f (x) must be an increasing function. Thus, {an } = {e f (xn ) } is increasing. b) Since ln x ≤ x − 1,     1 1 ln ak = k ln 1 + ≤k 1+ −1 =1 k k which implies that ak ≤ e for all k. Since {an } is increasing, e is an upper bound for {an }.

33. Suppose {an } is ultimately increasing, say an+1 ≥ an if

n ≥ N. Case I. If there exists a real number K such that an ≤ K for all n, then lim an = a exists by completeness. Case II. Otherwise, for every integer K , there exists n ≥ N such that an > K , and hence a j > K for all j ≥ n. Thus lim an = ∞. If {an } is ultimately decreasing, then either it is bounded below, and therefore converges, or else it is unbounded below, and therefore diverges to negative infinity.

34. If {|an |} is bounded then it is bounded above, and there

exists a constant K such that |an | ≤ K for all n. Therefore, −K ≤ an ≤ K for all n, and so {an } is bounded above and below, and is therefore bounded.

354 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 9.2 (PAGE 510)

35. Suppose limn→∞ |an | = 0. Given any ǫ > 0,

3.

there exists an integer N = N (ǫ) such that if n > N , then ||an | − 0| < ǫ. In this case |an − 0| = |an | = ||an | − 0| < ǫ, so limn→∞ an = 0.

36.

∞ X n=5

1 (2 + π )2n

1 (2 + π )10 1 = (2 + π )10 1 = (2 + π )10

=

a) “If lim an = ∞ and lim bn = L > 0, then lim an bn = ∞” is TRUE. Let R be an arbitrary, large positive number. Since lim an = ∞, and 2R L > 0, it must be true that an ≥ for n sufL ficiently large. Since lim bn = L, it must also be L that bn ≥ for n sufficiently large. Therefore 2 2R L an bn ≥ = R for n sufficiently large. Since L 2 R is arbitrary, lim an bn = ∞. b) “If lim an = ∞ and lim bn = −∞, then lim(an + bn ) = 0” is FALSE. Let an = 1 + n and bn = −n; then lim an = ∞ and lim bn = −∞ but lim(an + bn ) = 1.

4.

e) “If {|an |} converges, then {an } converges” is FALSE. Let an = (−1)n . Then limn→∞ |an | = limn→∞ 1 = 1, but limn→∞ an does not exist.

∞ X (−5)n n=2

c) “If lim an = ∞ and lim bn = −∞, then lim an bn = −∞” is TRUE. Let R be an arbitrary, large positive number. Since lim an = √∞ and lim bn =√ −∞, we must have an ≥ R and bn ≤ − R, for all sufficiently large n. Thus an bn ≤ −R, and lim an bn = −∞. d) “If neither {an } nor {bn } converges, then {an bn } does not converge” is FALSE. Let an = bn = (−1)n ; then lim an and lim bn both diverge. But an bn = (−1)2n = 1 and {an bn } does converge (to 1).

   2 ∞ X 1 1 5 =5 1+ + + ··· 103n 1000 1000 n=0 =

5.

6.

7.

8.

1 1 + + +··· (2 + π )12 (2 + π )14   1 1 + + · · · 1+ (2 + π )2 (2 + π )4 1 1 h i. · = 1 8 (2 + π ) (2 + π )2 − 1 1− (2 + π )2

82n

5

1 1− 1000

1.

(page 510)

9.

(−5)2 (−5)3 (−5)4 + + +··· 4 6 8 8 88   25 52 5 = 4 1− + 2 −··· 8 64 64 25 1 25 25 = 4 · = = . 5 64 × 69 4416 8 1+ 64

 2 ∞ X 1 1 1 + +··· = = 1 + n e e e n=0

1 · 3

1 1 1− 3

=

P∞

j =1 π

3 4

1 1− e

=

e . e−1

j/2 cos( j π )

P j j/2 diverges because = ∞ j =2 (−1) π does not exist.

P∞ 3 + 2n diverges to ∞ because n=1 2n+2 3 +1 1 3 + 2n 2n = > 0. lim = lim n→∞ n→∞ 2n+2 4 4

1 . 2

10.

∞ X 3 + 2n n=0

2. 3 − +

1

∞ k+3 ∞  k X X 8e3 2 8e4 2 3 = . = 8e = k−3 2 e e e−2 k=0 k=0 1− e

!  2 1 1 1 1 1 1 + + + ··· = 1+ + +··· 3 9 27 3 3 3 =

5000 . 999

=

lim j →∞ (−1) j π j/2

Section 9.2 Infinite Series

=

  ∞ X 3 3 1 n−1 3 − +··· = 3 − = 16 64 4 1 + n=1

1 4

=

3n+2

12 . 5

= =

∞  n ∞  n 1X 1 1X 2 + 3 n=0 3 9 n=0 3

1 · 3

1

1 1− 3

+

1 · 9

1

2 1− 3

=

1 1 5 + = . 2 3 6

355 Copyright © 2014 Pearson Canada Inc.

SECTION 9.2 (PAGE 510)

11. Since

1 1 = n(n + 2) 2



ADAMS and ESSEX: CALCULUS 8

 1 1 − , therefore n n+2

14. Since   1 1 2 1 1 = − + , n(n + 1)(n + 2) 2 n n+1 n+2

1 1 1 1 + + +··· + 1×3 2×4 3×5 n(n + 2)  1 1 1 1 1 1 1 1 1 = − + − + − + − + ··· 2 1 3 2 4 3 5 4 6  1 1 1 1 1 1 − + − + − + n−2 n n−1 n+1 n n+2   1 1 1 1 = 1+ − − . 2 2 n+1 n+2

sn =

Thus lim sn =

the partial sum is     1 2 1 1 1 2 1 sn = 1− + + − + + ··· 2 2 3 2 2 3 4     1 2 1 2 1 1 1 1 + − + − + + 2 n−1 n n+1 2 n n+1 n+2   1 1 1 1 = − + . 2 2 n+1 n+2 Hence,

∞ X 3 3 1 , and = . 4 n(n + 2) 4 n=1

n=1

12. Let ∞ X n=1



 1 1 − , the 2n − 1 2n + 1

    1 1 1 1 1 1− + − +··· sn = 2 3 2 3 5     1 1 1 1 1 1 + − + − 2 2n − 3 2n − 1 2 2n − 1 2n + 1   1 1 = 1− . 2 2n + 1 Hence,

16.

n=1

1 1 = lim sn = . (2n − 1)(2n + 1) 2

1 1 = (3n − 2)(3n + 1) 3



1 1 − 3n − 2 3n + 1

1 1 1 1 sn = + + +··· + 1 × 4 4 × 7 7 × 10 (3n − 2)(3n + 1)  1 1 1 1 1 1 1 − + − + − +··· = 3 1 4 4 7 7 10  1 1 1 1 + − + − 3n − 5 3n − 2 3n − 2 3n + 1   1 1 1 = 1− → . 3 3n + 1 3 P∞

n=1

1 1 = . (3n − 2)(3n + 1) 3

n n diverges to infinity since lim = 1 > 0. n+2 n+2 1 n

17. Since n −1/2 = √ ≥ n X

19.  , there-

∞ X n=1

18. ∞ X

Thus

1 1 1 1 > = · , therefore the partial sums 2n − 1 2n 2 n of the given series P exceed half those of the divergent harmonic series (1/2n). Hence the given series diverges to infinity.

1 1 1 1 = + + +···. (2n − 1)(2n + 1) 1×3 3×5 5×7

1 1 = (2n − 1)(2n + 1) 2 partial sum is

fore

1 1 = lim sn = . n(n + 1)(n + 2) 4

15. Since

Since

13. Since

∞ X

1 for n ≥ 1, we have n

n X 1 → ∞, k k=1 k=1 P −1/2 as n → ∞ (harmonic series). Thus n diverges to infinity.   ∞ X 1 1 2 1 = 2 + + + · · · diverges to infinity n+1 2 3 4 n=1 since it is just twice the harmonic series with the first term omitted. n −1 if n is odd sn = −1 + 1 − 1 + · · · + (−1)n = . P 0 n if n is even Thus lim sn does not exist, and (−1) diverges.

k −1/2 ≥

n(n + 1) , the given series 2 which converges to 2 by the result of

20. Since 1 + 2 + 3 + · · · + n = 2 n(n + 1) Example 3 of this section.

is

P∞

n=1

21. The total distance is "

#  2 3 3 2+2 2× +2× + ··· 4 4 " #  2 3 3 3 =2+2× 1+ + + ··· 2 4 4 =2 +

356 Copyright © 2014 Pearson Canada Inc.

3 1−

3 4

= 14 metres.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 9.3 (PAGE 520)

28. “If

2 m

29.

P

an and

P

bn both diverge, then so does 1 and (an + bn )” is FALSE. Let an = n P P 1 bn = − , then an = ∞ and bn = −∞ but n P P (an + bn ) = (0) = 0. P “If an ≥ c > 0 for all n, then an diverges to infinity” is TRUE. We have

P

sn = a1 + a2 + a3 + · · · + an ≥ c + c + c + · · · + c = nc,

30. Fig. 9.2.21

22. The balance at the end of 8 years is h i sn = 1000 (1.1)8 + (1.1)7 + · · · + (1.1)2 + (1.1)   (1.1)8 − 1 = 1000(1.1) ≈ $12, 579.48. 1.1 − 1

23. For n > N let sn =

n X j =1

a j , and Sn =

Then sn = Sn + C, where C =

n X

j =N P N −1 j =1 a j .

31.

aj .

and nc → ∞ as n → ∞. P P “If an diverges and {bn } is bounded, then an bn 1 1 diverges” is FALSE. Let an = and bn = . n n + 1 P Then an = ∞ and 0 ≤ bn ≤ 1/2. But P P 1 an bn = which converges by Example n(n + 1) 3. P P 2 “If an > 0 and an converges, then an converges” is TRUE.P Since an converges, therefore lim an = 0. Thus there exists N such that 0 < an ≤ 1 for n ≥ N . Thus 0 < an2 ≤ an for n ≥ N . n n X X If Sn = ak2 and sn = ak , then {Sn } is increasing k=N

k=N

and bounded above:

We have

Sn ≤ sn ≤

lim sn = lim Sn + C :

n→∞

n→∞

eitherPboth sides exist or neither does. Hence and ∞ n=N both converge or neither does.

P∞

n=1 an

Thus

25. If {an } is ultimately negative, then the series

1.

2.

P

P an converges, then 1/a Pn diverges to infinity” is FALSE. A counterexample is (−1)n /2n .

27. “If

P

∞ X

ak2 converges.

k=1

X 1 1 converges by comparison with since +1 n2 1 1 0< 2 < 2. n +1 n ∞ ∞ X X 1 n converges by comparison with since 4−2 3 n n n=1 n=1 X

n2

lim

P

an converge” is TRUE because s = 0 = 0, for every n, and so n P an = lim sn = 0. n k=0

ak2 converges, and so

ak < ∞.

Section 9.3 Convergence Tests for Positive Series (page 520)

an must either converge (if its partial sums are bounded below), or diverge to −∞ (if its partial sums are not bounded below).

26. “If an = 0 forPevery n, then

k=1

k=N

24. If {an } is ultimately positive, then the sequence {sn } of

partial sums of the series must be ultimately increasing. By Theorem 2, if {sn } is ultimately increasing, then either it is bounded above, and therefore convergent, or else it above and diverges to infinity. Since Pis not boundedP an = lim sn , an must either converge when {sn } converges and lim sn = s exists, or diverge to infinity when {sn } diverges to infinity.

∞ X

∞ X

3.



n 4 n −2   1 n3



= 1,

and 0 < 1 < ∞.

X n2 + 1 diverges to infinity by comparison with n3 + 1 X1 n2 + 1 1 , since 3 > . n n +1 n 357

Copyright © 2014 Pearson Canada Inc.

SECTION 9.3 (PAGE 520)

4.

∞ X n=1

ADAMS and ESSEX: CALCULUS 8

√

∞ X n 1 converges by comparison with 3/2 +n+1 n n=1

n2

since

lim



√

n n2 + n + 1   1 n 3/2



= 1,

and

0 < 1 < ∞.

13.

14.

X 1 X 1 . so sin n 2 converges by comparison with n2 ∞ X

1 converges by comparison with the geometric n +5 π n=8 ∞  n X 1 1 1 since 0 < n series < n. π π +5 π n=8

7. Since (ln n)3 < n for large n, infinity by comparison with ∞ X

11.

πn

∞ X n=1

lim

1 n 1/3

1

nπ 1− n π

= lim

a

dt = t ln t (ln ln t)2

Z

∞

ln ln a

du 0.

X 1 − (−1)n X 1 converges by comparison with , 4 n n4 n 2 1 − (−1) ≤ 4. since 0 ≤ n4 n

diverges to infinity.

= 1, the series

n 1/3 + n 5/3 = 1. 2 + n 5/3

n2 √ diverges to infinity since 1+n n

X 1 1 1 < n , the series conn + 1) 2 2 (n + 1) X 1 verges by comparison with the geometric series . 2n ∞ X n4 converges by the ratio test since n! n=1

17. Since

18.

2n (n

(n + 1)4   n+1 4 1 (n + 1)! = lim lim = 0. n n+1 n4 n!

19.

X 1 + n 4/3 diverges to infinity by comparison with the 2 + n 5/3 X 1 divergent p-series , since n 1/3 

∞

∞ X 2 1 + (−1)n 2 2 = 0 + √ +0 + √ +0 + √ +··· √ n 6 2 4 n=1 ∞ ∞ X 1 √ X 1 √ = 2 √ =2 2k k k=1 k=1

πn

1 + n 4/3 n→∞ 2 + n 5/3

1 converges by the integral test: n ln n(ln ln n)2

16. The series

X1 . n

πn = lim − nπ

lim

12.

15.

1 diverges to (ln n)3

1 converges by comparison with the geomet− nπ X 1 ric series . πn ∞ X 1+n 1+n diverges to infinity since lim = 1 > 0. 2 + n 2+n n=0 X

10.

X

1 diverges to infinity by comparison with the ln(3n) n=1 ∞ X 1 1 1 harmonic series since > for n ≥ 1. 3n ln(3n) 3n n=1

9. Since limn→∞

∞ X

Z

sin 1 = sin 1 ≤ 1 , n2 n2 n2

8.

1 diverges to infinity by the integral test, √ n ln n ln ln n n=3 since Z ∞ Z ∞ dt du √ = ∞. √ = u t ln t ln ln t 3 ln ln 3

n=2

5. Since sin x ≤ x for x ≥ 0, we have

6.

∞ X

X n! diverges to infinity by the ratio test, since n 2 en ρ = lim

20.

(n + 1)! n 2 en 1 n2 · = lim = ∞. (n + 1)2 en+1 n! e n+1

∞ X (2n)!6n converges by the ratio test since (3n)! n=1

 (2n + 2)!6n+1 (2n)!6n (3n + 3)! (3n)! (2n + 2)(2n + 1)6 = lim = 0. (3n + 3)(3n + 2)(3n + 1)

lim

n2 √ = ∞. 1+n n

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INSTRUCTOR’S SOLUTIONS MANUAL

21.

SECTION 9.3 (PAGE 520)

√ ∞ X n converges by the ratio test, since n ln n 3 n=2 √

3n ln n n+1 · √ n+1 3 ln(n + 1) n r ln n 1 1 n+1 · lim = < 1. = lim 3 n ln(n + 1) 3

ρ = lim

22.

We use the approximation s≈

1 = sn + 2

sn∗

− sn∗ |

|s

lim

23.

X (2n)! converges by the ratio test, since (n!)3 (2n + 2)! (n!)3 (2n + 2)(2n + 1) · = lim = 0 < 1. ((n + 1)!)3 (2n)! (n + 1)3

ρ = lim

24.

25.

∞ X 1 + n! diverges by comparison with the harmonic (1 + n)! n=1 ∞ X 1 + n! n! 1 1 since > = . series n+1 (1 + n)! (1 + n)! n+1 n=1

X

2n converges by the ratio test since 3n − n 3

ρ = lim

3n+1

27.



Z

∞ n

  1 1 1 − ≤ 2 3n 3 3(n + 1)3 3 1 (n + 1) − n 3 = 6 n 3 (n + 1)3 1 3n 2 + 3n + 1 7 = < 4. 6 n 3 (n + 1)3 6n

that is, if n 4 > 7000/6. Since 64 = 1296 > 7000/6, n = 6 will do. Thus   ∞ X 1 1 1 1 1 1 1 1 1 ∗ ≈ s = 1 + + + + + + + 6 n4 24 34 44 54 64 6 73 63 n=1 ≈ 1.082

with error less than 0.001 in absolute value.

1 is positive, continuous and decreasing x3 on [1, ∞), for any n = 1, 2, 3, . . ., we have

28. Since f (x) =

where sn =

Z ∞ n X 1 dx 1 and A = = . If n 3 3 2 k x 2n n k=1

1 sn∗ = sn + (An+1 + An ), then 2

  An − An+1 1 1 1 = − 2 4 n2 (n + 1)2 1 2n + 1 = < 0.001 4 n 2 (n + 1)2

|sn − sn∗ | ≤

  nn 1 1 n e = lim 1 + = < 1. π n n! π n π

f (x) = 1/x 4 is positive, continuous, and decreasing on [1, ∞). Let An =

.

sn + An+1 ≤ s ≤ sn + An

∞ X nn converges by the ratio test since π n n! n=1

(n + 1)n+1 lim (n+1) π (n + 1)!



We have used 3n 2 + 3n + 1 ≤ 7n 2 and n 3 (n + 1)3 > n 6 to obtain the last inequality. We will have |s − sn∗ | < 0.001 provided 7 < 0.001, 6n 4

2n+1 3n − n 3 · 3 2n − (n + 1)

n3 1 − n 3 2 3 −n 2 2 3n = lim = lim = < 1. 3 3 3 3 3 (n + 1) (n + 1) 3n − 1− n+1 3 3

26.

1 1 + 3 3(n + 1)3 3n

The error satisfies

∞ X n 100 2n √ converges by the ratio test since n! n=0

 (n + 1)100 2n+1 n 100 2n √ √ (n + 1)! n!   n + 1 100 1 = 0. = lim 2 √ n n+1



  R dx 1 1 = lim = 3. − 4 3 R→∞ x 3x 3n n

if n = 8. Thus, the error in the approximation s ≈ s8∗ is less than 0.001. 1 is positive, continuous and decreasing x 3/2 on [1, ∞), for any n = 1, 2, 3, . . ., we have

29. Since f (x) =

sn + An+1 ≤ s ≤ sn + An

359 Copyright © 2014 Pearson Canada Inc.

SECTION 9.3 (PAGE 520)

ADAMS and ESSEX: CALCULUS 8

Z ∞ n X 1 dx 2 and A = = √ . If n 3/2 3/2 k x n n k=1   1 1 1 sn∗ = sn + (An+1 + An ) = sn + √ + √ , then 2 n n+1 where sn =

31.

An − An+1   2 2 1 2 √ −√ = 2 n n+1 √ √ n+1− n 1 = √ √ = √ √ √ √ n n+1 n n + 1( n + n + 1) 1 < 3/2 < 0.001 2n

∗ has error less n = 22 will do. The approximation s ≈ s22 than 0.001. ∞ X 1 We have s = and k k! 2 k=1

|sn − sn∗ | ≤

sn = Then

0 < s − sn

1 1 1 + + + ··· 2n+1 (n + 1)! 2n+2 (n + 2)! 2n+3 (n + 3)!   1 1 1 = n+1 + 2 +··· 1+ 2 (n + 1)! 2(n + 2) 2 (n + 2)(n + 3) " #  2 1 1 1 < n+1 1+ + + ··· 2 (n + 1)! 2(n + 2) 2(n + 2) =

∗ is if n ≥ 63. Thus, the error in the approximation s ≈ s63 less than 0.001.

30. Again, we have sn + An+1 ≤ s ≤ sn + An where sn =

Pn

k=1

An = If

sn∗

Z

n

∞

k2

1 and +4

  ∞ n  dx 1 π 1 −1 x = tan = − tan−1 . 2 x +4 2 2 n 4 2 2

1 = sn + (An+1 + An ), then 2

An − An+1 ≤ 2      π 1 π 1 1 −1 n −1 n + 1 = − tan − + tan 2 4 2 2 4 2 2       1 1 n+1 n tan−1 − tan−1 = (a − b), = 4 2 2 4   n  n+1 where a = tan−1 and b = tan−1 . Now 2 2

=

1 · 2n+1 (n + 1)!

1

=

n+2 < 0.001 2n (n + 1)!(2n + 3)

1−

1 2(n + 2)

if n = 4. Thus, s ≈ s4 =

|sn − sn∗ |

tan a − tan b tan(a − b) = 1+ tan atan b n n+1 − 2 2   = n + 1 n  1+ 2 2 2 = 2 n + n+ 4  2 . ⇔ a − b = tan−1 n2 + n + 4

n X 1 1 1 1 1 = + 2 + 3 +··· + n . k 2 k! 2 2 2! 2 3! 2 n! k=1

32.

1 1 1 1 + 2 + 3 + 4 with 2 2 2! 2 3! 2 4!

error less than 0.001. ∞ X 1 We have s = and (2k − 1)! k=1 sn =

n X k=1

1 1 1 1 1 = + + + ··· + . (2k − 1)! 1! 3! 5! (2n − 1)!

Then 1 1 1 + + + ··· (2n + 1)! (2n + 3)! (2n + 5)!  1 1 = 1+ + (2n + 1)! (2n + 2)(2n + 3)  1 + ··· (2n + 2)(2n + 3)(2n + 4)(2n + 5)  1 1 < 1+ + (2n + 1)! (2n + 2)(2n + 3)  1 + · · · [(2n + 2)(2n + 3)]2  

0 < s − sn =

We want error less than 0.001:   1 1 2 (a − b) = tan−1 < 0.001 4 4 n2 + n + 4 2 ⇔ < tan 0.004 n2 + n + 4 ⇔ n 2 + n > 2 cot(0.004) − 4 ≈ 496.

360 Copyright © 2014 Pearson Canada Inc.

 1  (2n + 1)! 

1

   1 1− (2n + 2)(2n + 3) 1 4n 2 + 10n + 6 = < 0.001 (2n + 1)! 4n 2 + 10n + 5

=

INSTRUCTOR’S SOLUTIONS MANUAL

if n = 3. Thus, s ≈ s3 = 1 +

SECTION 9.3 (PAGE 520)

1 1 + = 1.175 with error 3! 5!

less than 0.001. ∞ n−1 X X 2k 2k and sn = . Thus We have s = (2k)! (2k)! k=0 k=0

33.

0 < s − sn =

2n

2n+1

2n · (2n)!

36. Let u = ln ln t, du = Z

n X 1 1 1 1 1 = + 2 + 3 + ··· + n . sn = k k 1 2 3 n k=1

∞ X

n=N

1 1 1 0 < s − sn = + + +··· (n + 1)n+1 (n + 2)n+2 (n + 3)n+3   1 1 1 1 + + < + · · · (n + 1)n+1 n + 1 (n + 1)2  

35.

1

 1  1− n+1

1 1 1 + 3 + 4 = 1.291 with 22 3 4

error less than 0.001. 1 Let f (x) = . Then f is decreasing on [1, ∞). 1 + x2 ∞ X Since f (n) is a right Riemann sum for

0

∞

1 n(ln n)(ln ln n) · · · (ln j n)(ln j +1 n) p

CASE I. Suppose σ < 1. Pick λ such that σ < λ < 1. Then there exists N such that (an )1/n ≤ λ for all n ≥ N . Therefore aN ≤ λN , Thus series

∞ X

a N +1 ≤ λ N +1 ,

a N +2 ≤ λ N +2 , . . . .

an converges by comparison with the geometric

n=N ∞ X

λn , and

n=N

∞ X

an also converges.

n=1

CASE II. Suppose σ > 1. Then (an )1/n ≥ 1, and an ≥ 1, for all sufficiently large values of n. ThereX fore lim an 6= 0 and an must diverge. Since an > 0 it diverges to infinity. 1 1 and bn = 2 . n n ln n Since lim n 1/n = 1 (because lim = 0), we have n 1/n 1/n lim(an ) = 1 and = 1. That is, σ =X 1 for X lim(bn ) both series. But an diverges to infinity, while bn converges. Thus the case σ = 1 provides no information on the convergence or divergence of a series.

38. Let an = 2n+1 /n n . Then lim

n=1

Z

du up

CASE III. Let an =

1 < 0.001 n(n + 1)n

if n = 4. Thus, s ≈ s4 = 1 +

ln ln a

Let σ = lim(an )1/n .

Then

=

∞

37. Let an > 0 for all n. (Let’s forget the “ultimately” part.)

if n = 4. Thus, s ≈ s4 with error less than 0.001. ∞ X 1 We have s = and k k k=1

1   (n + 1)n+1

a

Z

dt = t ln t (ln ln t) p

converges if and only if p > 1, where N is large enough that ln j N > 1.

1−

=

∞

dt and ln ln a > 0; then t ln t

will converge if and only if p > 1. Thus, ∞ X 1 will converge if and only if p > 1. n ln n(ln ln n) p n=3 Similarly,

1

2 (2n + 1)(2n + 2) 2n 4n 2 + 6n + 2 = · < 0.001 (2n)! 4n 2 + 6n

34.

∞ X 1 = f (n) converges by the integral test, and 2 1+n n=1 n=1 its sum is less than π/2.

2n+2

+ + +··· (2n)! (2n + 2)! (2n + 4)!  2n 2 = 1+ (2n)! (2n + 1)(2n + 2)  22 + + ··· (2n + 1)(2n + 2)(2n + 3)(2n + 4) # "  2 2 2 2n + ··· 1+ + < (2n)! (2n + 1)(2n + 2) (2n + 1)(2n + 2) =

∞ X

R π f (x) d x = lim tan−1 x = , R→∞ 2 0

n→∞

√ 2 × 21/n n an = lim = 0. n→∞ n

Since this limit is less than 1, root test.

P∞

n=1 an

converges by the

361 Copyright © 2014 Pearson Canada Inc.

SECTION 9.3 (PAGE 520)

39.

ADAMS and ESSEX: CALCULUS 8

n 2 ∞  X n converges by the root test of Exercise 31 n+1 n=1 since σ = lim



n→∞

40. Let an =

n n+1

n 2 1/n

1

1 = lim   = < 1. n→∞ e 1 n 1+ n

2n+1 . Then nn

2n+2 nn an+1 · = an (n + 1)n+1 2n+1 2 1 2 =  · n =  n n+1 1 n (n + 1) 1+ n+1 n 1 → 0 × = 0 as n → ∞. e P Thus ∞ n=1 an converges by the ratio test. (Remark: the question contained a typo. It was intended to ask that #33 be repeated, using the ratio test. That is a little harder.)

41. Trying to apply the ratio test to

ρ = lim

X 22n (n!)2 , we obtain (2n)!

22n+2 ((n + 1)!)2 (2n)! 4(n + 1)2 · 2n = 1. = lim 2 (2n + 2)! 2 (n!) (2n + 2)(2n + 1)

Thus the ratio test provides no information. However, 22n (n!)2 [2n(2n − 2) · · · 6 · 4 · 2]2 = (2n)! 2n(2n − 1)(2n − 2) · · · 3 · 2 · 1 2n − 2 4 2 2n · · · · · · · > 1. = 2n − 1 2n − 3 3 1 Since the terms exceed 1, the series diverges to infinity.

42. We have (2n)! 1 × 2 × 3 × 4 × · · · × 2n = 22n (n!)2 (2 × 4 × 6 × 8 × · · · × 2n)2 1 × 3 × 5 × · · · × (2n − 1) = 2 × 4 × 6 × · · · × (2n − 2) × 2n 3 5 7 2n − 1 1 1 = 1 × × × × ··· × × > . 2 4 6 2n − 2 2n 2n

an =

∞ X (2n)! diverges to infinity by comparison 2n (n!)2 2 n=1 ∞ X 1 with the harmonic series . 2n n=1

Therefore

43.

a) If n is a positive integer and k > 0, then 1 (1 + k)n ≥ 1 + nk > nk, so n < (1 + k)n . k b) Let s N =

 N  N X 1X 1+k n n < 2n k n=0 2 n=0

N 1X 1 1 − r N +1 rn = · , k n=0 k 1−r where r = (1 + k)/n. Thus

=

 1 + k N +1 1− 1 2 sn < · 1+k k 1− 2 !   2 1 + k N +1 2 = 1− . ≤ k(1 − k) 2 k(1 − k) 

∞ X 2 n ≤ . n 2 k(1 − k) n=0 Since the maximum value of k(1 − k) is 1/4 (at k = 1/2), the best upper bound we get for s by this method is s ≤ 8.  ∞ ∞  X j 1+k j 1 X c) s − sn = < 2j k j =n+1 2 j =n+1  n+1 1 1 1+k · = 1+k k 2 1− 2 (1 + k)n+1 G(k) = = n , k(1 − k)2n 2 (1 + k)n+1 where G(k) = . For minimum G(k), look k(1 − k) for a critical point:

Therefore, s =

k(1 − k)(n + 1)(1 + k)n − (1 + k)n+1 (1 − 2k) =0 k 2 (1 − k)2 (k − k 2 )(n + 1) − (1 + k)(1 − 2k) = 0 k 2 (n + 1) − k(n + 1) + 1 − k − 2k 2 = 0

(n − 1)k 2 − (n + 2)k + 1 = 0 p (n + 2) ± (n + 2)2 − 4(n − 1) k= 2(n − 1) √ n + 2 ± n2 + 8 = . 2(n − 1)

For given n, the √ upper bound is minimal if n + 2 − n2 + 8 k= (for n ≥ 2). 2(n − 1)

44. If s =

∞ X k=1

362 Copyright © 2014 Pearson Canada Inc.

ck =

∞ X k=1

1 , then we have + 1)

k 2 (k

sn + An+1 ≤ s ≤ sn + An

INSTRUCTOR’S SOLUTIONS MANUAL

where sn = Z

n X k=1

SECTION 9.4 (PAGE 526)

1 and + 1)

(b) Let Sn =

k 2 (k

∞

Z

∞

1 1 dx −1 = + 2 + + 1) x x x +1 n n ∞ 1 = − ln x − + ln(x + 1) x n ∞   1 1 = ln 1 + − x x  n 1 1 = − ln 1 + . n n

An =

If

sn∗

x 2 (x



= = ≤ =

dx

45.

1 1 − n . Since n 2 2 +1

2n + 1 − 2n 1 < n, 2n (2n + 1) 4

i=1

 1 1 + + · · · 4n+1 4 42 4 1 1 = f r ac14n+1 × = < 3 3 × 4n 1,000 

1

<

1+

provided 4n > 1,000/3. Thus n = 5 will P do (but n = 4 is insufficient). S5 approximates ∞ n=1 bn to within 0.001. P n (c) Since ∞ n=1 1/2 = 1, we have ∞ X n=1

2n

∞ ∞ X X 1 1 = − bn n + 1 n=1 2 n=1

≈1−

5 X

bn

n=1

     1 1 1 1 1 1 − − − − − 2 3 4 5 8 9     1 1 1 1 − − − − 16 17 32 33 ≈ 0.765 with error less than 0.001.

=1−

if n = 8. Thus,

with error less than 0.001. P n s= ∞ n=1 1/(2 + 1).

where bn =

we have ∞ X 0< bi − Sn = bn+1 + bn+2 + bn+3 + · · ·

An − An+1   2    1 1 1 1 1 − ln 1 + + ln 1 + − 2 n n n+1 n+1    2 1 n + 2n 1 + ln 2 n(n + 1) n 2 + 2n + 1    1 1 n 2 + 2n + −1 2 n(n + 1) n 2 + 2n + 1 1 < 0.001 2n(n + 1)2

∞ X 1 1 = 1 + s8∗ = 1 + s8 + (A9 + A8 ) 2 n 2 n=1   1 1 1 1 =1+ + 2 + 2 +··· + 2 + 2 2 (3) 3 (4) 8 (9)     10 9 1 1 1 − ln + − ln 2 9 9 8 8 = 1.6450

i=1 bi ,

0 < bn =

1 = sn + (An+1 + An ), then 2

|sn − sn∗ | ≤

Pn



Section 9.4 Absolute and Conditional Convergence (page 526) 1.

2.

(a) We have

X (−1)n converges by the alternating series test (since √ n the terms alternate in sign, decrease in size, and approach 0). However, the convergence is only conditional, since X 1 √ diverges to infinity. n ∞ X (−1)n converges absolutely since 2 n + ln n n=1 ∞ X (−1)n 1 1 converges. n 2 + ln n ≤ n 2 and n2 n=1

∞ X

1 1 1 1 0 < s − sn = = n+1 + n+2 + n+3 + · · · i +1 2 2 2 2 i=1   1 1 1 = n+1 1 + + 2 + · · · 2 2 2 1 1 if 2n > 1,000. = n < 2 1,000 Since 210 = 1,024, s10 will approximate s to within 0.001.

3.

4.

X (−1)n cos(nπ ) = converges (n + 1) ln(n + 1) (n + 1) ln(n + 1) by the alternating series test, but only conditionally since X 1 diverges to infinity (by the integral (n + 1) ln(n + 1) test).

X

∞ ∞ X X (−1)2n 1 = is a positive, convergent geometric n 2 2n n=1 n=1 series so must converge absolutely.

363 Copyright © 2014 Pearson Canada Inc.

SECTION 9.4 (PAGE 526)

5.

6.

ADAMS and ESSEX: CALCULUS 8

X (−1)n (n 2 − 1) diverges since its terms do not apn2 + 1 proach zero.

9.

10.

X 20n 2 − n − 1 converges by the alternating se(−1)n 3 n + n 2 + 33 ries test (the terms are ultimately decreasing in size, and approach zero), but the convergence is only conditional X 20n 2 − n − 1 since diverges to infinity by comparin 3 + n 2 + 33 X1 son with . n ∞ ∞ X X 100 cos(nπ ) 100(−1)n = converges by the alter2n + 3 2n + 3 n=1 n=1 nating series test but only conditionally since 100(−1)n 100 2n + 3 = 2n + 3

∞ X 100 diverges to infinity. and 2n +3 n=1

11. 12.

X

n! n! diverges since lim = ∞. n (−100) 100n

∞ ∞ X X sin(n + 21 )π (−1)n = converges by the alterln ln n ln ln n n=10 n=10 ∞ X 1 nating series test but only conditionally since ln ln n n=10 ∞ X 1 diverges to infinity by comparison with . n n=10 (ln ln n < n for n ≥ 10.)

k=1

k2

n X k k , and sn = (−1)k−1 2 , +1 k +1 k=1

n+1 < 0.001 (n + 1)2 + 1

if n = 999, because the series satisfies the conditions of the alternating series test. P∞ (−1)n are altern=0 (2n)! nating in sign and decreasing in size, the size of the error in the approximation s ≈ sn does not exceed that of the first omitted term:

14. Since the terms of the series s =

|s − sn | ≤

X 1 is a convergent geometric series. πn

∞ X −n diverges to −∞ since all terms are negative 2+1 n n=0 ∞ X n and diverges to infinity by comparison with 2+1 n n=0 ∞ X 1 . n n=0

∞ X (−1)k−1

|s − sn | <

X (−1)n converges absolutely, since, for n ≥ 1, nπ n (−1)n 1 nπ n ≤ π n , and

8.

then

∞ X (−2)n converges absolutely by the ratio test since n! n=1

(−2)n+1 n! 1 lim · = 2 lim = 0. n (n + 1)! (−2) n+1

7.

13. If s =

1 < 0.001 (2n + 2)!

1 1 1 + − ; four terms 2! 4! 6! will approximate s with error less than 0.001 in absolute value. if n = 3. Hence s ≈ 1 −

15. If s =

∞ n X X k k (−1)k−1 k , and sn = (−1)k−1 k , then 2 2 k=1 k=1

|s − sn | <

n+1 < 0.001 2n+1

if n = 13, because the series satisfies the conditions of the alternating series test from the second term on.

16. Since the terms of the series s =

17.

n n3 n=0 (−1)

P∞

n! are alternating in sign and ultimately decreasing in size (they decrease after the third term), the size of the error in the approximation s ≈ sn does not exceed that of the first omitted term (provided n ≥ 3): 3n+1 |s − sn | ≤ < 0.001 if n = 12. Thus twelve terms (n + 1)! will suffice to approximate s with error less than 0.001 in absolute value. X xn √ , we obtain Applying the ratio test to n+1 r √ x n+1 n + 1 n+1 = |x|. ρ = lim √ · = |x| lim n+2 xn n+2

Hence the series converges absolutely if |x| < 1, that is, if −1 < x < 1. The series converges conditionally for x = −1, but diverges for all other values of x.

18. Let an =

(x − 2)n . Apply the ratio test n 2 22n

(x − 2)n+1 n 2 22n |x − 2| ρ = lim × 1 or x < −1. In Exercise 36 of Sec1 tion 9.3 it was shown that an ≥ , so the given series 2n definitely diverges at x = 1 and may at most converge conditionally at x = −1. To see whether it does converge at −1, we write, as in Exercise 36 of Section 9.3, (2n)! 1 × 2 × 3 × 4 × · · · × 2n = 22n (n!)2 (2 × 4 × 6 × 8 × · · · × 2n)2 1 × 3 × 5 × · · · × (2n − 1) = 2 × 4 × 6 × · · · × (2n − 2) × 2n 1 3 2n − 3 2n − 1 = × × ··· × × 2 4 2n− 2  2n     1 1 1 1 = 1− 1− ··· 1 − 1− . 2 4 2n − 2 2n

an =

      1 1 1 ln an = ln 1 − + ln 1 − + · · · + ln 1 − 2 4 2n 1 1 1 ≤ − − −··· − 2 4 2n  1 1 1 1 + + ··· + → −∞ as n → ∞. =− 2 2 n

P 1 1 and qn = − . Then pn diverges 2n − 1 2n P to ∞ and qn diverges to −∞. Also, the alternating harmonic series is the sum of all the pn s and qn s in a specific order:

30. Let pn =

if and only if −e < x < e. If x = ±e, then, by (a), n n!e ln n = ln(n!) + ln en − ln n n n > (n ln n − n + 1) + n − n ln n = 1. n n!e ⇒ n > e. n ∞ X n=1

an converges absolutely if −e < x < e

and diverges elsewhere.

(2n + 2)(2n + 1) = |x|. 4(n + 1)2

Thus lim an = 0, and the given series converges conditionally at x = −1 by the alternating series test.

(n + 1)!x n+1 n n ρ = lim × (n + 1)n+1 n!x n |x| |x| = lim   = 0 and t = x − a, then x = t + a and   t ln x = ln(a + t) = ln a + ln 1 + a ∞ n X t = ln a + (−1)n−1 n (−a < t ≤ a) a n=1

lim x→0 csc x does not exist. π π and sin x = cos y. Let y = x − . Then x = y + 2 2 Therefore, using the result of Exercise 25, y2 5y 4 csc x = sec y = 1 + + + ··· 2 24   1 5  π 2 π 4 =1+ + +···. x− x− 2 2 24 2

33. 1 + x 2 + 34.

35.

36.

37.

3

x −

x4 2!

x9

+ +

x6 3!

+ · · · = ex

x 15

−

2

x 21

(for all x). +

x 27

3! × 4 5! × 16 7! × 64 9! × 256 " #  3  5 x3 1 x3 1 x3 =2 − + − ··· 2 3! 2 5! 2  3 x = 2 sin (for all x). 2

= ln a +

∞ X (x − a)n (−1)n−1 an n=1

Since the series converges to ln x on an interval of positive radius (a), centred at a, ln is analytic at a.

40. If f (x) =

−···

x4 x6 x2 + + + ··· 3! 5! 7! 1 e x − e−x = sinh x = x 2x if x 6= 0. The sum is 1 if x = 0.

41.

1+

1 1 1 + + + ··· 2"× 2! 4 × 3! 8 × 4! #  2  3 1 1 1 1 1 =2 + + +··· 2 2! 2 3! 2   = 2 e1/2 − 1 .



2

e−1/x , 0,

if x 6= 0; if x = 0;

then the Maclaurin series for f (x) is the identically zero series 0 + 0x + 0x 2 + · · · since f (k) (0) = 0 for every k. The series converges for every x, but converges to f (x) only at x = 0, since f (x) 6= 0 if x 6= 0. Hence, f cannot be analytic at 0. ! ! ∞ ∞ X X xn ym x y e e = n! m! n=0 m=0 e x+y =

1+

(0 < x < 2a).

= =

∞ k ∞ X X 1 X k! (x + y)k = x j y k− j k! k! j !(k − j )! k=0 j =0 k=0 ∞ ∞ X xj X y k− j j ! k= j (k − j )! j =0

(let k − j = m)

∞ ∞ X xj X ym = ex e y . j ! m! j =0 m=0

42. We want to prove that f (x) = Pn (x) + E n (x), where Pn

P(x) = 1 + x + x 2 .

is the nth-order Taylor polynomial for f about c and

a) The Maclaurin series for P(x) is 1 + x + (for all x).

x2 E n (x) =

b) Let t = x − 1, so x = t + 1. Then P(x) = P(t + 1) = 1 + t + 1 + (t + 1)2 = 3 + 3t + t 2 . The Taylor series for P(x) about 1 is 3 + 3(x − 1) + (x − 1)2 .

1 n!

Z

x c

(x − t)n f (n+1) (t) dt.

(a) The Fundamental Theorem of Calculus written in the form Z x f (x) = f (c) + f ′ (t) dt = P0 (x) + E 0 (x) c

38. If a 6= 0 and |x − a| < |a|, then 1 1 1 1 = = x a + (x − a) a 1+ x −a a   1 x −a (x − a)2 (x − a)3 = 1− + − + · · · . a a a2 a3

is the case n = 0 of the above formula. We now apply integration by parts to the integral, setting

374 Copyright © 2014 Pearson Canada Inc.

U = f ′ (t), dU = f ′′ (t) dt,

d V = dt, V = −(x − t).

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 9.6 (PAGE 545)

(We have broken our usual rule about not including a constant of integration with V . In this case we have included the constant −x in V in order to have V vanish when t = x.) We have t=x Z x + (x − t) f ′′ (t) dt f (x) = f (c) − f ′ (t)(x − t) c t=c Z x = f (c) + f ′ (c)(x − c) + (x − t) f ′′ (t) dt

and f (0) = 0, f ′ (0) = 1, f ′′ (0) = −1, f ′′′ (0) = 2,

f (4) (0) = −3!, . . . , f (n) (0) = (−1)n−1 (n − 1)!. Therefore, the Taylor Formula is −1 2 2 −3! 4 x + x3 + x +···+ 2! 3! 4! (−1)n−1 (n − 1)! n x + E n (x) n!

f (x) = x +

c

= P1 (x) + E 1 (x).

We have now proved the case n = 1 of the formula.

where Z 1 x (x − t)n f (n+1) (t) dt n! 0 Z (−1)n n! 1 x (x − t)n dt = n! 0 (1 + t)n+1 Z x (x − t)n = (−1)n dt. n+1 0 (1 + t)

(b) We complete the proof for general n by mathematical induction. Suppose the formula holds for some n = k: f (x) = Pk (x) + E k (x) Z 1 x = Pk (x) + (x − t)k f (k+1) (t) dt. k! c Again we integrate by parts. Let U= f

(k+1)

k

(t),

dU = f (k+2) (t) dt,

d V = (x − t) dt, −1 (x − t)k+1 . V = k +1

We have f (x) = Pk (x) + +

Z

x c

t=x 1 f (k+1) (t)(x − t)k+1 − k! k+1 t=c ! k+1 (k+2) (x − t) f (t) dt k+1

f (k+1) (c) = Pk (x) + (x − c)k+1 (k + 1)! Z x 1 + (x − t)k+1 f (k+2) (t) dt (k + 1)! c = Pk+1 (x) + E k+1 (x). Thus the formula is valid for n = k + 1 if it is valid for n = k. Having been shown to be valid for n = 0 (and n = 1), it must therefore be valid for every positive integer n for which E n (x) exists.

43. If f (x) = ln(1 + x), then −1 2 1 , f ′′ (x) = , f ′′′ (x) = , 1+x (1 + x)2 (1 + x)3 −3! (−1)n−1 (n − 1)! f (4) (x) = , . . . , f (n) = 4 (1 + x) (1 + x)n f ′ (x) =

E n (x) =

If 0 ≤ t ≤ x ≤ 1, then 1 + t ≥ 1 and |E n (x)| ≤

Z

0

x

(x − t)n dt =

x n+1 1 ≤ →0 n+1 n+1

as n → ∞. If −1 < x ≤ t ≤ 0, then x − t t − x 1 + t = 1 + t ≤ |x|,

t−x increases from 0 to −x = |x| as t in1+t creases from x to 0. Thus,

because

|E n (x)| <

1 1+x

Z

|x| 0

|x|n dt =

|x|n+1 →0 1+x

as n → ∞ since |x| < 1. Therefore, f (x) = x −

∞ X x2 x3 x4 xn + − + ··· = (−1)n−1 , 2 3 4 n n=1

for −1 < x ≤ 1.

44. We follow the steps outlined in the problem: Rj (a) Note that ln( j − 1) < j −1 ln x d x < ln j , j = 1, 2, . . .. For j = 0 the integral is improper but convergent. We have n ln n − n =

Z

0

n

ln x d x < ln(n!) <

Z

n+1

ln x d x

1

= (n + 1) ln(n + 1) − n − 1 < (n + 1) ln(n + 1) − n.

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SECTION 9.6 (PAGE 545)

ADAMS and ESSEX: CALCULUS 8

(b) If cn = ln(n!) − n +

1 2




or, equivalently,

ln n + n, then


n! − n + 12 ln n (n + 1)!
+ n + 32 ln(n + 1) − 1
1 − n + 21 ln n = ln n+1
+ n + 12 ln(n + 1) + ln(n + 1) − 1
n+1 = n + 12 ln −1 n 1
1 + 2n+1 = n + 21 ln − 1. 1 1 − 2n+1

22n (n!)2 √ = n→∞ (2n)! 2n + 1 lim

cn − cn+1 = ln

1+t t3 t5 = 2 t+ + +··· 1−t 3 5 −1 < t < 1. Thus

(c) ln

0 < cn − cn+1 = (2n + 1)

= = =

n! = lim ecn = ec n→∞ n n+1/2 e−n √ exists. It remains to show that ec = 2π. (d) The Wallis Product, 2n 2n π 22446 ··· = 13355 2n − 1 2n + 1 2

can be rewritten in the form 2n n!

√ = n→∞ 1 · 3 · 5 · · · (2n − 1) 2n + 1

√

=

ec e2c = . 2ec 2

22n+1/2 n 2n+1/2 e−2n ec2n 2n √ √ Thus ec /2 = π2, and ec = 2π , which completes the proof of Stirling’s Formula. n→∞

45. We have:

r

1<

(n + 1) ln(n + 1) − −n ln(n!) < . n ln n − n n ln n − n

We divide the numerator and denominator of the fraction on the right by n and take the limit as n → ∞ by l’Hˆopital’s rule:   1 1+ ln(n + 1) − 1 n lim n→∞ ln n − 1 1 (n + 1) 1 − 2 ln(n + 1) + n n n+1 = lim 1 n→∞ n ln(n + 1) = lim − + 1 = 1. n→∞ n ln(n!) = 1, and n ln n − n ln(n!) − (n ln n − n) =0 lim n→∞ ln(n!)

Hence lim

n→∞

so that the relative error in the approximation ln(n!) ≈ n ln −n approaches 0as n → ∞.

lim

lim

22n n 2n+1 e−2n e2cn

lim

Therefore,

1 1 + 2n + 1 3(2n + 1)3 !

n→∞

lim

Substituting n! = n n+1/2 e− necn and a similar expression for (2n)!, we obtain

n ln n − n < ln(n!) < (n + 1) ln(n + 1) − n.

These inequalities imply  that {cn } is de1 creasing and cn − 12n is increasing. Thus 1 {cn } is bounded below by c1 − 12 = 11 12 and so limn→∞ cn = c exists. Since ecn = n!n −(n+1/2) en , we have

n→∞

π . 2

(a) By part (a) of Exercise 44,

for

1 +··· −1 5(2n + 1)5   1 1 1 + + · · · 3 (2n + 1)2 (2n + 1)4 (geometric) 1 1 2 1 3(2n + 1) 1− (2n + 1)2 1 12(n 2 + n)   1 1 1 − . 12 n n+1

+ <

!

r

π , 2

(b) From Stirling’s √ Formula, ln(n!) ≈ ln 2π n + n ln n − n. For this approximation we have the following relative errors (using a calculator)   √ ln(10!) − ln 20π + 10 ln(10) − 10 ≈ 0.000552 ln(10!)   √ ln(20!) − ln 40π + 20 ln(20) − 20 ≈ 0.000098 ln(20!)

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 9.7 (PAGE 549)

For the modified Stirling formula, ln(n!) ≈ n ln n − n, these errors are 

 ln(10!) − 10 ln(10) − 10

≈ 0.137613

ln(20!)

≈ 0.057185

ln(10!)   ln(20!) − 20 ln(20) − 20

3.

e0.2 ≈ 1 + 0.2 +

(0.2)2 (0.2)n + ··· + = sn 2! n!

Error estimate: (0.2)n+1 (0.2)n+2 0 < e0.2 − sn = + +··· (n + 1)! (n + 2)!   (0.2)n+1 0.2 (0.2)2 ≤ 1+ + + · · · (n + 1)! n + 2 (n + 2)2 n+1 (0.2) 10n + 20 = · < 5 × 10−5 if n = 4. (n + 1)! 10n + 18 (0.2)2 (0.2)3 (0.2)4 e0.2 ≈ 1 + 0.2 + + + 2! 3! 4! ≈ 1.221400

Evidently, Stirling’s formula gives a significantly better approximation than the modified version.

4. We have Section 9.7 Applications of Taylor and Maclaurin Series (page 549)

1. If f (x) = sin x, then P5 (x) = x −

1 1 1 1 1 = e−1 = 1 − + − + − ··· e 1! 2! 3! 4! which satisfies the conditions for the alternating series test, and the error incurred in using a partial sum to approximate e−1 is less than the first omitted term in abso1 lute value. Now < 5 × 10−5 if n = 7, so (n + 1)!

x5 x3 + . 6 120

METHOD I. (using an alternating series bound)

1 1 1 1 1 1 1 ≈ − + − + − ≈ 0.36786 e 2 6 24 120 720 5040 | f (0.2) − P5 (0.2)| ≤

(0.2)7 < 2.6 × 10−9 . 7!

with error less than 5 × 10−5 in absolute value.

5. e1.2 = ee0.2 . From Exercise 1: e0.2 ≈ 1.221400,

(0.2)5 60 · ≈ 0.000003. Since 5! 58 e = 2.718281828 · · ·, it follows that e1.2 ≈ 3.3201094 · · ·, 1 . with error less than 3 × 0.000003 = 0.000009 < 20, 000 Thus e1.2 ≈ 3.32011 with error less than 1/20,000.

with error less than

METHOD II. (using Taylor’s Theorem) Since P5 (x) = P6 (x) (Maclaurin polynomials for sin have only odd degree terms) we are better off using the remainder E6. | f (0.2) − P5 (0.2)| = |E 6 (0.2)| =

| f (7) (s)| (0.2)7 , 7!

6. We have sin(0.1) = 0.1 −

for some s between 0 and 0.2. Now f (7) (x) = − cos x, so | f (0.2) − P5 (0.2)| <

(0.1)5 = 8.33 × 10−8 < 5 × 10−5 , therefore 5!

1 × (0.2)7 < 2.6 × 10−9 . 7!

2. If f (x) = ln x, then f ′ (x) = 1/x, f ′′ (x) = −1/x 2 ,

f ′′′ (x) = 2/x 3 , f (4) (x) = −6/x 4 , and f (5) (x) = 24/x 5 . If P4 (x) is the Taylor polynomial for f about x = 2, then for some s between 1.95 and 2 we have (using Taylor’s Theorem) 24 (0.05)5 · 5! s5 24(0.05)5 ≤ < 2.22 × 10−9 . (1.95)5 120

| f (1.95) − P4 (1.95)| =

Since

(0.1)5 (0.1)7 (0.1)3 + − + ···. 3! 5! 7!

sin(0.1) = 0.1 −

7.

(0.1)3 ≈ 0.09983 3!

with error less than 5 × 10−5 in absolute value. π 5π = cos cos 5◦ = cos 180 36 1  π 2 1  π 4 (−1)n  π 2n ≈1− + − ··· + 2! 36 4! 36 (2n)! 36  π 2n+2 1 |Error| < (2n + 2)! 36 1 < < 0.00005 if n = 1. (2n + 2)!92n+2 1  π 2 cos 5◦ ≈ 1 − ≈ 0.996192 2! 36 with error less than 0.00005.

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SECTION 9.7 (PAGE 549)

ADAMS and ESSEX: CALCULUS 8

55 π 5 < 0.00000005. Thus 5!1805 √ 0.996192 3(0.0871557) cos 65◦ ≈ − ≈ 0.42262 2 2

8. We have ln

    6 1 = ln 1 + 5 5       1 1 1 2 1 1 3 1 1 4 + − + ···. = − 5 2 5 3 5 4 5  n 1 < 5 × 10−5 if n = 6, therefore 5

1 Since n

ln

with error less than

with error less than 0.00005.

12. We have tan−1 (0.2) = 0.2 −

          1 1 1 2 1 1 3 1 1 4 1 1 5 6 + − + ≈ − 5 5 2 5 3 5 4 5 5 5 ≈ 0.18233

Since

ln(0.9) = ln(1 − 0.1)

(0.1)2 (0.1)3 (0.1)n − −··· − 2 3 n (0.1)n+1 (0.1)n+2 |Error| < + + ··· n+1 n+2 i (0.1)n+1 h < 1 + 0.1 + (0.1)2 + · · · n+1 (0.1)n+1 10 · < 0.00005 if n = 3. = n+1 9 (0.1)3 (0.1)2 ln(0.9) ≈ −0.1 − − ≈ −0.10533 2 3 with error less than 0.00005. ≈ −0.1 −

(0.2)7 < 5 × 10−5 , therefore 7 tan−1 (0.2) ≈ 0.2 −

with error less than 5 × 10−5 in absolute value.

9.

(0.2)5 (0.2)7 (0.2)3 + − +···. 3 5 7

13.

(0.2)3 (0.2)5 + ≈ 0.19740 3 5

with error less than 5 × 10−5 in absolute value. 1 1 1 + + ··· + with error less than cosh 1 ≈ 1 + 2! 4! (2n)!   1 1 1 1+ + + · · · (2n + 2)! (2n + 3)2 (2n + 3)4 1 1 = · < 0.00005 if n = 3. 1 (2n + 2)! 1− (2n + 3)2 Thus cosh 1 ≈ 1 + less than 0.00005.

1 1 1 + + ≈ 1.54306 with error 2 24 720

14. We have 10. We have

    1 3 = ln 1 + 2 2       1 1 1 2 1 1 3 1 1 4 = − + − + ···. 2 2 2 3 2 4 2   1 1 1 n < if n = 11, therefore Since n 2 20000 ln

π  sin 80◦ = cos 10◦ = cos 18 1  π 2 1  π 4 =1− + −···. 2! 18 4! 18 Since

1  π 4 < 5 × 10−5 , therefore 4! 18 1  π 2 sin 80 ≈ 1 − ≈ 0.98477 2! 18

ln

◦

11.

with error less than 5 × 10−5 in absolute value.   π 5π cos 65◦ = cos + 3 180 √ 1 3 5π 5π = cos − sin 2 180 2 180 From Exercise 5, cos(5π/180) ≈ 0.996192 with error less than 0.000003. Also sin

5π 5π 1 = − 180 180 3!



5π 180

3

≈ 0.0871557

15.

        1 1 1 2 1 1 3 1 1 10 3 ≈ − + −··· − 2 2 2 2 3 2 10 2 ≈ 0.40543

with error less than 5 × 10−5 in absolute value. Z x sin t I (x) = dt t 0  Z x t2 t4 t6 = 1− + − + · · · dt 3! 5! 7! 0 x3 x5 =x− + − ··· 3 × 3! 5 × 5! ∞ X x 2n+1 = (−1)n for all x. (2n + 1)(2n + 1)! n=0

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INSTRUCTOR’S SOLUTIONS MANUAL

16.

17.

18.

19.

SECTION 9.7 (PAGE 549)

et − 1 dt t 0  Z x t2 t3 t = + + + · · · dt 1+ 2! 3! 4! 0 x3 x4 x2 =x+ + + +··· 2! · 2 3! · 3 4! · 4 ∞ X xn = . n! · n n=1

J (x) =

x

Z

Z

21. From Exercise 15: x3 x5 + −··· 3!3 5!5 1 1 1 + − · · · + (−1)n I (1) ≈ 1 − 3!3 5!5 (2n + 1)!(2n + 1) 1 |Error| ≤ < 0.0005 if n = 2. (2n + 3)!(2n + 3) I (x) = x −

ln t K (x) = dt let u = t − 1 t −1 Z1 x ln(1 + u) = du u 0  Z x u u2 u3 = 1− + − + · · · du 2 3 4 0 x2 x3 x4 = x − 2 + 2 − 2 +··· 2 3 4 ∞ n+1 X x = (−1)n (−1 ≤ x ≤ 1) (n + 1)2 n=0 Z

22.

0

23.

x

tan−1 (t 2 ) dt t2 0  Z x t4 t8 t 12 1− + − + · · · dt = 3 5 7 0 x5 x9 x 13 =x− + − +··· 3 × 5 5 × 9 7 × 13 ∞ 4n+1 X x = (−1)n (−1 ≤ x ≤ 1) (2n + 1)(4n + 1) n=0

M(x) =

decimal places.

Z

x6 x 10 + −··· 3! 5! = lim lim 3 5 x→0 x→0 sinh x x x x+ + +··· 3! 5! 5 9 x x x− + −··· 3! 5! = lim = 0. 4 2 x→0 x x + + ··· 1+ 3! 5! sin(x 2 )

x2 −

L(0.5) = 0.5 −

(0.5)5 (0.5)9 (0.5)13 + − +···. 2! · 5 4! · 9 6! · 13

(0.5)4n+1 < 5 × 10−4 if n = 2, therefore (2n)! · (4n + 1) L(0.5) ≈ 0.5 −

rounded to 3 decimal places.

(0.5)5 ≈ 0.497 2! · 5

x8 x4 − + ··· 2! 4! lim = lim  2 2 x→0 (1 − cos x) x→0 x4 x2 − + ··· 1−1+ 2! 4! 1 2 + O(x ) = lim 2! = 2. x→0 1 + O(x 2 ) 4 1−1+

1 − cos(x 2 )

24. We have 2 x3 x4 + + · · · −1− = lim 2!4 3!6 4!8 lim x→0 x→0 x 2 − ln(1 + x 2 ) x x x − + −··· 2 3 4     4 2 2 x x x 1 1+ + + ··· 1 4 3 12 4 = lim =   = . 1 x→0 2 x4 x6 x8 − + − ··· 2 2 3 4 (e x

20. We have

Since

1 1 + ≈ 0.946 correct to three 3!3 5!5

x

cos(t 2 ) dt  Z x t8 t 12 t4 + − + · · · dt = 1− 2! 4! 6! 0 x5 x9 x 13 =x− + − + ··· 2! · 5 4! · 9 6! · 13 ∞ X x 4n+1 (−1)n = . (2n)! · (4n + 1) n=0

L(x) =

Thus I (1) ≈ 1 −

1+x

25.

x)2

 x2

+

2 sin 3x − 3 sin 2x 5x − tan−1 5x     33 x 3 23 x 3 2 3x − + · · · − 3 2x − + ··· 3! 3! = lim   x→0 53 x 3 5x − 5x − +··· 3 −9 + 4 + O(x 2 ) 5×3 3 =− = lim =− . x→0 125 125 25 + O(x 2 ) 3 lim

x→0

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SECTION 9.7 (PAGE 549)

ADAMS and ESSEX: CALCULUS 8

26. We have

3.

sin(sin x) − x x→0 x[cos(sin x) − 1]   1 1 sin x − sin3 x + sin5 x − · · · − x 3! 5! = lim h i 1 1 x→0 2 x 1 − sin x + sin4 x − · · · − 1 2! 4!   3 5 1 x3 1 x3 x− + ··· − + ··· + x− x − ··· −··· − x 3! 3! 3! 5! = lim h 1 2 i 4 x→0 1 x3 x − +··· + x− x −··· − ··· 2! 3! 4! 2 3 2 − x + higher degree terms 2 = lim 3! = 3! = . 1 3 1 x→0 3 − x + higher degree terms 2! 2!

  1 1 −  1 x 2 2  x 2 = 2 1 + 2 · 4 + 2! 4

lim

27.

sinh x − sin x x→0 cosh x − cos x    x3 + ··· − x − x+ 3! = lim    x→0 x2 + ··· − 1 − 1+ 2! 3 x + O(x 5 ) 3 = lim 2 = 0. x→0 x + O(x 4 )

    1 1 3 − −    x 3 2 2 2 + + · · ·  3! 4

=2+ =2+

4.

lim

 x3 +··· 3!  x2 + ··· 2!

Section 9.8 The Binomial Theorem and Binomial Series (page 554) √ 1 + x = (1 + x)1/2      1 1 x2 1 1 3 x3 x − + − − + ··· =1+ + 2 2 2 2! 2 2 2 3! ∞ 1 · 3 · 5 · · · (2n − 3) n x X (−1)n−1 =1+ + x 2 n=2 2n n! =1+

2.

∞ x X (2n − 2)! + (−1)n−1 2n−1 xn 2 n=2 2 (n − 1)!n!

∞ X x (2n − 1)! +2 (−1)n−1 4n−1 xn 4 2 n!(n − 1)! n=2

(−4 < x < 4).   x 2 −1/2 1 1 1 = √ = r 1 +  x 2 2 2 4 + x2 2 1+ 2       1  x 2 1 1 3  x 4 1 1+ − + − − + = 2 2 2 2! 2 2 2      1 1 3 5  x 6 + ··· − − − 3! 2 2 2 2 1 1 3×5 3 = − 4 x 2 + 7 x 4 − 10 x 6 + · · · 2 2 2 2! 2 3! ∞ 1 X 1 × 2 × 3 × · · · × (2n − 1) 2n (−1)n x = + 2 n=1 23n+1 n! (1 − x)−2

(−2)(−3) (−2)(−3)(−4) (−x)2 + (−x)3 + · · · 2! 3! ∞ X = 1 + 2x + 3x 2 + 4x 3 + · · · = nx n−1 (−1 < x < 1).

= 1 − 2(−x) +

n=1

6.

(−3)(−4) 2 (−3)(−4)(−5) 3 x + x +··· 2! 3! (3)(4) 2 (4)(5) 3 = 1 − 3x + x − x +··· 2 2 ∞ X (n + 2)(n + 1) n x (−1 < x < 1). = (−1)n 2 n=0

(1 + x)−3 = 1 − 3x +

(−1 < x < 1).

√ x 1 − x = x(1 − x)1/2   x2 1 1 (−1)2 x 3 =x− + − 2 2 2 2!    1 1 3 (−1)3 x 4 + − − +··· 2 2 2 3! ∞ x2 X 1 · 3 · 5 · · · (2n − 3) n+1 =x− − x 2 2n n! n=2

∞ X 1 · 3 · 5 · · · (2n − 3) n x (−1)n−1 +2 x 4 23n n! n=2

(−2 ≤ x ≤ 2).

5. 1.

r √ x 4+x =2 1+ 4 

7. Using the Maclaurin series for sin−1 x obtained in this

section, we have π cos−1 x = − sin−1 x 2 ∞ X 1 × 3 × 5 × · · · × (2n − 1) 2n+1 π = −x − x 2 2n n!(2n + 1) n=1

∞ x2 X (2n − 2)! =x− − (−1)n−1 2n−1 x n+1 2 2 (n − 1)!n! n=2

(−1 < x < 1).

π x3 3 5 −x − − x − ··· 2 6 40 for −1 < x < 1 =

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 9.8 (PAGE 554)

d 1 11. sinh−1 x = √ , we have, using the series 2 d x 1 + x √ for 1 + x 2 , Z x dt √ sinh−1 x = 1 + t2 0 # Z x" ∞ X n 1 × 3 × 5 × · · · × (2n − 1) 2n = 1+ t dt (−1) 2n n! 0 n=1

8. Since

=x+ =x−

9.

∞ X 1 × 3 × 5 × · · · × (2n − 1) 2n+1 (−1)n x 2n n!(2n + 1) n=1

x3 3 5 + x −··· 6 40

for −1 < x < 1 n  n  n! n! = 1, = 1. i) = = 0!n! n n!0! 0 ii) If 0 ≤ k ≤ n, then     n! n n n! + + = (k − 1)!(n − k + 1)! k!(n − k)! k−1 k   n! = k + (n − k + 1) k!(n − k + 1)!   n+1 (n + 1)! = . = k k!(n + 1 − k)! Pn

n

a n−k bk k holds for n = 1; it says a + b = a + b in this case. Suppose the formula holds for n = m, where m is some positive integer. Then

10. The formula (a + b)n =

(a + b)m+1 = (a + b) =

m   X m k=0

k

a

k=0

m   X m

k

k=0

m+1−k k

b +

k

=a =

m+1

+

m+1 X k=0

k=1

k

a

m+1−k k

 m+1 a m+1−k bk k

n   X n k=0

k

f (n−k) g (k) .

This holds for n = 1; it says ( f g)′ = f ′ g + f g ′ in this case. Suppose the formula holds for n = m, where m is some positive integer. Then

d ( f g)(m) dx m   d X m = f (m−k) g (k) d x k=0 k m   m   i X X m m f (m−k) g (k+1) = f (m+1−k) g (k) + k k k=0 k=0

( f g)(m+1) =

(replace k by k − 1 in the latter sum) m   m+1 X X m  m = f (m+1−k) g (k) + f (m+1−k) g (k) k k −1 k=0 k=1   m   X m m = f (m+1) g (0) + + k k −1 k=1 × f (m+1−k) g (k) + f (0) g (m+1) (by Exercise 9(i))

a m−k bk+1

(replace k by k − 1 in the latter sum) m   X X m  m m+1−k k m+1 = a b + a m+1−k bk k k − 1 k=0 k=1   m   X m m m+1 =a + + a m+1−k bk + bm+1 k k − 1 k=1 (by #13(i))  m  X m +1

( f g)(n) =

a m−k bk

m   X m k=0

Consider the Leibniz Rule:

b +b

m+1

(by #13(ii))

(by #13(i) again).

= f (m+1) g (0) +

 m  X m +1 k=1

k

f (m+1−k) g (k) + f (0) g (m+1)

(by Exercise 9(ii)) m+1 X m + 1 = f (m+1−k) g (k) k k=0

(by 9(i) again).

Thus the formula holds for n = m + 1. By induction it holds for all positive integers n.

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SECTION 9.8 (PAGE 554)

ADAMS and ESSEX: CALCULUS 8

Thus the Rule holds for n = m +1. By induction, it holds for all positive integers n.

which completes the induction and the proof.

12. As suggested in the hint, we assume that (x1 +x2 +· · ·+xn )k =

X

|m|=k

holds for some n ≥ 2 and all k, and we apply the Binomial theorem to  k (x1 + · · · + xm + xm+1 )k = (x1 + · · · + xn ) + xn+1 = =

k X j =0

k! k− j (x1 + · · · |xn ) j xn+1 j ! (k − j )!

j =0

X j! k! k− j xn+1 x1m 1 · · · xnm n . j ! (k − j )! m ! · · · m ! 1 n |m|= j

k X

m∗

Now let k − j = m n+1 and let = (m 1 , · · · m n , m n+1 ), so that |m ∗ | = j + (k − j ) = k. The result above then simplifies to (x1 + · · · + xn + xn+1 )k X = |m ∗ |=k

Section 9.9

k! x m 1 x m 2 · · · xnm n , m1! m2! · · · mn ! 1 2

1.

2π 3

( f1 f2 · · · fn )

=

X

|m|=k

k! m 1 !m 2 ! · · · m n !

holds for all n ≥ 2 and all positive integers k. The sum is taken over all multiindices m or order n satisfying |m| = k. Observe that the case n = 2 has been proved in Exercise 11. To complete the induction on n, we assume that the formula above holds for some n and all k. If u = f 1 f 2 · · · f n and v = f 1 f 2 · · · f n f n+1 = u f n+1 then the Product rule and the induction hypothesis show that v (k) = (u fn+1 )(k) = =

k X j =0

k X j =0

k! (k− j ) f j !(k − j )! n+1

k! (k− j ) u ( j ) f n+1 j !(k − j )! X

|m|= j

j! f (m 1 ) f 2(m 2 ) · · · f n(m n ) . m1! · · · mn ! 1

m∗

Now let = (m 1 , m 2 , . . . , m n , m n+1 ), where m n+1 = k − j to ensure that |m ∗ | = k. Then we have

2π 3





= sin(3t) = f (t).

= sin(3t + 2π )

= cos(3 + π t) = g(t).

3. h(t) = cos2 t = 21 (1 + cos 2t) has fundamental period π : h(t + π ) =

1 + cos(2t + 2π ) 1 + cos 2t) = = h(t). 2 2

4. Since sin 2t has periods π , 2π , 3π , . . . , and cos 3t 4π 6π 8π has periods 2π 3 , 3 , 3 = 2π , 3 , . . . , the sum k(t) = sin(2t) + cos(3t) has periods 2π , 4π , . . . . Its fundamental period is 2π .

5. Since f (t) = t is odd on (−π, π ) and has period 2π , its

cosine coefficients are 0 and its sine coefficients are given by bn =

2 2π

Z

π

−π

t sin(nt) dt =

2 π

Z

π

t sin(nt) dt.

0

This integral can be evaluated by a single integration by parts. Instead we used Maple to do the integral: 2 2 bn = − cos(nπ ) = (−1)n+1 . n n

k X

X k! j! (m ) f 1(m 1 ) · · · f n(m n ) f n+1n+1 j !m ! m ! · · · m ! n+1 1 n j =0 |m|= j X k! (m ) = f 1(m 1 ) · · · f n(m n ) f n+1n+1 , m ! · · · m !m ! 1 n n+1 ∗ |m |=k

v (k) =

= sin 3 t +

  g(t + 2) = cos 3 + π(t + 2) = cos(3 + π t + 2π )

13. We want to prove that (k)




g(t) = cos(3 + π t) has fundamental period 2 since cos t has fundamental period 2π :

k! m n+1 , x m 1 · · · xnm n xn+1 m 1 ! · · · m n ! m n+1 ! 1

completing the induction and thus the proof.

(page 560)

f (t) = sin(3t) has fundamental period 2π/3 since sin t has fundamental period 2π : f t+

2.

Fourier Series

The Fourier series of f is

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∞ X 2 (−1)n+1 sin(nt). n n=1

INSTRUCTOR’S SOLUTIONS MANUAL

6.

SECTION 9.9 (PAGE 560)



0 if 0 ≤ t < 1 , f has period 2. 1 if 1 ≤ t < 2 The Fourier coefficients of f are as follows: Z Z a0 1 2 1 2 1 = f (t) dt = dt = 2 2 0 2 1 2 Z 2 Z 2 an = f (t) cos(nπ t) dt = cos(nπ t) dt f (t) =

0

1

2 1 (n ≥ 1) = sin(nπ t) = 0, nπ 1 2 Z 2 1 bn = sin(nπ t) dt = − cos(nπ t) nπ 1 1 ( 2 1 − (−1)n =− = − nπ if n is odd nπ 0 if n is even The Fourier series of f is ∞   1 X 2 − sin (2n − 1)π t . 2 n=1 (2n − 1)π

7.

∞ ∞ 2 9 X 2nπ t 1 X 1 1 − cos + cos(2nπ t). 2 2 2 3 2π n=1 n 3 2π n=1 n 2

9. The even extension of h(t) = 1 on [0, 1] to [−1, 1] has

the value 1 everywhere. Therefore all the coefficients an and bn are zero except a0 , which is 2. The Fourier series is a0 /2 = 1.

10. The Fourier sine series of g(t) = π − t on [0, π ] has coefficients

2 bn = π



0 if −1 ≤ t < 0 , f has period 2. f (t) = t if 0 ≤ t < 1 The Fourier coefficients of f are as follows: Z Z a0 −1 1 1 1 1 = f (t) dt = t dt = 2 1 −1 2 0 4 Z 1 Z 1 an = f (t) cos(nπ t) dt = t cos(nπ t) dt −1 0  2 (−1)n − 1 = −2/(nπ ) if n is odd = 2 2 n π 0 if n is even Z 1 t sin(nπ t) dt bn = (−1)n

0

. nπ The Fourier series of f is ∞ ∞  1 2 X (−1)n 1 1 X − 2 sin(nπ t). cos (2n−1)π t)− 2 4 π n=1 (2n − 1) π n=1 n =−

8.

The latter expression was obtained using Maple to evaluate the integrals. If n = 3k, where k is an integer, then an = 0. For other integers n we have an = −9/(2π 2 n 2 ). Thus the Fourier series of f is

t if 0 ≤ t < 1 1 if 1 ≤ t < 2 , f has period 3. 3 − t if 2 ≤ t < 3 f is even, so its Fourier sine coefficients are all zero. Its cosine coefficients are Z 1 2 3 2 a0 2 = · f (t) dt = (2) = 2 2 3 0 3 3 Z 2nπ t 2 3 f (t) cos dt an = 3 0 3 Z 1 Z 2 2 2nπ t 2nπ t = t cos dt + cos dt 3 0 3 3 1  Z 3 2nπ t + (3 − t) cos dt 3  2  3 2nπ 4nπ = 2 2 cos − 1 − cos(2nπ ) + cos . 2n π 3 3 f (t) =

Z

π

(π − t) sin nt dt =

0

2 . n

The required Fourier sine series is ∞ X 2 sin nt. n n=1

11. The Fourier sine series of f (t) = t on [0, 1] has coefficients

bn = 2

Z

0

1

t sin(nπ t) dt = −2

(−1)n . nπ

The required Fourier sine series is

(

∞ X 2(−1)n sin(nπ t). nπ n=1

12. The Fourier cosine series of f (t) = t on [0, 1] has coefficients

a0 = 2

Z

an = 2

1

t dt =

0

Z

1 2

1

t cos(nπ t) dt ( 0 2(−1)n − 2 −4 = = 2 2 n π n2π 2 0

if n is even if n is odd.

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SECTION 9.9 (PAGE 560)

ADAMS and ESSEX: CALCULUS 8

The required Fourier cosine series is

and is proved similarly.

  ∞ cos (2n − 1)π t 1 4 X − . 2 π 2 n=1 (2n − 1)2

Review Exercises 9 (page 561) 1.

13. From Example 3, 2. ∞ X

π 4 + cos (2n − 1)π t = π − |t| 2 n=1 π(2n − 1)2 



3.

for −π ≤ t ≤ π . Putting t = π , we obtain ∞ π X 4 + (−1) = 0. 2 n=1 π(2n − 1)2

Thus

∞ X n=1

4.

5.

π π 1 π = · = . 2 4 8 (2n − 1)2

14. If f is even and has period T , then bn =

2 T

=

2 T

Z

T /2

f (t) sin

−T /2 Z 0

2nπ t dt T 2nπ t dt + T

f (t) sin

−T /2

T /2

Z

f (t) sin

0

 2nπ t dt . T

0

Z

Similarly, an =

2 T

2 = T =

4 T

(−1)n n 2 (−1)n = lim does not exist. n→∞ π n(n − π ) n→∞ 1 − (π/n) The sequence diverges (oscillates). √ an 1 Let a1 > 2 and an+1 = + . 2 an √ x 1 1 1 If f (x) = + , then f ′ (x) = − 2 > 0 if x > 2. 2 x 2 x √ √ √ √ Since f ( 2) = 2,√we have f (x) > 2 if x√> 2. Therefore, if √an > 2, then an+1 = f (an ) > 2. Thus an > 2 for all n ≥ 1, by induction. √ an > 2 ⇒ 2 < an2 ⇒ an2 + 2 < 2an2 lim

an2 + 2 < an ⇒ an+1 < an . 2an √ Thus {an } is decreasing and an > 2 for all n. √ Being decreasing and bounded below by 2, {an } must converge by the completeness axiom. Let √ limn→∞ an = a. Then a ≥ 2, and

  2nπ t (−dt) f (t) − sin T T /2  Z T /2 2nπ t dt + f (t) sin T 0  Z T /2  Z T /2 2 2nπ t 2nπ t = − f (t) sin dt + f (t) sin dt T T T 0 0 = 0. 2 T

ln n ln n = ∞. ≥ lim tan−1 n n→∞ π/2 The sequence diverges to infinity. lim

n→∞

⇒

In the first integral in the line above replace t with −t. Since f (−t) = f (t) and sine is odd, we get bn =

(−1)n en = 0. The sequence converges. n→∞ n!   n 100 + 2n π n 100 lim = π. = lim π + n→∞ n→∞ 2n 2n The sequence converges. lim

lim an+1 = lim

n→∞

1 a a= + . 2 a

n→∞



an 1 + 2 an



Thus a/2 = 1/a, so a 2 = 2, and limn→∞ an = a =

√

2.

6. By l’Hˆopital’s Rule,

Z

0

f (t) cos

−T /2 0

Z Z

T /2 T /2

2nπ t dt + T

Z

0

2nπ t dt. T

f (t) cos

0

2nπ t f (t) cos (−dt) + T

f (t) cos

T /2

Z

0

T /2

2nπ t dt T

2nπ t f (t) cos dt T

The corresponding result for an odd function f states that an = 0 and bn =

4 T

Z

T /2

f (t) sin 0



2nπ t dt, T

lim

x→∞



ln(x + 1) 1/(x + 1) x = lim = lim = 1. x→∞ x→∞ x + 1 ln x 1/x

Thus   ln(n + 1) lim ln ln(n+1)−ln ln n = lim ln = ln 1 = 0. n→∞ n→∞ ln n

7.

∞ X

  1 1 2−(n−5)/2 = 22 1 + √ + + · · · 2 2 n=1 √ 4 4 2 √ =√ . = 1 − (1/ 2) 2−1

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INSTRUCTOR’S SOLUTIONS MANUAL

8.

∞ X n=0

n ∞  4n−1 1X 4 = (π − 1)2n 4 n=0 (π − 1)2 =

1 · 4

1

=

4 (π − 1)2 since (π − 1)2 > 4.

9.

∞ X n=1

1 1 4

n2 −

=

∞ X n=1

1−

1 n−

= 2 − lim

N →∞

10.

∞ X n=1

1 9 4

n2 −

1 2

− 1

1 n+

N+

1 2

1 2

!

REVIEW EXERCISES 9 (PAGE 561)

14. (π − 1)2 , 4(π − 1)2 − 16

15.

! ∞ X 1 1 1 = − (telescoping) 3 n − 32 n + 32 n=1  1 1 1 1 1 = − + − 3 −1/2 5/2 1/2 7/2  1 1 1 1 + − + − + ··· 3/2 9/2 5/2 11/2   1 2 2 = −2 + 2 + = . 3 3 9

12.

32(n+1)+1 n! 9 · 2n+1 = lim = 0 < 1. n→∞ (n + 1)! n→∞ n + 1 3

16.

∞ X n + 2n converges by comparison with the convergent 1 + 3n n=1 ∞  n X 2 because geometric series 3 n=1

18.

∞ X

n √ converges by comparison with the (1 + n)(1 + n n) n=1 ∞ X 1 convergent p-series because 3/2 n n=1 n √ 1 (1 + n)(1 + n n)   = 1. lim = lim  1 n→∞ n→∞ 1 1 + 1 n 3/2 n n 3/2 + 1

∞ X

n! converges by comparison with the con(n + 2)! + 1 n=1 ∞ X 1 vergent p-series , because 2 n n=1 0≤

n + 2n n (n/2n ) + 1 lim 1 + 3n = lim = 1. n→∞ (2/3) n→∞ (1/3n ) + 1

13.

∞ 2n+1 X 3 converges by the ratio test, because n! n=1

lim

17.

∞ X n−1 must also converge. n3 n=1

(1 + 2n )(1

n2 √ 1 (1 + 2n )(1 + n n)   = 1. √ lim = lim  n→∞ n→∞ 1 1 n + 1 2n n 3/2 + 1 2n

(telescoping)

= 2.

n2

√ converges by comparison with + n n) ∞ √ X n the convergent series (which converges by the n 2 n=1 ratio test) because n=1

∞ X n−1 1 1 ≤ 2 for n ≥ 1 and converges, 3 2 n n n n=1

11. Since 0 ≤

∞ X

19.

n! 1 1 n! < = < 2. (n + 2)! + 1 (n + 2)! (n + 2)(n + 1) n

∞ X (−1)n−1 converges absolutely by comparison with 1 + n3 n=1 ∞ X 1 the convergent p-series , because n3 n=1 (−1)n−1 ≤ 1 . 0 ≤ 3 1 + n n3

∞ X (−1)n converges absolutely by comparison with the 2n − n n=1 ∞ X 1 convergent geometric series , because n 2 n=1 (−1)n 2n − n 1 lim = lim n = 1. 1 n→∞ n→∞ 1− n n 2 2 ∞ X (−1)n−1 converges by the alternating series test, but ln ln n n=1 ∞ X 1 the convergence is only conditional since ln ln n n=1 diverges to infinity by comparison with the divergent ∞ X 1 . (Note that ln ln n < n for all harmonic series n n=1 n ≥ 1.)

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REVIEW EXERCISES 9 (PAGE 561)

20.

21.

22.

ADAMS and ESSEX: CALCULUS 8

∞ X n 2 cos(nπ ) converges by the alternating series test 1 + n3 n=1 (note that cos(nπ ) = (−1)n ), but the convergence is only conditional because 2 n cos(nπ ) n2 1 1 + n 3 = 1 + n 3 ≥ 2n

∞ X 1 for n ≥ 1, and is a divergent harmonic series. 2n n=1 (x − 2)n+1 r n+1 √ |x − 2| |x − 2| n n+1 3 = lim lim = . n→∞ n→∞ (x − 2)n 3 n+1 3 √ 3n n

∞ X (x − 2)n |x − 2| < 1, that is, converges absolutely if √ n 3 3 n n=1 if −1 < x < 5, and diverges if x < −1 or x > 5. X (−1)n If x = −1 the series is √ , which converges n conditionally. X 1 If x = 5 the series is √ , which diverges (to ∞). n (5 − 2x)n+1 n+1 = lim |5 − 2x| n = |5 − 2x|. lim n n→∞ n→∞ (5 − 2x) n+1 n ∞ X (5 − 2x)n converges absolutely if |5 − 2x| < 1, that n n=1 is, if 2 < x < 3, and diverges if x < 2 or x > 3. X1 , which diverges. If x = 2 the series is n X (−1)n If x = 3 the series is , which converges condin tionally.

23. Let s =

∞ n X X 1 1 and sn = . Then 3 3 k k k=1 k=1 ∞

s≈

24.

Z ∞ dt dt < s − s < n 2 4 + t 4 + t2 n n+1 π 1 −1 n + 1 π 1 n sn + − tan < s < sn + − tan−1 . 4 2 2 4 2 2

sn∗ = sn +

Then s ≈ sn∗ with error satisfying 1 1 1 − 2 2 2n 2(n + 1)2

=

  π 1 n+1 n − tan−1 + tan−1 . 4 4 2 2

Then s ≈ sn∗ with error satisfying |s − sn∗ | <

  n+1 n 1 tan−1 − tan−1 . 4 2 2

This error is less than 0.001 if n ≥ 22. Hence s≈

25.

22 X k=1

  1 π 1 −1 23 + tan−1 (11) ≈ 0.6605 − tan + 4 + k2 4 4 2

with error less than 0.001. 1 1 =  x 3−x 3 1− 3 ∞  n ∞ X 1X x xn = = 3 n=0 3 3n+1 n=0

(−3 < x < 3).

get

  1 1 1 n 2 + (n + 1)2 sn∗ = sn + + s + = s + . n n 2 2(n + 1)2 2n 2 4n 2 (n + 1)2

|s − sn∗ | <

∞

Let

∞ X x x 2n+1 = 2 3−x 3n+1 n=0

Z



with error less than 0.001. ∞ n X X 1 1 Let s = and sn = . Then 2 4 + k 4 + k2 k=1 k=1 Z

27.

Let



1 1 1 1 1 1 1 1 + 3 + 3 + 3 + 3 + 3 + 3 + 3 3 1 2 3 4 5 6 7 8 64 + 81 ≈ 1.202 + 4(64)(81)

26. Replace x with x 2 in Exercise 25 and multiply by x to

∞ dt dt < s − sn < 3 t3 n+1 t n 1 1 sn + < s < sn + 2 . 2(n + 1)2 2n

Z

This error is less than 0.001 if n ≥ 8. Hence

2n + 1 . + 1)2

4n 2 (n

28.

√ √ (− 3 < x < 3).

  x2 ln(e + x 2 ) = ln e + ln 1 + e ∞ X x 2n = ln e + (−1)n−1 n ne n=1 1 − e−2x 1 = x x

1−1−

∞ X (−2x)n n=1

n!

∞ X 2n x n−1 = (−1)n−1 n! n=1

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√ √ (− e < x ≤ e). !

(for all x 6= 0).

INSTRUCTOR’S SOLUTIONS MANUAL

29.

x (1 + cos(2x)) 2 ! ∞ 2n X x n (2x) = 1+ (−1) 2 (2n)! n=0

x cos2 x =

=x+

30.

31.

REVIEW EXERCISES 9 (PAGE 561)

∞ X 22n−1 x 2n+1 (−1)n (2n)! n=1

34. Let u = x − (π/4), so x = u + (π/4). Then

(for all x).

 π π π sin x + = sin x cos + cos x sin 3 3 √ ∞ 3 ∞ 2n+1 3X 1X x x 2n = (−1)n + (−1)n 2 n=0 (2n + 1)! 2 n=0 (2n)! ! √ ∞ X 3x 2n x 2n+1 (−1)n + (for all x). = 2 (2n)! (2n + 1)! n=0 1 x −1/3 1+ 2   8  4 1  − − 1 1 x  3 3  x 2 = 1− + 2 3 8 2! 8     1 4 7  − − − 3 3 3  x 3 +··· + 3! 8 ∞ 1 · 4 · 7 · · · (3n − 2) n 1 X (−1)n x (−8 < x < 8). = + 2 n=1 2 · 3n · 8n · n!

   2 1 − 1 3 3 x2 (1 + x)1/3 = 1 + x + 3 2!     1 2 5 − − 3 3 3 3 + x + ··· 3! ∞ 2 · 5 · 8 · · · (3n − 4) n x X =1+ + (−1)n−1 x 3 n=2 3n n! (Remark: the series also converges at x = 1.)

33.

35.

1 1 1 1 = = · x π + (x − π ) π 1+ x −π π  n ∞ 1 X x − π = (−1)n π n=0 π =

∞ X n=0

(−1)n

(x − π )n π n+1

(0 < x < 2π ).

36.

37.

38.

(−1 < x < 1).

ex

2 +2x

2

= e x e2x

  4x 2 8x 3 = (1 + x 2 + · · ·) 1 + 2x + + + ··· 2! 3! 4 3 2 2 3 = 1 + 2x + 2x + x + x + 2x + · · · 3 10 3 2 x . P3 (x) = 1 + 2x + 3x + 3

(8 + x)−1/3 =

(Remark: Examining the ln of the absolute value of the nth term at x = 8 shows that this term → 0 as n → ∞. Therefore the series also converges at x = 8.)

32.

  π π sin x + cos x = sin u + + cos u + 4 4  1  = √ (sin u + cos u) + (cos u − sin u) 2 ∞ √ √ X u 2n = 2 cos u = 2 (−1)n (2n)! n=0 ∞   n √ X 2n π (−1) (for all x). = 2 x− (2n)! 4 n=0

sin(1 + x) = sin(1) cos x + cos(1) sin x     x2 x3 = sin(1) 1 − + · · · + cos(1) x − +··· 2! 3! sin(1) 2 cos(1) 3 x − x . P3 (x) = sin(1) + cos(1)x − 2 6 2 x3 + ··· (x − · · ·)4 3! cos(sin x) = 1 − + − ··· 2! 4!   1 x4 x4 =1− x2 − +··· + +··· 2 3 24 1 5 4 P4 (x) = 1 − x 2 + x . 2 24    1 1 − √ 1 2 2 1 + sin x = 1 + sin x + (sin x)2 2 2!     1 3 1 − − 2 2 2 + (sin x)3 3!      1 1 3 5 − − − 2 2 2 2 + (sin x)4 + · · · 4!    2 1 x3 1 x3 =1+ x− +··· − x− + ··· 2 6 8 6 1 5 + (x − · · ·)3 − (x − · · ·)4 + · · · 16 128 x x3 x2 x4 x3 5x 4 =1+ − − + + − + ··· 2 12 8 24 16 128 2 3 4 x x x x − + . P4 (x) = 1 + − 2 8 48 384 

x−

387 Copyright © 2014 Pearson Canada Inc.

REVIEW EXERCISES 9 (PAGE 561)

39. The series

ADAMS and ESSEX: CALCULUS 8

∞ X (−1)n x n is the Maclaurin series for cos x (2n)! n=0

44.

n=1 ∞ X

with x 2 replaced by x. For x > 0 the series therefore ∞ X √ |x|n represents cos x. For x < 0, the series is , (2n)! n=0 √ which is the Maclaurin series for cosh |x|. Thus the given series is the Maclaurin series for f (x) =

40. Since 1+



√ cos √x if x ≥ 0 cosh |x| if x < 0.

45.

41. ∞ X

∞ X

nx

n=0 ∞ X n=0

=

nx n =

∞ X n=0

∞ X

lim

x→0

(x − tan−1 x)(e2x − 1) 2x 2 − 1 + cos(2x)

nx

n−1

= lim

16x 4 4x 2 + −··· 2x 2 − 1 + 1 − 2! 4!   2 x4 + ··· 3  = 1.  = lim x→0 4 2 + ··· x 3

x (1 − x)2 1 π

1 1− π

2 +

1 1−

nx n =

x (1 − x)2

!  x3 x5 4x 2 − +··· 2x + + ··· 3 5 2!

x→0

1 π

47.

Z

1/2

e

−x 4

0



π π −1

2

.

as in Exercise 23

d x 1+x = = d x (1 − x)2 (1 − x)3

x(1 + x) (1 − x)3 n=0   1 1 1 + ∞ X n2 π(π + 1) π π =  . 3 = n π (π − 1)3 1 n=0 1− π

43.

46.

1 = (1 − x)2

π π = = + π −1 (π − 1)2

n 2 x n−1

n=0 ∞ X

0

x−x+

n=1

∞ X n+1 =  πn n=0

42.

f (2n−1) (0) = 0.

1 x = 1 − x n=0 n−1

x

Z

sin(t 2 ) dt  Z x t6 = t2 − + · · · dt 3! 0 x7 x3 − + ··· = 3 7 · 3!

S(x) =

n

∞ X

= sin x

x7 − ··· x3 − x3 + x 3 − 3S(x) 1 14 lim = lim = . 7 7 x→0 x→0 x x 14

for x near 0, we have, for n = 1, 2, 3, . . . (2n)! , n2

(2n − 1)!

(−1)n π 2n−1 = − sin π = 0 (2n − 1)! n=1   ∞ X 1 (−1)n π 2n−4 (−1)π 1 = 3 0− = 2. (2n − 1)! π 1! π n=2

∞ ∞ X X x 2n f (k) (0) k = x 2 n k! n=1 k=0

f (2n) (0) =

∞ X (−1)n−1 x 2n−1

n2 x n =

∞ X xn = − ln(1 − x) n n=1   ∞ X 1 1 = − ln 1 − = 1 − ln(e − 1). nen e n=1

∞ 1/2 X

(−x 4 )n dx n! 0 n=0 1/2 ∞ X (−1)n x 4n+1 = (4n + 1)n! 0 n=0

dx =

=

Z

∞ X n=0

(−1)n . 24n+1 (4n + 1)n!

The series satisfies the conditions of the alternating series test, so if we truncate after the term for n = k − 1, then the error will satisfy |error| ≤

1 . 24k+1 (4k + 1)k!

This is less than 0.000005 if 24k+1 (4k + 1)k! > 200, 000, which happens if k ≥ 3. Thus, rounded to five decimal places, Z

0

1/2

4

e−x d x ≈

388 Copyright © 2014 Pearson Canada Inc.

1 1 1 − + ≈ 0.49386. 2 · 1 · 1 32 · 5 · 1 512 · 9 · 2

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 9 (PAGE 562)

48. If f (x) = ln(sin x), then calculation of successive deriva-

The required Fourier series is, therefore,

tives leads to

f (5) (x) = 24 csc4 x cot x − 8 csc2 cot x. Observe that 1.5 < π/2 ≈ 1.5708, that csc x ≥ 1 and cot x ≥ 0, and that both functions are decreasing on that interval. Thus | f (5) (x)| ≤ 24 csc4 (1.5) cot(1.5) ≤ 2 for 1.5 ≤ x ≤ π/2. Therefore, the error in the approximation ln(sin 1.5) ≈ P4 (x),

−

2+π 4    ∞ X 2 cos (2n − 1)t n=1

π(2n − 1)2

1. If an > 0 and

0

∞ X 2 The series is sin(nt). n n=1

50.

⇒

an n−1 > an−1 n

coefficients

2 (π − t) sin(nt) dt = . n

a1 2 2a2 a1 a3 > > 3 3 a2 >

⇒

.. .

49. The Fourier sine series of f (t) = π − t on [0, π ] has π

 sin(2nt) . 2n

an+1 n for all n, then > an n+1 a2 1 > a1 2 a3 2 > a2 3

π 5 2 |error| ≤ 1.5 − ≤ 3 × 10−8 . 5! 2

Z

π(2n − 1)

+

Challenging Problems 9 (page 562)

where P4 is the 4th degree Taylor polynomial for f (x) about x = π/2, satisfies

2 bn = π

+

  (2 − π ) sin (2n − 1)t

2.

an >

⇒

a1 . n

(This can be P verified by induction.) Therefore ∞ n=1 an diverges by comparison with the harP 1 monic series ∞ n=1 . n P a) If sn = nk=1 v k for n ≥ 1, and s0 = 0, then v k = sk − sk−1 for k ≥ 1, and n X k=1

uk vk =

n X k=1

u k sk −

n X

u k sk−1 .

k=1

In the second sum on the right replace k with k + 1:



1 if −π < t ≤ 0 f (t) = has period 2π . Its Fourier t if 0 < t ≤ π coefficients are Z π 1 a0 = f (t) dt 2 2π −π Z 0  Z π 1 1 π = dt + t dt = + 2π −π 2 4 0 Z 0  Z π 1 an = cos(nt) dt + t cos(nt) dt π −π 0 Z 1 π = (1 + t) cos(nt) dt π 0  (−1)n − 1 −2/(π n 2 ) if n is odd = = 2 πn 0 if n is even  Z 0 Z π 1 t sin(nt) dt bn = sin(nt) dt + π −π 0 Z 1 π = (t − 1) sin(nt) dt π 0  1 + (−1)n (π − 1) (π − 2/(π n) if n is odd =− = −(1/n) if n is even. πn

n X k=1

uk vk = =

n X

k=1 n X k=1

u k sk −

n−1 X

u k+1 sk

k=0

(u k − u k+1 )sk − u 1 s0 + u n+1 sn

= u n+1 sn +

n X (u k − u k+1 )sk . k=1

b) If {u n } is positive and decreasing, and limn→∞ u n = 0, then

n X (u k − u k+1 ) = u 1 − u 2 + u 2 − u 3 + · · · + u n − u n+1 k=1

= u 1 − u n+1 → u 1 as n → ∞.

Thus

n X (u k − u k+1 ) is a convergent, positive, telek=1

scoping series.

If the partial sums sn of {v n } are bounded, say |sn | ≤ K for all n, then |(u n − u n+1 )sn | ≤ K (u n − u n+1 ),

389 Copyright © 2014 Pearson Canada Inc.

CHALLENGING PROBLEMS 9 (PAGE 562)

ADAMS and ESSEX: CALCULUS 8

P so ∞ n=1 (u n − u n+1 )sn is absolutely convergent (and therefore convergent) by the comparison test. Therefore, by part (a), ∞ X k=1

u k v k = lim

n→∞

n X u n+1 sn + (u k − u k+1 )sk k=1

!

∞ X = (u k − u k+1 )sk k=1

converges.

f′ k −

∞ n=1 (1/n) sin(nx)

f ′ (x)

are 0, so the series converges

∞ X

n=N +1

=

sin(nx) = =

h    i N cos (n − 21 )x − cos (n + 12 )x X 2 sin(x/2)  cos(x/2) − cos (N + 21 )x

n=1



2 sin(x/2)

.

P Therefore, the partial sums of ∞ n=1 sin(nx) are bounded. Since the sequence {1/n} is positive, decreasing, and has P limit 0, part (b) of Problem 2 shows that ∞ sin(nx)/n n=1 converges in this case too. Therefore the series converges for all x.

4. Let an be the nth integer that has no zeros in its decimal representation. The number of such integers that have m digits is 9m . (There are nine possible choices for each of the m digits.) Also, each such m-digit number is greater than 10m−1 (the smallest m-digit number). Therefore the sum of all the terms 1/an for which an has m digits is less than 9m /(10m−1 ). Therefore, ∞ X 1 9 0, and let Ik =

x

0

1 d V = 2 e−1/t dt t V = e−1/t

x

t k+1 e−1/t dt

0

Ik = x k+2 e−1/x − (k + 2)Ik+1 . Therefore,

n=2

+ (−1) N +1 N !

x

Z

t N −1 e−1/t dt.

0

The Maclaurin series for e−1/t does not exist. The function is not defined at t = 0. For x = 0.1 and N = 5, the approximation I =

Z

0.1 0

e−1/t dt ≈ e−10

5 X (−1)n (n − 1)!(0.1)n

n=2   = e−10 (0.1)2 − 2(0.1)3 + 6(0.1)4 − 24(0.1)5

≈ 0.00836e−10 has error E given by

E = (−1)6 5!

Z

0.1

t 4 e−1/t dt.

0

Since e−1/t ≤ e−10 for 0 ≤ t ≤ 0.1, we have |E| ≤ 120e−10

Z

0.1 0

t 4 dt ≈ 2.4 × 10−4 e−10 ,

For N = 10, the error estimate is |E| ≤ 10!e

−10

Z

0.1

0

t 9 dt ≈ 3.6 × 10−5 e−10 ,

which is about 0.4% of the size of I . For N = 20, the error estimate is |E| ≤ 20!e−10

e−1/t dt = I0 = x 2 e−1/x − 2I1   = x 2 e−1/x − 2 x 3 e−1/x − 3I2   = e−1/x [x 2 − 2!x 3 ] + 3! x 4 e−1/x − 4I3 0

N X (−1)n (n − 1)!x n

t k e−1/t dt

0

U= dU = (k + 2)t k+1 dt x Z k+2 −1/t − (k + 2) =t e

x

.. .

which is about 3% of the size of I .

t k+2

Z

  = e−1/x [x 2 − 2!x 3 + 3!x 4 ] − 4! x 5 e−1/x − 5I4 = e−1/x

× ( j + k)( j + k − 1) · · · ( j + k − i + 1)x j +k−i k   k ( j + k)! 1 X x j +k−i . m k− j (−n) j = k! j =0 j ( j + k − i )!

Z

ADAMS and ESSEX: CALCULUS 8

Z

0

0.1

t 19 dt ≈ 1.2 × 10−3 e−10 ,

which is about 15% of the size of I . Observe, therefore, that the sum for N = 10 does a better job of approximating I than those for N = 5 or N = 20.

392 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.1 (PAGE 569)

CHAPTER 10. VECTORS AND COORDINATE GEOMETRY IN 3-SPACE

8. If A = (1, 2, 3), B = (1, 3, 4), and C = (0, 3, 3), then | AB| = | AC| =

Section 10.1 Analytic Geometry in Three Dimensions (page 569) 1. The distance between (0, 0, 0) and (2, −1, −2) is p 22 + (−1)2 + (−2)2 = 3 units.

|BC| =

(0 − 1)2

+ (2

− 1)2

+ (−2 − 0)2

√ = 6 units.

4. The distance between (3, 8, −1) and (−2, 3, −6) is p

5.

(−2 − 3)2

+ (3

− 8)2

+ (−6 + 1)2

√ = 5 3 units.

a) The shortest distance from (x, y, z) to the x y-plane is |z| units.

p

(0 − 1)2

(0 − 1)2

+ (3 + (3

p

a = |BC| =

p

12 + 12 + 12 + · · · + 1 =

12.

2.

√

2. √

2. Its

√

n units.

z

2 z=2

y x

Fig. 10.1.12

13.

y ≥ −1 is the half-space consisting of all points on the plane y = −1 (which is perpendicular to the y-axis at (0, −1, 0)) and all points on the same side of that plane as the origin. z

(1 − 2)2 + (−3 + 1)2 + (1 + 1)2 = 3 √ (1 − 0)2 + (−3 − 1)2 + (1 + 2)2 = 26.

y=−1 −1

By the Cosine Law, x

6

2

z = 2 is a plane, perpendicular to the z-axis at (0, 0, 2).

(0 − 2)2 + (1 + 1)2 + (−2 + 1)2 = 3

a 2 = b2 + c2 − 2bc cos 6 A 26 = 9 + 9 − 18 cos 6 A 26 − 18 A = cos−1 ≈ 116.4◦ . −18

=

√

(1, 0, 0, . . . , 0). The distance between these points is p √ 02 + 12 + 12 + · · · + 12 = n − 1 units.

then

p

=

2

11. The point on the x1 -axis closest to (1, 1, 1, . . . , 1) is

7. If A = (2, −1, −1), B = (0, 1, −2), and C = (1, −3, 1),

b = | AC| =

+ (3

− 4)2

√

10. The distance from the origin to (1, 1, 1, . . . , 1) in Rn is

p √ | AB| = 32 + (−2)2 + 22 = 17 p √ | AC| = 22 + 42 + 12 = 21 p √ |BC| = (−1)2 + 62 + (−1)2 = 38.

p

+ (3

− 3)2

Thus the triangle ABC is equilateral with sides area is, therefore, r √ 1 √ 3 1 × 2× 2− = sq. units. 2 2 2

6. If A = (1, 2, 3), B = (4, 0, 5), and C = (3, 6, 4), then

c = | AB| =

− 3)2

| AB| = | AC| = |BC| =

b) p The shortest distance from (x, y, z) to the x-axis is y 2 + z 2 units.

Since | AB|2 + | AC|2 = 17 + 21 = 38 = |BC|2, the triangle ABC has a right angle at A.

− 2)2

√

9. IfA = (1, 1, 0), B = (1, 0, 1), and C = (0, 1, 1), then

p √ (1 + 1)2 + (1 + 1)2 + (1 + 1)2 = 2 3 units.

p

(1 − 1)2 + (3 − 2)2 + (4 − 3)2 =

p

All three sides being equal, the triangle is equilateral.

2. The distance between (−1, −1, −1) and (1, 1, 1) is

3. The distance between (1, 1, 0) and (0, 2, −2) is

p

y

Fig. 10.1.13

14.

z = x is a plane containing the y-axis and making 45◦ angles with the positive directions of the x- and z-axes.

393 Copyright © 2014 Pearson Canada Inc.

SECTION 10.1 (PAGE 569)

ADAMS and ESSEX: CALCULUS 8

z

z x 2 +z 2 =4

(1,0,1)

2

z=x

y x

y x

Fig. 10.1.14

Fig. 10.1.20

21. 15.

x + y = 1 is a vertical plane (parallel to the z-axis) passing through the points (1, 0, 0) and (0, 1, 0). z

z = y 2 is a “parabolic cylinder” — a surface all of whose cross-sections in planes perpendicular to the x-axis are parabolas. z

x+y=1 z=y 2

1 y x

1 y

x

Fig. 10.1.15

16.

Fig. 10.1.21

x 2 + y 2 + z 2 = 4 is a sphere centred at the origin and having radius 2 (i.e., all points at distance 2 from the origin).

22.

17. (x − 1)2 + (y + 2)2 + (z − 3)2 = 4 is a sphere of radius 2

p

z ≥ x 2 + y 2 represents every point whose distance above the x y-plane is not less than its horizontal distance from the z-axis. It therefore consists of all points inside and on a circular cone with axis along the positive z-axis, vertex at the origin, and semi-vertical angle 45◦ . z

with centre at the point (1, −2, 3).

18.

x 2 + y 2 + z 2 = 2z can be rewritten 45◦

x 2 + y 2 + (z − 1)2 = 1,

√

z=

y

x

and so it represents a sphere with radius 1 and centre at (0, 0, 1). It is tangent to the x y-plane at the origin.

x 2 +y 2

Fig. 10.1.22

z x 2 +y 2 +z 2 =2z

23.

(0,0,1)

x + 2y + 3z = 6 represents the plane that intersects the coordinate axes at the three points (6, 0, 0), (0, 3, 0), and (0, 0, 2). Only the part of the plane in the first octant is shown in the figure. z (0,0,2)

y

x

(0,3,0)

Fig. 10.1.18

19.

20.

y

y 2 +z 2 ≤ 4 represents all points inside and on the circular cylinder of radius 2 with central axis along the x-axis (a solid cylinder). x 2 + z 2 = 4 is a circular cylindrical surface of radius 2 with axis along the y-axis.

x

(6,0,0)

Fig. 10.1.23

24.



x =1 represents the vertical straight line in which the y=2 plane x = 1 intersects the plane y = 2.

394 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.1 (PAGE 569)

z

27. y=2 x=1

y x

(1,2,0)

Fig. 10.1.24

28. 25.



x =1 is the straight line in which the plane z = 1 y=z intersects the plane y = z. It passes through the points (1, 0, 0) and (1, 1, 1). z



x 2 + y2 + z2 = 4 is the circle in which the sphere of x 2 + y 2 + z 2 = 4z radius 2 centred at the origin intersects the sphere of radius 2 centred at (0, 0, 2). (The second equation can be rewritten x 2 + y 2 + (z − 2)2 = 4 for easier recognition.) Subtracting the equations of the two spheres we get z = 1, so the circle must lie in the plane z = 1 as well. Thus it is the same circle as in the previous exercise.  x 2 + y 2 + z 2 = 4 represents the two circles in x 2 + z2 = 1 which the cylinder x 2 + z 2 − 1 intersects the sphere x 2 + y 2 + z 2 = 4. Subtracting the two equations, we √ get y 2 = 3. Thus,√one circle lies in the plane y = 3 and has√centre (0, 3, 0) and the √ other lies in the plane y = − 3 and has centre (0, − 3, 0). Both circles have radius 1. z

x=1

(1,1,1)

z=y

2

1 √

y

3

x (1,0,0)

y

x

Fig. 10.1.25 Fig. 10.1.28

26.



x2

y2

29.

z2

+ + = 4 is the circle in which the horizontal z=1 plane z = 1 intersects the sphere of radius 2 centred at the origin. √The circle has centre (0, 0, 1) and radius √ 4 − 1 = 3.



x 2 + y 2 = 1 is the ellipse in which the slanted plane z=x z = x intersects the vertical cylinder x 2 + y 2 = 1. z

x 2 +y 2 =1

z

z=x

√ 3

(0,0,1) 1

z=1

y

2

x

y

Fig. 10.1.29

x x 2 +y 2 +z 2 =4

Fig. 10.1.26

30.

ny ≥ x is the quarter-space consisting of all points lying z≤y on or on the same side of the planes y = x and z = y as does the point (0, 1, 0).

395 Copyright © 2014 Pearson Canada Inc.

SECTION 10.1 (PAGE 569)

31.

ADAMS and ESSEX: CALCULUS 8



x 2 + y2 ≤ 1 represents all points which are inside or z≥y on the vertical cylinder x 2 + y 2 = 1, and are also above or on the plane z = y. z

35.

S = {(x, y) : x + y = 1} The boundary of S is S. The interior of S is the empty set. S is closed, but not bounded. There are points on the line x + y = 1 arbitrarily far away from the origin.

36.

S = {(x, y) : |x| + |y| ≤ 1} The boundary of S consists of all points on the edges of the square with vertices (±1, 0) and (0, ±1). The interior of S consists of all points inside that square. S is closed since it contains all its boundary points. It is bounded since all points in it are at distance not greater than 1 from the origin.

37.

S = {(x, y, z) : 1 ≤ x 2 + y 2 + z 2 ≤ 4} Boundary: the spheres of radii 1 and 2 centred at the origin. Interior: the region between these spheres. S is closed.

38.

S = {(x, y, z) : x ≥ 0, y > 1, z < 2} Boundary: the quarter planes x = 0, (y ≥ 1, z ≤ 2), y = 1, (x ≥ 0, z ≤ 2), and z = 2, (x ≥ 0, y ≥ 1). Interior: the set of points (x, y, z) such that x > 0, y > 1, z < 2. S is neither open nor closed.

39.

S = {(x, y, z) : (x − z)2 + (y − z)2 = 0} The boundary of S is S, that is, the line x = y = z. The interior of S is empty. S is closed.

40.

S = {(x, y, z) : x 2 + y 2 < 1, y + z > 2} Boundary: the part of the cylinder x 2 + y 2 = 1 that lies on or above the plane y + z = 2 together with the part of that plane that lies inside the cylinder. Interior: all points that are inside the cylinder x 2 + y 2 = 1 and above the plane y + z = 2. S is open.

z=y

x

y x 2 +y 2 =1

Fig. 10.1.31

32.



x 2 + y2 + z2 ≤ 1 p represents all points which are inside x 2 + y2 ≤ z or on the sphere of radius 1 centred at the origin and which are also inside or on the upper half of the circular cone with axis along the z-axis, vertex at the origin, and semi-vertical angle 45◦ . z

√

z=

x 2 +y 2

y

Section 10.2

x x 2 +y 2 +z 2 =1

1.

Fig. 10.1.32

33.

34.

2.

(page 578)

A = (−1, 2), B = (2, 0), C = (1, −3), D = (0, 4). −→ −→ (a) AB = 3i − 2j (b) B A = −3i + 2j −→ (c) AC = 2i − 5j

S = {(x, y) : 0 < x 2 + y 2 < 1} The boundary of S consists of the origin and all points on the circle x 2 + y 2 = 1. The interior of S is S, which is therefore open. S is bounded; all points in it are at distance less than 1 from the origin. S = {(x, y) : x ≥ 0, y < 0} The boundary of S consists of points (x, 0) where x ≥ 0, and points (0, y) where y ≤ 0. The interior of S consists of all points of S that are not on the y-axis, that is, all points (x, y) satisfying x > 0 and y < 0. S is neither open nor closed; it contains some, but not all, of its boundary points. S is not bounded; (x, −1) belongs to S for 0 < x < ∞.

Vectors

−→ (d) B D = −2i + 4j

−→ −→ −→ (e) D A = −i − 2j (f) AB − BC = 4i + j −→ −→ −→ (g) AC − 2 AB + 3C D = −7i + 20j 5 1 −→ −→ −→ AB + AC + AD = 2i − j (h) 3 3 u=i−j v = j + 2k a)

u + v = i + 2k u − v = i − 2j − 2k 2u − 3v = 2i − 5j − 6k

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.2 (PAGE 578)

√ 2 √ √ |v| = 1 + 4 = 5

b) |u| =

√

1+1=

Thus M N is parallel to and half as long as BC. C

1 c) uˆ = √ (i − j) 2 1 vˆ = √ (j + 2k) 5 d) u • v = 0 − 1 + 0 = −1

N

A

e) The angle between u and v is −1 cos−1 √ ≈ 108.4◦ . 10 f) The scalar projection of u in the direction of v is −1 u•v = √ . |v| 5

g) The vector projection of v along u is 1 (v • u)u = − (i − j). 2 |u| 2

3.

M

B

Fig. 10.2.5

6. We have −→ −→ −→ 1 −→ P Q = P B + B Q = 2 AB + − → −→ −→ −→ S R = S D + D R = 21 AD +

u = 3i + 4j − 5k v = 3i − 4j − 5k

1 −→ 2 BC = 1 −→ 2 DC =

→ 1− 2 AC; → 1− 2 AC.

−→ − → Therefore, P Q = S R. Similarly,

a)

u + v = 6i − 10k u − v = 8j 2u − 3v = −3i + 20j + 5k √ √ b) |u| = 9 + 16 + 25 = 5 2 √ √ |v| = 9 + 16 + 25 = 5 2

−→ −→ −→ 1 −→ Q R = QC + C R = 2 B D; −→ −→ − → −→ P S = P A + AS = 12 B D. −→ − → Therefore, Q R = P S. Hence, P Q RS is a parallelogram. Q

B

1 c) uˆ = √ (3i + 4j − 5k) 5 2 1 vˆ = √ (3i − 4j − 5k) 5 2 d) u • v = 9 − 16 + 25 = 18

P

e) The angle between u and v is 18 cos−1 ≈ 68.9◦ . 50 f) The scalar projection of u in the direction of v is u•v 18 = √ . |v| 5 2

g) The vector projection of v along u is (v • u)u 9 = (3i + 4j − 5k). |u|2 25

A

S

D

7. Let the parallelogram be ABC O. Take the origin at O. The position vector of the midpoint of O B is −→ −→ −→ −→ −→ OB OB + CB OC + O A = = . 2 2 2 The position vector of the midpoint of C A is −→ −→ −→ −→ C A −→ O A − OC OC + = OC + 2 2 −→ −→ OC + O A = . 2

5. Let the triangle be ABC. If M and N are the midpoints −−→ −→ of AB and AC respectively, then AM = 12 AB, and −→ 1 −→ AN = 2 AC. Thus −→ −→ −→ AC − AB BC −−→ −→ −−→ M N = AN − AM = = . 2 2

R

Fig. 10.2.6

4. If a = (−1, 1), B = (2, 5) and C = (10, −1), then

−→ −→ −→ −→ AB = 3i + 4j and BC = 8i − 6j. Since AB • BC = 0, −→ −→ therefore, AB ⊥ BC. Hence, △ ABC has a right angle at B.

C

Thus the midpoints of the two diagonals coincide, and the diagonals bisect each other.

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SECTION 10.2 (PAGE 578)

ADAMS and ESSEX: CALCULUS 8

9. Let i point east and j point north. Let the wind velocity B

C

be vwind = ai + bj. Now vwind = vwind rel car + vcar . When vcar = 50j, the wind appears to come from the west, so vwind rel car = λi. Thus ai + bj = λi + 50j,

A

O

Fig. 10.2.7

8. Let X be the point of intersection of the medians AQ and B P as shown. We must show that C X meets AB −→ −→ in the midpoint of AB. Note that P X = α P B and −→ −→ Q X = β Q A for certain real numbers α and β. Then   1 −→ −→ −→ 1 −→ −→ 1 −→ C X = C B + βQA = C B + β CB + BA 2 2 2 1 + β −→ −→ C B + β B A; = 2   1 −→ −→ −→ 1 −→ −→ 1 −→ C X = C A + α P B = C A + α C A + AB 2 2 2 1 + α −→ −→ = C A + α AB. 2

so a = λ and b = 50. When vcar = 100j, the wind appears to come from the northwest, so vwind rel car = µ(i-j). Thus ai + bj = µ(i − j) + 100j, so a = µ and b = 100 − µ. Hence 50 = 100 − µ, so µ = 50.√Thus a = b = 50. The wind is from the southwest at 50 2 km/h.

10. Let the x-axis point east and the y-axis north. The velocity of the water is vwater = 3i. If you row through the water with speed 5 in the direction making angle θ west of north, then your velocity relative to the water will be vboat

Thus, 1 + β −→ −→ 1 + α −→ −→ CB + βBA = C A + α AB 2 2 −→ 1 + α −→ 1 + β −→ (β + α) B A = CA− CB 2 2 1 + α −→ 1 + β −→ −→ −→ (β + α)(C A − C B) = CA− CB 2   2 1 + α −→ 1 + β −→ β +α− CA = β +α − C B. 2 2 −→ −→ Since C A is not parallel to C B, 1+β 1+α =β+α− =0 2 2 1 ⇒α=β= . 3

β +α−

Since α = β, x divides AQ and B P in the same ratio. By symmetry, the third median C M must also divide the other two in this ratio, and so must pass through X and M X = 13 MC.

rel water

= −5 sin θ i + 5 cos θ j.

Therefore, your velocity relative to the land will be vboat

rel land

= vboat rel water + vwater = (3 − 5 sin θ )i + 5 cos θ j.

To make progress in the direction j, choose θ so that 3 = 5 sin θ . Thus θ = sin−1 (3/5) ≈ 36.87◦ . In this case, your actual speed relative to the land will be 4 5 cos θ = × 5 = 4 km/h. 5 To row from A to B, head in the direction 36.87◦ west of north. The 1/2 km crossing will take (1/2)/4 = 1/8 of an hour, or about 7 12 minutes. B

vboat

rel water

A θ M

vwater

P

X

B

j i A

Q

C

Fig. 10.2.8

Fig. 10.2.10

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.2 (PAGE 578)

11. We use the notations of the solution to Exercise 4. You

1 now want to make progress in the direction ki + j, that 2 is, in the direction making angle φ = tan−1

1 2k

If this velocity is true easterly, then 100 750 sin θ = √ , 2 so θ ≈ 5.41◦ . The speed relative to the ground is

with vector i. Head at angle θ upstream of this direction. Since your rowing speed is 2, the triangle with angles θ and φ has sides 2 and 3 as shown in the figure. By the 3 2 Sine Law, = , so sin θ sin φ

100 750 cos θ − √ ≈ 675.9 km/h. 2 The time for the 1500 km trip is

1500 ≈ 2.22 hours. 675.9

y

3 3 1 q sin θ = sin φ = 2 2 2 k2 +

3

1 4

= √ . 2 4k 2 + 1

3 This is only possible if √ ≤ 1, that is, if 2 4k 2 + 1 √ 5 . k≥ 4 3 Head in the direction θ = sin−1 √ upstream of 2 4k 2 + 1 the direction of AC,√as shown in the figure. The trip is not possible if k < 5/4. B

750 θ x 100

C

k

Fig. 10.2.12

φ

13. The two vectors are perpendicular if their dot product is 3

2

1 2

zero: (2ti + 4j − (10 + t)k) • (i + tj + k) = 0 2t + 4t − 10 − t = 0 ⇒ t = 2.

θ

3i

The vectors are perpendicular if t = 2.

14. The cube with edges i, j, and k has diagonal i + j + k.

φ

The angle between i and the diagonal is

A Fig. 10.2.11

cos−1

12. Let i point east and j point north. If the aircraft heads in

i • (i + j + k) 1 = cos−1 √ ≈ 54.7◦ . √ 3 3

a direction θ north of east, then its velocity relative to the air is 750 cos θ i + 750 sin θ j.

z

The velocity of the air relative to the ground is 100 100 − √ i + − √ j. 2 2

θ

Thus the velocity of the aircraft relative to the ground is

i+j+k

i

y

x



   100 100 750 cos θ − √ i + 750 sin θ − √ j. 2 2

Fig. 10.2.14

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SECTION 10.2 (PAGE 578)

ADAMS and ESSEX: CALCULUS 8

15. The cube of Exercise 10 has six faces, each with 2 diagonals. The angle between i + j + k and the face diagonal i + j is cos−1

(i + j) • (i + j + k) 2 √ √ = cos−1 √ ≈ 35.26◦ . 6 2 3

Six of the face diagonals make this angle with i + j + k. The face diagonal i − j (and five others) make angle cos−1

(i − j) • (i + j + k) √ √ = cos−1 0 = 90◦ 2 3

with the cube diagonal i + j + k.

u•i u1 = . |u| |u| u2 u3 Similarly, cos β = and cos γ = . |u| |u| Thus, the unit vector in the direction of u is

16. If u = u 1 i + u 2 j + u 3 k, then cos α

uˆ =

u = cos αi + cos βj + cos γ k, |u|

ˆ 2 = 1. and so cos2 α + cos2 β + cos2 γ = |u|

17. If uˆ makes equal angles α = β = γ with √ the coordinate axes, then 3 cos2 α = 1, and cos α = 1/ 3. Thus uˆ =

ABC = cos−1 6 6

BC A = cos−1 C AB = cos−1

r is perpenducular to a. Thus the equation is satisfied by all points on the plane through the origin that is normal (perpendicular) to a.

21. If r • a = b, then the vector projection of r along a is the constant vector r•a a b = 2 a = r0 , |a| |a| |a|

−→ −→ B A • BC 4 = cos−1 √ √ ≈ 60.26◦ |B A||BC| 5 13 −→ −→ CB • CA 9 = cos−1 √ √ ≈ 37.87◦ |C B||C A| 10 13 −→ −→ AC • AB 1 = cos−1 √ √ ≈ 81.87◦ . | AC|| AB| 10 5

19. Since r − r1 = λr1 + (1 − λ)r2 − r1 = (1 − λ)(r1 − r2 ),

therefore r − r1 is parallel to r1 − r2 , that is, parallel to the line P1 P2 . Since P1 is on that line, so must P be on it. 1 1 If λ = , then r = (r1 + r2 ), so P is midway between 2 2 P1 and P2 . 2 2 1 If λ = , then r = r1 + r2 , so P is two-thirds of the 3 3 3 way from P2 towards P1 along the line. If λ = −1, the r = −r1 + 2r2 = r2 + (r2 − r1 ), so P is such that P2 bisects the segment P1 P. If λ = 2, then r = 2r1 − r2 = r1 + (r1 − r2 ), so P is such that P1 bisects the segment P2 P.

say.

Thus r • a = b is satisfied by all points on the plane through r0 that is normal to a.

In Exercises 22–24, u = 2i + j − 2k, v = i + 2j − 2k, and w = 2i − 2j + k. 22. Vector x = xi + yj + zk is perpendicular to both u and v if

u•x=0 v•x=0

⇔ ⇔

2x + y − 2z = 0 x + 2y − 2z = 0.

Subtracting these equations, we get x − y = 0, so x = y. The first equation now gives 3x = 2z. Now x is a unit vector if x 2 + y 2 + z 2 = 1, that is, if x 2 + x 2 + 94 x 2 = 1, √ or x = ±2/ 17. The two unit vectors are   2 2 3 x =± √ i+ √ j+ √ k . 17 17 17

23. Let x = xi + yj + zk. Then

i+j+k √ . 3

18. If A = (1, 0, 0), B = (0, 2, 0), and C = (0, 0, 3), then 6

20. If a 6= 0, then a • r = 0 implies that the position vector

x•u=9 x•v=4 x•w=6

⇔ ⇔ ⇔

2x + y − 2z = 9 x + 2y − 2z = 4 2x − 2y + z = 6.

This system of linear equations has solution x = 2, y = −3, z = −4. Thus x = 2i − 3j − 4k.

24. Since u, v, and w all have the same length (3), a vector x = xi + yj + zk will make equal angles with all three if it has equal dot products with all three, that is, if 2x + y − 2z = x + 2y − 2z 2x + y − 2z = 2x − 2y + z

⇔ ⇔

x =y=0 3y − 3z = 0.

Thus x = y = z. Two unit vectors satisfying this condition are   1 1 1 x=± √ i+ √ j+ √ k . 3 3 3

25. Let uˆ = u/|u| and vˆ = v/|v|.

Then uˆ + vˆ bisects the angle between u and v. A unit vector which bisects this angle is u v + uˆ + vˆ |u| |v| = u v |uˆ + vˆ | + |u| |v| |v|u + |u|v . = |v|u + |u|v

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.2 (PAGE 578)

30. Suppose |u| = |v| = |w| = 1, and u•v = u•w = v•w = 0, and let r = au + bv + ww. Then

u uˆ

r • u = au • u + bv • u + cw • u = a|u|2 + 0 + 0 = a.

uˆ + vˆ

vˆ

v

31.

Fig. 10.2.25

Similarly, r • v = b and r • w = c. w•a a, (the vector projection of w along a). Let u = |a|2 Let v = w − u. Then w = u + v. Clearly u is parallel to a, and v•a=w•a−

26. If u and v are not parallel, then neither is the zero vector, and the origin and the two points with position vectors u and v lie on a unique plane. The equation r = λu + µv (λ, µ real) gives the position vector of an arbitrary point on that plane.

27.

w•a a • a = w • a − w • a = 0, |a|2

so v is perpendicular to a. w

2

a) |u + v| = (u + v) • (u + v) = u•u+u•v+v•u+v•v

v

a u

= |u|2 + 2u • v + |v|2 .

b) If θ is the angle between u and v, then cos θ ≤ 1, so u • v = |u||v| cos θ ≤ |u||v|.

≤ |u|2 + 2|u||v| + |v|2

= (|u| + |v|)2 . Thus |u + v| ≤ |u| + |v|.

a) u, v, and u + v are the sides of a triangle. The triangle inequality says that the length of one side cannot exceed the sum of the lengths of the other two sides. b) If u and v are parallel and point in the same direction, (or if at least one of them is the zero vector), then |u + v| = |u| + |v|.

29. u =

3 5i

+

4 5 j,

a) |u| = u•v=

12 25

q

−

v= 9 25 12 25

+

16 25

v = s uˆ + t nˆ r = x uˆ + y nˆ . Since v is not parallel to u, we have t 6= 0. Thus ˆ and nˆ = (1/t)(v − s u) r = x uˆ +

y ˆ = λu + µv, (v − s u) t

where λ = (t x − ys)/(t|u|) and µ = y/t.

33. Let |a|2 − 4r st = K 2 , where K > 0. Now

4 3 5 i − 5 j,

w = k. q = 1, |v| = 16 25 +

32. Let nˆ be a unit vector that is perpendicular to u and lies in the plane containing the origin and the points U , V , and P. Then uˆ = u/|u| and nˆ constitute a standard basis in that plane, so each of the vectors v and r can be expressed in terms of them:

c) |u + v|2 = |u|2 + 2u • v + |v|2

28.

Fig. 10.2.31

9 25

|a|2 = a • a = (r x + sy) • (r x + sy)

= 1, |w| = 1,

= r 2 |x|2 + s 2 |y|2 + 2r sx • y

= 0, u • w = 0, v • w = 0.

K 2 = |a|2 − 4r sx • y = |r x − sy|2

b) If r = xi + yj + zk, then (r • u)u + (r • v)v + (r • w)w    3 4 3 4 = x+ y i+ j 5 5 5 5    4 3 4 3 x− y i − j + zk + 5 5 5 5 9x + 16x 16y + 9y = i+ j + zk 25 25 = xi + yj + zk = r.

(since x • y = t). ˆ for some unit vector u. ˆ Therefore r x − sy = K u, Since r x + sy = a, we have 2r x = a + K uˆ ˆ 2sy = a − K u. Thus x=

a + K uˆ , 2r

y=

a − K uˆ , 2s

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SECTION 10.2 (PAGE 578)

ADAMS and ESSEX: CALCULUS 8

p where K = |a|2 − 4r st, and uˆ is any unit vector. (The solution is not unique.)

The length of the cable between x = 0 and x = 10 m is Z 10 q 1 + sinh2 (ax) d x L=

34. The derivation of the equation of the hanging cable given

0

in the text needs to be modified by replacing W = −δgsj with W = −δgxj. Thus Tv = δgx, and the slope of the cable satisfies dy δgx = = ax dx H

=

Z

0

10

10 1 cosh(ax) d x = sinh(ax) a 0

1 = sinh(10a) ≈ 12.371 m. a

where a = δg/H . Thus y=

1 2 ax + C; 2

Section 10.3 (page 586)

the cable hangs in a parabola.

35. If y =

1. (i − 2j + 3k) × (3i + j − 4k) = 5i + 13j + 7k

1 cosh(ax), then y ′ = sinh(ax), so a

2. (j + 2k) × (−i − j + k) = 3i − 2j + k

Z x q s= 1 + sinh2 (au) du = cosh(au) du 0 0 x sinh(au) 1 = = a sinh(ax). a 0 Z

x

As shown in the text, the tension T at P has horizontal δg and vertical components that satisfy Th = H = and a δg sinh(ax). Hence Tv = δgs = a |T| =

36.

q

Th2

+

Tv2

50 =

0

45 q

1 + sinh2 (ax) d x =

1 sinh(45a). a

The equation sinh(45a) = 50a has approximate solution a ≈ 0.0178541. The vertical distance between the lowest point on the cable and the support point is  1 cosh(45a) − 1 ≈ 19.07 m. a 1 cosh(ax). a At the point P where x = 10 m, the slope of the cable is sinh(10a) = tan(55◦ ). Thus

37. The equation of the cable is of the form y =

a=

3. If A = (1, 2, 0), B = (1, 0, 2), and C = (0, 3, 1), then

−→ −→ AB = −2j+ 2k, AC = −i + j+ k, and the area of triangle ABC is −→ −→ | − 4i − 2j − 2k| √ | AB × AC| = = 6 sq. units. 2 2

4. A vector perpendicular to the plane containing the three given points is (−ai + bj) × (−ai + ck) = bci + acj + abk. A unit vector in this direction is

δg cosh(ax) = δgy. = a

1 The cable hangs along the curve y = cosh(ax), and a its length from the lowest point at x = 0 to the support tower at x = 45 m is 50 m. Thus Z

The Cross Product in 3-Space

bci + acj + abk √ . b2 c2 + a 2 c2 + a 2 b2 1p 2 2 b c + a 2 c2 + a 2 b2 . 2 A vector perpendicular to i + j and j + 2k is

The triangle has area

5.

±(i + j) × (j + 2k) = ±(2i − 2j + k), which has length 3. A unit vector in that direction is   2 2 1 ± i− j+ k . 3 3 3

6. A vector perpendicular to u = 2i − j − 2k and to

v = 2i − 3j + k is the cross product j k i u × v = 2 −1 −2 = −7i − 6j − 4k, 2 −3 1 √ which has length 101. A unit vector with positive k component that is perpenducular to u and v is

1 sinh−1 (tan(55◦ ) ≈ 0.115423. 10

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−1 1 u×v = √ (7i + 6j + 4k). √ 101 101

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.3 (PAGE 586)

7. Since u makes zero angle with itself, |u × u| = 0 and 8.

9.

10.

11.

u × u = 0. j k i u × v = u1 u2 u3 v1 v2 v3 j k i = − v 1 v 2 v 3 = −v × u. u1 u2 u3 i j k (u + v) × w = u 1 + v 1 u 2 + v 2 u 3 + v 3 w1 w2 w3 j k j k i i = u1 u2 u3 + v1 v2 v3 w1 w2 w3 w1 w2 w3 = u × w + v × w. j k i (tu) × v = tu 1 tu 2 tu 3 v v2 v3 1 j k i = t u 1 u 2 u 3 = t (u × v), v1 v2 v3 u × (tv) = −(tv) × u = −t (v × u) = t (u × v). u • (u × v) u u3 − u2 u1 = u 1 2 v1 v2 v3

u1 u 3 + u 3 v1 v3

= u1u2 v3 − u1v2u3 − u2 u1v3 + u 2 v 1 u 3 + u 3 u 1 v 2 − u 3 v 1 u 2 = 0, v • (u × v) = −v • (v × u) = 0.

13. Suppose that u + v + w = 0. Then u × v + v × v + w × v = 0 × v = 0. Thus u × v + w × v = 0. Thus u × v = −w × v = v × w. By symmetry, we also have v × w = w × u.

14. The base of the tetrahedron is a triangle spanned by v and w, which has area A=

1 |v × w|. 2

The altitude h of the tetrahedron (measured perpendicular to the plane of the base) is equal to the length of the projection of u onto the vector v × w (which is perpendicular to the base). Thus h=

|u • (v × w)| . |v × w|

The volume of the tetrahedron is V =

u 2 v2

1 1 Ah = |u • (v × w)| 3 6 1 u 1 u 2 u 3 = | v 1 v 2 v 3 |. 6 w w w 1 2 3 u×v

u h h

w

12. Both u = cos β i + sin β j and v = cos α i + sin α j are

unit vectors. They make angles β and α, respectively, with the positive x-axis, so the angle between them is |α − β| = α − β, since we are told that 0 ≤ α − β ≤ π . They span a parallelogram (actually a rhombus) having area

v Fig. 10.3.14

|u × v| = |u||v| sin(α − β) = sin(α − β). But i u × v = cos β cos α

j sin β sin α

k 0 = (sin α cos β − cos α sin β)k. 0

Because v is displaced counterclockwise from u, the cross product above must be in the positive k direction. Therefore its length is the k component. Therefore sin(α − β) = sin α cos β − cos α sin β.

15. The tetrahedron with vertices (1, 0, 0), (1, 2, 0), (2, 2, 2), and (0, 3, 2) is spanned by u = 2j, v = i + 2j + 2k, and w = −i + 3j + 2k. By Exercise 14, its volume is 4 1 0 2 0 V = | 1 2 2 | = cu. units. 6 −1 3 2 3

16. Let the cube be as shown in the figure. The required parallelepiped is spanned Its volume is a a V = | 0 a a 0

by ai + aj, aj + ak, and ai + ak. 0 a | = 2a 3 cu. units. a 403

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SECTION 10.3 (PAGE 586)

ADAMS and ESSEX: CALCULUS 8

z

But v × w is perpendicular to both v and w, so (0,a,a)

u • (v × w) = 0 + 0 + ν(v × w) • (v × w).

(a,0,a)

If u • (v × w) = 0, then ν = 0, and u = λv + µw. y x (a,a,0)

21.

Fig. 10.3.16

17. The points A = (1, 1, −1), B = (0, 3, −2),

C = (−2, 1, 0), and D = (k, 0, 2) are coplanar if −→ −→ −→ ( AB × AC) • AD = 0. Now j k i −→ −→ AB × AC = −1 2 −1 = 2i + 4j + 6k. −3 0 1 Thus the four points are coplanar if

2(k − 1) + 4(0 − 1) + 6(2 + 1) = 0,

18.

that is, if k = −6. u1 u2 u3 u • (v × w) = v 1 v 2 v 3 w1 w2 w3 v1 v2 v3 = − u1 u2 u3 w1 w2 w3 v1 v2 v3 = w1 w2 w3 u1 u2 u3 = v • (w × u) = w • (u × v) (by symmetry).

19. If u • (v × w) 6= 0, and x = λu + µv + νw, then x • (v × w) = λu • (v × w) + µv • (v × w) + νw • (v × w) = λu • (v × w). Thus λ=

x • (v × w) . u • (v × w)

Since u • (v × w) = v • (w × u) = w • (u × v), we have, by symmetry,

u = i + 2j + 3k v = 2i − 3j w=j−k u × (v × w) = u × (3i + 2j + 2k) = −2i + 7j − 4k (u × v) × w = (9i + 6j − 7k) × w = i + 9j + 9k. u × (v × w) lies in the plane of v and w; (u × v) × w lies in the plane of u and v.

22. u • v × w makes sense in that it must mean u • (v × w).

((u • v) × w makes no sense since it is the cross product of a scalar and a vector.)

u × v × w makes no sense. It is ambiguous, since (u × v) × w and u × (v × w) are not in general equal.

23. As suggested in the hint, let the x-axis lie in the direction of v, and let the y-axis be such that w lies in the x yplane. Thus v = v 1 i,

w = w1 i + w2 j.

Thus v × w = v 1 w2 i × j = v 1 w2 k, and u × (v × w) = (u 1 i + u 2 j + u 3 k) × (v 1 w2 k) = u 1 v 1 w2 i × k + u 2 v 1 w2 j × k = −u 1 v 1 w2 i − u 1 v 1 w2 j. But (u • w)v − (u • v)w = (u 1 w1 + u 2 w2 )v 1 i − u 1 v 1 (w1 i + w2 j) = u 2 v 1 w2 i − u 1 v 1 w2 j. Thus u × (v × w) = (u • w)v − (u • v)w.

24. If u, v, and w are mutually perpendicular, then v × w is parallel to u, so u × (v × w) = 0. In this case, u • (v × w) = ±|u||v||w|; the sign depends on whether u and v × w are in the same or opposite directions.

25. Applying the result of Exercise 23 three times, we obtain x • (w × u) µ= , u • (v × w)

x • (u × v) ν= . u • (v × w)

20. If v × w 6= 0, then (v × w) • (v × w) 6= 0. By the previous exercise, there exist constants λ, µ and ν such that

u × (v × w) + v × (w × u) + w × (u × v) = (u • w)v − (u • v)w + (v • u)w − (v • w)u + (w • v)u − (w • u)v = 0.

u = λv + µw + ν(v × w).

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SECTION 10.4 (PAGE 594)

26. If a = −i + 2j + 3k and x = xi + yj + zk, then j k i a × x = −1 2 3 x y z = (2z − 3y)i + (3x + z)y − (y + 2x)k = i + 5j − 3k, provided 2z − 3y = 1, 3x + z = 5, and −y − 2x = −3. This system is satisfied by x = t, y = 3 − 2t, z = 5 − 3t, for any real number t. Thus

3. The plane through the origin having normal i − j + 2k has equation x − y + 2z = 0.

4. The plane passing through (1, 2, 3), parallel to the plane 3x + y − 2z = 15, has equation 3z + y − 2z = 3 + 2 − 6, or 3x + y − 2z = −1.

5. The plane through (1, 1, 0), (2, 0, 2), and (0, 3, 3) has normal (i − j + 2k) × (i − 2j − 3k) = 7i + 5j − k. It therefore has equation

x = ti + (3 − 2t)j + (5 − 3t)k gives a solution of a × x = i + 5j − 3k for any t. These solutions constitute a line parallel to a.

27. Let a = −i + 2j + 3k and b = i + 5j. If x is a solution of a × x = b, then

7(x − 1) + 5(y − 1) − (z − 0) = 0, or 7x + 5y − z = 12.

6. The plane passing through (−2, 0, 0), (0, 3, 0), and (0, 0, 4) has equation x y z + + = 1, −2 3 4

a • b = a • (a × x) = 0. However, a • b 6= 0, so there can be no such solution x.

28. The equation a × x = b can be solved for x if and only if a • b = 0. The “only if” part is demonstrated in the previous solution. For the “if” part, observe that if a • b = 0 and x0 = (b × a)/|a|2 , then by Exercise 23, a × x0 =

1 (a • a)b − (a • b)a a × (b × a) = = b. |a|2 |a|2

The solution x0 is not unique; as suggested by the example in Exercise 26, any multiple of a can be added to it and the result will still be a solution. If x = x0 + ta, then a × x = a × x0 + ta × a = b + 0 = b.

or 6x − 4y − 3z = −12.

7. The normal n to a plane through (1, 1, 1) and (2, 0, 3) must be perpendicular to the vector i − j + 2k joining these points. If the plane is perpendicular to the plane x + 2y − 3z = 0, then n must also be perpendicular to i + 2j − 3k, the normal to this latter plane. Hence we can use n = (i − j + 2k) × (i + 2j − 3k) = −i + 5j + 3k. The plane has equation −(x − 1) + 5(y − 1) + 3(z − 1) = 0, or x − 5y − 3z = −7.

8. Since (−2, 0, −1) does not lie on x − 4y + 2z = −5, the required plane will have an equation of the form

Section 10.4 Planes and Lines 1.

(page 594)

a) x 2 + y 2 +z 2 = z 2 represents a line in 3-space, namely the z-axis. b) x + y + z = x + y + z is satisfied by every point in 3-space. c) x 2 + y 2 + z 2 = −1 is satisfied by no points in (real) 3-space.

2. The plane through (0, 2, −3) normal to 4i − j − 2k has

2x + 3y − z + λ(x − 4y + 2z + 5) = 0 for some λ. Thus −4 + 1 + λ(−2 − 2 + 5) = 0, so λ = 3. The required plane is 5x − 9y + 5z = −15.

9. A plane through the line x + y = 2, y − z = 3 has equation of the form

equation

4(x − 0) − (y − 2) − 2(z + 3) = 0,

x + y − 2 + λ(y − z − 3) = 0. This plane will be perpendicular to 2x + 3y + 4z = 5 if (2)(1) + (1 + λ)(3) − (λ)(4) = 0,

or 4x − y − 2z = 4.

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SECTION 10.4 (PAGE 594)

ADAMS and ESSEX: CALCULUS 8

that is, if λ = 5. The equation of the required plane is x + 6y − 5z = 17.

or in scalar parametric form, x = −1 + 2t,

10. Three distinct points will not determine a unique plane

then they are coplanar if, for example,

h i (r2 − r1 ) • (r3 − r1 ) × (r4 − r1 ) = 0 (or if they satisfy any similar such condition that asserts that the tetrahedron whose vertices they are has zero volume).

13.

x +1 y z−1 = = . 2 −1 7

17. A line parallel to the line with equations

11. If the four points have position vectors ri , (1 ≤ i ≤ 4),

12.

z = 1 + 7t,

or in standard form

through them if they all lie on a straight line. If the points have position vectors r1 , r2 , and r3 , then they will all lie on a straight line if (r2 − r1 ) × (r3 − r1 ) = 0.

y = −t,

x + y + z = λ is the family of all (parallel) planes normal to the vector i + j + k. x + λy + λz = λ is the family of all planes containing the line of intersection of the planes x = 0 and y + z = 1, except the plane y + z = 1 itself. All these planes pass through the points (0, 1, 0) and (0, 0, 1).

14. The distance from the planes p λx + 1 − λ2 y = 1 √ to the origin is 1/ λ2 + 1 − λ2 = 1. Hence the equation represents the family of all vertical planes at distance 1 from the origin. All such planes are tangent to the cylinder x 2 + y 2 = 1.

x + 2y − z = 2,

2x − y + 4z = 5

is parallel to the vector (i + 2j − k) × (2i − j + 4k) = 7i − 6j − 5k. Since the line passes through the origin, it has equations r = 7ti − 6tj − 5tk x = 7t, y = −6t, x y z = = 7 −6 −5

z = −5t

(vector parametric) (scalar parametric) (standard form).

18. A line parallel to x + y = 0 and to x − y + 2z = 0

is parallel to the cross product of the normal vectors to these two planes, that is, to the vector (i + j) × (i − j + 2k) = 2(i − j − k).

Since the line passes through (2, −1, −1), its equations are, in vector parametric form r = (2 + t)i − (1 + t)j − (1 + t)k,

15. The line through (1, 2, 3) parallel to 2i − 3j − 4k has equations given in vector parametric form by

or in scalar parametric form r = (1 + 2t)i + (2 − 3t)j + (3 − 4t)k, or in scalar parametric form by x = 1 + 2t,

y = 2 − 3t,

z = 3 − 4t,

x = 2 + t,

y = −(1 + t),

z = −(1 + t),

or in standard form x − 2 = −(y + 1) = −(z + 1).

or in standard form by y−2 z−3 x −1 = − . 2 −3 −4

19. A line making equal angles with the positive directions

16. The line through (−1, 0, 1) perpendicular to the plane 2x − y + 7z = 12 is parallel to the normal vector 2i − j + 7k to that plane. The equations of the line are, in vector parametric form, r = (−1 + 2t)i − tj + (1 + 7t)k,

of the coordinate axes is parallel to the vector i + j + k. If the line passes through the point (1, 2, −1), then it has equations r = (1 + t)i + (2 + t)j + (−1 + t)k (vector parametric) x = 1 + t, y = 2 + t, z = −1 + t (scalar parametric) x −1= y−2= z+1 (standard form).

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.4 (PAGE 594)

20. The line r = (1 − 2t)i + (4 + 3t)j + (9 − 4t)k has standard form

x −1 y−4 z−9 = = . −2 3 −4

21. The line

(

22. The line



26. The distance from (0, 0, 0) to x + 2y + 3z = 4 is √

x = 4 − 5t has standard form y = 3t z=7 x −4 y = , −5 3

(Other similar answers are possible.)

4 = √ units. 14

27. The distance from (1, 2, 0) to 3x − 4y − 5z = 2 is

z = 7.

|3 − 8 − 0 − 2| 7 = √ units. √ 5 2 32 + 42 + 52

x − 2y + 3z = 0 is parallel to the vector 2x + 3y − 4z = 4

(i − 2j + 3k) × (2i + 3j − 4k) = −i + 10j + 7k. We need a point on this line. Putting z = 0, we get x − 2y = 0,

4 12 + 22 + 32

2x + 3y = 4.

The solution of this system is y = 4/7, x = 8/7. A possible standard form for the given line is

28. A vector parallel to the line x + y + z = 0, 2x − y −5z = 1 is

a = (i + j + k) × (2i − j − 5k) = −4i + 7j − 3k. We need a point on this line: if z = 0 then x + y = 0 and 2x − y = 1, so x = 1/3 and y = −1/3. The position vector of this point is

8 4 y− 7 = 7 = z, −1 10 7

x−

r1 =

though, of course, this answer is not unique as the coordinates of any point on the line could have been used.

1 1 i − j. 3 3

The distance from the origin to the line is |r1 × a| |i + j + k| s= = = √ |a| 74

23. The equations       

x = x1 + t (x2 − x1 )    y = y1 + t (y2 − y1 )    z = z 1 + t (z 2 − z 1 )

24. The point on the line corresponding to t = −1 is the

point P3 such that P1 is midway between P3 and P2 . The point on the line corresponding to t = 1/2 is the midpoint between P1 and P2 . The point on the line corresponding to t = 2 is the point P4 such that P2 is the midpoint between P1 and P4 .

25. Let ri be the position vector of Pi (1 ≤ i ≤ 4). The

3 units. 74



x + 2y = 3 contains the points (1, 1, 1) and y + 2z = 3 1 (3, 0, 3/2), so is parallel to the vector 2i − j + k, or to 2 4i − 2j +nk. x +y+z =6 The line contains the points (−5, 11, 0) x − 2z = −5 and (−1, 5, 2), and so is parallel to the vector 4i−6j+2k, or to 2i − 3j + k. Using the values

29. The line

certainly represent a straight line. Since (x, y, z) = (x1 , y1 , z 1 ) if t = 0, and (x, y, z) = (x2 , y2 , z 2 ) if t = 1, the line must pass through P1 and P2 .

r

r1 = i + j + k r2 = −i + 5j + 2k

a1 = 4i − 2j + k a2 = 2i − 3j + k,

we calculate the distance between the two lines by the formula in Section 10.4 as

line P1 P2 intersects the line P3 P4 in a unique point if the four points are coplanar, and P1 P2 is not parallel to P3 P4 . It is therefore sufficient that (r2 − r1 ) × (r4 − r3 ) 6= 0, and h i (r3 − r1 ) • (r2 − r1 ) × (r4 − r3 ) = 0.

|(r1 − r2 ) • (a1 × a2 )| |a1 × a2 | |(2i − 4j − k) • (i − 2j − 8k)| = |i − 2j − 8k| 18 = √ units. 69

s=

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SECTION 10.4 (PAGE 594)

ADAMS and ESSEX: CALCULUS 8

y+3 z−1 = passes through the point 2 4 (2, −3, 1), and is parallel to a = i + 2j + 4k. The plane 2y − z = 1 has normal n = 2j − k. Since a • n = 0, the line is parallel to the plane. The distance from the line to the plane is equal to the distance from (2, −3, 1) to the plane 2y − z = 1, so is

30. The line x − 2 =

D=

| − 6 − 1 − 1| 8 = √ units. √ 4+1 5

3.

2(x 2 − 2x + 1) + 2(y 2 + 4y + 4) + 2(z 2 − 6z + 9) = −27 + 2 + 8 + 18 1 (x − 1)2 + (y + 2)2 + (z − 3)2 = 2 √ This is a sphere with radius 1/ 2 and centre (1, −2, 3).

4.

(x + 2)2 (y − 1)2 z2 + + =1 2 2 4 2 (4/3)2 This is an ellipsoid with centre (−2, 1, 0) and semi-axes 4, 2, and 4/3.

x y-plane passing through (x0 , y0 ). Therefore, in 3-space the pair of equations

32.

z = z0

5.

represents all straight lines in the plane z = z 0 which pass through the point (x0 , y0 , z 0 ). x − x0 y − y0 √ = = z − z 0 represents all lines through λ 1 − λ2 (x0 , y0 , z 0 ) parallel to the vectors a=

x 2 + 4y 2 + 9z 2 + 4x − 8y = 8

(x + 2)2 + 4(y − 1)2 + 9z 2 = 8 + 8 = 16

31. (1 − λ)(x − x0 ) = λ(y − y0 ) represents any line in the

(1 − λ)(x − x0 ) = λ(y − y0 ),

2x 2 + 2y 2 + 2z 2 − 4x + 8y − 12z + 27 = 0

z = x 2 + 2y 2 represents an elliptic paraboloid with vertex at the origin and axis along the positive z-axis. Crosssections in planes z = k > 0 are ellipses with semi-axes √ √ k and k/2. z z=x 2 +2y 2

p 1 − λ2 i + λj + k.

All such lines are generators of the circular cone (z − z 0 )2 = (x − x0 )2 + (y − y0 )2 , so the given equations specify all straight lines lying on that cone.

33. The equation

y

x

(A1 x + B1 y + C1 z + D1 )(A2 x + B2 y + C2 z + D2 ) = 0 is satisfied if either A1 x + B1 y + C1 z + D1 = 0 or A2 x + B2 y + C2 z + D2 = 0, that is, if (a, y, z) lies on either of these planes. It is not necessary that the point lie on both planes, so the given equation represents all the points on each of the planes, not just those on the line of intersection of the planes.

Section 10.5 Quadric Surfaces 1.

Fig. 10.5.5

6.

z = x 2 − 2y 2 represents a hyperbolic paraboloid. z z=x 2 −2y 2

(page 598)

x 2 + 4y 2 + 9z 2 = 36

x2 y2 z2 + + =1 62 32 22 This is an ellipsoid with centre at the origin and semiaxes 6, 3, and 2.

2.

x 2 + y 2 + 4z 2 = 4 represents an oblate spheroid, that is, an ellipsoid with its two longer semi-axes equal. In this case the longer semi-axes have length 2, and the shorter one (in the z direction) has length 1. Cross-sections in planes perpendicular to the z-axis between z = −1 and z = 1 are circles.

y x

Fig. 10.5.6

7.

x 2 − y 2 − z 2 = 4 represents a hyperboloid of two sheets with vertices at (±2, 0, 0) and circular cross-sections in planes x = k, where |k| > 2.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 10.5 (PAGE 598)

z

z x 2 −y 2 −z 2 =4 x 2 +4z 2 =4

x

y y

x

Fig. 10.5.7

Fig. 10.5.10

8. −x 2 + y 2 + z 2 = 4 represents a hyperboloid of one sheet,

11.

with circular cross-sections in all planes perpendicular to the x-axis.

x 2 − 4z 2 = 4 represents a hyperbolic cylinder with axis along the y-axis. z

z −x 2 +y 2 +z 2 =4

y

y

x

x

x

y

x 2 −4z 2 =4

Fig. 10.5.11 Fig. 10.5.8

12. 9.

z = x y represents a hyperbolic paraboloid containing the x- and y-axes.

y = z 2 represents a parabolic cylinder with vertex line along the x-axis. z

z

y=z 2 y y x

z=xy x

Fig. 10.5.12 Fig. 10.5.9

10.

x2

4z 2

+ = 4 represents an elliptic cylinder with axis along the y-axis.

13.

  1 2 1 x = z 2 +z = z + − represents a parabolic cylinder 2 4 with vertex line along the line z = −1/2, x = −1/4.

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SECTION 10.5 (PAGE 598)

ADAMS and ESSEX: CALCULUS 8

z z

(z−1)2 =(x−2)2 +(y−3)2 +4

x=z 2 +z

(2,3,1) y

y x x

Fig. 10.5.13

14.

Fig. 10.5.16

x 2 = y 2 + 2z 2 represents an elliptic cone with vertex at the origin and axis along the x-axis.

17.

z

x 2 =y 2 +2z 2



x 2 + y2 + z2 = 4 represents the circle of intersection of x +y+z =1 a sphere and a plane. The circle lies in the plane x + y + z = 1,√and has centre (1/3, 1/3, 1/3) and radius √ 4 − (3/9) = 11/3. z



x+y+z=1

x

1 1 1 3,3,3



y x 2 +y 2 +z 2 =4

x

y

Fig. 10.5.14

15. (z − 1)2 = (x − 2)2 + (y − 3)2 represents a circular

Fig. 10.5.17

cone with axis along the line x = 2, y = 3, and vertex at (2, 3, 1) z (z−1)2 =(x−2)2 +(y−3)2

18.



x 2 + y2 = 1 is the ellipse of intersection of the plane z=x+y z = x + y and the circular cylinder x 2 + y 2 = 1. The centre of the ellipse √ is at the √ origin, √ and the ends of the major axis are ±(1/ 2, 1/ 2, 2). z

(2,3,1) y

z=x+y

x

Fig. 10.5.15

x 2 +y 2 =1

x

16. (z − 1)2 = (x − 2)2 + (y − 3)2 + 4 represents a hyperboloid of two sheets with centre at (2, 3, 1), axis along the line x = 2, y = 3, and vertices at (2, 3, −1) and (2, 3, 3).

410 Copyright © 2014 Pearson Canada Inc.

Fig. 10.5.18

y

INSTRUCTOR’S SOLUTIONS MANUAL

19.

SECTION 10.6 (PAGE 602)



z 2 = x 2 + y 2 is the parabola in which the plane z =1+x z = 1 + x intersects the circular cone z 2 = x 2 + y 2 . (It is a parabola because the plane is parallel to a generator of the cone, namely the line z = x, y = 0.) The vertex of the parabola is (−1/2, 0, 1/2), and its axis is along the line y = 0, z = 1 + x. z

z 2 =x 2 +y 2

y

Fig. 10.5.19 

x 2 + 2y 2 + 3z 2 = 6 is an ellipse in the plane y=1 y = 1. Its projection onto the x z-plane is the ellipse x 2 + 3z 2 = 4. One quarter of the ellipse is shown in the figure. z

y=1

3

x 2 +2y 2 +3z 2 =6

Fig. 10.5.20

22.

x2 y2 z2 + − =1 a2 b2 c2 2 2 x z y2 − = 1 − a2 c2  b2  x z x z y y + − = 1+ 1− a c a c b b   x z y  + =λ 1+ a c  b Family 1:  λ x − z = 1− y. a c b   x z y  + =µ 1− a c b .  Family 2:  µ x − z = 1+ y. a c b

z = xy

Family 1:



z = λx λ = y.

in ellipses with semi-axes 1 in the y direction and 1/ 2 in the x direction. Tilting the plane in the x direction will cause the shorter semi-axis to increase in length. The plane z = cx intersects the cylinder in an ellipse with √ principal axes √ through the points (0, ±1, 0) and (±1/ 2, 0, ±c/ 2). The semi-axes will be equal (and the ellipse will be a circle) if (1/2) + (c2 /2) = 1, that is, if c = ±1. Thus cross-sections of the cylinder perpendicular to the vectors a = i ± k are circular.

c2 x 2 + 2ckx + k 2 = 2x 2 + y 2

(2 − c2 )x 2 − 2ckx + y 2 = k 2  2 ck c2 k 2 2k 2 (2 − c2 ) x − + y 2 = k2 + = 2 2 2−c 2−c 2 − c2 (x − x0 )2 y2 + 2 = 1, 2 a b where x0 =

2k 2 2k 2 ck , a2 = , and b2 = . 2 2 2 2−c (2 − c ) 2 − c2

(x0 − a, 0, c(x0 − a) + k) and (x0 , −b, cx0 + k) y

21.

23. The cylinder 2x 2 + y 2 = 1 intersects horizontal planes √

As in the previous exercise, z = cx + k intersects the cylinder (and hence the cone) in an ellipse with principal axes joining the points

√ x √ 6

z = µy µ = x.

z 2 = 2x 2 + y 2 on the cylinder

x

√ 2

n

24. The plane z = cx + k intersects the elliptic cone

z=1+x

20.

Family 2:

to to

(x0 + a, 0, c(x0 + a) + k), (x0 , b, cx0 + k).

The centre of this ellipse is (x0 , 0, cx0 + k). The ellipse is a circle if its two semi-axes have equal lengths, that is, if a 2 + c2 a 2 = b2 , that is, 2k 2 2k 2 = , 2 2 (2 − c ) 2 − c2 √ or 1 + c2 = 2 − c2 . Thus √ c = ±1/ 2. A√vector normal to the plane z = ±(x/ 2) + k is a = i ± 2k. (1 + c2 )

Section 10.6 Cylindrical and Spherical Coordinates (page 602) 1. Cartesian: (2, −2, √ 1);

Cylindrical: [2 2, −π/4, 1]; Spherical: [3, cos−1 (1/3), −π/4].

2. Cylindrical: √[2, π/6, −2];

√ Cartesian: ( 3, 1, −2]; Spherical: [2 2, 3π/4, π/6].

3. Spherical: [4,√ π/3, 2π/3];

√ Cartesian: (− 3, 3, −2); Cylindrical: [2 3, 2π/3, 2].

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SECTION 10.6 (PAGE 602)

ADAMS and ESSEX: CALCULUS 8

4. Spherical: [1, φ, θ ]; Cylindrical: [r, π/4, r ].

4.

√ x = sin φ cos θ = r cos π/4 = r/ 2 √ y = sin φ sin θ = r sin π/4 = r/ 2 z = cos φ = r. Thus x = y, θ = π/4,√and r = sin φ = cos φ. Hence φ = π/4, √ so r = 1/ 2. Finally: x = y = 1/2, z = 1/ 2. √ Cartesian: (1/2, 1/2, 1/ 2).

5.

5. θ = π/2 represents the half-plane x = 0, y > 0. 6. φ = 2π/3 represents the lower half of the right-circular cone with vertex at the origin, axis along the z-axis, and semi-vertical angle π/3. Its Cartesian equation is p z = − (x 2 + y 2 )/3.

7. φ = π/2 represents the x y-plane. 8.

R = 4 represents the sphere of radius 4 centred at the origin.

9. r = 4 represents the circular cylinder of radius 4 with axis along the z-axis.

10.

R = z represents the positive half of the z-axis.

11.

R = r represents the x y-plane.

12.

R = 2x represents the half-cone with vertex at the origin, axis along the positive x-axis, and semi-vertical angle p π/3. Its Cartesian equation is x = (y 2 + z 2 )/3.

6.

13. If R = 2 cos φ, then R 2 = 2R cos φ, so 2

2

2

2



w y

b d

1 0 T AA =  0 0  4 3 = 2 1  1 0 A2 =  0 0  1 0 = 0 0

1 1 0 0 3 3 2 1 1 1 0 0 2 1 0 0



x= T

xx =

2

x + y + z = 2z

x + y + z 2 − 2z + 1 = 1

x 2 + y 2 + (z − 1)2 = 1.



a c

x z

x y z

!

,

x y z

!

=

1 1 1 0 2 2 2 1 1 1 1 0 3 2 1 0



a p q

(x, y, z) = x y z

!

aw + cx ay + cz

 1 1 11  1 1 1 1  1 1  1 1  1 1 10 10 1 0  4 3  2 1

A=

xT x = (x, y, z)

Thus R = 2 cos φ represents the sphere of radius 1 centred at (0, 0, 1).



p b r x2 xy xz

bw + d x by + dz

0 1 1 1

0 0 1 1

 0 0  0 1

1 1 0 0

1 1 1 0

 1 1 1 1

! q r c ! xy xz 2 y yz yz z 2

= (x 2 + y 2 + z 2 )

! ! a p q x x Ax = (x, y, z) p b r y q r c z ! ax + py + qz = (x, y, z) px + by + r z qx + r y + cz = ax 2 + by 2 + cz 2 + 2 px y + 2qx z + 2r yz T

14. r = 2 cos θ ⇒ x 2 + y 2 = r 2 = 2r cos θ = 2x, or

(x − 1)2 + y 2 = 1. Thus the given equation represents the circular cylinder of radius 1 with axis along the vertical line x = 1, y = 0.

Section 10.7 A Little Linear Algebra (page 611) 1.

2. 3.

3 0 −2 1 1 2 −1 1 −1 ! 1 1 1 0 1 1 0 0 1   a b w c d y

! ! 1 6 7 = 5 −3 0 −2 1 1 ! ! 1 1 1 1 2 3 0 1 1 = 0 1 2 0 0 1 0 0 1    x aw + by ax + bz = z cw + d y cx + dz

!

2 3 0

7.

2 4 1 −2

3 −1 0 2 1 4 0 2 1 = −3 1 −1 1 0 −1 1 −2 0 1 0 0 1   1 1 4 1 = −3 −2 −1 −2 1 −2 1 = 6(3) + 3(6) = 36

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INSTRUCTOR’S SOLUTIONS MANUAL

8.

SECTION 10.7 (PAGE 611)

1 1 2 1 −2 0 3 −3 1 = −2 2 −3 1 = −2 0 0 1 = −2 5

1 1 3 4 2 4 2 −2 1 1 1 1 1 4 3 4 +2 1 2 3 −3 −2 2 −2 1 1 1 − 4 1 2 3 3 −3 2 1 1 1 1 1 1 2 +2 0 1 3 5 1 0 −6 −5 1 1 1 2 − 4 0 1 0 −6 −1 1 1 2 3 2 −4 + 2 −6 −1 −6 −5 1

= −2(−9) + 2(13) − 4(11) = 0

9.

10.

a11 0 0 . . . 0

a12 a22 0 .. .

a13 a23 a33 .. .

a1n a2n a3n .. . 0 0 · · · ann a22 a23 · · · a2n 0 a33 · · · a3n = a11 . .. .. .. . . . .. 0 0 · · · ann a33 · · · a3n . .. .. = a11 a22 .. . . 0 · · · ann = a11 a22 a33 · · · ann ··· ··· ··· .. .

(or use induction on n) 1 1 x y = y − x. If

1 f (x, y, z) = x 2 x

so A = y − x and

1 x 1 x2 1 . . . n−1 x1

11. Let A =

1 x2 x22 .. .

1 x3 x32 .. .

x2n−1

x3n−1



a c

1 xn xn2 .. .

··· ··· ··· .. . ···

Y = (x j − xi ). 1≤i< j ≤n n−1

xn

  b ℓ ,B= d n

  m w , C= p y

 x . Then z

   aℓ + bn am + bp w x (AB)C = cℓ + dn cm + d p y z   aℓw + bnw + amy + bpy aℓx + bnx + amz + bpz = cℓw + dnw + cmy + d py cℓx + dnx + cmz + d pz    a b ℓw + my ℓx + mz A(BC) = c d nw + py nx + pz   aℓw + amy + bnw + bpy aℓx + amz + bnx + bpz = cℓw + cmy + dnw + d py cℓx + cmz + dnx + d pz Thus (AB)C = A(BC).

12. If A =



  b a , then AT = d b

a c

 c , and d

det(A) = ad − bc = det(AT ). We generalize this by induction. Suppose det(B T )=det(B) for any (n − 1) × (n − 1) matrix, where n ≥ 3. Let a

11

1 y y2

1 z , z2

then f is a polynomial of degree 2 in z. Since f (x, y, x) = 0 and f (x, y, y) = 0, we must have f (x, y, z) = A(z − x)(z − y) for some A independent of z. But 1 Ax y = f (x, y, 0) = x 2 x

Generalization:

1 y y2

1 0 = x y(y − x), 0

 a21 A=  .. . an1

a12 a22 .. .

a1n  a2n  ..   . · · · ann ··· ··· .. .

an2

be an n × n matrix. If det(A) is expanded in minors about the first row, and det(A T ) is expanded in minors about the first column, the corresponding terms in these expansions are equal by the induction hypothesis. (The (n − 1) × (n − 1) matrices whose determinants appear in one expansion are the transposes of those in the other expansion.) Therefore det(A T )=det(A) for any square matrix A.

13. Let A =



f (x, y, z) = (y − x)(z − x)(z − y).

a c

b d



and B =

AB =





w y

aw + by cw + d y

 x . Then z ax + bz cx + dz



.

413 Copyright © 2014 Pearson Canada Inc.

SECTION 10.7 (PAGE 611)

ADAMS and ESSEX: CALCULUS 8

Therefore,

18.

det(A)det(B) = (ad − bc)(wz − x y) = adwz − ad x y − bcwz + bcx y det(AB) = (aw + by)(cx + dz) − (ax + bz)(cw + d y) = awcx + awdz + bycx + bydz − axwc − axd y − bzcw − bzd y = adwz − ad x y − bcwz + bcx y = det(A)det(B).

14. If Aθ =



cos θ − sin θ



cos(−θ ) sin(−θ ) − sin(−θ ) cos(−θ )

A−θ =

1 Let A = −1 2 AA−1 = I we

a−g=1 −a + d = 0 2a + d + 3g = 0



=

0 1





cos θ sin θ

− sin θ cos θ



Aθ A−θ =

A−1 = 

,



1 0

= I. A

Thus A−θ = (Aθ )−1 .

15. If D = ad − bc, we have a c

b d



−b   ad − bc D D = a cd − cd D D

 d  D −c D

−ab + ab   1 D  −cb + ad = 0 D

with a similar calculation for the product with a reversed order of factors.

16. Since D = detB = x y 2 − x 2 y = x y(y − x), B is

c−i = 0 −c + f = 0 2c + f + 3i = 1.

− 61

1 2 1 2

1 6 1 6 1 6

5 6

− 21

− 61



.

B−1

y2  D = −x 2 D

  y −y   x(y − x) D = −x x y(y − x) D



1 Let A = 0 0 AA−1 = I we

! 1 1 1 1 , A−1 = 0 1 must have

a+d +g =1 d+g=0 g=0

a d g

0 1

b+e+h = 0 e+h = 1 h=0

A

=

1 0 0

−1 0 1 −1 0 1

!

.

=

−2 1 13

!

.

!

= A−1

−2 1 13

!

=

1 2 3

!

,

20. If A is the matrix of Exercises 16 and 17 then det(A) = 6. By Cramer’s Rule, 1 −2 x= 1 6 13 1 1 y = −1 6 2 1 1 z = −1 6 2

! c f . Since i

c+ f +i = 0 f +i = 0 i = 1.

!

so x = 1, y = 2, and z = 3.

y(y − x)

Thus a = 1, d = g = 0, h = 0, e = 1, b = −1, i = 1, f = −1, c = 0, and so −1

x y z

−1  x(y − x)   1

b e h

x y z

Thus 

nonsingular, and therefore invertible, provided x 6= 0, y 6= 0, and x 6= y. Using the formula for the inverse of a general 2 × 2 matrix, we have

17.

! c f . Since i

19. The given system of equations is

and



b−h = 0 −b + e = 1 2b + e + 3h = 0

b e h

Solving these three systems of equations, we get

 sin θ , then cos θ 

a d g

! 0 −1 1 0 , A−1 = 1 3 must have

21.

0 −1 6 1 0 = =1 6 1 3 −2 −1 12 1 0 = =2 6 13 3 0 −2 18 1 1 = = 3. 6 1 13

 1 1 1 1 1 −1  1 1 A=  1 1 −1 −1 1 −1 −1 −1 0 2 0 0 2 0 0 0 det(A) = 0 0 0 2 1 −1 −1 −1 2 0 0 0 2 =8 = −2 0 2 0 = −4 1 −1 1 −1 −1

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

INSTRUCTOR’S SOLUTIONS MANUAL

1 1 0 1 1 4 1 1 −1 x1 = 8 6 1 −1 −1 2 −1 −1 −1 0 1 1 1 1 4 0 0 −2 = 8 6 2 0 0 2 0 0 0 2 1 1 1 4 1 = − 0 0 −2 = − 8 2 0 0 8 0 1 1 1 x2 = 8 1 1 2 1 0 = 8 0 0 2 4 = 6 8 2

22.

0 4 6 2 0 4 6 2 2 0 0

SECTION 10.7 (PAGE 611)

Thus, G ◦ F is represented by the matrix GF.  −1 1 . Use Theorem 8. D1 = −1 < 0, 1 −2 −1 1 = 1 > 0. Thus A is negative definite. D2 = 1 −2

1 =1 −2

1 1 1 −1 −1 −1 −1 −1 0 1 2 −1 0 −1 0 −1 −1 −4 6 −1 = 2 −1 = 2 −1 8 −1

1 1 0 1 1 1 1 4 −1 x3 = 8 1 1 6 −1 1 −1 2 −1 0 2 0 1 1 0 0 4 −1 = 8 0 0 6 −1 2 −2 2 −1 2 0 1 2 4 4 −1 = − 0 4 −1 = − = −1 8 0 6 −1 8 6 −1 x4 = −(x1 + x2 + x3 ) = −2.     x1 a b Let F(x1 , x2 ) = F , where F = .  x2  c d y1 p q , where G = . Let G(y1 , y2 ) = G y2 r s If y1 = ax1 + bx2 and y2 = cx1 + d x2 , then G ◦ F(x1 , x2 ) = G(y1 , y2 )    p q ax1 + bx2 = r s cx1 + d x2   pax1 + pbx2 + qcx1 + qd x2 = r ax1 + r bx2 + scx1 + sd x2    pa + qc pb + qd x1 = r a + sc r b + sd x2     p q a b x1 = r s c d x2   x1 = GF . x2

23. A =



24. A =

1 2 0

! 2 0 1 0 . Use Theorem 8. 0 1

D1 = 1 > 0, 1 D3 = 2 0

1 D2 = 2 2 1 0

Thus A is indefinite.

25. A =

2 1 1

2 = −3 < 0, 1

0 0 = −3 < 0. 1

! 1 1 2 1 . Use Theorem 8. 1 2 D1 = 2 > 0, 2 D3 = 1 1

2 D2 = 1

1 = 3 > 0, 2

1 1 2 1 = 4 > 0. 1 2

Thus A is positive definite.

26.

1 1 A= 1 1 0 0 use Theorem

! 0 1 1 = 0, we cannot 0 . Since D2 = 1 1 1 8. The corresponding quadratic form is

Q(x, y, z) = x 2 + y 2 + 2x y + z 2 = (x + y)2 + z 2 , which is positive semidefinite. (Q(1, −1, 0) = 0.). Thus A is positive semidefinite.

27. A =

1 0 1

! 0 1 1 −1 . Use Theorem 8. −1 1 D1 = 1 > 0,

1 D2 = 0

1 0 D3 = 0 1 1 −1

Thus A is indefinite.

0 = 1 > 0, 1

1 −1 = −1 < 0. 1

415 Copyright © 2014 Pearson Canada Inc.

SECTION 10.7 (PAGE 611)

28. A =

2 0 1

ADAMS and ESSEX: CALCULUS 8

! 0 1 4 11 . Use Theorem 8. −1 1 D1 = 2 > 0, 2 D3 = 0 1

2 D2 = 0

0 = 8 > 0, 4

0 1 4 11 = 2 > 0. −1 1

>

a := DotProduct(U,(V W),conjugate=false):

&x

>

b := DotProduct(V,(W U),conjugate=false):

&x

>

&x

c := DotProduct(W,(U V),conjugate=false): >

simplify(a-b); simplify(a-c);

Thus A is positive definite.

Section 10.8 Using Maple for Vector and Matrix Calculations (page 620) It is assumed that the Maple package LinearAlgebra has been loaded for all the calculations in this section.

1. We use the result of Example 9 of Section 10.4. > > > >

r1 := : v1 := : r2 := : v2 := : v1xv2 := v1 &x v2: dist := abs((r2-r1).v1xv2)/Norm(v1xv2,2);

0 0

4. These calculations verify the identity: >

U := Vector[row](3,symbol=u): V := Vector[row](3,symbol=v): > > >

LHS := (U &x V) &x (U &x W): RHS := (DotProduct(U,(V &x W),conjugate=false))*U: >

di st := 2

W := Vector[row](3,symbol=w):

simplify(LHS-RHS);

The distance between the two lines is 2 units.

2. The plane P through the origin containing the vectors v1 = i − 2j − 3k and v2 = 2i + 3j + 4k has normal n = v1 × v2 . >

[0, 0, 0]

5. sp := (U,V) -> DotProduct(U,Normalize(V,2),conjugate=false)

n := &x ; n := [1, −10, 7]

The angle between v = i − j + 2k and n (in degrees) is >

angle := evalf((180/Pi)*VectorAngle(n,));

6. vp := (U,V) -> DotProduct(U,Normalize(V,2), conjugate=false)*Normalize(V,2) 7. ang := (u,v) -> evalf((180/Pi)*VectorAngle(U,V))

angvn := 33.55730975 Since this angle is acute, the angle between v and the plane P is its complement. >

9. VolT := (U,V,W)->(1/6)*abs(DotProduct(U,(V &x W), conjugate=false))

angle := 90 - angvn; angle := 56.44269025

10. dist:=(A,B)->Norm(A-B,2)

3. These calculations verify the identity: >

U := Vector[row](3,symbol=u): V := Vector[row](3,symbol=v): >

8. unitn := (U,V)->Normalize((U &x V),2)

>

dist(,); 5

W := Vector[row](3,symbol=w): 11. We use LinearSolve.

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INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 10 (PAGE 621)

> A := Matrix([[1,2,3,4,5], > [6,-1,6,2,-3],[2,8,-8,-2,1], > [1,1,1,1,1],[10,-3,3,-2,2]]): > X :=

16.

LinearSolve(A,,free=t);  

> A := Matrix([[1,1/2,1/3], > [1/2,1/3,1/4],[1/3,1/4,1/5]]): > Ainv := MatrixInverse(A): >

Digits := 10: evalf(Eigenvalues(A));

1  0    X :=  −1   3  2

"

The solution is u = 1, v = 0, x = −1, y = 3, z = 2.

>

12. We use LinearSolve.

1.408318927 − 4 10−11 I 0.00268734034 − 5.673502694 10−10 I 0.1223270659 + 5.873502694 10−10 I

evalf(Eigenvalues(Ainv)); "

> B := Matrix([[1,1,1,1,0], > [1,0,0,1,1],[1,0,1,1,0], > [1,1,1,0,1],[0,1,0,1,-1]]): > X := 11 − 2t5 2     X :=  −2 + t5    −1 + t5 t5

14.

15.

> B := Matrix([[1,1,1,1,0], > [1,0,0,1,1],[1,0,1,1,0], > [1,1,1,0,1],[0,1,0,1,-1]]): > Digits := 5: evalf(Eigenvalues(B));   0  3.3133 − 0.0000053418I     0.8693 + 0.0000073520I    −1.2728 − 0.0000025143I −7 −1.9098 + 5.041 10 I

#

Although they appear in different orders, each eigenvalue of A−1 is the reciprocal of an eigenvalue of A. This is to be expected since A−1 x = λx

There is a one-parameter family of solutions: u = 11−2t, v = 2, x = −2 + t, y = −1 + t, z = t, for arbitrary t. > A := Matrix([[1,2,3,4,5], > [6,-1,6,2,-3],[2,8,-8,-2,1], > [1,1,1,1,1],[10,-3,3,-2,2]]): > Determinant(A); −935

372.1151279 − 2 10−9 I 0.710066409 − 5.096152424 10−8 I 8.174805711 + 5.296152424 10−8 I

The small imaginary parts are due to round-off errors in the solution process. The eigenvalues are real since the matrix and its inverse are real and symmetric.

LinearSolve(B,,free=t);  

13.

#

≡

(1/λ)x = Ax.

Review Exercises 10 (page 621) 1.

x + 3z = 3 represents a plane parallel to the y-axis and passing through the points (3, 0, 0) and (0, 0, 1).

2.

y − z ≥ 1 represents all points on or below the plane parallel to the x-axis that passes through the points (0, 1, 0) and (0, 0, −1).

3.

x + y + z ≥ 0 represents all points on or above the plane through the origin having normal vector i + j + k.

4.

x − 2y − 4z = 8 represents all points on the plane passing through the three points (8, 0, 0), (0, −4, 0), and (0, 0, −2).

5.

The tiny imaginary parts are due to roundoff error in the calculations. They should all be 0. Since B is a real, symmetric matrix, its eigenvalues are all real. The eigenvalues, rounded to 5 decimal places are 0, 3.3133, 0.8693, −1.2728, and −1.9098.

y = 1 + x 2 + z 2 represents the circular paraboloid obtained by rotating about the y-axis the parabola in the x y-plane having equation y = 1 + x 2 .

6.

y = z 2 represents the parabolic cylinder parallel to the x-axis containing the curve y = z 2 in the yz-plane.

> A := Matrix([[1,1/2,1/3], > [1/2,1/3,1/4],[1/3,1/4,1/5]]): > Ainv := MatrixInverse(A); " # 9 −36 30 Ai nv := −36 192 −180 30 −180 180

7.

x = y 2 − z 2 represents the hyperbolic paraboloid whose intersections with the x y- and x z-planes are the parabolas x = y 2 and x = −z 2 , respectively.

8.

z = x y is the hyperbolic paraboloid containing the x- and y-axes that results from rotating the hyperbolic paraboloid z = (x 2 − y 2 )/2 through 45◦ about the z-axis.

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REVIEW EXERCISES 10 (PAGE 621)

9.

10.

11.

12.

ADAMS and ESSEX: CALCULUS 8

x 2 + y 2 + 4z 2 < 4 represents the interior of the circular ellipsoid (oblate spheroid) centred at the origin with semi-axes 2, 2, and 1 in the x, y, and z directions, respectively. x 2 + y 2 − 4z 2 = 4 represents a hyperboloid of one sheet with circular cross-sections in planes perpendicular to the z-axis, and asymptotic to the cone obtained by rotating the line x = 2z about the z-axis. x2

y2

4z 2

− − = 0 represents an elliptic cone with axis along the x-axis whose cross-sections in planes x = k are ellipses with semi-axes |k| and |k|/2 in the y and z directions, respectively. x 2 − y 2 − 4z 2 = 4 represents a hyperboloid of two sheets asymptotic to the cone of the previous exercise.

13. (x − z)2 + y 2 = 1 represents an elliptic cylinder with

oblique axis along the line z = x in the x z-plane, having circular cross-sections of radius 1 in horizontal planes z = k.

14. (x − z)2 + y 2 = z 2 represents an elliptic cone with oblique axis along the line z = x in the x z-plane, having circular cross-sections of radius |k| in horizontal planes z = k. The z-axis lies on the cone.

15.

x + 2y = 0, z = 3 together represent the horizontal straight line through the point (0, 0, 3) parallel to the vector 2i − j.

16.

x + y + 2z = 1, x + y + z = 0 together represent the straight line through the points (−1, 0, 1) and (0, −1, 1).

17.

18.

x 2 + y 2 + z 2 = 4, x + y + z = 3 together represent the circle in which the sphere of radius 2 centred at the origin intersects the plane through (1, 1, 1) √ with normal i + j + k. Since this plane lies at distance 3 from the √ origin, the circle has radius 4 − 3 = 1.

x 2 + z 2 ≤ 1, x − y ≥ 0 together represent all points that lie inside or on the circular cylinder of radius 1 and axis along the y-axis and also either on the vertical plane x − y = 0 or on the side of that plane containing the positive x-axis.

19. The given line is parallel to the vector a = 2i − j + 3k. The plane through the origin perpendicular to a has equa-

21. A plane perpendicular to x − y +z = 0 and 2x + y −3z = 2 has normal given by the cross product of the normals of these two planes, that is, by i 1 2

If the plane also passes through (2, −1, 1), then its equation is 2(x − 2) + 5(y + 1) + 3(z − 1) = 0, or 2x + 5y + 3z = 2.

22. The plane through A = (−1, 1, 0), B = (0, 4, −1) and C = (2, 0, 0) has normal i −→ −→ AC × AB = 3 1

b = (2 − 1)i + (−1 − 0)j + (1 − (−1))k = i − j + 2k. If it is also parallel to the vector a in the previous solution, then it is normal to j k i a × b = 2 −1 3 = i − j − k. 1 −1 2

The plane has equation (x − 1) − (y − 0) − (z + 1) = 0, or x − y − z = 2.

j k −1 0 = i + 3j + 10k. 3 −1

Its equation is (x −2)+3y +10z = 0, or x +3y +10z = 2.

23. A plane containing the line of intersection of the planes x + y + z = 0 and 2x + y − 3z = 2 has equation 2x + y − 3z − 2 + λ(x + y + z − 0) = 0. This plane passes through (2, 0, 1) if −1 + 3λ = 0. In this case, the equation is 7x + 4y − 8z = 6.

24. A plane containing the line of intersection of the planes x + y + z = 0 and 2x + y − 3z = 2 has equation 2x + y − 3z − 2 + λ(x + y + z − 0) = 0. This plane is perpendicular to x − 2y − 5z = 17 if their normals are perpendicular, that is, if 1(2 + λ) − 2(1 + λ) − 5(−3 + λ) = 0, or 9x + 7y − z = 4.

25. The line through (2, 1, −1) and (−1, 0, 1) is parallel to

the vector 3i + j − 2k, and has vector parametric equation

tion 2x − y + 3z = 0.

20. A plane through (2, −1, 1) and (1, 0, −1) is parallel to

j k −1 1 = 2i + 5j + 3k. 1 −3

r = (2 + 3t)i + (1 + t)j − (1 + 2t)k.

26. A vector parallel to the planes x − y = 3 and

x + 2y + z = 1 is (i − j) × (i + 2j + k) = −i − j + 3k. A line through (1, 0, −1) parallel to this vector is y z+1 x −1 = = . −1 −1 3

27. The line through the origin perpendicular to the plane 3x − 2y + 4z = 5 has equations x = 3t, y = −2t, z = 4t.

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INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 10 (PAGE 621)

28. The vector   a = (1 + t)i − tj − (2 + 2t)k − 2si + (s − 2)j − (1 + 3s)k = (1 + t − 2s)i − (t + s − 2)j − (1 + 2t − 3s)k

32. The tetrahedron with vertices A = (1, 2, 1),

B = (4, −1, 1), C = (3, 4, −2), and D = (2, 2, 2) has volume

joins points on the two lines and is perpendicular to both lines if a • (i − j − 2k) = 0 and a • (2i + j − 3k) = 0, that is, if

1 −→ −→ −→ 1 |( AB × AC) • AD| = |(9i + 9j + 12k) • (i + k)| 6 6 7 9 + 12 = cu. units. = 6 2

1 + t − 2s + t + s − 2 + 2 + 4t − 6s = 0 2 + 2t − 4s − t − s + 2 + 3 + 6t − 9s = 0, or, on simplification,

33. The inverse of A satisfies 6t − 7s = −1 7t − 14s = −7.

This system has solution t = 1, s = 1. We would expect to use a as a vector perpendicular to both lines, but, as it happens, a = 0 if t = s = 1, because the two given lines intersect at (2, −1, −4). A nonzero vector perpendicular to both lines is j k i 1 −1 −2 = 5i − j + 3k. 2 1 −3

Thus the required line is parallel to this vector and passes through (2, −1, −4), so its equation is r = (2 + 5t)i − (1 + t)j + (−4 + 3t)k.

1 2  3 4

0 1 2 3



 0 a 0 e  0 i 1 m

a = 1, b = 0, c = 0, d = 0,

  d 1 h  0 = l 0 p 0

0 1 0 0

0 0 1 0

 0 0 . 0 1

= 0, = 0, = 1, = 0,

4a + 3e + 2i + m = 0. 4b + 3 f + 2 j + n = 0. 4c + 3g + 2k + o = 0. 4d + 3h + 2l + p = 1.

These systems have solutions a = 1, b = 0, c = 0, d = 0,

e f g h

= −2, = 1, = 0, = 0,

i j k l

= 1, = −2, = 1, = 0,

m = 0, n = 1, o = −2, p = 1.

Thus

(Any permutation of the subscripts 1, 2, and 3 in the above equation will do as well.)

A−1

30. The points with position vectors r1 , r2 , r3 , and r4 are coplanar if the tetrahedron having these points as vertices has zero volume, that is, if h i (r2 − r1 ) × (r3 − r1 ) • (r4 − r1 ) = 0. (Any permutation of the subscripts 1, 2, 3, and 4 in the above equation will do as well.)

31. The triangle with vertices A = (1, 2, 1), B = (4, −1, 1), k 0 | −3

c g k o

2a + e = 0, 3a + 2e + i 2b + f = 1, 3b + 2 f + j 2c + g = 0, 3c + 2g + k 2d + h = 0, 3d + 2h + l

collinear if the triangle having these points as vertices has zero area, that is, if

and C = (3, 4, −2) has area j 1 −→ −→ 1 i | AB × AC| = | 3 −3 2 2 2 2

b f j n

Expanding the product on the left we get four systems of equations:

29. The points with position vectors r1 , r2 , and r3 are

(r2 − r1 ) × (r3 − r1 ) = 0.

0 0 1 2

√ 3 34 1 = |9i + 9j + 12k| = sq. units. 2 2

1 2 1

34. Let A = Then

1 0 0 0  −2 1 = 1 −2 1 0 1 −2 

! 1 1 1 0 ,x = 0 −1

Ax = b

⇔

 0 0 . 0 1

! x1 x2 , and b = x3

! b1 b2 . b3

x1 + x2 + x3 = b1 2x1 + x2 = b2 x1 − x3 = b3 .

The sum of the first and third equations is 2x1 + x2 = b1 + b3 , which is incompatible with the second equation unless b2 = b1 + b3 , that is, unless b • (i − j + k) = 0.

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CHALLENGING PROBLEMS 10 (PAGE 621)

If b satisfies this condition then there will be a line of solutions; if x1 = t, then x2 = b2 − 2t, and x3 = t − b3 , so ! t x = b2 − 2t t − b3

35.

ADAMS and ESSEX: CALCULUS 8

3. The triangle with vertices (x1 , y1 , 0), (x2 , y2 , 0), and (x3 , y3 , 0), has two sides corresponding to the vectors (x2 − x1 )i + (y2 − y1 )j and (x3 − x1 )i + (y3 − y1 )j. Thus the triangle has area given by i 1 A = | x2 − x1 2 x −x

is a solution for any t. ! 3 −1 1 A = −1 1 −1 . We use Theorem 8. 1 −1 2 D1 = 3 > 0,

3 D2 = −1

Thus A is positive definite.

(page 621)

1. If d is the distance from P to the line AB, then d is the altitude of the triangle AP B measured perpendicular to the base AB. Thus the area of the triangle is −→ (1/2)d| B A| = (1/2)d|r A − r B |. On the other hand, the area is also given by −→ −→ (1/2)| P A × P B| = (1/2)|(r A − r P ) × (r B − r P )|. Equating these two expressions for the area of the triangle and solving for d we get d=

|(r A − r P ) × (r B − r P )| . |r A − r B |

1

k 0 | 0

1 = |[(x2 − x1 )(y3 − y1 ) − (x3 − x1 )(y2 − y1 )]k| 2

−1 = 2 > 0, 3

3 −1 1 D3 = −1 1 −1 = 2 > 0. 1 −1 2

Challenging Problems 10

3

j y2 − y1 y3 − y1

1 |x2 y3 − x2 y1 − x1 y3 − x3 y2 + x3 y1 + x1 y2 | 2 1 x1 y1 1 = | x2 y2 1 |. 2 x 3 y3 1 =

4.

a) Let Q 1 and Q 2 be the points on lines L 1 and L 2 , respectively, that are closest together. As observed in −−−→ Example 9 of Section 1.4, Q 1 Q 2 is perpendicular to both lines. Therefore, the plane P1 through Q 1 having normal −−−→ Q 1 Q 2 contains the line L 1 . Similarly, the plane P2 −−−→ through Q 2 having normal Q 1 Q 2 contains the line L 2 . These planes are parallel since they have the same normal. They are different planes because Q 1 6= Q 2 (because the lines are skew). b) Line L 1 through (1, 1, 0) and (2, 0, 1) is parallel to i − j + k, and has parametric equation r1 = (1 + t)i + (1 − t)j + tk. Line L 2 through (0, 1, 1) and (1, 2, 2) is parallel to i + j + k, and has parametric equation r2 = si + (1 + s)j + (1 + s)k. Now r2 − r1 = (s − t − 1)i + (s + t)j + (1 + s − t)k.

2. By the formula for the vector triple product given in Exercise 23 of Section 1.3, (u × v) × (w × x) = [(u × v) • x]w − [(u × v) • w]x (u × v) × (w × x) = −(w × x) × (u × v) = −[(w × x) • v]u + [(w × x) • u]v. In particular, if w = u, then, since (u × v) • u = 0, we have (u × v) × (u × x) = [(u × v) • x]u, or, replacing x with w,

To find the points Q 1 on L 1 and Q 2 on L 2 for −−−→ which Q 1 Q 2 is perpendicular to both lines, we solve (s − t − 1) − (s + t) + (1 + s − t) = 0 (s − t − 1) + (s + t) + (1 + s − t) = 0. Subtracting these equations gives s + t = 0, so t = −s. Then substituting into either equation gives 2s − 1 + 1 + 2s = 0, so s = −t = 0. Thus Q = (1, 1, 0) and Q 2 = (0, 1, 1), and −−−→ 1 Q 1 Q 2 = −i + k. The required planes are x − z = 1 (containing L 1 ) and x − z = −1 (containing L 2 ).

5. This problem is similar to Exercise 28 of Section 1.3. (u × v) × (u × w) = [(u × v) • w]u.

The equation a×x = b has no solution x unless a•b = 0. If this condition is satisfied, then x = x0 + ta is a solution for any scalar t, where x0 = (b × a)/|a|2 .

420 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.1 (PAGE 629)

CHAPTER 11. VECTOR FUNCTIONS AND CURVES

9. Position: r = 3 cos ti + 4 cos tj + 5 sin tk

Velocity: v = √−3 sin ti − 4 sin tj + 5 cos tk Speed: v = 9 sin2 t + 16 sin2 t + 25 cos2 t = 5 Acceleration : a = −3 cos ti − 4 cos tj − 5 sin tk = −r Path: the circle of intersection of the sphere x 2 + y 2 + z 2 = 25 and the plane 4x = 3y.

Section 11.1 Vector Functions of One Variable (page 629)

10. Position: r = 3 cos ti + 4 sin tj + tk

Velocity: v = √−3 sin ti + 4 cos tj + k √ Speed: v = 9 sin2 t + 16 cos2 t + 1 = 10 + 7 cos2 t Acceleration : a = −3 cos ti − 4 sin tj = tk − r Path: a helix (spiral) wound around the elliptic cylinder (x 2 /9) + (y 2 /16) = 1.

1. Position: r = i + tj

Velocity: v = j Speed: v = 1 Acceleration : a = 0 Path: the line x = 1 in the x y-plane.

11. Position: r = aet i + bet j + cet k

2. Position: r = t 2 i + k

Velocity: v = 2ti Speed: v = 2|t| Acceleration : a = 2i Path: the line z = 1, y = 0.

12.

3. Position: r = t 2 j + tk

Velocity: v = √2tj + k Speed: v = 4t 2 + 1 Acceleration : a = 2j Path: the parabola y = z 2 in the plane x = 0.

+ aω(2 cos ωt − ω sin ωt)j − (b/t 2 )k Path: a spiral on the surface x 2 + y 2 = a 2 e z/b .

t 13. Position: r = e−tcos(et )i + e−t sin(et )j −e k

4. Position: r = i + tj + tk

Velocity: v = − e−t cos(et ) + sin(et ) i   − e−t sin(et ) − cos(et ) j − et k √ Speed: v = 1 + e−2t + e2t  Acceleration: a = (e−t − et ) cos(et ) + sin(et ) i   + (e−t − et ) sin(et ) − cos(et ) j − et k p Path: a spiral on the surface z x 2 + y 2 = −1.

Velocity: v = √j + k Speed: v = 2 Acceleration : a = 0 Path: the straight line x = 1, y = z.

5. Position: r = t 2 i − t 2 j + k

Velocity: v =√2ti − 2tj Speed: v = 2 2t Acceleration: a = 2i − 2j Path: the half-line x = −y ≥ 0, z = 1.

14. Position: r = a cos t sin ti + a sin2 tj + a cos tk

 a a = sin 2ti + 1 − cos 2t j + a cos tk 2 2 Velocity: v =√ a cos 2ti + a sin 2tj − a sin tk Speed: v = a 1 + sin2 t Acceleration: a = −2a sin 2ti + 2a cos 2tj − a cos tk Path: the path lies on the sphere x 2 + y 2 + z 2 = a 2 , on the surface defined in terms of spherical polar coordinates by φ = θ , on the circular cylinder x 2 + y 2 = ay, and on the parabolic cylinder ay + z 2 = a 2 . Any two of these surfaces serve to pin down the shape of the path.

6. Position: r = ti + t 2 j + t 2 k

Velocity: v = √i + 2tj + 2tk Speed: v = 1 + 8t 2 Acceleration: a = 2j + 2k Path: the parabola y = z = x 2 .

7. Position: r = a cos ti + a sin tj + ctk

Velocity: v = √−a sin ti + a cos tj + ck Speed: v = a 2 + c2 Acceleration: a = −a cos ti − a sin tj Path: a circular helix.

8. Position: r = a cos ωti + bj + a sin ωtk

Velocity and acceleration: v=a=r √ Speed: v = et a 2 + b2 + c2 x y z Path: the half-line = = > 0. a b c Position: r = at cos ωti + at sin ωtj + b ln tk Velocity: v = a(cos ωt − ωt sin ωt)i + a(sin ωt + ωt cos ωt)j + (b/t)k p Speed: v = a 2 (1 + ω2 t 2 ) + (b2 /t 2 ) Acceleration: a = −aω(2 sin ωt + ω cos ωt)i

15. The position of the particle is given by r = 5 cos(ωt)i + 5 sin(ωt)j,

Velocity: v = −aω sin ωti + aω cos ωtk Speed: v = |aω| Acceleration: a = −aω2 cos ωti − aω2 sin ωtk Path: the circle x 2 + z 2 = a 2 , y = b.

where ω = π to ensure that r has period 2π/ω = 2 s. Thus d 2r a = 2 = −ω2 r = −π 2 r. dt At (3, 4), the acceleration is −3π 2 i − 4π 2 j.

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SECTION 11.1 (PAGE 629)

ADAMS and ESSEX: CALCULUS 8

16. When its x-coordinate is x, the particle is at position

Thus

r = xi + (3/x)j, and its velocity and speed are

d 2u −2 du −16u = 4u = . dt 2 (1 + 2u 2 )2 dt (1 + 2u 2 )3

dr dx 3 dx = i− 2 j dt dt x dt r dx 9 v = 1 + 4 . dt x v=

The particle is at (3, 3, 2) when u = 1. At this point du/dt = 2/3 and d 2 u/dt 2 = −16/27, and so

We know that d x/dt > 0 since the particle is moving to the right. When x = √ 2, we have 10 = v = (d x/dt) 1 + (9/16) = (5/4)(d x/dt). Thus d x/dt = 8. The velocity at that time is v = 8i − 6j.

2 (3i + 6uj + 6u 2 k) = 2i + 4j + 4k 3  2 −16 2 a= (6j + 12k) (3i + 6j + 6k) + 27 3 8 = (−2i − j + 2k). 9 v=

17. The particle moves along the curve z = x 2 , x + y = 2, in the direction of increasing y. Thus its position at time t is r = (2 − y)i + yj + (2 − y)2 k, where y is an increasing function of time t. Thus

20.

i dy h −i + j − 2(2 − y)k dt q dy v= 1 + 1 + 4(2 − y)2 = 3 dt v=

since the speed 3. When y = 1, we have √ is √ d y/dt = 3/ 6 = 3/2. Thus v=

r

3 (−i + j − 2k). 2



dx dt

2

.

The left side is 3 when x = 1, so 3(d 2 x/dt 2 ) + 48 = 3, and d 2 x/dt 2 = −15 at that point, and the acceleration there is

r = xi + x 2 j + x 3 k, i dx h i + 2xj + 3x 2 k . Since dt dz/dt = 3x 2 d x/dt = 3, when x = 2 we have 12 d x/dt = 3, so d x/dt = 1/4. Thus

so its velocity is v =

a = −15(i − 2j + 2k) + 9(−2j + 2k) = −15i + 12j − 12k.

21.

1 v = i + j + 3k. 4

r = 3ui + 3u 2 j + 2u 3 k du v= (3i + 6uj + 6u 2 k) dt  2 d 2u du a = 2 (3i + 6uj + 6u 2 k) + (6j + 12uk). dt dt Since u is increasing and the speed of the particle is 6, 6 = |v| = 3

r = xi − x 2 j + +x 2 k dx v= (i − 2xj + 2xk) dt  2 d2x dx a = 2 (i − 2xj + 2xk) + (−2j + 2k). dt dt √ dx √ dx Thus |v| = 1 + 4x 4 + 4x 4 = 1 + 8x 4 , dt dt since x is increasing. At (1, −1, 1), x = 1 and |v| = 9, so d x/dt = 3, and the velocity at that point is v = 3i − 6j + 6k. Now p d d2x 16x 3 |v| = 1 + 8x 4 2 + √ dt dt 1 + 8x 4

18. The position of the object when its x-coordinate is x is

19.

du 2 = , and dt 1 + 2u 2

du p du 1 + 4u 2 + 4u 4 = 3(1 + 2u 2 ) . dt dt

d d 2 |v| = v • v = 2v • a. dt dt If v • a > 0 then the speed v = |v| is increasing. If v • a < 0 then the speed is decreasing.

22. If u(t) = u 1 (t)i + u 2 (t)j + u 3 (t)k

v(t) = v 1 (t)i + v 2 (t)j + v 3 (t)k then u • v = u 1 v 2 + u 2 v 2 + u 3 v 3 , so d du 1 dv 1 du 2 dv 2 u•v= v1 + u1 + v2 + u2 dt dt dt dt dt du 3 dv 3 + v3 + u3 dt dt du dv = •v+u• . dt dt

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INSTRUCTOR’S SOLUTIONS MANUAL

23.

24.

25.

SECTION 11.1 (PAGE 629)

a11 a12 a13 a21 a22 a23 a31 a32 a33 dh = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 dt i − a11 a23 a32 − a12 a21 a33 − a13 a22 a31

d dt

31.

′ ′ ′ = a11 a22 a33 + a11 a22 a33 + a11 a22 a33 ′ ′ ′ + a12 a23 a31 + a12 a23 a31 + a12 a23 a31 ′ ′ ′ + a13 a21 a32 + a13 a21 a32 + a13 a21 a32 ′ ′ ′ − a11 a23 a32 − a11 a23 a32 − a11 a23 a32 ′ ′ ′ − a12 a21 a33 − a12 a21 a33 − a12 a21 a33 ′ ′ ′ − a13 a22 a31 − a13 a22 a31 − a13 a22 a31 ′ ′ ′ a11 a12 a13 a11 a12 a13 ′ ′ ′ a23 = a21 a22 a23 + a21 a22 a31 a32 a33 a31 a32 a33 a11 a12 a13 + a21 a22 a23 ′ ′ ′ a31 a32 a33

32.

33. Since

28.

29.

30.

dr = v(t) = 2r(t) and r(0) = r0 , we have dt r(t) = r(0)e2t = r0 e2t , dr dv =2 = 4r0 e2t . a(t) = dt dt

d d 2 |r| = r • r = 2r • v = 0 implies that |r| is constant. dt dt Thus r(t) lies on a sphere centred at the origin.

34.

d d |r − r0 |2 = (r − r0 ) • (r − r0 ) dt dt dr =0 = 2(r − r0 ) • dt implies that |r − r0 | is constant. Thus r(t) lies on a sphere centred at the point P0 with position vector r0 . above.) Thus r is moving farther away from the origin. If r • v < 0 then r is moving closer to the origin.   d 2 u d 2 u du d 3 u d du d 2 u × 2 = 2 × 2 + × 3 dt dt dt dt dt dt dt du d 3 u = × 3. dt dt   d u • (v × w) dt = u′ • (v × w) + u • (v′ × w) + u • (v × w′ ).  d u × (v × w) dt = u′ × (v × w) + u × (v′ × w) + u × (v × w′ ).    d du d 2 u u× × 2 dt dt dt    2  du du d 2 u d u d 2u = × × 2 +u× × dt dt dt dt 2 dt 2   3 du d u +u× × 3 dt dt     du du d 2 u du d 3 u = × × 2 +u× × 3 . dt dt dt dt dt

The path is the half-line from the origin in the direction of r0 . v  0 r = r0 cos ωt + sin ωt ω dr = −ωr0 sin ωt + v0 cos ωt dt d 2r = −ω2 r0 cos ωt − ωv0 sin ωt = −ω2 r dt 2 dr = v0 . r(0) = r0 , dt t=0

Observe that r • (r0 × v0 ) = 0 for all t. Therefore the path lies in a plane through the origin having normal N = r0 × v0 . Let us choose our coordinate system so that r0 = ai (a > 0) and v0 = ωbi + ωcj (c > 0). Therefore, N is in the direction of k. The path has parametric equations

26. If r • v > 0 then |r| is increasing. (See Exercise 16

27.

i dh (u + u′′ ) • (u × u′ ) dt = (u′ + u′′′ ) • (u × u′ ) + (u + u′′ ) • (u′ × u′ ) + (u + u′′ ) • (u × u′′ ) ′′′ = u • (u × u′ ). i dh (u × u′ ) • (u′ × u′′ ) dt = (u′ × u′ ) • (u′ × u′′ ) + (u × u′′ ) • (u′ × u′′ ) + (u × u′ ) • (u′′ × u′′ ) + (u × u′ ) • (u′ × u′′′ ) = (u × u′′ ) • (u′ × u′′ ) + (u × u′ ) • (u′ × u′′′ ).

x = a cos ωt + b sin ωt y = c sin ωt. The curve is a conic section since it has a quadratic equation:   1 by 2 y 2 x − + 2 = 1. a2 c c Since the path is bounded (|r(t)| ≤ |r0 | + (|v0 |/ω)), it must be an ellipse. If r0 is perpendicular to v0 , then b = 0 and the path is the ellipse (x/a)2 + (y/c)2 = 1 having semi-axes a = |r0 | and c = |v0 |/ω.

35.

d 2r dr = −gk − c dt 2 dt dr r(0) = r0 , = v0 . dt t=0 423

Copyright © 2014 Pearson Canada Inc.

SECTION 11.1 (PAGE 629)

Let w = ect

ADAMS and ESSEX: CALCULUS 8

dr . Then dt

2. Let v(t) be the speed of the tank car at time t seconds. d 2r

dw dr = cect + ect 2 dt dt dt dr dr = cect − ect gk − cect dt dt = −ect gk Z ect w(t) = − ect gk dt = − gk + C. c

The mass of the car at time t is m(t) = M − kt kg. At full power, the force applied to the car is F = Ma (since the motor can accelerate the full car at a m/s2 ). By Newton’s Law, this force is the rate of change of the momentum of the car. Thus i dh (M − kt)v = Ma dt dv − kv = Ma (M − kt) dt dv dt = Ma + kv M − kt 1 1 1 ln(Ma + kv) = − ln(M − kt) + ln C k k k C . Ma + kv = M − kt

g Put t = 0 and get v0 = − k + C, so c ect

g dr = w = v0 + (1 − ect )k dt c dr g −ct = e v0 − (1 − e−ct )k dt c   e−ct g e−ct r=− v0 − t+ k+D c c c 1 g r0 = r(0) = − v0 − 2 k + D. c c

At t = 0 we have v = 0, so Ma = C/M. Thus C = M 2 a and kv =

Thus we have r = r0 +

g 1 − e−ct v0 − 2 (ct + e−ct − 1)k. c c

The speed of the tank car at time t (before it is empty) is v(t) =

The limit of this solution, as c → 0, is calculated via l’Hˆopital’s Rule: t − te−ct te−ct − gk lim c→0 c→0 1 2c t 2 e−ct = r0 + v0 t − gk lim c→0 2 1 2 = r0 + v0 t − gt k, 2

lim r(t) = r0 + v0 lim

c→0

which is the solution obtained in Example 4.

Mat m/s. M − kt

dr = k × r, r(0) = i + k. dt Let r(t) = x(t)i + y(t)j + z(t)k. Then x(0) = z(0) = 1 and y(0) = 0. Since k • (dr/dt) = k • (k × r) = 0, the velocity is always perpendicular to k, so z(t) is constant: z(t) = z(0) = 1 for all t. Thus

3. Given:

dx dy dr i+ j= = k × r = xj − yi. dt dt dt

Section 11.2 Some Applications of Vector Differentiation (page 636)

Separating this equation into components,

1. It was shown in the text that v(T ) − v(0) = − ln

M2a Makt − Ma = . M − kt M − kt



 m(0) ve . m(T )

If v(0) = 0 and v(T ) = −ve then ln(m(0)/m(T )) = 1 and m(T ) = (1/e)m(0). The rocket must therefore e−1 of its initial mass to accelerate to burn fraction e the speed of its exhaust gases. (1/e2 )m(0),

Similarly, if v(T ) = −2ve , then m(T ) = so e2 − 1 the rocket must burn fraction of its initial mass to e2 accelerate to twice the speed of its exhaust gases.

dx = −y, dt

dy = x. dt

Therefore, d2x dy =− = −x, dt 2 dt and x = A cos t + B sin t. Since x(0) = 1 and y(0) = 0, we have A = 1 and B = 0. Thus x(t) = cos t and y(t) = sin t. The path has equation

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r = cos ti + sin tj + k.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.2 (PAGE 636)

Remark: This result also follows from comparing the given differential equation with that obtained for circular motion in the text. This shows that the motion is a rotation with angular velocity k, that is, rotation about the z-axis with angular speed 1. The initial value given for r then forces r = cos ti + sin tj + k.

which is pointing towards the ground.

6. We use the fixed and rotating frames as described in the text. Assume the satellite is in an orbit in the plane spanned by the fixed basis vectors I and K. When the satellite passes overhead an observer at latitude 45◦ , its position is I+K R=R √ , 2 where R is the radius of the earth, and since it circles the earth in 2 hours, its velocity at that point is

4. First observe that   d dr |r − b|2 = 2(r − b) • = 2(r − b) • a × (r − b) = 0, dt dt so |r − b| is constant; for all t the object lies on the sphere centred at the point with position vector b having radius r0 − b. Next, observe that

I−K V = πR √ . 2 The angular velocity of the earth is Ω = (π/12)K. The rotating frame with origin at the observer’s position has, at the instant in question, its basis vectors satisfying

  d (r − r0 ) • a = a × (r − b) • a = 0, dt so r − r0 ⊥ a; for all t the object lies on the plane through r0 having normal a. Hence the path of the object lies on the circle in which this plane intersects the sphere described above. The angle between r − b and a must therefore also be constant, and so the object’s speed |dr/dt| is constant. Hence the path must be the whole circle.

1 1 I = −√ j + √ k 2 2 J=i 1 1 K = √ j + √ k. 2 2 As shown in the text, the velocity v of the satellite as it appears to the observer is given by V = v + Ω × R. Thus

5. Use a coordinate system with origin at the observer, i

v =V−Ω×R R pi πR = √ (I − K) − K × √ (I + K) 12 2 2 πR πR = √ (I − K) − √ J 2 12 2 πR = −π Rj − √ i. 12 2

pointing east, and j pointing north. The angular velocity of the earth is 2π/24 radians per hour northward: Ω=

π j. 12

Because the earth is rotating west to east, the true north to south velocity of the satellite will appear to the observer to be shifted to the west by π R/12 km/h, where R is the radius of the earth in kilometres. Since the satellite circles the earth at a rate of π radians/h, its velocity, as observed at the moving origin, is v R = −π Rj −

πR i. 12

 π R/12 = tan−1 (1/12) ≈ 4.76◦ πR with the southward direction. Thus the satellite appears to the observer to be moving in a direction 4.76◦ west of south. v R makes angle tan−1



v makes

√ ! √ π R/12 2 angle = tan−1 (1/(12 2) ≈ 3.37◦ πR with the southward direction. Thus the satellite appears to the observer to be moving in a direction 3.37◦ west of south. tan−1

The apparent Coriolis force is

The apparent Coriolis force is −2Ω × v R = −

  π πR πR K × √ (I − K − √ J 12 2 12 2   π2R 1 =− √ J+ I 12 6 2   2 π R 1 = − √ i + √ (−j + k) . 6 2 12 2

−2Ω × v = −2

  2π πR π2R j × −π Rj − i =− k, 12 12 72

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SECTION 11.2 (PAGE 636)

ADAMS and ESSEX: CALCULUS 8

7. The angular velocity of the earth is Ω, pointing due

to the east and west of north.

north. For a particle moving with horizontal velocity v, the tangential and normal components of the Coriolis force C, and of Ω, are related by CT = −2Ω N × v,

Section 11.3 (page 643)

C N = −2ΩT × v.

At the north or south pole, ΩT = 0 and Ω N = Ω. Thus C N = 0 and CT = −2Ω × v. The Coriolis force is horizontal. It is 90◦ east of v at the north pole and 90◦ west of v at the south pole.

1. On the first quadrant part of the circle x 2 + y 2 = a 2 p we have x = a 2 − y 2 , 0 ≤ y ≤ a. The required parametrization is

At the equator, Ω N = 0 and ΩT = Ω. Thus CT = 0 and C N = −2Ω × v. The Coriolis force is vertical.

r = r(y) =

8. We continue with the same notation as in Example 4. Since j points northward at the observer’s position, the angle µ between the direction vector of the sun, S = cos σ I + sin σ J and north satisfies

a 2 − y 2 i + yj,

(0 ≤ y ≤ a).

we have y = a 2 − x 2 , 0 ≤ x ≤ a. The required parametrization is r = r(x) = xi +

For the sun, θ = 0 and at sunrise and sunset we have, by Example 4, cos θ = − tan σ/ tan φ, so that tan σ + sin σ sin φ tan φ cos2 φ = sin σ + sin σ sin φ sin φ sin σ = . sin φ

q

2. On the first quadrant part of the circle x 2 + y 2 = a 2 √

cos µ = S • j = − cos σ cos φ cos θ + sin σ sin φ.

cos µ = cos σ cos φ

Curves and Parametrizations

p a 2 − x 2 j,

(0 ≤ x ≤ a).

3. From the figure we see that π π , 0≤θ ≤ 2 2  π x = a cos θ = a cos φ − = a sin φ 2  π y = a sin θ = a sin φ − = −a cos φ. 2

φ=θ+

9. At Vancouver, φ = 90◦ − 49.2◦ = 40.8◦ . On June

21st, σ = 23.3◦ . Ignoring the mountains and the rain, by Example 4 there will be   tan 23.3◦ 24 −1 − cos ≈ 16 π tan 40.8◦

The required parametrization is π

r = a sin φi − a cos φj,

hours between sunrise and sunset. By Exercise 8, the sun will rise and set at an angle   sin 23.3◦ cos−1 ≈ 52.7◦ sin 40.8◦

2

 ≤φ≤π .

y a (x,y)

to the east and west of north.

10. At Ume˚a, φ = 90◦ − 63.5◦ = 26.5◦ . On June 21st,

φ

σ = 23.3◦ . By Example 4 there will be   24 −1 tan 23.3◦ cos − ≈ 20 π tan 26.5◦

hours between sunrise and sunset. By Exercise 8, the sun will rise and set at an angle   sin 23.3◦ cos−1 ≈ 27.6◦ sin 26.5◦

θ a

Fig. 11.3.3

4.

s s π s 0≤ ≤ x = a sin , y = a cos , a a  a 2 s s aπ  r = a sin i + a cos j, 0≤s≤ . a a 2

426 Copyright © 2014 Pearson Canada Inc.

x

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.3 (PAGE 643)

y

2 2 2 a) If t = x, √ then z = 1 + t, so 1 + 2t + t = t + y , and y = ± 1 + 2t. Two parametrizations are needed to get the whole parabola, one for y ≤ 0 and one for y ≥ 0.

s

a

(x,y)

b) If t = y, then x 2 + t 2 = z 2 = 1 + 2x + x 2 , so 2x + 1 = t 2 , and x = (t 2 − 1)/2. Thus z = 1 + x = (t 2 + 1)/2. The whole parabola is parametrized by

s a

a

x

r=

Fig. 11.3.4

5.

z = x 2 , z = 4y 2 . If t = y, then z = 4t 2 , so x = ±2t. The curve passes through (2, −1, 4) when t = −1, so x = −2t. The parametrization is r = −2ti + tj + 4t 2 k.

6.

z = x 2 , x + y + z = 1. If t = x, then z = t 2 and y = 1 − t − t 2 . The parametrization is r = ti + (1 − t − t 2 )j + t 2 k.

7.

z = x + y, x 2 + y 2 = 9. One possible parametrization is r = 3 cos ti + 3 sin tj + 3(cos t + sin t)k. p x + y = 1, z = 1 − x 2 − y 2 . If x = t, then y = 1 − t and p p z = 1 − t 2 − (1 − t)2 = 2(t − t 2 ). One possible parametrization is

8.

r = ti + (1 − t)j +

9.

p

2 2 2 c) If t = z, then √ x = t − 1 and t = t − 2t + 1 + y , so y = ± 2t − 1. Again two parametrizations are needed to get the whole parabola.

12. By symmetry, the centre of the circle C of intersection of the plane x + y + z = 1 and the sphere x 2 + y 2 + z 2 = 1 must lie on the plane and must have its three coordinates equal. Thus the centre has position vector r0 =

Any vector v that satisfies v • (i + j + k) = 0 is parallel to the plane x + y + z = 1 containing C. One such vector is v1 = i − j. A second one, perpendicular to v1 , is

z = x 2 + y 2, 2x − 4y − z − 1 = 0. These surfaces intersect on the vertical cylinder

v2 = (i + j + k) × (i − j) = i + j − 2k. Two perpendicular unit vectors that are parallel to the plane of C are

that is

(x − 1)2 + (y + 2)2 = 4.

i−j vˆ 1 = √ , 2

One possible parametrization is x = 1 + 2 cos t y = −2 + 2 sin t z = −1 + 2(1 + 2 cos t) − 4(−2 + 2 sin t) = 9 + 4 cos t − 8 sin t r = (1 + 2 cos t)i − 2(1 − sin t)j + (9 + 4 cos t − 8 sin t)k.

10.

yz + x = 1, x z − x = 1. One possible parametrization is x = t, z = (1 + t)/t, and y = (1 − t)/z = (1 − t)t/(1 + t), that is, t − t2 1+t r = ti + j+ k. 1+t t

11.

z 2 = x 2 + y 2 , z = 1 + x.

1 (i + j + k). 3

Since C passes through the point (0, 0, 1), its radius is s r       1 2 2 1 2 1 2 0− . + 0− + 1− = 3 3 3 3

2(t − t 2 )k.

x 2 + y 2 = 2x − 4y − 1,

t2 − 1 t2 + 1 i + tj + k. 2 2

13.

vˆ 2 =

i + j − 2k . √ 6

Thus one possible parametrization of C is r 2 r = r0 + (cos t vˆ 1 + sin t vˆ 2 ) 3 i + j + k cos t sin t = + √ (i − j) + (i + j − 2k). 3 3 3 r = t 2 i + t 2 j + t 3 k, (0 ≤ t ≤ 1) p p v = (2t)2 + (2t)2 + (3t 2 )2 = t 8 + 9t 2 Z 1 p Length = t 8 + 9t 2 dt Let u = 8 + 9t 2 0 du = 18t dt √ √ 17 1 2 3/2 17 17 − 16 2 = u = units. 18 3 27 8 427

Copyright © 2014 Pearson Canada Inc.

SECTION 11.3 (PAGE 643)

14.

r = ti + λt 2 j + t 3 k, (0 ≤ t ≤ T ) p p v = 1 + (2λt)2 + 9t 4 = (1 + 3t 2 )2 √ if 4λ2 = 6, that is, if λ = ± 3/2. In this case, the length of the curve is s(T ) =

15.

16.

ADAMS and ESSEX: CALCULUS 8

T

Length =

Z

T

=

Z

1

Z

T

0

18. One-eighth of the curve C lies in the first octant. That part can be parametrized

dr dt dt s

1 z = √ sin t, (0 ≤ t ≤ π/2) 2 1 1 1 − cos2 t − sin2 t = √ sin t. 2 2

x = cos t, r

(1 + 3t 2 ) dt = T + T 3 .

y=

4a 2 t 2 + b2 +

1

The curve is called a conical helix because it is a spiral lying on the cone x 2 + y 2 = z 2 .

c2 t2

Since the first octant part of C lies in the plane y = z, it must be a quarter of a circle of radius 1. Thus the length of all of C is 8 × (π/2) = 4π units. If you wish to use an integral, the length is

dt units.

If b2 = rthen Z 4ac T c 2 Length = 2at + dt t 1 Z T c = 2at + dt t 1 = a(T 2 − 1) + c ln T units. a x = a cos t sin t = sin 2t, 2 a y = a sin2 t = (1 − cos 2t), 2 z = bt. The curve is a circular helix lying on the cylinder

8

Z

π/2

0

=8

Z

r

sin2 t +

1 1 cos2 t + cos2 t dt 2 2

π/2

dt = 4π units.

0

z

x 2 + y2 + z2 = 1 x 2 + 2z 2 = 1

 a 2 a2 . x2 + y − = 2 4

C

Its length, from t = 0 to t = T , is L=

T

Z

y

0 p = T a 2 + b2 units.

17.

L=

=2

2π

0

Z

p 2 + t 2 dt

t=2π

Fig. 11.3.18

19. If C is the curve

r = t cos ti + t sin tj + tk, 0 ≤ t ≤ 2π v = (cos t − t sin t)i + (sin t + t cos t)j + k p p v = |v| = (1 + t 2 ) + 1 = 2 + t 2 . The length of the curve is Z

x

p a 2 cos2 2t + a 2 sin2 2t + b2 dt

√ Let t = 2 tan θ √ dt = 2 sec2 θ dθ

3

sec θ dθ

x = et cos t,

(0 ≤ t ≤ 2π ),

z = t,

then the length of C is

L= =

t=0

  t=2π = sec θ tan θ + ln | sec θ + tan θ | !t=0 √ √ 2π t 2 + t2 2 + t2 t = √ + ln +√ 2 2 2 0 p p √  = π 2 + 4π 2 + ln 1 + 2π 2 + 2π units.

y = et sin t,

Z

2π

0

Z

2π

0

Z

2π

s 

dx dt

2

+



dy dt

2

+



dz dt

2

dt

p e2t (cos t − sin t)2 + e2t (sin t + cos t)2 + 1 dt p 2e2t + 1 dt

Let 2e2t + 1 = v 2 2e2t dt = v dv  Z t=2π 2 Z t=2π  v dv 1 = = 1 + dv v2 − 1 v2 − 1 t=0 t=0  t=2π  1 v − 1 = v + ln 2 v + 1 t=0 =

0

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.3 (PAGE 643)

√ 2π p √ 1 2e2t + 1 − 1 4π = 2e + 1 − 3 + ln √ 2 2e2t + 1 + 1 0 √ 2π p √ 2e2t + 1 − 1 = 2e4π + 1 − 3 + ln √ 2et 0  p p √ 4π 4π = 2e + 1 − 3 + ln 2e + 1 − 1 √ − 2π − ln( 3 − 1) units.

The centreline of the cable is wound around a cylinder of 2a radius a +b and must rise a vertical distance in one cos θ revolution. The figure below shows the cable unwound from the spool and inclined at angle θ . The total length of spool required is the total height H of the cable as shown in that figure. 2a cos θ

Remark: This answer appears somewhat different from that given in the answers section of the text. The two are, however, equal. Somewhat different simplifications were used in the two.

20.

L a

r = t 3i + t 2j

a

v = 3t 2 i + 2tj p p v = |v| = 9t 4 + 4t 2 = |t| 9t 2 + 4

The length L between t = −1 and t = 2 is Z 0 Z 2 p p L= t 9t 2 + 4 dt. (−t) 9t 2 + 4 dt +

θ 2π(a + b) one revolution

Fig. 11.3.22 Observe that tan θ =

0

−1

Making the substitution u = 9t 2 + 4 in each integral, we obtain Z 13  Z 40 1 L= u 1/2 du + u 1/2 du 18 4 4  1  3/2 3/2 = 13 + 40 − 16 units. 27

21. r1 = ti + tj, (0 ≤ t ≤ 1) represents the straight line segment from the origin to (1, 1) in the x y-plane.

r2 = (1 − t)i + (1 + t)j, (0 ≤ t ≤ 1) represents the straight line segment from (1, 1) to (0, 2). Thus C = C1 + C2 is the 2-segment polygonal line from the origin to (1, 1) and then to (0, 2).

L sin θ

2a cos θ

sin θ = cos θ =

2a 1 × . Therefore cos θ 2π(a + b)

a π(a + b) s

a2 1− 2 = π (a + b)2

p

π 2 (a + b)2 − a 2 . π(a + b)

The total length of spool required is H = L sin θ + 2a cos θ   p a L + 2 π 2 (a + b)2 − a 2 units. = π(a + b)

23. r = Ati + Btj + Ctk.

The arc length from the point where t = 0 to the point corresponding to arbitrary t is

22. (Solution due to Roland Urbanek, a student at Okanagan College.) Suppose the spool is vertical and the cable windings make angle θ with the horizontal at each point.

s = s(t) =

Z tp 0

A2 + B 2 + C 2 du =

p

A2 + B 2 + C 2 t.

√ Thus t = s/ A2 + B 2 + C 2 . The required parametrization is Asi + Bsj + Csk r= √ . A2 + B 2 + C 2 H

v = et i + √2j + e−t k v = |v| = e2t + 2 + e−2t = et + e−t . The arc length from the point where t = 0 to the point corresponding to arbitrary t is

θ 2a b Fig. 11.3.22

√

24. r = et i + √2tj − e−t k

a s = s(t) =

Z

0

t

(eu + e−u ) du = et − e−t = 2 sinh t.

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SECTION 11.3 (PAGE 643)

Thus t =

sinh−1 (s/2)

s+

ADAMS and ESSEX: CALCULUS 8

! √ s2 + 4 , 2

= ln √ s + s2 + 4 and et = . The required parametrization is 2 ! √ √ s + s2 + 4 2k s + s2 + 4 √ i+ 2 ln j− r= √ . 2 2 s + s2 + 4

28. If r = r(t) has nonvanishing velocity v = dr/dt on [a, b], then for any t0 in [a, b], the function s = g(t) =

25.

3

|v(u)| du,

on [a, b], by the Fundamental Theorem of Calculus. Hence g is invertible, and defines t as a function of arc length s: t = g −1 (s) ⇔ s = g(t).

1p 2 9a + 16b2 sin2 t = K sin2 t 2 1p 2 where K = 9a + 16b2 2 r r s s Therefore sin t = , cos t = 1 − , K K 2s cos 2t = 1 − 2 sin2 t = 1 − . K The required parametrization is =

Then

  r = r2 (s) = r g −1 (s)

is a parametrization of the curve r = r(t) in terms of arc length.

   s 3/2 2s s 3/2 k i+a +b 1− r=a 1− K K K 

1p 2 for 0 ≤ s ≤ K , where K = 9a + 16b2 . 2 √ r = 3t cos ti + 3t sin tj + 2 2t 3/2 k, (t ≥ 0)

√ √ v = 3(cos t − t sin t)i + 3(sin t + t cos t)j + 3 2 tk p v = |v| = 3 1 + t 2 + 2t = 3(1 + t)   Z t t2 s= 3(1 + u) du = 3 t + 2 0 r 2s 2s 2 Thus t +2t = , so t = −1+ 1 + since t ≥ 0. The 3 3 required parametrization is the given one with t replaced √ by −1 + 1 + (2s)/3.

Section 11.4 Curvature, Torsion, and the Frenet Frame (page 651) 1.

Tˆ = 2.

and so Z b Z d  b d d  d r1 (t) dt = r2 (u) du. r2 u(t) du dt = dt dt du du a a c

i − 4tj + 9t 2 k v = √ . v 1 + 16t 2 + 81t 4

r = a sin ωti + a cos ωtk v = aω cos ωti − aω sin ωtk, v = |aω| h i Tˆ = sgn(aω) cos ωti − sin ωtk .

3.

r = cos t sin ti + sin2 t + cos tk 1 1 = sin 2ti + (1 − cos 2t)j + cos tk 2 2 v = cos 2ti + sin 2tj − sin tk p v = |v| = 1 + sin2 t   1 Tˆ = √ cos 2ti + sin 2tj − sin tk . 1 + sin2 t

4.

r = a cos ti + b sin tj + tk v = −a sin ti + b cos tj + k p v = a 2 sin2 t + b2 cos2 t + 1 v −a sin ti + b cos tj + k Tˆ = = √ . v a 2 sin2 t + b2 cos2 t + 1

r1 (t) = r2 u(t) , where u is a function from [a, b] to [c, d], having u(a) = c and u(b) = d. We assume u is differentiable. Since u is one-to-one and orientationpreserving, du/dt ≥ 0 on [a, b]. By the Chain Rule: d d du r1 (t) = r2 (u) , dt du dt

r = ti − 2t 2 j + 3t 3 k

v = i − 4tj + 9t 2 k p v = 1 + 16t 2 + 81t 4

27. As claimedin the  statement of the problem,

Z

t0

ds = g ′ (t) = |v(t)| > 0 dt

0

26.

t

which gives the (signed) arc length s measured from r(t0 ) along the curve, is an increasing function:

π r = a cos ti + a sin tj + b cos 2tk, 0 ≤ t ≤ 2 v = −3a cos2 t sin ti + 3a sin2 t cos tj − 4b sin t cos tk p v = 9a 2 + 16b2 sin t cos t Z tp s= 9a 2 + 16b2 sin u cos u du 3

Z

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.5 (PAGE 657)

ˆ dT ˆ = 0, so = κN ds dr ˆ ˆ ˆ T(s) = T(0) is constant. This says that = T(0), so ds ˆ r = T(0)s + r(0), which is the vector parametric equation of a straight line.

Section 11.5 Curvature and Torsion for General Parametrizations (page 657)

5. If κ(s) = 0 for all s, then

1. For y = x 2 we have κ(x) =

6. If τ (s) = 0 for all s, then

d Bˆ ˆ = 0, so B(s) ˆ ˆ = −τ N = B(0) is constant. Therefore, ds  d  dr ˆ ˆ ˆ ˆ • B(s) = T(s) • B(s) = 0. r(s) − r(0) • B(s) = ds ds

It follows that 

√ Hence κ(0) = 2 and κ( 2) √ = 2/27. The radii of curvature at x = 0 and x = 2 are 1/2 and 27/2, respectively.

2. For y = cos we have κ(x) =

   ˆ ˆ r(s) − r(0) • B(0) = r(s) − r(0) • B(s) =0

r=

3.

1 1 cos Csi + sin Csj C C

a = (2/t 3 )j

8. The circular helix C1 given by

v × a = (4/t 3 )i + (4/t 3 )k At (2, 1, −2), that is, at t = 1, we have κ = κ(1) =

v = 3t 2 i + 2tj + k a = 6ti + 2j v(1) = 3i + 2j + k, a(1) = 6i + 2j v(1) × a(1) = −2i + 6j − 6k √ √ 4 + 36 + 36 2 19 κ(1) = = (9 + 4 + 1)3/2 143/2 √ At t = 1 the radius of curvature is 143/2 /(2 19).

has curvature and torsion given by b , a 2 + b2

by Example 3. if a curve C has constant curvature κ(s) = C > 0, and constant torsion τ (s) = T 6= 0, then we can choose a and b so that a = C, a 2 + b2

b = T. a 2 + b2

C T , and b = 2 .) By C2 + T 2 C + T2 Theorem 3, C is itself a circular helix, congruent to C1 . (Specifically, a =

r = t 3 i + t 2 j + tk

4.

τ (s) =

√ |v × a| 4 2 . = v3 27

√ Thus the radius of curvature is 27/(4 2).

r = a cos ti + a sin tj + btk

a , a 2 + b2

r = 2ti + (1/t)j − 2tk

v = 2i − (1/t 2 )j − 2k

is parametrized in terms of arc length, and has curvature C and torsion 0. (See Examples 2 and 3.) If curve C has constant curvature κ(s) = C and constant torsion τ (s) = 0, then C is congruent to C1 by Theorem 3. Thus C must itself be a circle (with radius 1/C).

κ(s) =

|d 2 y/d x 2 | | cos x| = . (1 + (d y/d x)2 )3/2 (1 + sin2 x)3/2

Hence κ(0) = 1 and κ(π/2) = 0. The radius of curvature at x = 0 is 1. The radius of curvature at x = π/2 is infinite.

for all s. This says that r(s) lies in the plane through ˆ r(0) having normal B(0).

7. The circle C1 given by

|d 2 y/d x 2| 2 = . 2 3/2 (1 + (d y/d x) ) (1 + 4x 2 )3/2

5.

r = ti + t 2 j + 2k v = i + 2tj a = 2j v × a = 2k At (1, 1, 2), where t = 1, we have √ Tˆ = v/|v| = (i + 2j)/ 5

Bˆ = (v × a)/|v × a| = k

√

ˆ = Bˆ × T ˆ = (−2i + j)/ 5. N

431 Copyright © 2014 Pearson Canada Inc.

SECTION 11.5 (PAGE 657)

6.

ADAMS and ESSEX: CALCULUS 8

r = ti + t 2 j + tk v = i + 2tj + k a = 2j v × a = −2i + 2k At (1, 1, 1), where t = 1, we have √ Tˆ = v/|v| = (i + 2j + k)/ 6

8.

√

Bˆ = (v × a)/|v × a| = −(i − k)/ 2 √

ˆ = Bˆ × T ˆ = −(i − j + k)/ 3. N

9. 7.

t3 t2 j+ k 2 3 v = i + tj + t 2 k da a = j + 2tk, = 2k dt 2 v × a = t i − 2tj + k p v = |v| = 1 + t 2 + t 4 , da =2 (v × a) • dt r = ti +

|v × a| =

p 1 + 4t 2 + t 4

v i + tj + t 2 k = √ v 1 + t2 + t4 v × a t 2 i − 2tj + k Bˆ = =√ |v × a| 1 + 4t 2 + t 4 3 4 3 ˆ =B ˆ × Tˆ = −(2tp + t)i + (1 − t )j + (t + 2t)k N (1 + t 2 + t 4 )(1 + 4t 2 + t 4 ) √ |v × a| 1 + 4t 2 + t 4 κ= = 3 v (1 + t 2 + t 4 )3/2 da (v × a) • 2 dt = τ = . 2 |v × a| 1 + 4t 2 + t 4

Tˆ =

10.

r = et cos ti + et sin tj + et k v = et (cos t − sin t)i + et (sin t + cos t)j + et k a = −2et sin ti + 2et cos tj + et k da = −2et (cos t + sin t)i + 2et (cos t − sin t)j + et k dt v × a = e2t (sin t − cos t)i − e2t (cos t + sin t)j + 2e2t k √ √ v = |v| = 3et , |v × a| = 6e2t da (v × a) • = 2e3t dt v (cos t − sin t)i + (cos t + sin t)j + k Tˆ = = √ v 3 v×a (sin t − cos t)i − (cos t + sin t)j + 2k ˆ B= √ = |v × a| 6 (cos t + sin t)i − (cos t − sin t)j ˆ N = Bˆ × Tˆ = − √ 2 √ 2 |v × a| = t κ= 3e v3 da (v × a) • dt = 1 . τ= 2 |v × a| 3et √ r = (2 + 2 cos t)i + (1 − sin t)j + (3 + sin t)k √ v = − 2 sin ti − cos tj + cos tk p √ v = 2 sin2 t + cos2 t + cos2 t = 2 √ a = − 2 cos ti + sin tj − sin tk da √ = 2 sin ti + cos tj − cos tk dt √ √ v × a = − 2j − 2k 1 |v × a| 2 κ= = √ =√ v3 2 2 2 √ √ da (v × a) • = − 2 cos t + 2 cos t = 0 dt τ = 0. √ Since κ = 1/ 2 is constant, and τ = 0, the √ curve is a circle. Its centre is (2, 1, 3) and its√radius is 2. It lies ˆ in a plane with normal j + k(= − 2B).

r = xi + sin xj dx dx v= i + cos x j = k(i + cos xj) dt dt p v = k 1 + cos2 x dx a = −k sin x j = −k 2 sin xj dt v × a = −k 3 sin xk |v × a| | sin x| κ= = . v3 (1 + cos2 x)3/2

432 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.5 (PAGE 657)

The tangential and normal components of acceleration are

13. The ellipse is the same one considered in Exercise 16, so its curvature is

dv k k 2 cos x sin x dx = √ =−√ 2 cos x)(− sin x) dt dt 2 1 + cos2 x 1 + cos2 x 2 k | sin x| . v 2κ = √ 1 + cos2 x

11.

(a 2

14.

1 1 At t = π/4 we have v = j − √ k, a = −2i − √ k, 2 2 √ da 1 1 = −4j + √ k, v × a = − √ i + 2j + 2k, dt 2 2 √ da = −3 2. (v × a) • dt Thus 1 √ Tˆ = √ ( 2j − k) 3 √ 1 Bˆ = √ (−i + 2j + 2 2k) 13 √ ˆ = − √1 (6i + j + 2k) N 39 √ √ 2 39 6 2 κ= , τ =− . 9 13

By Example 2, the curvature of y = x 2 at (1, 1) is κ=

2 2 = √ . 2 3/2 (1 + 4x ) 5 5 x=1

Thus the magnitude of the normal √ acceleration of the bead at that point is v 2 κ = 2v 2 /(5 5). The rate of change of the speed, dv/dt, is the tangential component of the acceleration, and is due entirely to the tangential component of the gravitational force since there is no friction: dv ˆ = g cos θ = g(−j) • T, dt where θ is the angle between Tˆ and −j. (See the figure.) Since the √ slope of y = x 2 at (1, 1) is 2, √ we have ˆ T = −(i + 2j)/ 5, and therefore dv/dt = 2g/ 5. y

y = x2

ˆ v2κN

(1, 1) θ

r = a cos ti + b sin tj v = −a sin ti + b cos tj a = −a cos ti − b sin tj v × a = abk p v = a 2 sin2 t + b2 cos2 t.

−gj dv ˆ T dt

dv (a 2 − b2 ) sin t cos t = √ , dt a 2 sin2 t + b2 cos2 t which is zero if t is an integer multiple of π/2, that is, at the ends of the major and minor axes of the ellipse. The normal component of acceleration is |v × a| ab . = √ 3 2 2 v a sin t + b2 cos2 t

x

Fig. 11.5.14

The tangential component of acceleration is

v 2κ = v 2

sin2 t

If a > b > 0, then the maximum curvature occurs when sin t = 0, and is a/b2 . The minimum curvature occurs when sin t = ±1, and is b/a 2 .

r = sin t cos ti + sin2 tj + cos tk v = cos 2ti + sin 2tj − sin tk a = −2 sin 2ti + 2 cos 2tj − cos tk da = −4 cos 2ti − 4 sin 2tj + sin tk. dt da At t = 0 we have v = i, a = 2j − k, = −4i, dt da v × a = j + 2k, (v × a) • = 0. dt √ √ ˆ = (2j − k)/ 5, Thus√Tˆ = i, Bˆ = (j + 2k)/ 5, N κ = 5, and τ = 0.

12.

ab + b2 cos2 t)3/2 ab =  3/2 . (a 2 − b2 ) sin2 t + b2

κ=

15. Curve: r =

xi + e x j.

Velocity: v = i + e x j. Speed: v = Acceleration: a = e x j. We have v × a = e x k,

√

1 + e2x .

|v × a| = e x .

ex . Therefore, the radius (1 + e2x )3/2 2x 3/2 (1 + e ) of curvature is ρ = . ex The curvature is κ =

433 Copyright © 2014 Pearson Canada Inc.

SECTION 11.5 (PAGE 657)

ADAMS and ESSEX: CALCULUS 8

The unit normal is −e x i + j

ˆ =B ˆ × Tˆ = (v × a) × v = √ N . |(v × a) × v| 1 + e2x

for some right-handed basis {i1 , j1 , k1 }, and some constant vector r0 . Example 3 of Section 2.4 provides values ˆ ˆ ˆ for T(0), N(0), and B(0), which we can equate to the given values of these vectors:

The centre of curvature is

1 1 ˆ i = T(0) = √ j1 + √ k1 2 2 ˆ j = N(0) = −i1 1 1 ˆ k = B(0) = − √ j1 + √ k1 . 2 2

ˆ rc = r + ρ N

  1 = xi + e x j + (1 + e2x ) −i + x j e = (x − 1 − e2x )i + (2e x + e−x )j.

This is the equation of the evolute.

16. The curve with polar equation r = f (θ ) is given para-

Solving these equations for i1 , j1 , and k1 in terms of the given basis vectors, we obtain

metrically by

i1 = −j 1 j1 = √ i − 2 1 k1 = √ i + 2

r = f (θ ) cos θ i + f (θ ) sin θ j. Thus we have   v = f ′ (θ ) cos θ − f (θ ) sin θ i   + f ′ (θ ) sin θ + f (θ ) cos θ j   a = f ′′ (θ ) cos θ − 2 f ′ (θ ) sin θ − f (θ ) cos θ i   + f ′′ (θ ) sin θ + 2 f ′ (θ ) cos θ − f (θ ) sin θ j r  2  2 f ′ (θ ) + f (θ ) v = |v| = h  2  2 i v × a = 2 f ′ (θ ) + f (θ ) − f (θ ) f ′′ (θ ) k. The curvature is, therefore,  2  2 |2 f ′ (θ ) + f (θ ) − f (θ ) f ′′ (θ )| . h 2  2 i3/2 f ′ (θ ) + f (θ )

17. If r = a(1 − cos θ ), then r ′ = a sin θ , and r ′′ = a cos θ . By the result of Exercise 20, the curvature of this cardioid is 1 2 2 κ=  3/2 × 2a sin θ a 2 sin2 θ + a 2 (1 − cos θ )2 + a 2 (1 − cos θ )2 − a 2 (cos θ − cos2 θ ) 3a 2 (1 − cos θ ) 3 =  . 3/2 = √ 2 2ar 2a 2 (1 − cos θ )

18. By Exercise 8 of Section 2.4, the required curve must be a circular helix with parameters a = 1/2 (radius), and b = 1/2. Its equation will be r=

1 1 1 cos ti1 + sin tj1 + tk1 + r0 2 2 2

1 √ k 2 1 √ k. 2

Therefore r(t) =

t + sin t cos t t − sin t j+ √ i− √ k + r0 . 2 2 2 2 2

1 We also require that r(0) = i, so r0 = i + j. The 2 required equation is, therefore, r(t) =

19. Given that



 t − sin t t + sin t 1 − cos t j+ √ +1 i+ √ k. 2 2 2 2 2

dr = c × r(t), we have dt

d 2 d |r| = r • r = 2r • (c × r) = 0 dt dt  d dr r(t) − r(0) • c = • c = (c × r) • c = 0. dt dt   Thus |r(t)| = |r(0)| is constant, and r(t) − r(0) • c = 0 is constant. Thus r(t) lies on the sphere centred at the origin with radius |r(0)|, and also on the plane through r(0) with normal c. The curve is the circle of intersection of this sphere and this plane.

20. For r = a cos ti + a sin tj + btk, we have, by Example 3 of Section 2.4,

ˆ = − cos ti − sin tj, N

434 Copyright © 2014 Pearson Canada Inc.

κ=

a . a 2 + b2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.5 (PAGE 657)

The centre of curvature rc is given by ˆ =r+ rc = r + ρ N

Therefore the evolute has equation

1ˆ N. κ

r = 2 cos ti + sin tj − =

Thus the evolute has equation

3 cos3 i − 3 sin3 tj. 2

23. We require that

r = a cos ti + a sin tj + btk

a 2 + b2 − (cos ti + sin tj) a b2 b2 = − cos ti − sin tj + btk. a a

f ′′ (1) = 0, f (−1) = 0.

′

′′

As in Example 5, we try a polynomial of degree 5. However, here it is clear that an odd function will do, and we need only impose the conditions at x = 1. Thus we try

21. The parabola y = x 2 has curvature

f (x) = Ax + Bx 3 + C x 5

f ′ (x) = A + 3Bx 2 + 5C x 4

2 , (1 + 4x 2 )3/2

f ′′ (x) = 6Bx + 20C x 3 .

by Exercise 18. The normal at (x, x 2 ) is perpendicular to the tangent, so has slope −1/(2x). Since the unit normal points upward (the concave side of the parabola), we have −2xi + j ˆ =√ N . 1 + 4x 2 Thus the evolute of the parabola has equation (1 + 4x 2 )3/2 r = xi + x 2 j + 2

f ′ (1) = 0, f (−1) = 0,

f (1) = 1, f (−1) = −1,

The evolute is also a circular helix.

κ=

3 sin2 t + 1 (cos ti + 2 sin tj) 2



−2xi + j √ 1 + 4x 2 1 + 4x 2 j = xi + x 2 j − (1 + 4x 2 )xi + 2   1 j. = −4x 3 i + 3x 2 + 2



The conditions at x = 1 become A A

+ +

B 3B 6B

+ + +

C 5C 20C

1 0 0.

= = =

This system has solution A = 15/8, B = −5/4, and C = 3/8. Thus 15 5 3 x − x3 + x5 8 4 8

f (x) = is one possible solution.

y

y=1

(1,1)

y= f (x)

22. For the ellipse r = 2 cos ti + sin tj, we have v = −2 sin ti + cos tj a = −2 cos ti − sin tj v × a = 2k p p v = 4 sin2 t + cos2 t = 3 sin2 t + 1.

2 The curvature is κ = , so the radius of 2 (3 sin t + 1)3/2 2 3/2 (3 sin t + 1) curvature is ρ = . We have 2 −2 sin ti + cos tj √ , 3 sin2 t + 1 ti + 2 sin tj ˆ = − cos . N √ 3 sin2 t + 1

Tˆ =

Bˆ = k

x y=−1 (−1,−1)

Fig. 11.5.23

24. We require f (0) = 1, f (−1) = 1,

f ′ (0) = 0, f (−1) = 0, ′

f ′′ (0) = −1, f (−1) = 0. ′′

The condition f ′′ (0) = −1 follows from the fact that d2 p 2 1 − x = −1. 2 dx x=0 435

Copyright © 2014 Pearson Canada Inc.

SECTION 11.5 (PAGE 657)

ADAMS and ESSEX: CALCULUS 8

26. After loading the LinearAlgebra and VectorCalculus pack-

As in Example 5, we try

ages, issue the following commands: f (x) = A + Bx + C x 2 + Dx 3 + E x 4 + F x 5

f ′ (x) = B + 2C x + 3Dx 2 + 4E x 3 + 5F x 4 f ′′ = 2C + 6Dx + 12E x 2 + 20F x 3 .

The required conditions force the coefficients to satisfy the system of equations A− B+C − D+E −F =1 B − 2C + 3D − 4E + 5F = 0 2C − 6D + 12E − 20F = 0 A=1 B=0 2C = −1 which has solution A = 1, B = 0, C = −1/2, D = −3/2, E = −3/2, F = −1/2. Thus we can use a track section in the shape of the graph of

>

R := t -> ; > assume(t::real): > interface(showassumed=0): > V := t -> diff(R(t),t): > A := t -> diff(V(t),t): > v := t -> Norm(V(t),2): > VxA := t -> V(t) &x A(t): > vxa := t -> Norm(VxA(t),2): > Ap := t -> diff(A(t),t): > Curv := t -> > simplify(vxa(t)/(v(t))ˆ3): > Tors := t -> simplify( > (VxA(t).Ap(t))/(vxa(t))ˆ2): > Curv(t); Tors(t); This leads to the values √

1 3 3 1 1 f (x) = 1 − x 2 − x 3 − x 4 − x 5 = 1 − x 2 (1 + x)3 . 2 2 2 2 2 y (−1,1) y=1

y= f (x)

x

(cos(t)2

2 p + 1) 2 cos(t)2 + 2

and 0

for the curvature and torsion, respectively. Maple doesn’t seem to recognize that the curvature simplifies to 1/(cos2 t + 1)3/2 . The torsion is zero because the curve is lies in the plane z = x. It is the ellipse in which this plane intersects the ellipsoid 2x 2 + y 2 + 2z 2 = 4. The maximum and minimum values of the curvature are 1 and 1/23/2 , respectively, at the ends of the major and minor axes of the ellipse.

27. After loading the LinearAlgebra and VectorCalculus packx 2 +y 2 =1

Fig. 11.5.24

25. Given: a(t) = λ(t)r(t) + µ(t)v(t), v × a 6= 0. We have v × a = λv × r + µv × v = λv × r da = λ′ r + λv + µ′ v + µa dt = λ′ r + (λ + µ′ )v + µ(λr + µv) = (λ′ + µλ)r + (λ + µ′ + µ2 )v.

Since v × r is perpendicular to both v and r, we have (v × a) •

da = 0. dt

Thus the torsion τ (t) of the curve is identically zero. It remains zero when expressed in terms of arc length: τ (s) = 0. By Exercise 6 of Section 2.4, r(t) must be a plane curve.

ages, issue the following commands: > > > > > > > > > > > > > >

R := t -> ; assume(t::real): interface(showassumed=0): V := t -> diff(R(t),t): A := t -> diff(V(t),t): v := t -> Norm(V(t),2): VxA := t -> V(t) &x A(t): vxa := t -> Norm(VxA(t),2): Ap := t -> diff(A(t),t): Curv := t -> simplify(vxa(t)/(v(t))ˆ3): Tors := t -> simplify( (VxA(t).Ap(t))/(vxa(t))ˆ2): Curv(t); Tors(t);

This leads to the values

436 Copyright © 2014 Pearson Canada Inc.

p

cos(t)2 + 2 − 2 cos(t) (3 − 2 cos(t))3/2

INSTRUCTOR’S SOLUTIONS MANUAL

and

SECTION 11.6 (PAGE 666)

1 − 2 cos(t)2 + sin(t)2 − 2 cos(t) + 1

29. After loading the LinearAlgebra and VectorCalculus packages, issue the following commands: >

R := t -> ; > assume(t::real): > interface(showassumed=0): > V := t -> diff(R(t),t): > A := t -> diff(V(t),t): > v := t -> Norm(V(t),2): > VxA := t -> V(t) &x A(t): > vxa := t -> Norm(VxA(t),2): > Ap := t -> diff(A(t),t): > Curv := t -> > simplify(vxa(t)/(v(t))ˆ3): > Tors := t -> simplify( > (VxA(t).Ap(t))/(vxa(t))ˆ2): > Curv(t); Tors(t);

for the curvature and torsion, respectively. Each of these formulas can be simplified somewhat: √ 2 − 2 cos t + cos2 t Curv(t) = (3 − 2 cos t)3/2 −1 Tors(t) = . 2 − 2 cos t + cos2 t

Since 3 − 2 cos t > 0 and 2 − 2 cos t + cos2 t = 1 + (1 − cos t)2 > 0 for all t, the curvature and torsion are both continuous for all t. The curve appears to be some sort of helix (but not a circular one) with central axis along the line x = z, y = 1.

28. After loading the LinearAlgebra and VectorCalculus packages, issue the following commands:

This leads to the values

>

R := t -> ; > assume(t::real): > interface(showassumed=0): > V := t -> diff(R(t),t): > A := t -> diff(V(t),t): > v := t -> Norm(V(t),2): > VxA := t -> V(t) &x A(t): > vxa := t -> Norm(VxA(t),2): > Ap := t -> diff(A(t),t): > Curv := t -> > simplify(vxa(t)/(v(t))ˆ3): > Tors := t -> simplify( > (VxA(t).Ap(t))/(vxa(t))ˆ2): > Curv(t); Tors(t); > simplify(%,trig); The last line simplifies the rather complicated expression that Tors(t) returns by applying some trigonometric identities. The values for the curvature and torsion are p 17 + 60 cos(t)2 + 48 cos(t)4 Curv(t) =
3/2 4 cos(t)2 + 1 Tors(t) =

12 cos t (2 cos(t)2

+ 3) . 17 + 60 cos(t)2 + 48 cos(t)4

Plotting the curvature as a function of t, (plot(Curv(t),t=-2*Pi..2*Pi)), shows that the minimum curvature occurs at t = 0 (and any integer multiple of π ). The minimum curvature is √ 125/53/2 = 1. The command simplify(Norm(R(t),2)); gives output 1, indicating that the curve lies on the sphere x 2 + y 2 + z 2 = 1.

p 2 cos(t)2 + cos t + 1 Curv(t) =
3/2 5 − cos(t)2 + 2 cos t 1 Tors(t) = 2(cos(t)2 ) + cos t + 1 This appears to be an elliptical helix with central axis along the line x = y = z − 1.

30. evolute := R -> (t -> R(t)+TNBFrame(R)[2](t) *(1/Curvature(R)(t))); 31. tanline := R -> ((t,u) -> R(t)+TNBFrame(R)[1](t)*u); Section 11.6 Kepler’s Laws of Planetary Motion (page 666)

1.

ℓ 1 + ǫ cos θ r = ℓ − ǫx

r=

H⇒

r + ǫx = ℓ

x 2 + y 2 = r 2 = ℓ2 − 2ℓǫx + ǫ 2 x 2

(1 − ǫ 2 )x 2 + 2ℓǫx + y 2 = ℓ2  2 ℓǫ ℓ2 ǫ 2 ℓ2 2 2 2 (1 − ǫ ) x + + y = ℓ + = 1 − ǫ2 1 − ǫ2 1 − ǫ2  2 ℓǫ x+ y2 1 − ǫ2  2 = 1. 2 +  ℓ ℓ √ 1 − ǫ2 1 − ǫ2

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SECTION 11.6 (PAGE 666)

ADAMS and ESSEX: CALCULUS 8

2. Position: r = r rˆ = kˆr.

ˆ speed: v = k θ˙ . Velocity: v = k r˙ˆ = k θ˙ θ; ˙ ˆ Acceleration: k θ¨ θˆ + k θ˙ θˆ = −k θ˙ 2 rˆ + k θ¨ θ. Radial component of acceleration: −k θ˙ 2 . Transverse component of acceleration: k θ¨ = v˙ (the rate of change of the speed).

Halley’s comet be TH = 76 years and a H km respectively. Similar parameters for the earth’s orbit are TE = 1 year and a E = 150 × 106 km. By Kepler’s third law TH2

a 3H

=

TE2 a 3E

.

Thus

3. Position: on the curve r = eθ . ˙ eθ θ.

Radial velocity: r˙ = ˙ Transverse √ velocity: r θ˙ = eθ θ. √ θ ˙ Speed v = 2e√θ = 1 H⇒ θ˙ = (1/ 2)e−θ . −θ −2θ Thus θ¨ = −(1/ 2)e θ˙ = −e /2. √ Radial velocity = transverse velocity = 1/ 2. Radial acceleration: r¨ − r θ˙ 2 = eθ θ˙ 2 + eθ θ¨ − eθ θ˙ 2 = eθ θ¨ = −e−θ /2. Transverse acceleration: r θ¨ + 2˙r θ˙ = −(e−θ )/2 + e−θ = e−θ /2.

¨ 4. Path: r = θ .pThus r˙ = θ˙ , r¨ = θ. √

˙ 2 = θ˙ 1 + r 2 . Speed: v = (˙r )2 + (r θ) Transverse acceleration = 0 (central force). Thus r θ¨ + 2˙r θ˙ = 0, or θ¨ = −2θ˙ 2 /r . Radial acceleration: r¨ − r θ˙ 2 = θ¨ − r θ˙ 2   2 (2 + r 2 )v 2 =− + r θ˙ 2 = − . r r (1 + r 2 ) The magnitude of the acceleration is, therefore, (2 + r 2 )v 2 . r (1 + r 2 )

5. For a central force, r 2 θ˙ = h (constant), and the acceleration is wholly radial, so

a H = 150 × 106 × 762/3 ≈ 2.69 × 109 . The major axis of Halley’s comet’s orbit is 2a H ≈ 5.38 × 109 km.

7. The period and semi-major axis of the moon’s orbit around the earth are TM ≈ 27 days,

a M ≈ 385, 000 km.

The satellite has a circular orbit of radius a S and period TS = 1 day. (If the orbit is in the plane of the equator, the satellite will remain above the same point on the earth.) By Kepler’s third law, TS2 a 3S

=

TM2 a 3M

.

Thus a S = 385, 000 × (1/27)2/3 ≈ 42, 788. The satellite’s orbit should have radius about 42,788 km, and should lie in the equatorial plane.

8. The period T (in years) and radius R (in km) of the asteroid’s orbit satisfies 2 Tearth T2 12 = = . 3 R3 (150 × 106 )3 Rearth Thus the radius of the asteroid’s orbit is R ≈ 150 × 106 T 2/3 km.

9. If R is the radius and T is the period of the asteroid’s circular orbit, then almost stopping the asteroid causes it to drop into a very eccentric elliptical orbit with major axis approximately R. (Thus, a = R/2.) The period Te of the new elliptical orbit satisfies

|a| = |¨r − r θ˙ 2 |. For r = θ −2 , we have r˙ = −2θ −3 θ˙ = −2θ −3

6. Let the period and the semi-major axis of the orbit of

h = −2hθ. r2

(R/2)3 1 Te2 = = . 2 T R3 8

√ Thus Te = T/(2 2). The time the asteroid will √ take to fall into the sun is half of Te . Thus it is T/(4 2).

Thus r¨ = −2h θ˙ = −2h 2 /r 2 . The speed v is given by v 2 = r˙ 2 + r 2 θ˙ 2 = 4h 2 θ 2 + (h 2 /r 2 ). R

Since the speed is v 0 when θ√= 1 (and so r = 1), we have v 02 = 5h 2 , and h = v 0 / 5. Hence the magnitude of the acceleration at any point on the path is   h2 h2 v2 2 1 |a| = −2 2 − r 4 = 0 + . r r 5 r2 r3

438 Copyright © 2014 Pearson Canada Inc.

Fig. 11.6.9

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 11.6 (PAGE 666)

10. At perihelion, r = a − c = (1 − ǫ)a.

At aphelion r = a + c = (1 + ǫ)a. Since r˙ = 0 at perihelion and aphelion, the speed is v = r θ˙ at each point. Since r 2 θ˙ = h is constant over the orbit, v = h/r . Therefore v perihelion =

h , a(1 − ǫ)

v aphelion =

It follows that 1 = 4 − 4ǫ, so ǫ = 3/4. The new elliptical orbit has eccentricity ǫ = 3/4.

h . a(1 + ǫ)

R c

If v perihelion = 2v aphelion then

a

S

2h h = . a(1 − ǫ) a(1 + ǫ) Hence 1 + ǫ = 2(1 − ǫ), and ǫ = 1/3. The eccentricity of the orbit is 1/3.

11. The orbital speed v of a planet satisfies (by conservation

Fig. 11.6.13

of energy) v2 k − =K 2 r

(total energy).

If v is constant so must be r , and the orbit will therefore be circular.

12. Since r 2 θ˙ = h = constant for the planet’s orbit, and since the speed is v = r θ˙ at perihelion and at aphelion (the radial velocity is zero at these points), we have r p v p = ra v a , where the subscripts p and a refer to perihelion and aphelion, respectively. Since r p /ra = 8/10, we must have v p /v a = 10/8 = 1.25. Also, rp =

ℓ ℓ = , 1 + ǫ cos 0 1+ǫ

ra =

ℓ ℓ = . 1 + ǫ cos π 1−ǫ

14. As in Exercise 12, r P v P = r A v A , where r A = ℓ/(1 − ǫ)

and r P = ℓ/(1 + ǫ), ǫ being the eccentricity of the orbit. Thus vP rA 1+ǫ . = = vA rP 1−ǫ Solving this equation for ǫ in terms of v P and v A , we get vP − vA . ǫ= vP + vA By conservation of energy the speed v at the ends of the minor axis of the orbit (where r = a) satisfies v2 v2 v2 k k k − = P − = A − . 2 a 2 rP 2 rA The latter equality shows that

Thus ℓ/(1+ǫ) = (8/10)ℓ/(1−ǫ), and so 10−10ǫ = 8+8ǫ. Hence 2 = 18ǫ. The eccentricity of the orbit is ǫ = 1/9.

v 2P − v 2A = 2k

13. Let the radius of the circular orbit be R, and let the parameters of the new elliptical orbit be a and c, as shown in the figure. Then R = a + c. At the moment of the collision, r does not change (r = R), but the speed r θ˙ is cut in half. Therefore θ˙ is cut in half, and so h = r 2 θ˙ is cut in half. Let H be the value of r 2 θ˙ for the circular orbit, and let h be the value for the new elliptical orbit. Thus h = H/2. We have R=

H2 , k

a=

Similarly, c = ǫa =

v 2 = v 2P + 2k



=

4kǫ . ℓ



1 1 − a rP



 2k  1 − ǫ 2 − (1 + ǫ) ℓ 2kǫ 2 = vP − (1 + ǫ) ℓ   2 v − v 2A vP − vA = v 2P − P 1+ 2 vP + vA vP − vA 2 (2v P ) = v P v A . = vP − 2 = v 2P +

ǫR , so 4(1 − ǫ 2 ) (1 + ǫ)R R = . 4(1 − ǫ 2 ) 4(1 − ǫ)

1 1 − rP rA

Using this result and the parameters of the orbit given in the text, we obtain

h2 H2 R = = . k(1 − ǫ 2 ) 4k(1 − ǫ 2 ) 4(1 − ǫ 2 )

R =c+a =



Thus v =

√ v P v A.

439 Copyright © 2014 Pearson Canada Inc.

SECTION 11.6 (PAGE 666)

ADAMS and ESSEX: CALCULUS 8

15. Since the radial line from the sun to the planet sweeps out equal areas in equal times, the fraction of the planet’s period spend on the same side of the minor axis as the sun is equal to the shaded area in the figure to the total area of the ellipse, that is,

Thus a =

k . By Kepler’s third law, −2K 4π 2 3 4π 2 a = k k

T2 =

1 2 π ab

1 − 12 (2bc) π ab − ǫab 1 ǫ = 2 = − , π ab π ab 2 π where ǫ = c/a is the eccentricity of the orbit.

2π Thus T = √ k

v2 2 − 0 r0 k

!−3/2



k −2K

3

.

. y

a

b

a

b

A

a

c

S

P

c

x

Fig. 11.6.16 Fig. 11.6.15

16. By conservation of energy, we have

17. Let r1 (s) and r2 (s) be the distances from the point P = r(s) on the ellipse E to the two foci. (Here s denotes arc length on E, measured from any convenient point.) By symmetry

  1 2 h2 k − r˙ + 2 = −K r 2 r where K is a constant for the orbit (the total energy). The term in the parentheses is v 2 , the square of the speed. Thus 1 k 1 k − v 2 = −K = − v 02 , r 2 r0 2 where r0 and v 0 are the given distance and speed. We evaluate −K at perihelion. The parameters of the orbit are ℓ=

h2 , k

a=

h2 , k(1 − ǫ 2 )

At perihelion P we have

h2 b= √ , k 1 − ǫ2

r = a − c = (1 − ǫ)a =

c = ǫa.

Z

E

r1 (s) ds =

Z

E

r2 (s) ds.

But r1 (s) + r2 (s) = 2a for any s. Therefore, Z

E

r1 (s) ds +

Z

E

r2 (s) ds =

Z

E

2a ds = 2ac(E).

R Hence E r1 (s) ds = ac(E), and

Z 1 r1 (s) ds = a. c(E) E

h2 . k(1 + ǫ)

˙ By Since r˙ = 0 at perihelion, the speed there is v = r θ. Kepler’s second law, r 2 θ˙ = h, so v = h/r = k(1 + ǫ)/ h. Thus k v2 −K = − r 2 k2 1 k2 = 2 (1 + ǫ) − (1 + ǫ)2 h 2 h2 h i k2 = 2 (1 + ǫ) 2 − (1 + ǫ) 2h k2 k = 2 (1 − ǫ 2 ) = . 2h 2a

440 Copyright © 2014 Pearson Canada Inc.

y P r2

r1

F2

F1

E

Fig. 11.6.17

x

INSTRUCTOR’S SOLUTIONS MANUAL

18. Start with r¨ −

h2 r3

=−

SECTION 11.6 (PAGE 666)

Note that r → ∞ as θ → θ0 +

k . r2

orbits in this case.

d2u − ω2 u = 0, where dθ 2 ω2 = (k − h 2 )/ h 2 . This has solution u = Aeωθ + Be−ωθ . Since u → 0 or ∞ as θ → ∞, the corresponding solution r = 1/u cannot be both bounded and bounded away from zero. (Note that θ˙ = h/r 2 ≥ K > 0 for any orbit which is bounded away from zero, so we can be sure θ → ∞ on such an orbit.) CASE II. If k > h 2 the DE is

1 Let r (t) = , where θ = θ (t). Since r 2 θ˙ = h u(θ ) (constant), we have 1 du ˙ du h du θ = −r 2 = −h u 2 dθ dθ r 2 dθ d 2u ˙ h2 d 2u d 2u r¨ = −h 2 θ = − 2 2 = −h 2 u 2 2 . dθ r dθ dθ r˙ = −

Thus −h 2 u 2

d 2u = 0, which has dθ 2 solutions u = Aθ + B, corresponding to CASE III. If k = h 2 the DE is

d 2u − h 2 u 3 = −ku 2 , or dθ 2

r=

d 2u k +u = 2. 2 dθ h

Thus, the only possible orbits which are bounded away from zero and infinity (i.e., which do not escape to infinity or plunge into the sun) in a universe with an inverse cube gravitational attraction are some circular orbits for which h 2 = k. Such orbits cannot be considered “stable” since even slight loss of energy would result in decreased h and the condition h 2 = k would no longer be satisfied. Now aren’t you glad you live in an inverse square universe?

 k  1 + ǫ cos(θ − θ ) 0 h2

is a solution for any choice of the constants ǫ and θ0 . Expressing the solution in terms of r , we have r=

h 2 /k , 1 + ǫ cos(θ − θ0 )

20. Since then

which is an ellipse if |ǫ| < 1.

19. For inverse cube attraction, the equation of motion is r¨ −

h2 r3

=−

k , r3

21.

where = h is constant, since the force is central. Making the same change of variables used in Exercise 18, we obtain

or

1 k = v 2 − K by conservation of energy, if K < 0, r 2 k ≥ −K > 0, r

k . The orbit is, therefore, bounded. K ℓ r= , (ǫ > 1). 1 + ǫ cos θ See the following figure. Vertices: At V1 , θ = 0 and r = ℓ/(1 + ǫ). At V2 , θ = π and r = ℓ/(1 − ǫ) = −ℓ/(ǫ − 1). Semi-focal separation:   ℓ ℓǫ 1 ℓ + = 2 . c= 2 1+ǫ 1−ǫ ǫ −1 so r ≤ −

r 2 θ˙

−h 2 u 2

1 . Aθ + B

Such orbits are bounded away from zero and infinity only if A = 0, in which case they are circular.

This is the DE for simple harmonic motion with a constant forcing term (nonhomogeneous term) on the righthand side. It is easily verified that u=

π . There are no bounded 2ω

d2u − h 2 u 3 = −ku 3 , dθ 2

d 2u k − h2 − u = 0. dθ 2 h2

The centre is (c, 0). Semi-transverse axis:

There are three cases to consider. d 2u + ω2 u = 0, where CASE I. If k < the DE is dθ 2 ω2 = (h 2 − k)/ h 2 . This has solution u = A cos ω(θ − θ0 ). Thus 1 r= . A cos ω(θ − θ0 )

a=

h2

ℓǫ ℓ ℓ − = 2 . ǫ2 − 1 ǫ + 1 ǫ −1

Semi-conjugate axis: b=

p ℓ c2 − a 2 = √ . 2 ǫ −1

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SECTION 11.6 (PAGE 666)

ADAMS and ESSEX: CALCULUS 8

Direction of asymptotes (see figure): θ = tan−1

for all points on the orbit. At perihelion, ℓ , ǫ+1 h h(ǫ + 1) . v = v p = r p θ˙ = = rp ℓ

b a 1 = cos−1 = cos−1 . a c ǫ

r = r p = c − a = (ǫ − 1)a =

y

Since h 2 = kℓ, we have 2 v∞ = v 2p − c θ

2k h2 (ǫ + 1)2 − (ǫ + 1) ℓ2 ℓ i kh 2 = (ǫ + 1) − 2(ǫ + 1) ℓ k k = (ǫ 2 − 1) = . ℓ a

b

=

F2

θ

V1

F1

a

C

x

V2

2k rp

2 = k. Thus av ∞

If D is the perpendicular distance from the sun S to an asymptote of the orbit (see the figure) then

Fig. 11.6.21

D = c sin θ = ǫa sin θ = a

22. By Exercise 17, the asymptotes make angle θ = cos−1 (1/ǫ) with the transverse axis, as shown in the figure. The angle of deviation δ satisfies 2θ + δ = π , so π δ θ = − , and 2 2

=a Therefore

sin θ cos θ

δ cos(δ/2) = a cot . sin(δ/2) 2

2 Dv ∞ v2 a δ δ = ∞ cot = cot . k k 2 2

δ sin θ = cos . 2

δ cos θ = sin , 2

Review Exercises 11 (page 668) y

1. Given that a • r = 0 and a • v = 0, we have d |r(t) − tv(t)|2 dt     = 2 r(t) − tv(t) • v(t) − v(t) − ta(t)   = 2 r(t) − tv(t) • a(t) = 0 − 0 = 0.

(c,0) D 2θ

θ S

rp

a

x δ

2. r = t cos ti + t sin tj + (2π − t)k, (0 ≤ t ≤ p2π ) is a conical helix wound around the cone z = 2π − x 2 + y 2 starting at the vertex (0, 0, 2π ), and completing one revolution to end up at (2π, 0, 0). Since v = (cos t − t sin t)i + (sin t + t cos t)j − k,

Fig. 11.6.22

the length of the curve is

By conservation of energy, v2

2 v∞

k − = constant = 2 r 2

L=

Z

0

2π

! √ p p 2π + 2 + 4π 2 2 2 2 + t dt = π 2 + 4π +ln √ 2

units.

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REVIEW EXERCISES 11 (PAGE 668)

3. The position of the particle at time t is

Hence the acceleration is

r = xi + x 2 j + 32 x 3 k, where x is an increasing function of t. Thie velocity is v=

If the particle is moving to the left, so that d x/dt < 0, a similar calculation shows that at t = 3 its acceleration is

 dx  i + 2xj + 2x 2 k . dt

√ √ 3+4 2 (i + 2 2j) + 2j. a=− 9

Since the speed is 6, we have 6=

dx dx p 1 + 4x 2 + 4x 4 = (2x 2 + 1) , dt dt

so that d x/dt = 6/(2x 2 + 1). The particle is at (1, 1, 23 ) when x = 1. At this time its velocity is

5.

d2x 6 dx 144x =− (4x) =− dt 2 (2x 2 + 1)2 dt (2x 2 + 1)3 d2x a = 2 (i + 2xj + 2x 2 k) dt   dx dx dx + 2 j + 4x k . dt dt dt At x = 1, we have

=

ticle are given by r = xi + x 2 j dx v= (i + 2xj), dt

p dx v = 1 + 4x 2 dt  2 2 d x dx a = 2 (i + 2xj) + 2 j. dt dt Let us assume that the particle is moving to the right, so that d x/dt > 0. Since the speed is t, we have dx t =√ dt 1 + 4x 2 √ 4t x dx 1 + 4x 2 − √ 2 d2x 1 + 4x dt . = 2 dt 1 + 4x 2 √ If the particle is at ( 2, 2) at t = 3, then d x/dt = 1 at that time, and √ d2x 3−4 2 = . dt 2 9

√

2tj + e−t k

t −t 6. Tangential acceleration: dv/dt √= e −e .

8 (−2i − j + 2k). 3

4. The position, velocity, speed, and acceleration of the par-

√

v = e i + 2j − e−t k a = et i + e−t k da = et i − e−t k dt √ √ v × a = 2e−t i − 2j − 2et k p v = e2t + 2 + e−2t = et + e−t √ |v × a| = 2(et + e−t ) √ |v × a| 2 κ= = t 3 v (e + e−t )2 da √ (v × a) • 2 dt = τ = = κ. |v × a|2 (et + e−t )2

Also

16 (i + 2j + 2k) + 2(4j + 8k) 3

r = et i + t

v(1) = 2(i + 2j + 2k).

a(1) = −

√ √ 3−4 2 (i + 2 2j) + 2j. 9

a=

7.

Normal acceleration: v 2 κ = 2. Since v = 2 cosh t, the minimum speed is 2 at time t = 0. Z s Z s kt 2 kt 2 sin cos dt, y(s) = dt, we have For x(s) = 2 2 0 0 dx ks 2 = cos , ds 2

dy ks 2 = sin , ds 2

so that the speed is unity:

v=

s 

dx ds

2

+



dy ds

2

= 1.

Since x(0) = y(0) = 0, the arc length along the curve, measured from the origin, is s. Also, ks 2 ks 2 i + sin j 2 2 ks 2 ks 2 a = −ks sin i + ks cos j 2 2 v × a = ksk. v = cos

Therefore the curvature at position s is κ = |v × a|/v 3 = ks.

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REVIEW EXERCISES 11 (PAGE 668)

ADAMS and ESSEX: CALCULUS 8

8. If r = e−θ , and θ˙ = k, then r˙ = −e−θ θ˙ = −kr , and r¨ = k 2 r . Since r = r rˆ , we have

12. Let P be the point with position vector r(t)

v = r˙ rˆ + r θ˙ θˆ = −kr rˆ + kr θˆ ˙ θˆ a = (¨r − r θ˙ 2 )ˆr + (r θ¨ + 2˙r θ)

ˆ = (k 2 r − k 2 r )ˆr + (0 − 2k 2 r )θˆ = −2k 2 r θ.

9.

r = a(t − sin t)i + a(1 − cos t)j v = a(1 − cos t)i + a sin tj p v = a 1 − 2 cos t + cos2 t + sin2 t √ √ t = a 2 1 − cos t = 2a sin if 0 ≤ t ≤ 2π . 2 The length of the cycloid from t = 0 to t = T ≤ 2π is s(T ) =

10.

Z

T 0

This is the same cycloid as given by r(t) but translated π a units to the right and 2a units downward.

  t T 2a sin dt = 4a 1 − cos units. 2 2

   s  t ⇒ t = 2cos−1 1 − s = 4a 1 − cos = t (s). 2 4a

The required arc length parametrization of the cycloid is     r = a t (s) − sin t (s) i + a 1 − cos t (s) j.

on the cycloid. By Exercise 9, the arc O P has length 4a − 4a cos(t/2), and so P Q has length 4a - arc O P = 4a cos(t/2) units. Thus t ˆ −→ P Q = 4a cos T(t) 2  t t t = 4a cos sin i + cos j 2 2 2 = 2a sin ti + 2a(1 + cos t)j. It follows that Q has position vector −→ rQ = r + P Q = a(t − sin t)i + a(1 − cos t)j + 2a sin ti + 2a(1 + cos t)j = a(t + sin t)i + a(1 + cos t + 2)j (let t = u + π ) = a(u − sin u + π )i + a(1 − cos u + 2)j. Thus r Q (t) represents the same cycloid as r(t), but translated π a units to the left and 2a units upward. From Exercise 11, the given cycloid is the evolute of its involute. y Q A

11. From Exercise 9 we have (1 − cos t)i + sin tj v = v 2 sin(t/2) t t = sin i + cos j 2 2 1 t 1 t cos i − sin j ˆ 1 dT d Tˆ 2 2 2 = = 2 t ds v dt 2a sin 2   1 t = cot i − j 4a 2 dT 1 ˆ κ(t) = = ds 4a sin(t/2)

P

Tˆ (t) =

O

x

Fig. R-11.12

13. The position vector of P is given by r = R sin φ cos θ i + R sin φ sin θ j + R cos φk. Mutually perpendicular unit vectors in the directions of increasing R, φ and θ can be found by differentiating r with respect to each of these coordinates and dividing the resulting vectors by their lengths. They are

1 d Tˆ ˆ rC (t) = r(t) + ρ(t)N(t) = r(t) + (κ(t))2 ds   2 2 16a sin (t/2) t = r(t) + cot i − j 4a 2 t t t = r(t) + 4a cos sin i − 4a sin2 j 2 2 2 = a(t − sin t)i + a(1 − cos t)j + 2a sin ti − 2a(1 − cos t)j = a(t + sin t)i − a(1 − cos t)j (let t = u − π ) = a(u − sin u − π )i + a(1 − cos u − 2)j.

dr = sin φ cos θ i + sin φ sin θ j + cos φk dR 1 dr = cos φ cos θ i + cos φ sin θ j − sin φk φˆ = R dφ 1 dr θˆ = = − sin θ i + cos θ j. R sin φ dθ ρˆ =

ˆ is right-handed. This is the reason for ˆ θ} The triad{ρ, ˆ φ, ordering the spherical polar coordinates (R, φ, θ ) rather than (R, θ, φ).

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CHALLENGING PROBLEMS 11 (PAGE 669)

Challenging Problems 11

14. By Kepler’s Second Law the position vector r from the origin (the sun) to the planet sweeps out area at a constant rate, say h/2: dA h = . dt 2 ˙ so r 2 θ˙ = h, and As observed in the text, d A/dt = r 2 θ/2, ˆ = r 2 θ˙ rˆ × θˆ = hk = h r × v = (r rˆ ) × (˙r rˆ + r θ˙ θ)

1.

(page 669)

a) The angular velocity Ω of the earth points northward in the direction of the earth’s axis; in terms of the basis vectors defined at a point P at 45◦ north latitude, it points in the direction of j + k: j+k Ω= √ , 2

is a constant vector.

=

2π rad/s. 24 × 3,600

15. By Exercise 14, r × r˙ = r × v = h is constant, so, by Newton’s second law of motion,

r × F(r) = mr × r¨ = m

b) If v = −vk, then d (r × r˙ ) = 0. dt

√ 2v aC = 2Ω × v = − √ (j + k) × k = − 2vi. 2

Thus F(r) is parallel to r, and therefore has zero transverse component: c) If r(t) = x(t)i + y(t)j + z(t)k is the position of the falling object at time t, then r(t) satisfies the DE

F(r) = − f (r)ˆr

d 2r dr = −gk + 2Ω × dt 2 dt

for some scalar function f (r).

16. By Exercise 15, F(r) = m(¨r − r θ˙ 2 )ˆr = − f (r)ˆr. We are given that r = ℓ/(1 + ǫ cos θ ). Thus

ℓ (−ǫ sin θ )θ˙ (1 + ǫ cos θ )2 ǫℓ sin θ = θ˙ (1 + ǫ cos θ )2 ǫ sin θ 2 ˙ hǫ r θ= sin θ = ℓ ℓ 2 hǫ h ǫ cos θ r¨ = (cos θ )θ˙ = . ℓ ℓr 2

and the initial conditions r(0) = 100k, r′ (0) = 0. If we use the approximation

r˙ = −

dz dr ≈ k, dt dt which is appropriate since  is much smaller than g, then dr √ dz 2Ω × ≈ 2 i. dt dt Breaking the DE into its components, we get

It follows that h2 h 2 ǫ cos θ r¨ − r θ˙ 2 = − ℓr 2 r3   ℓ h2 h2 = − 2, = 2 ǫ cos θ − ℓr r ℓr (because (ℓ/r ) = 1 + ǫ cos θ ). Hence f (r) =

√ dz d2x = 2 , 2 dt dt

This says that the magnitude of the force on the planet is inversely proportional to the square of its distance from the sun. Thus Newton’s law of gravitation follows from Kepler’s laws and the second law of motion.

d2z = −g. dt 2

Solving these equations (beginning with the last one), using the initial conditions, we get

z(t) = 100 −

mh 2 . ℓr 2

d2 y = 0, dt 2

gt 2 , 2

y(t) = 0,

gt 3 x(t) = − √ . 3 2

Since g ≈ 9.8 m/s2 , the time of fall is t=

s

200 ≈ 4.52, g

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CHALLENGING PROBLEMS 11 (PAGE 669)

ADAMS and ESSEX: CALCULUS 8

at which time we have

3. x ≈−

2.

2π 9.8 √ (4.52)3 ≈ −0.0155 m. 24 × 3,600 3 2

The object strikes the ground about 15.5 cm west of P.  dv  = k × v − 32k dt  v(0) = 70i

a) If v = v 1 i+v 2 j+v 3 k, then k×v = v 1 j−v 2 i. Thus the initial-value problem breaks down into component equations as   dv 1 = −v 2 dt  v 1 (0) = 70

  dv 2 = v 1 dt  v 2 (0) = 0

  dv = ωv × k, dt  v(0) = v0 a)

b) If w(t) = v(t) − (v0 • k)k, then w • k = 0 by part (a). Also, using the result of Exercise 23 of Section 1.3, we have d 2w d2v dv = =ω × k = ω2 (v × k) × k dt 2 dt 2 h dt i = −ω2 (k • k)v − (k • v)k h i = −ω2 v − (v0 • k)k = −ω2 w, the equation of simple harmonic motion. Also, w(0) = v0 − (v0 • k)k w′ (0) = ωv0 × k.

dz = v 3 = −32t ⇒ z = −16t 2 . dt d 2 v1 dv 2 = −v 1 (the equation of simple =− dt 2 dt harmonic motion), so Also,

v 1 (t) = A cos t + B sin t,

dx = v 1 = 70 cos t dt x(t) = 70 sin t

c) Solving the above initial-value problem for w, we get w = A cos(ωt) + B sin(ωt), A = w(0) = v0 − (v0 • k)k, ωB = w′ (0) = ω × k.

v 2 (t) = A sin t − B cos t.

Since v 1 (0) = 70, v 2 (0) = 0, x(0) = 0, and y(0) = 0, we have dy = v 2 = 70 sin t dt y(t) = 70(1 − cos t).

At time t seconds after it is thrown, the ball is at position r = 70 sin ti + 70(1 − cos t)j − 16t 2 k. c) At t = 1/5 s, the ball is at about (13.9, 1.40, −0.64). If it had been thrown without the vertical spin, its position at time t would have been

qB m

d dv (v • k) = • k = ω(v × k) • k = 0. dt dt Thus v • k = constant = v0 • k. d 2 dv Also, |v| = 2 • v = 2ω(v × k) • v = 0, dt dt so |v| = constant = |v0 | for all t.

  dv 3 = −32 dt  v 3 (0) = 0.

b) If r = xi+yj+zk denotes the position of the baseball t s after it is thrown, then x(0) = y(0) = z(0) = 0 and we have

ω=

Therefore, v(t) = w(t) + (v0 • k)k h i = v0 − (v0 • k)k cos(ωt) + (v0 × k) sin(ωt) + (v0 • k)k.

d) If dr/dt = v and r(0) = 0, then r(t) =

v0 − (v0 • k)k sin(ωt) ω   v0 × k + 1 − cos(ωt) + (v0 • k)tk. ω

Since the three constant vectors

2

r = 70ti − 16t k,

v0 − (v0 • k)k , ω

so its position at t = 1/5 s would have been (14, 0, −0.64). Thus the spin has deflected the ball approximately 1.4 ft to the left (as seen from above) of what would have been its parabolic path had it not been given the spin.

where and

v0 × k , and (v0 • k)k ω

are mutually perpendicular, and the first two have the same length because

446 Copyright © 2014 Pearson Canada Inc.

|v0 − (v0 • k)k| = |v0 | sin θ = |v0 × k|,

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 11 (PAGE 669)

where θ is the angle between v0 and k, the curve r(t) is generally a circular helix with axis in the z direction. However, it will be a circle if v0 • k = 0, that is, if v0 is horizontal, and it will be a straight line if v0 × k = 0, that is, if v0 is vertical.

5.

a) The curve BC D is the graph of an even function; a fourth degree polynomial with terms of even degree only will enable us to match the height, slope, and curvature at D, and therefore also at C. We have f (x) = ax 4 + bx 2 + c

4. The arc length element on x = a(θ − sin θ ),

f ′ (x) = 4ax 3 + 2bx

y = a(cos θ − 1) is (for θ ≤ π ) p ds = a (1 − cos θ )2 + sin2 θ dθ p = a 2(1 − cos θ ) dθ = 2a sin(θ/2) dθ.

f ′′ (x) = 12ax 2 + 2b. At D we have x = 2, so we need

If the bead slides downward from rest at height y(θ0 ) to height y(θ ), its gravitational potential energy has decreased by h i mg y(θ0 ) − y(θ ) = mga(cos θ0 − cos θ ).

Since there is no friction, all this potential energy is converted to kinetic energy, so its speed v at height y(θ ) is given by 1 2 mv = mga(cos θ0 − cos θ ), 2 √ and so v = 2ga(cos θ0 − cos θ ). The time required for the bead to travel distance ds at speed v is dt = ds/v, so the time T required for the bead to slide from its starting position at θ = θ0 to the lowest point on the wire, θ = π , is Z θ =π Z π ds 1 ds T = = dθ v θ =θ θ0 v dθ s 0Z 2a π sin(θ/2) = √ dθ g θ0 cos θ0 − cos θ s Z 2a π sin(θ/2) p = dθ 2 g θ0 2 cos (θ0 /2) − 2 cos2 (θ/2) Let u = cos(θ/2) du = − 21 sin(θ/2) dθ r Z cos(θ0 /2) a du p =2 2 g 0 cos (θ0 /2) − u 2   cos(θ0 /2) r a −1 u =2 sin g cos(θ0 /2) 0 √ = π ag which is independent of θ0 . y θ = θ0 starting point

2 = f (2) = 16a + 4b + c 1 = f ′ (2) = 32a + 4b 0 = f ′′ (2) = 48a + 2b. These equations yield a = −1/64, b = 3/8, c = 3/4, so the curved track BC D is the graph of 1 (−x 4 + 24x 2 + 48). y = f (x) = 64 b) Since we are ignoring friction, the speed √ v of the car during its drop is given by v = 2gs, where s is the vertical distance dropped. (See the previous solution.) At B the car√has dropped about 7.2 m, so its speed there is v ≈ 2(9.8)(7.2) √ ≈ 11.9 m/s. At C the car has dropped 10 − (c/ 2) ≈ 9.47 m, so its speed there is v = 13.6 m/s. At D the car has dropped 10 m, so its speed is v = 14.0 m/s. c) At C we have x = 0, f ′ (0) = 0, and f ′′ (0) = 2b = 3/4. Thus the curvature of the track at C is | f ′′ (0)| 3 κ= = . (1 + ( f ′ (0))2 )3/2 4 The normal about 14g). √ dv 2g = √ dt 2 s

acceleration√is v 2 κ ≈ 138.7 m/s2 (or Since v = 2gs, we have √ √ ds 2g 19.6 = √ v≈ √ (13.6) ≈ 9.78 m/s2 , dt 2 s 2 9.47

so the total acceleration has magnitude approximately p (138.7)2 + (9.78)2 ≈ 139 m/s2 , which is again about 14g. A y

E

vertical section horizontal section √ g = (g/ 2)(i − j)

x B

θ =π

D

(−2, 2)

(2, 2) C x Fig. C-11.5

Fig. C-11.4

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CHALLENGING PROBLEMS 11 (PAGE 669)

6.

ADAMS and ESSEX: CALCULUS 8

a) At time t, the hareis at P = (0, vt) and the fox is at Q = x(t), y(t) , where x and y are such that the slope d y/d x of the fox’s path is the slope of the line P Q:

c) If u = d y/d x, then u = 0 and y = 0 when x = a, and p du x = 1 + u2 Z dx Z du dx √ = Let u = tan θ 2 x 1+u du = sec2 θ dθ Z sec θ dθ = ln x + ln C

y − vt dy = . dx x

b) Since

ln(tan θ + sec θ ) = ln(C x) p u + 1 + u 2 = C x.

d dy d2 y d x = , we have dt d x d x 2 dt

Since u = 0 when x = a, we have C = 1/a.

  d x d2 y d y − vt = dt d x 2 dt x   dy dx x − v − (y − vt) dt dt = 2 x   1 dy dx 1 dx = − v − 2 (y − vt) x d x dt x dt dx v 1 dx 1 − − 2 (y − vt) = 2 (y − vt) x dt x x dt v =− . x

p x 1 + u2 = − u a x2 2xu 1 + u2 = 2 − + u2 a a 2xu x2 = 2 −1 a a x a dy =u= − dx 2a 2x x2 a y= − ln x + C1 . 4a 2

d2 y v . =− 2 dx d x/dt Since the fox’s speed is also v, we have Thus x



dx dt

2

+



dy dt

2

Since y = 0 when x a a C1 = − + ln a, so 4 2

= v2.

y=

s

(d y/dt)2 1+ = (d x/dt)2

s

1+



dy dx

2

,

7. and so the fox’s path y = y(x) satisfies the DE

d2 y x = dx2

s

1+



dy dx

2

x 2 − a2 a x − ln 4 2 a

is the path of the fox.

Also, the fox is always running to the left (towards the y-axis from points where x > 0), so d x/dt < 0. Hence

v   = dx − dt

= a, we have

a) Since you are always travelling northeast √ at speed v, you are always moving north at rate v/ 2. Therefore you will reach the north pole in finite time

.

448 Copyright © 2014 Pearson Canada Inc.

T =

π a/2 πa √ =√ . v/ 2 2v

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 11 (PAGE 669)

b) Since your velocity at any point has a northward √ component v/ 2, and progress northward is measured along a circle of radius a (a meridian), your colatitude φ(t) satisfies a

As θ = 0 at t = 0, we have C = 0, and so

dφ v = −√ . dt 2

  vt vt θ (t) = ln sec √ + tan √ . a 2 a 2

Since φ(0) = π/2, it follows that φ(t) =

vt π − √ . 2 a 2

Since √ your velocity also has an eastward component v/ 2 measured along a parallel of latitude that is a circle of radius a sin φ, your longitude coordinate θ satisfies dθ v (a sin φ) = √ dt 2   v dθ vt = √ cos √ dt a 2  a 2 Z vt v θ= √ sec √ dt a 2 a 2  vt vt + C. = ln sec √ + tan √ a 2 a 2

√ c) As t → T = π a/( 2v), the expression for θ (t) → ∞, so your path spirals around the north pole, crossing any meridian infinitely often.

449 Copyright © 2014 Pearson Canada Inc.

SECTION 12.1 (PAGE 677)

ADAMS and ESSEX: CALCULUS 8

z

CHAPTER 12. PARTIAL DIFFERENTIATION (2,0,2)

(2,3,2)

Section 12.1 Functions of Several Variables (page 677) 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

z=x

x+y f (x, y) = . x−y The domain consists of all points in the x y-plane not on the line x = y. √ f (x, y) = x y. Domain is the set of points (x, y) for which x y ≥ 0, that is, points on the coordinate axes and in the first and third quadrants. x . + y2 The domain is the set of all points in the x y-plane except the origin. f (x, y) =

Fig. 12.1.11

12.

x2

f (x, y) = sin x, 0 ≤ x ≤ 2π, 0 ≤ y ≤ 1 z

z = sin x

xy f (x, y) = 2 . x − y2 The domain consists of all points not on the lines x = ±y. p f (x, y) = 4x 2 + 9y 2 − 36. The domain consists of all points (x, y) lying on or outside the ellipse 4x 2 + 9y 2 = 36.

2π x 1

p f (x, y) = 1/ x 2 − y 2 . The domain consists of all points in the part of the plane where |x| > |y|. f (x, y) = ln(1 + x y). The domain consists of all points satisfying x y > −1, that is, points lying between the two branches of the hyperbola x y = −1.

y

Fig. 12.1.12

13.

z = f (x, y) = y 2 z

f (x, y) = sin−1 (x + y). The domain consists of all points in the strip −1 ≤ x + y ≤ 1.

z = y2

x yz . + y2 + z2 The domain consists of all points in 3-dimensional space except the origin. f (x, y, z) =

y

3 x

x2

y

e xyz f (x, y, z) = √ . x yz The domain consists of all points (x, y, z) where x yz > 0, that is, all points in the four octants x > 0, y > 0, z > 0; x > 0, y < 0, z < 0; x < 0, y > 0, z < 0; and x < 0, y < 0, z > 0. z = f (x, y) = x

x

Fig. 12.1.13

14.

f (x, y) = 4 − x 2 − y 2 , (x 2 + y 2 ≤ 4, x ≥ 0, y ≥ 0)

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.1 (PAGE 677)

z z 4

z =4−

x2

−

y2

y

2 x

y

2

x

Fig. 12.1.14

Fig. 12.1.17

18. 15.

z = f (x, y) =

p

x 2 + y2 z

f (x, y) = 6 − x − 2y z

z=

p

x 2 + y2

6

z = 6 − x − 2y y 3

y

6

x

x

Fig. 12.1.18 Fig. 12.1.15

19. 16.

f (x, y) = x − y = C, a family of straight lines of slope 1. y

f (x, y) = 4 − x 2

c=−3 z

c=−2

z =4−

x2

c=−1 c=0 x c=1 c=2 y

c=3

x

x−y=c Fig. 12.1.16

Fig. 12.1.19

20. 17.

z = f (x, y) = |x| + |y|

f (x, y) = x 2 + 2y 2 = C, a family of similar ellipses centred at the origin.

451 Copyright © 2014 Pearson Canada Inc.

SECTION 12.1 (PAGE 677)

y

ADAMS and ESSEX: CALCULUS 8

23.

x 2 + 2y 2 = c

x−y = C, a family of straight lines through x+y the origin, but not including the origin. f (x, y) =

c=16

c=9

y c=−1

c=4

c=−.5 c=1 c=0 x c=.5 c=1

x c=2

c=−2

Fig. 12.1.20

21.

x−y x+y

f (x, y) = x y = C, a family of rectangular hyperbolas with the coordinate axes as asymptotes.

=c

Fig. 12.1.23

y c=9 c=4 c=1

24. c=0 x c=−1 c=−4

y = C. x 2 + y2
1 2 = 4C1 2 of circles This is the family x 2 + y − 2C passing through the origin and having centres on the y-axis. The origin itself is, however, not on any of the level curves. f (x, y) =

y

c=−9

c=1

xy = c

c=2

Fig. 12.1.21

22.

c=3

x2 f (x, y) = = C, a family of parabolas, y = x 2 /C, y with vertices at the origin and vertical axes.

x

y c=0.5

c=1

c=−3

c=−2 c=2 c=−1

x2 y

Fig. 12.1.24

x

=c

c=−2

c=−0.5

Fig. 12.1.22

c=−1

25.

f (x, y) = xe−y = C. x This is the family of curves y = ln . C

452 Copyright © 2014 Pearson Canada Inc.

y x 2 +y 2

=c

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.1 (PAGE 677)

y

500 c=−1

500

600

c=1

A

C

c=2

c=−2

N

B W

x

E S

400

c=4

c=−4

300 200 100

xe−y = c

Fig. 12.1.28

Fig. 12.1.25

26.

f (x, y) =

r

f (x, y) =

r

1 1 − x2 = C ⇒ y = 2 . y x + C2 y

29. The graph of the function whose level curves are as 1 − x2 = C y

shown in part (a) of Figure 12.1.29 is a plane containing the y-axis and sloping uphill to the right. It is consistent with, say, a function of the form f (x, y) = y. (a) (b) y y

C = 0.8 C=1

C=10

C =2 x

x

x

Fig. 12.1.26

27. The landscape is steepest at B where the level curves are C=−5

closest together.

(c)

y

C=0

C=5 C=5

(d)

y

C=3

500 A

x

500

600 C

x

N

B W

E

C=0

S 400

C=−5

Fig. 12.1.29 300 200 100

Fig. 12.1.27

30. The graph of the function whose level curves are as

28. C is a “pass” between two peaks to the east and west. The land is level at C and rises as you move to the east or west, but falls as you move to the north or south.

shown in part (b) of Figure 12.1.29 is a cylinder parallel to the x-axis, rising from height zero first steeply and then more and more slowly as y increases. It is p consistent with, say, a function of the form f (x, y) = y + 5. 453

Copyright © 2014 Pearson Canada Inc.

SECTION 12.1 (PAGE 677)

(a)

ADAMS and ESSEX: CALCULUS 8

(b)

y

(a)

y

(b)

y

C=10

x

C=−5

(c)

y

x

C=0

C=5 C=5

(d)

y

C=10

x

C=−5

y

(c)

C=3

x

x

C=0

C=0

C=5 C=5

y

x

(d)

y

C=3

x

x

C=0

C=−5

Fig. 12.1.30

C=−5

Fig. 12.1.32

31. The graph of the function whose level curves are as

33. The curves y = (x − C)2 are all horizontally shifted

shown in part (c) of Figure 12.1.29 is an upside down circular cone with vertex at height 5 on the z-axis and base circle in the x y-plane. It is consistent with, say, a p function of the form f (x, y) = 5 − x 2 + y 2 . (a) (b) y y

versions of the parabola y = x 2 , and they all lie in the half-plane y ≥ 0. Since each of these curves intersects all of the others, they cannot be level curves of a function f (x, y) defined in y ≥ 0. To be a family of level curves of a function f (x, y) in a region, the various curves in the family cannot intersect one another in that region.

C=10

x

x

34. 4z 2 = (x − z)2 + (y − z)2 .

If z = c > 0, we have (x − c)2 + (y − c)2 = 4c2 , which is a circle in the plane z = c, with centre (c, c, c) and radius 2c. y

C=−5

(c)

y

C=0

C=5 C=5

(d)

c=3

y

C=3 c=2

x

x c=1

C=0

C=−5

Fig. 12.1.31 x

32. The graph of the function whose level curves are as shown in part (d) of Figure 12.1.29 is a cylinder (possibly parabolic) with axis in the yz-plane, sloping upwards in the direction of increasing y. It is consistent with, say, a function of the form f (x, y) = y − x 2 .

454 Copyright © 2014 Pearson Canada Inc.

(x − c)2 + (y − c)2 = 4c2

Fig. 12.1.34

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.1 (PAGE 677)

z The graph of the function z = z(x, y) ≥ 0 defined by the given equation is (the upper half of) an elliptic cone with axis along the line x = y = z, and circular cross-sections in horizontal planes.

35.

z=

1 1 + x 2 + y2

a) f (x, y) = p C is x 2 + y 2 = C 2 implies that f (x, y) = x 2 + y 2 .

b) f (x, y) = C is x 2 + y 2 = C 4 implies that f (x, y) = (x 2 + y 2 )1/4 .

y

c) f (x, y) = C is x 2 + y 2 = C implies that f (x, y) = x 2 + y 2 .

x

d) f (x, y) = C√is x 2 + y 2 = (ln C)2 implies that f (x, y) = e

Fig. 12.1.43

x 2 +y 2 .

36. If the level surface f (x, y, z) = C is the plane y z x + + = 1, C3 2C 3 3C 3 that is, x +

44.

z z=

y z + = C 3 , then 2 3

cos x 1 + y2

 y z 1/3 . f (x, y, z) = x + + 2 3

37.

38.

39.

40.

41.

42.

43.

y

f (x, y, z) = x 2 + y 2 + z 2 . The level surface f (x, y, z) = c > 0 is a sphere of radius √ c centred at the origin.

−5 ≤ x ≤ 5,

x

f (x, y, z) = x + 2y + 3z. The level surfaces are parallel planes having common normal vector i + 2j + 3k. f (x, y, z) = x 2 + y 2 . The level √ surface f (x, y, z) = c > 0 is a circular cylinder of radius c with axis along the z-axis.

−5 ≤ y ≤ 5

Fig. 12.1.44

45.

z

x 2 + y2 f (x, y, z) = . z2 The equation f (x, y, z) = c can be rewritten x 2 + y 2 = C 2 z 2 . The level surfaces are circular cones with vertices at the origin and axes along the z-axis.

z=

f (x, y, z) = |x| + |y| + |z|. The level surface f (x, y, z) = c > 0 is the surface of the octahedron with vertices (±c, 0, 0), (0, ±c, 0), and (0, 0, ±c). (An octahedron is a solid with eight planar faces.)

y 1 + x 2 + y2

y x

f (x, y, z, t) = x 2 + y 2 + z 2 + t 2 . The “level hypersurface” √ f (x, y, z, t) = c > 0 is the “4-sphere” of radius c centred at the origin√in R4 . That is, it consists of all points in R4 at distance c from the origin.

Fig. 12.1.45

46.

455 Copyright © 2014 Pearson Canada Inc.

SECTION 12.1 (PAGE 677)

ADAMS and ESSEX: CALCULUS 8

z z=

(x 2

x − 1)2 + y 2

3.

x 2 + y2 does not exist. y

lim

(x,y)→(0,0)

If (x, y) → (0, 0) along x = 0, then

y

4. x Fig. 12.1.46

5. 47.

If (x, y) → (0, 0) along y = x 2 + y2 = 1 + x 2 → 1. y x Let f (x, y) = 2 . x + y2 Then | f (x, 0)| = |1/x| → ∞ as x → 0. But | f (0, y)| = 0 → 0 as y → 0. Thus lim(x,y)→(0,0) f (x, y) does not exist.

cos(x y) cos π = = −1 1 − x − cos y 1 − 1 − cos π

lim

(x,y)→(1,π )

z

6. z = xy y

7.

x 2 (y − 1)2 = 0, because x 2 + (y − 1)2 2 x (y − 1)2 ≤ x2 0 ≤ 2 x + (y − 1)2

lim

(x,y)→(0,1)

and x 2 → 0 as (x, y) → (0, 1). y3 y2 x 2 + y 2 ≤ x 2 + y 2 |y| ≤ |y| → 0 as (x, y) → (0, 0). Thus

x

8.

Fig. 12.1.47

lim

(x,y)→(0,0) x 2

sin 0 sin(x − y) = = 0. cos(x + y) cos 0

lim

(x,y)→(0,0)

y3 = 0. + y2

sin(x y) . x 2 + y2 Now f (0, y) = 0/x 2 = 0 → 0 as x → 0. sin x 2 1 However, f (x, x) = → as x → 0. 2x 2 2 Therefore lim f (x, y) does not exist.

9. Let f (x, y) =

48. The graph is asymptotic to the coordinate planes. z

z=

x 2 + y2 = y → 0. y 2 x , then

1 xy

(x,y)→(0,0)

10. The fraction is not defined at points of the line y = 2x

and so cannot have a limit at (1, 2) by Definition 4. However, if we use the extended Definition 6, then, cancelling the common factor 2x − y, we get

y −4 ≤ x ≤ 4 −4 ≤ y ≤ 4

x

Section 12.2 Limits and Continuity (page 682)

2.

lim

(x,y)→(2,−1)

lim

(x,y)→(0,0)

x y + x 2 = 2(−1) + 22 = 2

q

x 2 + y2 = 0

(x,y)→(1,2)

11.

Fig. 12.1.48

1.

lim

1 2x 2 − x y x = . = lim (x,y)→(1,2) 2x + y 4x 2 − y 2 4

x 2 ≤ x 2 + y 4 . Thus lim

(x,y)→(0,0)

x 2 y2 ≤ y 2 → 0 as y → 0. Thus x 2 + y4

x 2 y2 = 0. x 2 + y4

x 2 y2 = 0. 2x 4 + y 4 x 2 y2 x4 1 If x = y 6= 0, then = 4 = . 4 4 4 2x + y 2x + x 3 x 2 y2 Therefore lim does not exist. (x,y)→(0,0) 2x 4 + y 4

12. If x = 0 and y 6= 0, then

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INSTRUCTOR’S SOLUTIONS MANUAL

13.

f (x, y) =

SECTION 12.2 (PAGE 682)

x 2 + y2 − x 3 y3 x 3 y3 =1− 2 . But 2 2 x +y x + y2

f (x, y) may not be continuous at (a, b) even if f u (t) is continuous at t = 0 for every unit vector u. A counterexample is the function f of Example 4 in this section. Here a = b = 0. The condition that each f u should be continuous is the condition that f should be continuous on each straight line through (0, 0), which it is if we extend the domain of f to include (0, 0) by defining f (0, 0) = 0. (We showed that f (x, y) → 0 as (x, y) → (0, 0) along every straight line.) However, we also showed that lim(x,y)→(0,0) f (x, y) did not exist.

3 3 x y x2 3 3 = x 2 + y 2 x 2 + y 2 |x y | ≤ |x y | → 0

as (x, y) → (0, 0). Thus

lim

(x,y)→(0,0)

f (x, y) = 1 − 0 = 1.

Define f (0, 0) = 1.

14. For x 6= y, we have f (x, y) =

x 3 − y3 = x 2 + x y + y 2. x−y

The latter expression has the value 3x 2 at points of the line x = y. Therefore, if we extend the definition of f (x, y) so that f (x, x) = 3x 2 , then the resulting function will be equal to x 2 + x y + y 2 everywhere, and so continuous everywhere.

15.

On the other hand, if f (x, y) is continuous at (a, b), then f (x, y) → f (a, b) if (x, y) approaches (a, b) in any way, in particular, along the line through (a, b) parallel to u. Thus all such functions f u (t) must be continuous at t = 0.

18. Since |x| ≤

x−y x−y = . x 2 − y2 (x − y)(x + y) Since f (x, y) = 1/(x + y) at all points off the line x = y and so is defined at some points in any neighbourhood of (1, 1), it approaches 1/(1 + 1) = 1/2 as (x, y) → (1, 1); If we define f (1, 1) = 1/2, then f becomes continuous at (1, 1). Similarly, f (x, y) can be defined to be 1/(2x) at any point on the line x = y except the origin, and becomes continuous at such points.

f (x, y) =

2x y x 2 + y2 0

lim

(x,y)→(0,0)

f (x, y) =

lim g(x) = lim f (x, y) = f (a, b). x →a

x→a

20.

k xy = . ax 2 + bx y + cy 2 a + bk + ck 2

sin x sin3 y cannot be defined at (0, 0) 1 − cos(x 2 + y 2 ) so as to become continuous there, because f (x, y) has no limit as (x, y) → (0, 0). To see this, observe that f (x, 0) = 0, so the limit must be 0 if it exists at all. However, f (x, y) =

y=b

f (x, x) =

Similarly, h(y) = f (a, y) is continuous at y = b.

17.

x m yn = 0. (x 2 + y 2 ) p

Thus f (x, y) has different constant values along different rays from the origin unless a = c = 0 and b 6= 0. If this condition is not satisfied, lim(x,y)→(0,0) f (x, y) does not exist. If the condition is satisfied, then lim(x,y)→(0,0) f (x, y) = 1/b does exist.

if (x, y) = (0, 0).

If f (x, y) is continuous at (a, b), then g(x) = f (x, b) is continuous at x = a because

x 2 + y 2 , we have

19. Suppose (x, y) → (0, 0) along the ray y = kx. Then

if (x, y) 6= (0, 0)

Let a = b = 0. If g(x) = f (x, 0) and h(y) = f (0, y), then g(x) = 0 for all x, and h(y) = 0 for all y, so g and h are continuous at 0. But, as shown in Example 3 of Section 3.2, f is not continuous at (0, 0).

p

The expression on the right → 0 as (x, y) → (0, 0), provided m + n > 2 p. In this case

16. Let f be the function of Example 3 of Section 3.2: (

x 2 + y 2 and |y| ≤

x m y n (x 2 + y 2 )(m+n)/2 = (x 2 + y 2 )− p+(m+n)/2 . (x 2 + y 2 ) p ≤ (x 2 + y 2 ) p

f (x, y) =

However there is no way to define f (x, −x) so that f becomes continuous on y = −x, since | f (x, y)| = 1/|x + y| → ∞ as y → −x.

p

f u (t) = f (a + tu, b + tv), where u = ui + vj is a unit vector.

sin4 x sin4 x = 1 − cos(2x 2 ) 2 sin2 (x 2 )

which approaches 1/2 as x → 0 by l’Hˆopital’s Rule or by using Maclaurin series.

457 Copyright © 2014 Pearson Canada Inc.

SECTION 12.2 (PAGE 682)

21.

ADAMS and ESSEX: CALCULUS 8

23. The graph of a function f (x, y) that is continuous on

z z=

region R in the x y-plane is a surface with no breaks or tears in it and that intersects each line parallel to the zaxis through a point (x, y) of R at exactly one point.

2x y + y2

x2

Section 12.3 (page 689)

y

x

1.

f (x, y) = x − y + 2, f 1 (x, y) = 1 = f 1 (3, 2), f 2 (x, y) = −1 = f 2 (3, 2).

2.

f (x, y) = x y + x 2 , f 1 (x, y) = y + 2x, f 2 (x, y) = x, f 1 (2, 0) = 4, f 2 (2, 0) = 2.

3.

f (x, y, z) = x 3 y 4 z 5 , f 1 (x, y, z) = 3x 2 y 4 z 5 , f 2 (x, y, z) = 4x 3 y 3 z 5 , f 3 (x, y, z) = 5x 3 y 4 z 4 ,

Fig. 12.2.21 The graphing software is unable to deal effectively with the discontinuity at (x, y) = (0, 0) so it leaves some gaps and rough edges near the z-axis. The surface lies between a ridge of height 1 along y = x and a ridge of height −1 along y = −x. It appears to be creased along the z-axis. The level curves are straight lines through the origin.

f 1 (0, −1, −1) = 0, f 2 (0, −1, −1) = 0, f 3 (0, −1, −1) = 0.

xz , y+z z 1 g1 (x, y, z) = , g1 (1, 1, 1) = , y+z 2 −x z 1 g2 (x, y, z) = , g (1, 1, 1) = − , 2 (y + z)2 4 xy 1 g3 (x, y, z) = , g3 (1, 1, 1) = . (y + z)2 4

4.

g(x, y, z) =

5.

z = tan−1

22. The graphing software is unable to deal effectively with the discontinuity at (x, y) = (0, 0) so it leaves some gaps and rough edges near the z-axis. The surface lies between a ridge along y = x 2 , z = 1, and a ridge along y = −x 2 , z = −1. It appears to be creased along the z-axis. The level curves are parabolas y = kx 2 through the origin. One of the families of rulings on the surface is the family of contours corresponding to level curves. z

Partial Derivatives

y

x  y  y − 2 =− 2 2 x x + y2 y 1+ 2 x   1 x ∂z 1 = = 2 2 ∂y x x + y2 y 1+ 2 x 1 1 ∂z ∂z = − , =− . ∂ x (−1,1) 2 ∂ y (−1,1) 2 ∂z = ∂x

1

∂w yze xyz = , ∂x 1 + e xyz x ze xyz x ye xyz ∂w ∂w = = , , ∂y 1 + e xyz ∂z 1 + e xyz ∂w ∂w = 0, = −1, At (2, 0, −1): ∂x ∂y

6. w = ln(1 + e xyz ),

x

2x 2 y z= 4 x + y2

y

7.

Fig. 12.2.22

∂w = 0. ∂z

√ f (x, y) = sin(x y), π  √ √ f 1 (x, y) = y cos(x y), f 1 , 4 = −1, 3  π  x π √ ,4 = − . f 2 (x, y) = √ cos(x y), f 2 2 y 3 24

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INSTRUCTOR’S SOLUTIONS MANUAL

8.

f (x, y) = p

1 x2

y2

SECTION 12.3 (PAGE 689)

14.

,

+ 1 2 x f 1 (x, y) = − (x + y 2 )−3/2 (2x) = − 2 , 2 (x + y 2 )3/2 y By symmetry, f 2 (x, y) = − 2 , (x + y 2 )3/2 3 4 f 1 (−3, 4) = , f 2 (−3, 4) = − . 125 125

9. w = x y ln z ,

∂w ∂w = y ln z x y ln z−1 , = 2e, ∂x ∂ x (e,2,e) ∂w ∂w = ln x ln z x y ln z , = e2 , ∂y ∂ y (e,2,e) ∂w y ∂w y ln z = ln x x , = 2e. ∂z z ∂z (e,2,e)

10. If g(x1 , x2 , x3 , x4 ) =

x1 − x22 x3 + x42

11.

12.

13.

f (x, y) = cos

x y

1 f (π, 4) = √ 2

x 1 1 f 1 (x, y) = − sin f 1 (π, 4) = − √ y y 4 2 x x π f 2 (x, y) = 2 sin f 2 (π, 4) = √ y y 16 2 The tangent plane at x = π , y = 4 is

, then

1 x3 + x42 −2x2 g2 (x1 , x2 , x3 , x4 ) = x3 + x42 x 2 − x1 g3 (x1 , x2 , x3 , x4 ) = 2 (x3 + x42 )2 (x 2 − x1 )2x4 g4 (x1 , x2 , x3 , x4 ) = 2 (x3 + x42 )2 g1 (x1 , x2 , x3 , x4 ) =

15.

x−y , f (1, 1) = 0, x+y (x + y) − (x − y) 1 f 1 (x, y) = , f 1 (1, 1) = (x + y)2 2 (x + y)(−1) − (x − y) 1 f 2 (x, y) = , f 2 (1, 1) = − . (x + y)2 2 Tangent plane to z = f (x, y) at (1,1) has equation x −1 y−1 z= − , or 2z = x − y. 2 2 Normal line: 2(x − 1) = −2(y − 1) = −z. f (x, y) =

1 3 2 g2 (3, 1, −1, −2) = − 3 2 g3 (3, 1, −1, −2) = − 9 8 g4 (3, 1, −1, −2) = . 9

  1 1 π z = √ 1 − (x − π ) + (y − 4) , 4 16 2

g1 (3, 1, −1, −2) =

  2x 3 − y 3 f (x, y) = x 2 + 3y 2 if (x, y) 6= (0, 0)  0 if (x, y) = (0, 0) 2h 3 − 0 =2 f 1 (0, 0) = lim h→0 h(h 2 + 0) 1 −k 3 − 0 f 2 (0, 0) = lim =− . k→0 k(0 + 3k 2 ) 3   x 2 − 2y 2 if x 6= y f (x, y) =  x−y 0 if x = y f (h, 0) − f (0, 0) h−0 f 1 (0, 0) = lim = lim = 1, h→0 h→0 h h f (0, k) − f (0, 0) 2k f 2 (0, 0) = lim = lim = 2. k→0 k→0 k k f (x, y) = x 2 − y 2 f (−2, 1) = 3 f 1 (x, y) = 2x f 1 (−2, 1) = −4 f 2 (x, y) = −2y f 2 (−2, 1) = −2 Tangent plane: z = 3 − 4(x + 2) − 2(y − 1), or 4x + 2y + z = −3. y−1 z−3 x +2 = = . Normal line: −4 −2 −1

√ or 4x − π y + 16 2z = 16. Normal line: √  √  √ 16 2 −4 2(x − π ) = (y − 4) = − z − (1/ 2) . π

16.

17.

f (x, y) = e xy , f 1 (x, y) = ye xy , f 2 (x, y) = xe xy , f (2, 0) = 1, f 1 (2, 0) = 0, f 2 (2, 0) = 2. Tangent plane to z = e xy at (2,0) has equation z = 1+2y. Normal line: x = 2, y = 2 − 2z. x x 2 + y2 (x 2 + y 2 )(1) − x(2x) y2 − x 2 f 1 (x, y) = = 2 2 2 2 (x + y ) (x + y 2 )2 2x y f 2 (x, y) = − 2 (x + y 2 )2 3 4 1 , f 2 (1, 2) = − . f (1, 2) = , f 1 (1, 2) = 5 25 25 The tangent plane at x = 1, y = 2 is f (x, y) =

z=

1 3 4 + (x − 1) − (y − 2), 5 25 25

or 3x − 4y − 25z = −10. y−2 5z − 1 x −1 = = . Normal line: 3 −4 −125

18.

2

2

2

f (x, y) = ye−x , f 1 = −2x ye−x , f 2 = e−x , f (0, 1) = 1, f 1 (0, 1) = 0, f 2 (0, 1) = 1. Tangent plane to z = f (x, y) at (0, 1) has equation z = 1 + 1(y − 1), or z = y. Normal line: x = 0, y + z = 2.

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SECTION 12.3 (PAGE 689)

19.

ADAMS and ESSEX: CALCULUS 8

f (x, y) = ln(x 2 + y 2 ) f (1, −2) = ln 5 2x 2 f 1 (x, y) = 2 f 1 (1, −2) = 2 x +y 5 4 2y f (1, −2) = − 2 f 2 (x, y) = 2 2 5 x +y The tangent plane at (1, −2, ln 5) is

23.

2 4 z = ln 5 + (x − 1) − (y + 2), 5 5 or 2x − 4y − 5z = 10 − 5 ln 5. x −1 y+2 z − ln 5 Normal line: = = . 2/5 −4/5 −1

20.

21.

24.

2x y , f (0, 2) = 0 x 2 + y2 (x 2 + y 2 )2y − 2x y(2x) 2y(y 2 − x 2 ) f 1 (x, y) = = 2 2 2 (x + y ) (x 2 + y 2 )2 2 2 2x(x − y ) f 2 (x, y) = (by symmetry) (x 2 + y 2 )2 f 1 (0, 2) = 1, f 2 (0, 2) = 0. Tangent plane at (0, 2): z = x. Normal line: z + x = 0, y = 2. f (x, y) =

f (x, y) = tan−1 f 1 (x, y) =

1

π , f (1, −1) = − , 4 y y − 2 =− 2 , x x + y2

y2 x2   1 x 1 = 2 f 2 (x, y) = , x + y2 y2 x 1+ 2 x 1 f 1 (1, −1) = f 2 (1, −1) = . The tangent plane is 2 π 1 1 π 1 z = − + (x − 1) + (y + 1), or z = − + (x + y). 4 2 2 4 2 π Normal line: 2(x − 1) = 2(y + 1) = −z − . 4

22.

f (x, y) =

1+

q

1 + x 3 y2 3x 2 y 2

f (2, 1) = 3 f 1 (2, 1) = 2 8 f 2 (2, 1) = 3

f 1 (x, y) = p 2 1 + x 3 y2 2x 3 y f 2 (x, y) = p 2 1 + x 3 y2 Tangent plane: z = 3 + 2(x − 2) + 83 (y − 1), or 6x + 8y − 3z = 11. x −2 y−1 z−3 Normal line: = = . 2 8/3 −1

2

2

z = x ye−(x +y )/2 ∂z 2 2 2 2 2 2 = ye−(x +y )/2 − x 2 ye−(x +y )/2 = y(1 − x 2 )e−(x +y )/2 ∂x ∂z 2 2 = x(1 − y 2 )e−(x +y )/2 (by symmetry) ∂y The tangent planes are horizontal at points where both of these first partials are zero, that is, points satisfying y(1 − x 2 ) = 0

and

x(1 − y 2 ) = 0.

These points are (0, 0), (1, 1), (−1, −1), (1, −1) and (−1, 1). At (0,0) the tangent plane is z = 0. At (1, 1) and (−1, −1) the tangent plane is z = 1/e. At (1, −1) and (−1, 1) the tangent plane is z = −1/e.

y x

z = x 4 − 4x y 3 + 6y 2 − 2 ∂z = 4x 3 − 4y 3 = 4(x − y)(x 2 + x y + y 2 ) ∂x ∂z = −12x y 2 + 12y = 12y(1 − x y). ∂y The tangent plane will be horizontal at points where both first partials are zero. Thus we require x = y and either y = 0 or x y = 1. If x = y and y = 0, then x = 0. If x = y and x y = 1, then x 2 = 1, so x = y = ±1. The tangent plane is horizontal at the points (0, 0), (1, 1), and (−1, −1).

∂z ∂z = e y and = xe y . ∂x ∂y ∂z ∂z Thus x = xe y = . ∂x ∂y

25. If z = xe y , then

26.

x+y , x−y ∂z (x − y)(1) − (x + y)(1) −2y = = , ∂x (x − y)2 (x − y)2 ∂z (x − y)(1) − (x + y)(−1) 2x = = . ∂y (x − y)2 (x − y)2 Therefore ∂z ∂z 2x y 2x y +y =− x + = 0. 2 ∂x ∂y (x − y) (x − y)2 z=

27. If z =

p

x 2 + y 2 , then

∂z x = p , and 2 ∂x x + y2

∂z y = p . Thus 2 ∂y x + y2

460 Copyright © 2014 Pearson Canada Inc.

x

∂z ∂z x 2 + y2 +y = p = z. ∂x ∂y x 2 + y2

INSTRUCTOR’S SOLUTIONS MANUAL

28. w = x 2 + yz,

∂w ∂w = 2x, == z, ∂x ∂y

SECTION 12.3 (PAGE 689)

∂w = y. ∂z

Thus X = Y = have

Therefore

−t =

∂w ∂w ∂w x +y +z ∂x ∂y ∂z = 2(x 2 + yz) = 2w.

∂w 2x 1 , then =− 2 , ∂x x 2 + y2 + z2 (x + y 2 + z 2 )2 2y ∂w 2z ∂w =− 2 , and =− 2 . ∂y (x + y 2 + z 2 )2 ∂z (x + y 2 + z 2 )2 Thus

29. If w =

30.

31.

32.

33.

∂w ∂w x 2 + y2 + z2 ∂w +y +z = −2 2 = −2w. ∂x ∂y ∂z (x + y 2 + z 2 )2 f (x 2

35. If Q = (X, Y, Z) is the point on the surface z = x 2 + 2y 2 that is closest to P = (0, 0, 1), then

y 2 ),

z= + ∂z ∂z ′ 2 = f (x + y 2 )(2x), = f ′ (x 2 + y 2 )(2y). ∂x ∂y ∂z ∂z Thus y −x = 2x y f ′ (x 2 + y 2) − 2x y f ′ (x 2 + y 2) = 0. ∂x ∂y

−→ P Q = Xi + Y j + (Z − 1)k must be normal to the surface at Q, and hence must be −→ parallel to n = 2Xi + 4Y j − k. Hence P Q = tn for some real number t, so

z = f (x 2 − y 2 ), ∂z ∂z = f ′ (x 2 − y 2 )(2x), = f ′ (x 2 − y 2 )(−2y). ∂x ∂y ∂z ∂z Thus y +x = (2x y − 2x y) f ′ (x 2 − y 2 ) = 0. ∂x ∂y

X = 2t X,

f (x + h, y, z) − f (x, y, z) h f (x, y + k, z) − f (x, y, z) f 2 (x, y, z) = lim k→0 k f (x, y, z + ℓ) − f (x, y, z) f 3 (x, y, z) = lim ℓ→0 ℓ   At a, b, c, f (a, b, c) the graph of w = f (x, y, z) has tangent hyperplane h→0

w = f (a, b, c) + f 1 (a, b, c)(x − a) + f 2 (a, b, c)(y − b) + f 3 (a, b, c)(z − c).

36.

must be normal to the surface at Q, and hence must be −→ parallel to n = 2Xi + 2Y j − k. Hence P Q = tn for some real number t, so X − 1 = 2t X,

Y − 1 = 2tY,

Z = −t.

Z − 1 = −t.

the √ closest point to (0, 0, 1) on z = x 2 + 2y 2 is (0, 3/8, 3/4), and the distance from (0, 0, 1) to that √ surface is 7/4 units.

34. If Q = (X, Y, Z) is the point on the surface z = x 2 + y 2 −→ P Q = (X − 1)i + (Y − 1)j + Zk

Y = 4tY,

If X √ 6= 0, then√t = 1/2, so Y = 0, Z =√ 1/2, and X = Z =√1/ 2. The distance from (1/ 2, 0, 1/2) to (0, 0, 1) is 3/2 units. If Y 6= t = 1/4, so X = 0, Z = 3/4,√and √ 0, then√ Y = Z/2 √ = 3/8. The distance from (0, 3/8, 3/4) to (0, 0, 1) is 7/4 units. If X = Y = 0, then Z = 0 (and t = 1). The distance from (0, 0, 0) to (0, 0, 1) is 1 unit. Since √ √ 7 3 < < 1, 4 2

f 1 (x, y, z) = lim

that is closest to P = (1, 1, 0), then

2 . (1 − 2t)2

1 Evidently this equation is satisfied by t = − . Since the 2 left and right sides of the equation have graphs similar to those in Figure 12.18(b) (in the text), the equation has 1 only this one real solution. Hence X = Y = , and so 2 1 Z= . 2 2 The distance from (1,
1, 0) to z√= x is the distance from 1 1 1 (1, 1, 0) to 2 , 2 , 2 , which is 3/2 units.

= 2x 2 + yz + yz

x

1 , and, since Z = X 2 + Y 2 , we must 1 − 2t

2x y if (x, y) 6= (0, 0), f (0, 0) = 0 x 2 + y2 0−0 f (h, 0) − f (0, 0) f 1 (0, 0) = lim = lim =0 h→0 h→0 h h f (0, k) − f (0, 0) 0−0 f 2 (0, 0) = lim = lim =0 k→0 k→0 h k Thus f 1 (0, 0) and f 2 (0, 0) both exist even though f is not continuous at (0, 0) (as shown in Example 2 of Section 3.2). f (x, y) =

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SECTION 12.3 (PAGE 689)

37.

f (x, y) =

(

(x 3 + y) sin 0

f 1 (0, 0) = lim

h→0

38.

1 h



h 3 sin

1 2 x + y2 1 h2



ADAMS and ESSEX: CALCULUS 8

if (x, y) 6= (0, 0)

40.

if (x, y) = (0, 0)

39.

0 = 0. h5

Also, f 2 (0, 0, 0) = lim

k→0

which has no limit as x → 0.) Thus f 1 (x, y) has no limit at (0, 0) and is not continuous there.   x 3 − y3 f (x, y) = x 2 + y 2 if (x, y) 6= (0, 0)  0 if (x, y) = (0, 0). If (x, y) 6= (0, 0), then (x 2 + y 2 )3x 2 − (x 3 − y 3 )2x (x 2 + y 2 )2 4 2 x + 3x y 2 + 2x y 3 = (x 2 + y 2)2 2 (x + y 2 )(−3y 2 ) − (x 3 − y 3 )2y f 2 (x, y) = (x 2 + y 2 )2 4 2 y + 3x y 2 + 2x 3 y =− . (x 2 + y 2 )2

f 1 (x, y, z) =

Section 12.4 (page 694)

x 3 y3 ≤ |x| + |y|, | f (x, y)| ≤ 2 + x + y2 x 2 + y2

Higher-Order Derivatives

1.

z = x 2 (1 + y 2 ) ∂z ∂z = 2x(1 + y 2 ), = 2x 2 y, ∂x ∂y ∂2z ∂2z = 2(1 + y 2 ), = 2x 2 , 2 ∂x ∂ y2 ∂2z ∂2z = 4x y = . ∂ y∂ x ∂ x∂ y

2.

f (x, y) = x 2 + y 2 , f 1 (x, y) = 2x, f 2 (x, y) = 2y, f 11 (x, y) = f 22 (x, y) = 2, f 12 (x, y) = f 21 (x, y) = 0.

3.

w = x 3 y3z3, ∂w ∂w = 3x 2 y 3 z 3 , = 3x 3 y 2 z 3 , ∂x ∂y ∂ 2w ∂ 2w = 6x y 3 z 3 , = 6x 3 yz 3 , 2 ∂x ∂ y2 ∂ 2w ∂2w = 9x 2 y 2 z 3 = , ∂ x∂ y ∂ y∂ x ∂ 2w ∂ 2w = 9x 2 y 3 z 2 = , ∂ x∂z ∂z∂ x ∂2w ∂ 2w = 9x 3 y 2 z 2 = . ∂ y∂z ∂z∂ y

Also, at (0, 0),

Neither f 1 nor f 2 has a limit at (0, 0) (the limits along x = 0 and y = 0 are different in each case), so neither function is continuous at (0, 0). However, f is continuous at (0, 0) because

(y 4 + z 4 − 3x 4 )y 2 z , (x 4 + y 4 + z 4 )2

which has no limit as (x, y, z) → (0, 0, 0) along the line x = y = z.

f 1 (x, y) =

−k 3 f 2 (0, 0) = lim = −1. k→0 k · k 2

0 = 0. k5

f is not continuous at (0, 0, 0); it has different limits as (x, y, z) → (0, 0, 0) along x = 0 and along x = y = z. None of f 1 , f 2 , and f 3 is continuous at (0, 0, 0) either. For example,

  2x 4 + 2x 2 1 1 1 1 cos = − 1 + cos 2 , 4x 4 2x 2 2 x2 2x

which → 0 as (x, y) → (0, 0).

if (x, y, z) 6= (0, 0, 0) if (x, y, z) = (0, 0, 0).

h→0

1 (x 3 + y)2x 1 − cos 2 . x 2 + y2 (x 2 + y 2 )2 x + y2

h3 f 1 (0, 0) = lim = 1, h→0 h · h 2

x y 2z + y4 + z4

f 3 (0, 0, 0) = f 1 (0, 0, 0) = lim

The first term on the right → 0 as (x, y) → (0, 0), but the second term has no limit at (0, 0). (It is 0 along x = 0, but along x = y it is −



x4

0 By symmetry we have

1 = lim h 2 sin 2 = 0 h→0 h   1 1 k sin 2 f 2 (0, 0) = lim k→0 k k 1 = lim sin 2 does not exist. k→0 k If (x, y) 6= (0, 0), then f 1 (x, y) = 3x 2 sin

f (x, y, z) =

 

462 Copyright © 2014 Pearson Canada Inc.

∂w = 3x 3 y 3 z 2 , ∂z ∂2w = 6x 3 y 3 z, ∂z 2

INSTRUCTOR’S SOLUTIONS MANUAL

4.

SECTION 12.4 (PAGE 694)

q z = 3x 2 + y 2 , ∂z ∂z 3x y , , =p =p 2 2 2 ∂x ∂ y 3x + y 3x + y 2 p 3x 3x 2 + y 2 (3) − 3x p 2 ∂ z 3y 2 3x 2 + y 2 = = , 2 2 2 2 ∂x 3x + y (3x + y 2 )3/2 p y 3x 2 + y 2 − y p ∂2z 3x 2 3x 2 + y 2 = = , 2 2 2 2 ∂y 3x + y (3x + y 2 )3/2 ∂2z 3x y ∂2z = =− . ∂ x∂ y ∂ y∂ x (3x 2 + y 2 )3/2

5.

z = xe y − ye x , ∂z ∂z = e y − ye x , = xe y − e x , ∂x ∂y ∂2z ∂2z = −ye x , = xe y , 2 ∂x ∂ y2 ∂2z ∂2z = e y − ex = . ∂ y∂ x ∂ x∂ y

6.

f (x, y) = ln(1 + sin(x y)) y cos(x y) x cos(x y) f 1 (x, y) = , f 2 (x, y) = 1 + sin(x y) 1 + sin(x y) f 11 (x, y)

9.

10.

11.

(1 + sin(x y))(−y 2 sin(x y)) − (y cos(x y))(y cos(x y)) (1 + sin(x y))2 2 y =− 1 + sin(x y) x2 f 22 (x, y) = − (by symmetry) 12. 1 + sin(x y) f 12 (x, y) = (1 + sin(x y))(cos(x y) − x y sin(x y)) − (y cos(x y))(x cos(x y)) (1 + sin(x y))2 cos(x y) − x y = f 21 (x, y). = 1 + sin(x y) =

7. A function f (x, y, z) of three variables can have 33 = 27 partial derivatives of order 3. Of these, ten can have different values, namely f 111 , f 222 , f 333 , f 112 , f 122 , f 223 , f 233 , f 113 , f 133 , and f 123 . For f (x, y, z) = xe xy cos(x z), we have f 133

8.

 ∂  3 xy −x e cos(x z) = f 313 = f 331 = ∂x = −(3x 2 + x 3 y)e xy cos(x z) + x 3 ze xy sin(x z). 2

2

f (x, y) = A(x − y ) + Bx y, f 1 = 2 Ax + By, f 2 = −2 Ay + Bx, f 11 = 2 A, f 22 = −2 A, Thus f 11 + f 22 = 0, and f is harmonic.

13.

f (x, y) = 3x 2 y − y 3 , f 1 (x, y) = 6x y, f 11 (x, y) = 6y, f 2 (x, y) = 3x 2 − 3y 2 , f 22 (x, y) = −6y. Thus f 11 + f 22 = 0 and f is harmonic. Also g(x, y) = x 3 − 3x y 2 is harmonic. x f (x, y) = 2 x + y2 y2 − x 2 x 2 + y 2 − 2x 2 = 2 f 1 (x, y) = 2 2 2 (x + y ) (x + y 2 )2 2x y f 2 (x, y) = − 2 (x + y 2 )2 (x 2 + y 2 )2 (−2x) − (y 2 − x 2 )2(x 2 + y 2 )(2x) f 11 (x, y) = (x 2 + y 2 )4 3 2 2x − 6x y = 2 (x + y 2 )3 (x 2 + y 2 )2 (2x) − 2x y2(x 2 + y 2 )(2y) f 22 (x, y) = − (x 2 + y 2 )4 3 2 −2x + 6x y = . (x 2 + y 2 )3 Evidently f 11 (x, y) + f 22 (x, y) = 0 for (x, y) 6= (0, 0). Hence f is harmonic except at the origin. 2x 2y , f2 = 2 2 +y x + y2 2 2 2 2 (x + y )(2) − 2x(2x) 2(y − x ) f 11 = = 2 (x 2 + y 2 )2 (x + y 2 )2 2(x 2 − y 2 ) f 22 = 2 (by symmetry) (x + y 2 )2 Thus f 11 + f 22 = 0 (everywhere except at the origin), and f is harmonic. y , (x 6= 0). f (x, y) = tan−1 x  1 y  y f 1 (x, y) = − =− 2 , 2 2 x x + y2 y 1+ 2 x   x 1 1 = 2 f 2 (x, y) = , 2 x x + y2 y 1+ 2 x 2x y 2x y f 11 = 2 , f 22 = − 2 . 2 2 (x + y ) (x + y 2 )2 Thus f 11 + f 22 = 0 and f is harmonic. f (x, y) = ln(x 2 + y 2 ),

w = e3x+4y sin(5z), w1 = 3w, w2 = 4w, 3x+4y

f1 =

x2

w11 = 9w,

w22 = 16w,

w3 = 5e cos(5z), w33 = −25w. Thus w11 + w22 + w33 = (9 + 16 − 25)w = 0, and w is harmonic in 3-space.

14. Let g(x, y, z) = z f (x, y). Then g1 (x, y, z) = z f1 (x, y), g2 (x, y, z) = z f2 (x, y), g3 (x, y, z) = f (x, y),

g11 (x, y, z) = z f11 (x, y) g22 (x, y, z) = z f22 (x, y) g33 (x, y, z) = 0.

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SECTION 12.4 (PAGE 694)

ADAMS and ESSEX: CALCULUS 8

Thus g11 + g22 + g33 = z( f 11 + f 22 ) = 0 and g is harmonic because f is harmonic. This proves (a). The proofs of (b) and (c) are similar.

This does not contradict Theorem 1 since the partials F12 and F21 are not continuous at (0, 0). (Observe, for instance, that F12 (x, x) = 0, while F12 (x, 0) = 2 for x 6= 0.)

If h(x, y, z) = f (ax + by, cz), then h 11 = a 2 f 11 , h 22 = b2 f 11 and h 33 = c2 f 22 . If a 2 + b2 = c2 and f is harmonic then h 11 + h 22 + h 33 = c2 ( f11 + f 22 ) = 0,

15.

u(x, t) = t −1/2 e−x /4t   ∂u 1 1 2 = − t −3/2 + t −5/2 x 2 e−x /4t ∂t 2 4 1 −3/2 −x 2 /4t ∂u = − xt e ∂x 2   ∂ 2u 1 −3/2 1 −5/2 2 −x 2 /4t = − t + t x e ∂x2 2 4 ∂u = . ∂t

18.

u(x, y, t) = t −1 e−(x

so h is harmonic. ∂u ∂v ∂u ∂v Since = , = − , and the second partials of ∂x ∂y ∂y ∂x u are continuous, we have ∂ ∂v ∂ ∂v ∂ 2u ∂ 2u = = = − . ∂x2 ∂x ∂y ∂y ∂x ∂ y2 ∂ 2u ∂2u + = 0, and u is harmonic. The proof that ∂x2 ∂ y2 v is harmonic is similar.

2

17.

Thus

16. Let f (x, y) =

(

2x y x 2 + y2 0

if (x, y) 6= (0, 0) if (x, y) = (0, 0).

For (x, y) 6= (0, 0), we have (x 2 + y 2 )2y − 2x y(2x) 2y(y 2 − x 2 ) = (x 2 + y 2 )2 (x 2 + y 2 )2 2 2 2x(x − y ) f 2 (x, y) = (by symmetry). (x 2 + y 2 )2

f 1 (x, y) =

2 +y 2 )/4t

∂u 1 x 2 + y 2 −(x 2 +y 2 )/4t 2 2 = − 2 e−(x +y )/4t + e ∂t t 4t 3 ∂u x 2 2 = − 2 e−(x +y )/4t ∂x 2t ∂ 2u 1 −(x 2 +y 2 )/4t x 2 −(x 2 +y 2 )/4t = − e + e ∂x2 2t 2 4t 3 2 2 ∂ u 1 y 2 2 2 2 = − 2 e−(x +y )/4t + 3 e−(x +y )/4t ∂ y2 2t 4t ∂ 2u ∂u ∂ 2u = Thus + . ∂t ∂x2 ∂ y2

Let F(x, y) = (x 2 − y 2 ) f (x, y). Then we calculate F1 (x, y) = 2x f (x, y) + (x 2 − y 2 ) f 1 (x, y)

2y(y 2 − x 2 )2 (x 2 + y 2 )2 F2 (x, y) = −2y f (x, y) + (x 2 − y 2 ) f2 (x, y)

19. For

∂ 2u ∂u ∂2u ∂2u = + + the solution is ∂t ∂x2 ∂ y2 ∂z 2

= 2x f (x, y) −

2x(x 2

2 +y 2 +z 2 )/4t

,

y 2 )2

− (x 2 + y 2 )2 2(x 6 + 9x 4 y 2 − 9x 2 y 4 − y 6 ) F12 (x, y) = = F21 (x, y). (x 2 + y 2 )3 = −2y f (x, y) +

u(x, y, z, t) = t −3/2 e−(x

For the values at (0, 0) we revert to the definition of derivative to calculate the partials: F(h, 0) − F(0, 0) F1 (0, 0) = lim = 0 = F2 (0, 0) h→0 h F1 (0, k) − F1 (0, 0) −2k(k 4 ) F12 (0, 0) = lim = lim = −2 k→0 k→0 k(k 4 ) k F2 (h, 0) − F2 (0, 0) 2h(h 4 ) F21 (0, 0) = lim = lim =2 h→0 h→0 h(h 4 ) h

which is verified similarly to the previous Exercise.

20. u(x, y) is biharmonic ⇔

∂ 2u ∂ 2u + is harmonic ∂x2 ∂ y2

 2  ∂2 ∂2 ∂ u ∂ 2u + 2 + 2 =0 ⇔ ∂x2 ∂y ∂x2 ∂y ∂4u ∂ 4u ∂4u ⇔ +2 2 2 + 4 =0 ∂x4 ∂x ∂y ∂y 

by the equality of mixed partials.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.5 (PAGE 704)

21. If u(x, y) = x 4 − 3x 2 y 2 , then ∂ 2u ∂x2 ∂2u ∂ y2 ∂ 4u ∂x4 4 ∂ u ∂ x 2∂ y2 ∂4u ∂ y4 ∂ 4u ∂x4

by the equality of mixed partials.

∂ (4x 3 − 6x y 2 ) = 12x 2 − 6y 2 ∂x ∂ = (−6x 2 y) = −6x 2 ∂y ∂ = (24x) = 24 ∂x ∂ = (−12x) = −12 ∂x

If u(x, y, z) is harmonic then the functions xu(x, y, z), yu(x, y, z), and zu(x, y, z) are all biharmonic. The proof is almost identical to that given in Exercise 22.

=

27. > f := x*y/(xˆ2+yˆ2);

=0 +2

∂ 4u ∂ 4u + 4 = 24 − 24 = 0. 2 2 ∂x ∂y ∂y ∂ 2u

∂x2 v(x, y) = xu(x, y), then

+

∂ 2u ∂ y2

= 0. If

  ∂2v ∂ ∂u ∂u ∂2u = u + x =2 +x 2 2 ∂x ∂x ∂x ∂x ∂x   ∂2v ∂ ∂u ∂2u = x =x 2 ∂y ∂y ∂ y2 ∂y  2  ∂2v ∂u ∂2v ∂u ∂ u ∂ 2u +x + 2 =2 + 2 =2 . ∂x2 ∂y ∂x ∂x2 ∂y ∂x

∂2 ∂2 + ∂x2 ∂ y2



∂ ∂u = ∂x ∂x



∂ 2u ∂ 2u + ∂x2 ∂ y2



=

∂ (0) = 0. ∂x

∂ 2v ∂ 2v + 2 is harmonic, and so v is biharmonic. 2 ∂x ∂y The proof that w(x, y) = yu(x, y) is biharmonic is similar.

Thus

23. By Example 3, e x sin y is harmonic. Therefore xe x sin y is biharmonic by Exercise 22.

24. By Exercise 11, ln(x 2 + y 2 ) is harmonic (except at the 25.

26.

origin). Therefore y ln(x 2 + y 2 ) is biharmonic by Exercise 22. x By Exercise 10, 2 is harmonic (except at the orix + y2 xy gin). Therefore 2 is biharmonic by Exercise 22. x + y2 ∂ 2u ∂ 2u ∂ 2u + + is harmonic u(x, y, z) is biharmonic ⇔ ∂ x 2 ∂ y 2 ∂z 2

 2  ∂2 ∂2 ∂ u ∂ 2u ∂ 2u ∂2 ⇔ + 2+ 2 + 2 + 2 =0 ∂x2 ∂y ∂z ∂x2 ∂y ∂z   ∂ 4u ∂4u ∂4u ∂ 4u ∂ 4u ∂ 4u ⇔ + + + 2 + + =0 ∂x4 ∂ y4 ∂z 4 ∂ x 2∂ y2 ∂ x 2 ∂z 2 ∂ y 2 ∂z 2 

>

xy + y2

simplify(diff(f,x$4) + 2*diff(f,x$2,y$2) + diff(f,y$4)); 0

Section 12.5

The Chain Rule

(page 704)

1. If w = f (x, y, z) where x = g(s, t), y = h(s, t), and ∂w = f 1 (x, y, z)g2 (s, t) + f 2 (x, y, z)h 2 (s, t) ∂t + f 3 (x, y, z)k2 (s, t).

2. If w = f (x, y, z) where x = g(s), y = h(s, t) and z = k(t), then

∂w = f 2 (x, y, z)h 2 (s, t) + f 3 (x, y, z)k ′ (t). ∂t

3. If z = g(x, y) where y = f (x) and x = h(u, v), then

Since u is harmonic, so is ∂u/d x: 

x2

z = k(s, t), then

Thus u is biharmonic.

22. If u is harmonic, then

f :=

∂z = g1 (x, y)h 1 (u, v) + g2 (x, y) f ′ (x)h 1 (u, v). ∂u

4. If w = f (x, y) where x = g(r, s), y = h(r, t), r = k(s, t) and s = m(t), then

h  dw = f 1 (x, y) g1 (r, s) k1 (s, t)m ′ (t) dt  i + k2 (s, t) + g2 (r, s)m ′ (t) h  + f 2 (x, y) h 1 (r, t) k1 (s, t)m ′ (t)  i + k2 (s, t) + h 2 (r, t) .

5. If w = f (x, y, z) where x = g(y, z) and y = h(z), then h i dw = f 1 (x, y, z) g1 (y, z)h ′ (z) + g2 (y, z) dz + f 2 (x, y, z)h ′ (z) + f 3 (x, y, z) ∂w = f 2 (x, y, z)h ′ (z) + f 3 (x, y, z) ∂z x ∂w = f 3 (x, y, z). ∂z x,y 465

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SECTION 12.5 (PAGE 704)

6. If u = then

p

ADAMS and ESSEX: CALCULUS 8

x 2 + y 2 , where x = est and y = 1 + s 2 cos t,

Method II.   z = t t + ln(et + t 2 ) e2t     ∂z et + 2t t 2 2t 2t = t + ln(e + t ) e + te 1+ t ∂t e + t2   2t t 2 + 2te t + ln(e + t )   y + 2t = x y2 + t y2 1 + + 2t x y 2 . y + t2

Method I. ∂u x y = p sest + p (−s 2 sin t) 2 2 2 ∂t x +y x + y2 xsest − ys 2 sin t p = . x 2 + y2 Method II.

9. p u = e2st + (1 + s 2 cos t)2

∂u 2se2st − 2s 2 sin t (1 + s 2 cos t) p = ∂t 2 e2st + (1 + s 2 cos t)2 x 2 s − ys 2 sin t = p . x 2 + y2

10. 11. 12.

u v Method I.

7. If z = tan−1 , where u = 2x + y and v = 3x − y, then ∂z ∂z ∂u ∂z ∂v = + ∂x ∂u ∂ x  ∂v ∂ x   1 1 1 −u = (2) + (3) u2 v u2 v 2 1+ 2 1+ 2 v v 2v − 3u 5y = 2 =− . u + v2 13x 2 − 2x y + 2y 2

13.

∂ f (2x, 3y) = 2 f 1 (2x, 3y). ∂x ∂ f (2y, 3x) = 3 f 2 (2y, 3x). ∂x ∂ f (y 2 , x 2 ) = 2x f 2 (y 2 , x 2 ). ∂x  ∂  f y f (x, t), f (y, t) ∂y   = f (x, t) f 1 y f (x, t), f (y, t)   + f 1 (y, t) f 2 y f (x, t), f (y, t) .

T = e−t z, where z = f (t).

∂T ∂ T dz dT = + = −e−t f (t) + e−t f ′ (t). dt ∂t ∂z dt dT = 0. The temdt perature is rising with respect to depth at the same rate at which it is falling with respect to time.

If f (t) = et , then f ′ (t) = et and

14. If E = f (x, y, z, t), where x = sin t, y = cos t and z = t,

Method II.

then the rate of change of E is

2x + y 3x − y ∂z 1 (3x − y)(2) − (2x + y)(3) = ∂x (3x − y)2 (2x + y)2 1+ (3x − y)2 −5y −5y = = . (3x − y)2 + (2x + y)2 13x 2 − 2x y + 2y 2 z = tan−1

8. If z = t x y 2 , where x = t + ln(y + t 2 ) and y = et , then Method I.   dz ∂z ∂z ∂ x ∂x ∂y = + + dt ∂t ∂ x ∂t ∂ y ∂t ∂z ∂ y + ∂ y ∂t   y + 2t 2 = x y + t y2 1 + + 2t x y 2 . y + t2

dE ∂E ∂E ∂E ∂E = cos t − sin t + + . dt ∂x ∂y ∂z ∂t

15.

z = f (x, y), where x = 2s + 3t and y = 3s − 2t.  ∂2z ∂  a) = 2 f (x, y) + 3 f (x, y) 1 2 ∂s 2 ∂s = 2(2 f11 + 3 f 12 ) + 3(2 f21 + 3 f 22 ) = 4 f 11 + 12 f 12 + 9 f 22 b)

c)

∂2z ∂2z ∂ = = (2 f1 + 3 f 2 ) ∂s∂t ∂t∂s ∂t = 2(3 f11 − 2 f 12 ) + 3(3 f21 − 2 f 22 ) = 6 f 11 + 5 f 12 − 6 f 22 ∂2z ∂ = (3 f1 − 2 f 2 ) ∂t 2 ∂t = 3(3 f11 − 2 f 12 ) − 2(3 f21 − 2 f 22 ) = 9 f 11 − 12 f 12 + 4 f 22

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INSTRUCTOR’S SOLUTIONS MANUAL

16. Let u =

x2

SECTION 12.5 (PAGE 704)

x y , v=− 2 . Then + y2 x + y2

∂u y2 − x 2 = 2 ∂x (x + y 2)2 ∂u 2x y =− 2 ∂y (x + y 2 )2

∂v 2x y = 2 ∂x (x + y 2 )2 y2 − x 2 ∂v = 2 . ∂y (x + y 2 )2

− sin s f 2 + t cos2 s f 12 − t sin s cos s f 22 = cos s f 1 − sin s f 2 + t cos s sin s( f 11 − f 22 ) + t (cos2 s − sin2 s) f 12 ,

where all partials of f are evaluated at (t sin s, t cos s).

18.

  2  2  2 ∂u 2 1 ∂u ∂v ∂v + = 2 + = ∂x ∂y ∂x ∂y (x + y 2 )2 ∂u ∂v ∂u ∂v + = 0, ∂x ∂x ∂y ∂y 

Finally, u is harmonic by Exercise 10 of Section 3.4, and, by symmetry, so is v. Thus ∂2 ∂x2

f (u, v) +

∂2 ∂ y2

f (u, v) = 0

∂2 ∂3 f (2x + 3y, x y) = (3 f 1 + x f2 ) 2 ∂ x∂ y ∂ x∂ y ∂ = (9 f11 + 3x f 12 + 3x f21 + x 2 f 22 ) ∂x ∂ (9 f11 + 6x f 12 + x 2 f 22 ) = ∂x = 18 f 111 + 9y f 112 + 6 f 12 + 12x f121 + 6x y f 122 + 2x f 22 + 2x 2 f 221 + x 2 y f 222

= 18 f 111 + (12x + 9y) f 112 + (6x y + 2x 2 ) f 122 + x 2 y f 222 + 6 f 12 + 2x f 22 ,

we have

because we are given that f is harmonic, that is, f 11 (u, v) + f 22 (u, v) = 0.

is harmonic for

 ∂2 ∂  f (x, y) = sin s f 1 (x, y) + cos s f 2 (x, y) ∂s∂t ∂s = cos s f 1 + t sin s cos s f 11 − t sin2 s f 12

Noting that

∂2 ∂2 f (u, v) + 2 f (u, v) 2 ∂x ∂y "   2 # ∂u ∂u 2 + = f 11 ∂x ∂y "   2 # ∂v 2 ∂v + f 22 + ∂x ∂y   ∂u ∂v ∂u ∂v + 2 f 12 + ∂x ∂x ∂y ∂y  2  2   ∂ v ∂ u ∂ 2u ∂ 2v + f1 + + f + 2 ∂x2 ∂ y2 ∂x2 ∂ y2  2  2   2 2 ∂ u ∂ v ∂ u ∂ v = f1 + 2 + f2 + 2 , ∂x2 ∂y ∂x2 ∂y



17. If x = t sin s and y = t cos s, then

We have ∂ ∂u ∂v f (u, v) = f 1 (u, v) + f 2 (u, v) ∂x ∂x ∂x ∂u ∂v ∂ f (u, v) = f 1 (u, v) + f 2 (u, v) ∂y ∂y ∂y  2 ∂2 ∂u ∂v ∂u ∂ 2u + f f (u, v) = f + f 12 11 1 ∂x ∂x ∂x ∂x2 ∂x2  2 ∂ 2v ∂u ∂v ∂v + f2 2 + f 21 + f 22 ∂x ∂x ∂x ∂x  2 2 2 ∂ ∂u ∂v ∂u ∂ u + f 12 f (u, v) = f 11 + f1 2 ∂ y2 ∂y ∂y ∂y ∂y  2 ∂ 2v ∂u ∂v ∂v + f2 2 . + f 21 + f 22 ∂y ∂y ∂y ∂y



y x ,− 2 2 2 x +y x + y2 (x, y) 6= (0, 0). and f

where all partials are evaluated at (2x + 3y, x y).

19.

20.

∂2 ∂ f (y 2 , x y, −x 2 ) = (y f 2 − 2x f 3 ) ∂ y∂ x ∂y = f 2 + 2y 2 f 21 + x y f 22 − 4x y f 31 − 2x 2 f 32 , where all partials are evaluated at (y 2 , x y, −x 2 ).

∂3 ∂2 2 2 f (s − t, s + t ) = (2s f1 + f 2 ) ∂t 2 ∂s ∂t 2 ∂ = (−2s f 11 + 4st f 12 − f 21 + 2t f 22 ) ∂t ∂ = (−2s f 11 + (4st − 1) f12 + 2t f 22 ) ∂t = 2s f 111 − 4st f 112 + 4s f 12 − (4st − 1) f 121

+ 2t (4st − 1) f 122 + 2 f 22 − 2t f 221 + 4t 2 f 222

= 2s f 111 + (1 − 8st) f 112 + 4t (2st − 1) f 122 + 4t 2 f 222 + 4s f 12 + 2 f 22 , where all partials are evaluated at (s 2 − t, s + t 2 ).

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SECTION 12.5 (PAGE 704)

ADAMS and ESSEX: CALCULUS 8

21. Let g(x, y) = f (u, v), where u = u(x, y), v = v(x, y).

Therefore we have

Then

g1 (x, y) = f 1 (u, v)u 1 (x, y) + f 2 (u, v)v 1 (x, y) g2 (x, y) = f 1 (u, v)u 2 (x, y) + f 2 (u, v)v 2 (x, y)

g11 (x, y) = f 1 (u, v)u 11 (x, y) + f 11 (u, v)(u 1 (x, y))2 + f 12 (u, v)u 1 (x, y)v 1 (x, y) + f 2 (u, v)v 11 (x, y)

+ f 21 (u, v)u 1 (x, y)v 1 (x, y) + f 22 (u, v)(v 1 (x, y))2

g22 (x, y) = f 1 (u, v)u 22 (x, y) + f 11 (u, v)(u 2 (x, y))2 + f 12 (u, v)u 2 (x, y)v 2 (x, y) + f 2 (u, v)v 22 (x, y)

+ f 21 (u, v)u 2 (x, y)v 2 (x, y) + f 22 (u, v)(v 2 (x, y))2 g11 (x, y) + g22 (x, y) = f 1 (u, v)[u 11 (x, y) + u 22 (x, y)] + f 2 (u, v)[v 11 (x, y) + v 22 (x, y)] + [(u 1 (x, y))2 + (u 2 (x, y))2 ] f11 (u, v)

+ [(v 1 (x, y))2 + (v 2 (x, y))2 ] f22 (u, v) + 2[u 1 (x, y)v 1 (x, y) + u 2 (x, y)v 2 (x, y)] f 12 (u, v).

The first two terms on the right are zero because u and v are harmonic. The next two terms simplify to [(v 1 )2 + (v 2 )2 ][ f 11 + f 22 ] = 0 because u and v satisfy the Cauchy-Riemann equations and f is harmonic. The last term is zero because u and v satisfy the Cauchy-Riemann equations. Thus g is harmonic. ∂r ∂r x = 2x, so = . ∂x ∂x r y ∂r z 1 ∂r = and = . If u = , then Similarly, ∂y r ∂z r r

22. If r 2 = x 2 + y 2 + z 2 , then 2r

∂z ∂z ∂z = es cos t + es sin t ∂s ∂x ∂y ∂z ∂z ∂z = −es sin t + es cos t ∂t ∂x ∂y ∂z ∂z ∂2z = es cos t + es sin t ∂s 2 ∂x ∂y   ∂2z ∂2z + es cos t es cos t 2 + es sin t ∂x ∂ y∂ x   2z ∂ ∂2z + es sin t es cos t + es sin t 2 ∂ x∂ y ∂y 2 ∂ z ∂z ∂z = −es cos t − es sin t ∂x ∂y ∂t 2   ∂2z ∂2z − es sin t −es sin t 2 + es cos t ∂ y∂ x ∂x   2 ∂ z ∂2z + es cos t −es sin t + es cos t 2 . ∂ x∂ y ∂y It follows that  2  ∂ z ∂2z ∂2z ∂2z 2s 2 2 + = e (cos t + sin t) + ∂s 2 ∂t 2 ∂x2 ∂ y2  2  2 ∂ z ∂ z = (x 2 + y 2 ) + 2 . ∂x2 ∂y

∂u 1 ∂r x =− 2 =− 3 ∂x r ∂x r 3x 2 − r 2 ∂2u 1 3x x = . = − + ∂x2 r3 r4 r r5 Similarly, ∂ 2u 3y 2 − r 2 = , 2 ∂y r5

∂2u 3z 2 − r 2 = . 2 ∂z r5

24. If x = r cos θ and y = r sin θ , then r 2 = x 2 + y 2 and

Adding these three expressions, we get ∂ 2u ∂x2

+

∂ 2u ∂ y2

+

∂ 2u ∂z 2

∂r ∂r x = 2x, so = = cos θ , and ∂x ∂x r ∂r y = = sin θ . Also similarly, ∂y r tan θ = y/x. Thus 2r

= 0,

so u is harmonic except at r = 0.

23. If x = es cos t and y = es sin t, then ∂x = es cos t ∂s ∂x = −e x sin t ∂t

sec2 θ

∂y = es sin t ∂s ∂y = es cos t. ∂t

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∂θ y =− 2 ∂x x ∂θ y =− 2 ∂x x + y2 sin θ =− r

sec2 θ

∂θ 1 = ∂y x ∂θ x = 2 ∂x x + y2 cos θ = . r

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.5 (PAGE 704)

Now ∂u ∂u ∂r ∂u ∂θ ∂u sin θ ∂u = + = cos θ − ∂x ∂r ∂ x ∂θ ∂ x ∂r r ∂θ ∂u ∂r ∂u ∂θ ∂u cos θ ∂u ∂u = + = sin θ + ∂y ∂r ∂ y ∂θ ∂ y ∂r r ∂θ     ∂ 2u ∂ ∂u ∂ 2u sin θ ∂ 2 u = cos θ + cos θ cos θ − ∂x ∂r r ∂θ ∂r ∂x2 ∂r 2     ∂ sin θ ∂u sin θ ∂ 2u sin θ ∂ 2 u − − cos θ − ∂x r ∂θ r ∂r ∂θ r ∂θ 2 2 2 2 sin θ cos θ ∂u ∂ u sin θ ∂u + + cos2 θ 2 = r ∂r r2 ∂θ ∂r 2 sin θ cos θ ∂ 2 u sin2 θ ∂ 2 u − + r ∂r ∂θ r 2 ∂θ 2     2 ∂ u ∂ ∂u ∂2u cos θ ∂ 2 u = sin θ + sin θ sin θ 2 + ∂ y2 ∂y ∂r ∂r r ∂θ ∂r     ∂ cos θ ∂u cos θ ∂ 2u cos θ ∂ 2 u + + sin θ + ∂y r ∂θ r ∂r ∂θ r ∂θ 2 2 2 2 sin θ cos θ ∂u ∂ u cos θ ∂u − + sin2 θ 2 = r ∂r r2 ∂θ ∂r cos2 θ ∂ 2 u 2 sin θ cos θ ∂ 2 u + + . r ∂r ∂θ r 2 ∂θ 2

27. If f (x1 , · · · , xn ) is positively homogeneous of degree k and has continuous partial derivatives of second order, then n X

i, j =1

x1 x j f i j (x1 , · · · , xn ) = k(k − 1) f (x1 , · · · , xn ).

Proof: Differentiate f (t x1 , · · · , t xn ) = t k f (x1 , · · · , xn ) twice with respect to t: n X

xi f i (t x1 , · · · , t xn ) = kt k−1 f /xn

i=1 n X

i, j =1

xi x j f i j (t x1 , · · · , t xn ) = k(k − 1)t k−2 f (x1 , · · · , xn ),

and then put t = 1.

28. If f (x1 , · · · , xn ) is positively homogeneous of degree k

and has continuous partial derivatives of mth order, then n X

Therefore 1 ∂ 2u ∂ 2u ∂ 2u ∂2u 1 ∂u + 2 + 2 = 2 + , 2 r ∂r ∂x ∂y ∂r r ∂θ 2

i 1 ,...,i m =1

xi1 · · · xim f i1 ...im (x1 , · · · , xn )

as was to be shown.

25. If u = r 2 ln r , where r 2 = x 2 + y 2 , then, since ∂r/∂ x = x/r and ∂r/∂ y = y/r , we have ∂u ∂x ∂2u ∂x2 ∂ 2u ∂ y2 ∂2u ∂x2

x = x(1 + 2 ln r ) r 2x 2 = 1 + 2 ln r + 2 r 2y 2 = 1 + 2 ln r + 2 (similarly) r ∂ 2u 2(x 2 + y 2 ) + 2 = 2 + 4 ln r + = 4 + 4 ln r. ∂y r2 = (2r ln r + r )

The constant 4 is harmonic, and so is 4 ln r by Exercise ∂ 2u ∂ 2u 11 of Section 3.4. Therefore + is harmonic, and ∂ x 2 ∂ y2 so u is biharmonic.

26.

f (t x, t y) = t k f (x, y)

x f 1 (t x, t y) + y f 2 (t x, t y) = kt k−1 f (x, y)   x x f 11 (t x, t y) + y f 12 (t x, t y)   + y x f 21 (t x, t y) + y f 22 (t x, t y)

29.

The proof is identical to those of Exercises 26 or 27, except that you differentiate m times before putting t = 1.   2x y(x 2 − y 2 ) if (x, y) 6= (0, 0) F(x, y) = 2 2  x +y 0 if (x, y) = (0, 0) a) For (x, y) 6= (0, 0), F(x, y) =

x 2 f 11 (x, y)+2x y f 12 (x, y)+ y 2 f 22 (x, y) = k(k−1) f (x, y).

2x y(x 2 − y 2) 2x y(y 2 − x 2 ) =− = −F(y, x). 2 2 x +y x 2 + y2

Since 0 = −0, this holds for (x, y) = (0, 0) also. b) For (x, y) 6= (0, 0), ∂ ∂ F(x, y) = − F(y, x) = −F2 (y, x) ∂x ∂x ∂ ∂ F12 (x, y) = F1 (x, y) = − F2 (y, x) = −F21 (y, x). ∂y ∂y F1 (x, y) =

c) If (x, y)

= k(k − 1)t k−2 f (x, y) Put t = 1 and get

= k(k − 1) · · · (k − m + 1) f (x1 , · · · , xn ).

(0, 0), 2y(x 2 − y 2 ) ∂ x 2 − y2 then F1 (x, y) = + 2x y . 2 2 x +y ∂ x x 2 + y2 Thus F1 (0, y) = −2y + 0 = −2y for y 6= 0. This result holds for y = 0 also, since F1 (0, 0) = limh→0 (0 − 0)/ h = 0.. 6=

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SECTION 12.5 (PAGE 704)

ADAMS and ESSEX: CALCULUS 8

d) By (b) and (c), F2 (x, 0) = −F1 (0, x) = 2x, and F21 (0, 0) = 2.

30.

a) Since F12 (x, y) = −F21 (y, x) for (x, y) 6= (0, 0), we have F12 (x, x) = −F21 (x, x) for x 6= 0. However, all partial derivatives of the rational function F are continuous except possibly at the origin. Thus F12 (x, x) = F21 (x, x) for x 6= 0. Therefore, F12 (x, x) = 0 for x 6= 0. b) F12 cannot be continuous at (0, 0) because its value there (which is −2) differs from the value of F21 (0, 0) (which is 2). Alternatively, F12 (0, 0) is not the limit of F12 (x, x) as x → 0.

and hence

By Exercise 38, w(r, s) = f (r ) + g(s), where f and g are arbitrary twice differentiable functions. Hence the original differential equation has solution u(x, t) = f (x + ct) + g(x − ct).

34. By Exercise 39, the DE u t = c2 u xx has solution u(x, t) = f (x + ct) + g(x − ct),

31. If ξ = x + ct, η = x, and v(ξ, η) = v(x + ct, x) = u(x, t), then

∂u ∂v ∂ξ ∂v = =c ∂t ∂ξ ∂t ∂ξ ∂v ∂ξ ∂v ∂η ∂v ∂v ∂u = + = + . ∂x ∂ξ ∂ x ∂η ∂ x ∂ξ ∂η

for arbitrary sufficiently smooth functions f and g. The initial conditions imply that p(x) = u(x, 0) = f (x) + g(x) q(x) = u t (x, 0) = c f ′ (x) − cg ′ (x).

∂u ∂u If u satisfies = c , then v satisfies ∂t ∂x c

∂v ∂v ∂v =c +c , ∂ξ ∂ξ ∂η

that is,

∂v = 0. ∂η

Thus v is independent of η, so v(ξ, η) = f (ξ ) for an arbitrary differentiable function f of one variable. The original differential equation has solution

Integrating the second of these equations, we get f (x) − g(x) =

1 p(x) + 2 1 g(x) = p(x) − 2 f (x) =

32. If w(r ) = f (r ) + g(s), where f and g are arbitrary twice differentiable functions, then

∂ ′ ∂2w = g (s) = 0. ∂r ∂s ∂r

Z

x

q(s) ds,

a

Z x 1 q(s) ds 2c a Z x 1 q(s) ds. 2c a

Thus the solution to the initial-value problem is p(x + ct) + p(x − ct) 1 + u(x, t) = 2 2c

33. If r = x + ct, s = x − ct, and

w(r, s) = w(x + ct, x − ct) = u(x, t), then ∂w ∂w −c ∂r ∂s 2w ∂ 2w ∂ ∂ 2w + c2 2 = c2 2 − 2c2 ∂r ∂r ∂s ∂s ∂w ∂w = + ∂r ∂s ∂2w ∂ 2w ∂ 2w = +2 + . 2 ∂r ∂r ∂s ∂s 2 =c

∂ 2u ∂2u = c2 2 , then w satisfies 2 ∂t ∂x  2   2  ∂ w ∂2w ∂2w ∂ 2w ∂ 2w 2 ∂ w c2 − 2 + = c + 2 + ∂r 2 ∂r ∂s ∂s 2 ∂r 2 ∂r ∂s ∂s 2 If u satisfies

1 c

where a is a constant. Solving the two equations for f and g we obtain

u(x, t) = f (x + ct).

∂u ∂t ∂ 2w ∂t 2 ∂u ∂x ∂ 2w ∂x2

∂2w = 0. ∂r ∂s

Z

x+ct

q(s) ds. x−ct

35. > f := u(r*cos(t),r*sin(t)): > simplify( diff(f,r$2) + (1/r)*diff(f,r) > +(1/rˆ2)*diff(f,t$2)); D2,2 (u)(r cos(t), r sin(t)) + D1,1 (u)(r cos(t), r sin(t)) which confirms the identity.

36. > g := f(x/(xˆ2+yˆ2),y/(xˆ2+yˆ2)): > simplify(diff(g,x$2)+diff(g,y$2));     D1,1 ( f )

y x y x , + D2,2 ( f ) , x 2 + y2 x 2 + y2 x 2 + y2 x 2 + y2 . (x 2 + y 2 )2

If f is harmonic, then the numerator is zero so g is harmonic.

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SECTION 12.6 (PAGE 714)

37. > simplify(diff(diff( > f(yˆ2,x*y,-xˆ2),x),y));

2.

f 1 (x, y) = −

where all terms are evaluated at (y 2 , x y, −x 2 ).

38. > simplify(diff(diff > (f(sˆ2-t,s+tˆ2),s),t$2));

where all terms are evaluated at (s 2 − t, s + t 2 ).

3.

39. > z := u(x,y): > x := exp(s)*cos(t): y := exp(s)*sin(t): > simplify( > (diff(z,s$2)+diff(z,t$2))/(xˆ2+yˆ2)); 4. D2,2 (u)(es cos t, es sin t) + D1,1 (u)(es cos t, es sin t), which confirms the identity in Exercise 23.

40. > u := (x,t) -> (p(x-c*t)+p(x+c*t))/2 > +(1/((2*c))*int(q(s),x=xc*t..x+c*t): > simplify(diff(u(x,t),t$2) > -cˆ2*diff(u(x,t),x$2));

5.

0 >

simplify(u(x,0)); p(x)

f (3, 3) =

π 4

x2

f (x, y) = sin(π x y + ln y), f (0, 1) = 0 f 1 (x, y) =  π y cos(π xy + ln y), f 1 (0, 1) = π 1 f 2 (x, y) = π x + cos(π x y + ln y), f 2 (0, 1) = 1 y f (0.01, 1.05) ≈ f (0, 1) + 0.01 f 1 (0, 1) + 0.05 f 2 (0, 1) = 0 + 0.01π + 0.05 ≈ 0.081416 24 x 2 + x y + y2 −24(2x + y) −24(x + 2y) f 1 (x, y) = 2 , f 2 (x, y) = 2 (x + x y + y 2 )2 (x + x y + y 2 )2 f (2, 2) = 2, f 1 (2, 2) = −1, f 2 (2, 2) = −1 f (x, y) =

f (2.1, 1.8) ≈ f (2, 2) + 0.1 f 1 (2, 2) − 0.2 f 2 (2, 2) = 2 − 0.1 + 0.2 = 2.1 √ f (x, y, z) = x + 2y + 3z, f (2, 2, 1) = 3 1 1 f 1 (x, y, z) = √ , f 2 (x, y, z) = √ 2 x + 2y + 3z x + 2y + 3z 3 f 3 (x, y, z) = √ 2 x + 2y + 3z

f (1.9, 1.8, 1.1) ≈ f (2, 2, 1) − 0.1 f 1 (2, 2, 1) − 0.2 f 2 (2, 2, 1) + 0.1 f 3 (2, 2, 1) 0.1 0.2 0.1 − + ≈ 2.967 =3− 6 3 2

>

simplify(subs(t=0,diff(u(x,t),t))); q(x) so u satisfies the PDE and initial conditions given in Exercise 34.

6. Section 12.6 Linear Approximations, Differentiability, and Differentials (page 714)

y x

1 y f 1 (3, 3) = − 2 +y 6 x 1 f 2 (x, y) = 2 f 2 (3, 3) = 6 x + y2 f (3.01, 2.99) = f (3 + 0.01, 3 − 0.01) ≈ f (3, 3) + 0.01 f 1 (3, 3) − 0.01 f 2 (3, 3) π 0.01 0.01 π 0.01 = − − = − 4 6 6 4 3 ≈ 0.7820648

2y 2 D1,2 ( f ) + x y D2,2 ( f ) + D2 ( f ) −4x y D1,3 ( f ) − 2x 2 D2,3 ( f )

2s D1,1,1 ( f ) − 8st D1,1,2 ( f ) + 8st 2 D1,2,2 ( f ) + 4s D1,2 ( f ) +D1,1,2 ( f ) − 4t D1,2,2 ( f ) + 4t 2 D2,2,2 ( f ) + 2D2,2 ( f )

f (x, y) = tan−1

2

f (x, y) = xe y+x f (2, −4) = 2 2 f 1 (x, y) = e y+x (1 + 2x 2 ) f 1 (2, −4) = 9 2 f 2 (2, −4) = 2 f 2 (x, y) = xe y+x f (2.05, −3.92) ≈ f (2, −4) + 0.05 f 1 (2, −4) + 0.08 f 2 (2, −4) = 2 + 0.45 + 0.16 = 2.61

7. We have dz = 2xe3y d x + 3x 2 e3y d y. 1.

f (x, y) = x 2 y 3

f 1 (x, y) = 2x y 3

f 2 (x, y) = 3x 2 y 2

If x = 3 and y = 0, then z = 9. If also d x = 0.05 and d y = −0.02, then dz = 6(0.05) + 27(−0.02) = −0.24. Thus z ≈ 0 − 0.24 = 8.76.

f (3, 1) = 9 f 1 (3, 1) = 6 f 2 (3, 1) = 27

f (3.1, 0.9) = f (3 + 0.1, 1 − 0.1) ≈ f (3, 1) + 0.1 f 1 (3, 1) − 0.1 f 2 (3, 1) = 9 + 0.6 − 2.7 = 6.9

2s s2 d x − 2 dt. t t If s = t = 2, and ds = −dt = 0.1, then g(s, t) = 2 and dg = 2(0.1) − 1(−0.1) = 0.3. Thus g(2.1, 1.9) ≈ 2 + 0.3 = 2.3.

8. We have dg =

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SECTION 12.6 (PAGE 714)

ADAMS and ESSEX: CALCULUS 8

x d x + y d y + z dz

9. We have d F = p

. x 2 + y + 2 + z2 Also F(1, 2, 2) = 3. If d x = dz = −0.3 and d y = 0.6,  1 then d F = 1(−0.3) + 2(0.6) + 2(−0.3) = 0.1. Thus 3 F(0.9, 2.2, 1.9) ≈ 3 + 0.1 = 3.1.

10. We have 

 du = sin(x + y) + x cos(x + y) d x + x cos(x + y) d y. If x = y = π/2, d x  = 1/20, d y = 1/30, then u = 0  and   π 1  π 1 π and du = 0 − + − =− . − 2 20 2 30 120 1 π 1 π + , y = − , we have Therefore, at x = 2 20 2 30 π u≈− . 120

11. If the edges are x, y, and z, and

dx dy dz 1 = = = , x y z 100

then a) V = x yz ⇒ d V = yz d x + x z d y + x y dz dV dx dy dz 3 ⇒ = + + = . V x y z 100 The volume can be in error by about 3%.

1 x y sin θ 2 1 1 1 d A = y sin θ d x + x sin θ d y + x y cos θ dθ 2 2 2 dA dx dy = + + cot θ dθ. A x y A=

For x = 224, y = 158, θ = 64◦ = 64π/180, d x = d y = 0.4, and dθ = 2◦ = 2π/180, we have 0.4 0.4 2π dA = + + (cot 64◦ ) ≈ 0.0213. A 224 158 180 The calculated area of the plot can be in error by a little over 2%.

15. From the figure we have h = s tan θ h = (s + x) tan φ =



 h + x tan φ. tan θ

Solving the latter equation for h, we obtain

c) D 2 = x 2 + y 2 + z 2 ⇒ 2D d D = 2x d x + 2y d y + 2z dz

We calculate the values of h and its first partials at x = 100, θ = 50◦ , φ = 35◦ :

x2

y2

z2

dD 1 dx dy dz = 2 + 2 + 2 = . D D x D y D z 100

The diagonal can be in error by about 1%. V = 13 πr 2 h ⇒ d V = 32 πr h dr + 31 πr 2 dh. If r = 25 ft, h = 21 ft, and dr = dh = 0.5/12 ft, then dV =

0.5 π (2 × 25 × 21 + 252 ) ≈ 73.08. 3 12

The calculated volume can be in error by about 73 cubic feet.

13.

y m and θ radians, then its area A satisfies

b) A = x y ⇒ d A = y d x + x d y dx dy 2 dA = + = . ⇒ A x y 100 The area of a face can be in error by about 2%.

⇒

12.

14. If the sides and contained angle of the triangle are x and

h=

x tan φ tan θ . tan θ − tan φ

h ≈ 170 ∂h tan φ tan θ = ≈ 1.70 ∂x tan θ − tan φ (tan θ − tan φ) sec2 θ − tan θ sec2 θ ∂h = x tan φ ∂θ (tan θ − tan φ)2 2 2 x tan φ sec θ =− ≈ −491.12 (tan θ − tan φ)2 x tan2 θ sec2 φ ∂h = ≈ 876.02. ∂φ (tan θ − tan φ)2 Thus dh ≈ 1.70 d x − 491 dθ + 876 dφ. For d x = 0.1 m and |dθ | = |dφ| = 1◦ = π/180, the largest value of dh will come from taking dθ negative and dφ positive:

√ S = πr r 2 + h 2 , so  p  πr 2 πr h d S = π r 2 + h2 + √ dr + √ dh 2 2 r +h r 2 + h2 p  252 + 25 × 21 0.5 =π 252 + 212 + √ ≈ 8.88. 12 252 + 212 The surface area can be in error by about 9 square feet.

dh ≈ (1.70)(0.1) + (491 + 876)

π ≈ 24.03. 180

The calculated height of the tower is 170 m and can be in error by as much as 24 m. The calculation of the height is most sensitive to the accuracy of the measurement of φ.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.6 (PAGE 714)

20.

g(r, s, t) = Dg(r, s, t) =

h Dg(1, 3, 3) =

−0.01 g(0.99, 3.02, 2.97) ≈ g(1, 3, 3) + Dg(1, 3, 3) 0.02 −0.03 ! ! ! 3 −0.04 2.96 = 3 + −0.09 = 2.91 0 0.30 0.30

φ

θ s

x

A

B

Fig. 12.6.15

16.

f (a + h, b + k) − f (a, b) − h f 1 (a, b) − k f2 (a, b) √ h 2 + k2 approaches 0 as (h, k) → (0, 0). Since the denominator of this fraction approaches zero, the numerator must also approach 0 or the fraction would not have a limit. Since the terms h f 1 (a, b) and k f2 (a, b) both approach 0, we must have

dx 1 = . Similarly, x 100

2 dz 3 dy = and = . Therefore y 100 z 100

19.

g ′ (t) = h f 1 (a + th, b + tk) + k f2 (a + th, b + tk). If h and k are small enough that (a + h, b + k) belongs to the disk referred to in the statement of the problem, then we can apply the (one-variable) Mean-Value Theorem to g(t) on [0, 1] and obtain

f(r, θ ) = (r cos θ, r sin θ )   cos θ −r sin θ Df(r, θ ) = sin θ r cos θ

g(1) = g(0) + g ′ (θ ),

f(R, φ, θ ) = (R sin φ cos θ, R sin φ sin θ, R cos φ) ! sin φ cos θ R cos φ cos θ −R sin φ sin θ Df(R, φ, θ ) = sin φ sin θ R cos φ sin θ R sin φ cos θ cos φ −R sin φ 0 f(x, y, z) = Df(x, y, z) = Df(2, 2, 1) =







x 2 + yz 2 y − x ln z



4 0



2x − ln z 1 4

[ f (a + h, b + k) − f (a, b)] = 0.

22. Let g(t) = f (a + th, b + tk). Then

and w decreases by about 4%.

18.

lim

(h,k)→(0,0)

Thus f is continuous at (a, b).

1w dw 2 + 6 − 12 4 ≈ = =− , w w 100 100

17.

!

21. If f is differentiable at (a, b), then

x 2 y3 ∂w 2x y 3 2w = 4 = 4 z ∂x z x 3x 2 y 2 3w 4x 2 y 3 4w ∂w ∂w = = =− 5 =− . ∂y z4 y ∂z x z ∂w ∂w ∂w dw = dx + dy + dz ∂x ∂y ∂z dw dx dy dz =2 +3 −4 . w x y z w=

Since x increases by 1%, then

! r 2s 2 r t s2 − t2 ! 2r s r 2 0 2 2r t 0 r 0 2s −2t ! 6 1 0 6 0 1 0 6 −6

z 2y

2 −2

for some θ satisfying 0 < θ < 1, i.e., f (a + h, b + k) = f (a, b) + h f 1 (a + θ h, b + θ k) + k f2 (a + θ h, b + θ k).

23. Apply Taylor’s Formula: y −x/z

g(1) = g(0) + g ′ (0) +



−0.02 f(1.98, 2.01, 1.03) ≈ f(2, 2, 1) + Df(2, 2, 1) 0.01 0.03       6 −0.01 5.99 = + = 4 −0.02 3.98

g ′′ (θ ) 2!

for some θ between 0 and 1 to g(t) = f (a + th, b + tk). We have !

g ′ (t) = h f 1 (a + th, b + tk) + k f2 (a + th, b + tk) g ′ (0) = h f 1 (a, b) + k f2 (a, b)

g ′′ (t) = h 2 f 11 (a + th, b + tk) + 2hk f12 (a + th, b + tk) + k 2 f 22 (a + th, b + tk).

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SECTION 12.6 (PAGE 714)

ADAMS and ESSEX: CALCULUS 8

26. To maximize g(x) = px − f (x) we look for a criti-

Thus f (a + h, b + k) = f (a, b) + h f 1 (a, b) + k f2 (a, b)  1 2 + h f 11 (a + θ h, b + θ k) + 2hk f 12 (a + θ h, b + θ k) 2  + k 2 f 22 (a + θ h, b + θ k)

1 f = f (a + h, b + k) − f (a, b) d f = h f 1 (a, b) + k f2 (a, b) 1 f − d f 1 ≤ h 2 f 11 (a + θ h, b + θ k) + 2hk f12 (a + θ h, b + θ k) 2 2 + k f 22 (a + θ h, b + θ k) ≤ K (h 2 + k 2 )

(since 2hk ≤ h 2 + k 2 ),

for some K depending on f , and valid in some disk h 2 + k 2 ≤ R 2 of positive radius R.

24. Since E(S, V , N1 , . . . , Nn ) is homogeneous of degree 1, Euler’s Theorem tells us that ∂E ∂E ∂E ∂E +V + N1 + · · · + Nn ∂S ∂V ∂ N1 ∂ Nn = ST − PV + µ1 N1 + · · · + µn Nn .

E=S

Now calclate the differential of this equation: d E = S d T + T d S − P d V − V d P + µ1 d N1 + N1 dµ1 + · · · + µn d Nn + Nn dµn . Subtracting Gibbs equation d E = T d S − P d V + µ1 d N1 + · · · + µn d Nn

cal point. Thus we want 0 = g ′ (x) = p − f ′ (x), so that the critical point of g satisfies f ′ (x) = p. Since g ′′ (x) = f ′′ (x) > 0 for all x, f ′ is increasing (and so invertible), and the critical point must provide a maximum value for g. Now f ′ (x) = p can be solved for x as a function of p and so the maximum must satisfy g(x( p)) = px( p) − f (x( p)). Comparing this equation with f ∗ ( p) = px − f (x) shows that f ∗ ( p) must be that maximum value of g.

27. If f (x) = x 2 , then f ′ (x) = 2x = p. Thus x = p/2, and f ∗ ( p) = px − f (x) = 2x 2 − x 2 = x 2 =

p2 . 4

28. If f (x) = x 4 , then f ′ (x) = 4x 3 = p, so that x = ( p/4)1/3 . Thus

f ∗ ( p) = x p − f (x) = 4x 4 − x 4 = 3x 4 = 3

29. If f (x) = ln(2 + 3x), then p = f ′ (x) = x=

3 − 2p . Thus 3p

 p 4/3 4

3 , so that 2 + 3x

2p f ( p) = x p − ln(2 + 3x) = 1 − − ln 3 ∗

  3 . p

30. Fixed numbers mean d Ni = 0 for all i . dG = d E − T d S − S d T + P d V + V d P = T dS − P dV − T dS − S dT + P dV + V d P = −S d T + V d P. Since G depends only on T and P, dG =

from the above result leaves us with the Gibbs-Duhem equation

∂G ∂G dT + d P. ∂T ∂P

We conclude that, 0 = S d T − V d P + N1 dµ1 + · · · + Nn dµn ,

∂G = −S ∂T

showing that the intensive variables are not independent.

25. By direct calculation ∂E 2 = E ⇒ E= ∂S 3N k 2 ∂ 2 ∂E P=− = −E V 3 V−3 = ∂V ∂V 2 ⇒ PV = E = N kT. 3

T =

∂G = V. ∂P

and

31. From the Gibbs equation 3 N kT, 2 2E 3V

.

T d S = d E + Pd V − µ1 d N1 − · · · − µn d Nn , from which we conclude that ∂S 1 = , ∂E T

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∂S P = , ∂V T

and

∂S µi =− . ∂ Ni T

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.7 (PAGE 725)

Therefore we have

3. 

1 T



1 dE T E P µ1 µn = 2 dT + dV − d N1 − · · · − d Nn . T T T T

d8 = d S − E d

−

Thus 8 = 8(T, V , N1 , · · · , Nn ).

32. We start with the differential of H = X∂H i

dqi +

∂H d pi ∂ pi



P

i

pi q˙i − L.

∂qi  X ∂L ∂L = dqi − d q˙i . q˙i d pi + pi d q˙i − ∂qi ∂ q˙i i

(a) Since each variable q˙i on the right side has been replaced by pi on the left side, the conjugate pairs for this Legendre transformation are the pairs { pi , q˙i }, (1 ≤ i ≤ n). Also, since there are no differentials d q˙i on the left side, the coefficients of these differentials on the right side ∂L must vanish. Thus = pi , for 1 ≤ i ≤ n. ∂ q˙i

4.

5.

(b) Comparing the coefficients of the remaining differentials on the left and right sides and using the principle of least action, we have ∂L ∂H =− = − p˙ i ∂qi ∂qi 1 2 2 (q ∂H ∂p =

(c) H = q˙ =

and

∂H = q˙i . ∂ pi

6.

+ p2 ), thus p˙ = − ∂∂qH = −q and

p. Thus q¨ = p˙ = −q or q¨ + q = 0.

Section 12.7 Gradients and Directional Derivatives (page 725) 1.

f (x, y) = x 2 − y 2 ,

f (2, −1) = 3. ∇ f (2, −1) = 4i + 2j. Tangent plane to z = f (x, y) at (2, −1, 3) has equation 4(x − 2) + 2(y + 1) = z − 3, or 4x + 2y − z = 3. Tangent line to f (x, y) = 3 at (2, −1) has equation 4(x − 2) + 2(y + 1) = 0, or 2x + y = 3. x−y f (x, y) = , f (1, 1) = 0. x+y 2yi − 2xj ∇f = , (x + y)2 1 ∇ f (1, 1) = (i − j). Tangent plane to z = f (x, y) 2 at (1, 1, 0) has equation 21 (x − 1) − 21 (y − 1) = z, or x − y − 2z = 0. Tangent line to f (x, y) = 0 at (1, 1) has equation 1 1 2 (x − 1) − 2 (y − 1), or x = y.

∇ f (x, y) = 2xi − 2yj,

2.

x , x 2 + y2 (x 2 + y 2 )(1) − x(2x) y2 − x 2 f 1 (x, y) = = , (x 2 + y 2 )2 (x 2 + y 2 )2 2x y f 2 (x, y) = − 2 . (x + y 2 )2   1 2 2 ∇ f (x, y) = 2 (y − x )i − 2x yj , (x + y 2 )2 4 3 i − 25 j. ∇ f (1, 2) = 25 f (x, y) =

Tangent plane to z = f (x, y) at (1, 2, 15 ) has equation 4 1 3 (x − 1) − (y − 2) = z − , or 3x − 4y − 25z = −10. 25 25 5 Tangent line to f (x, y) = 1/5 at (1, 2) has equation 3 4 (x − 1) − (y − 2) = 0, or 3x − 4y = −5. 25 25 f (x, y) = e xy , ∇ f = ye xy i + xe xy j, ∇ f (2, 0) = 2j. Tangent plane to z = f (x, y) at (2, 0, 1) has equation 2y = z − 1, or 2y − z = −1. Tangent line to f (x, y) = 1 at (2, 0) has equation y = 0. 2xi + 2yj , x 2 + y2 ∇ f (1, −2) = 25 i − 45 j. Tangent plane to z = f (x, y) at 2 4 (1, −2, ln 5) has equation (x − 1) − (y + 2) = z − ln 5, 5 5 or 2x − 4y − 5z = 10 − 5 ln 5. Tangent line to f (x, y) = ln 5 at (1, −2) has equation 4 2 (x − 1) − (y + 2) = 0, or x − 2y = 5. 5 5 p f (x, y) = 1 + x y 2 , f (2, −2) = 3. y 2 i + 2x yj ∇ f (x, y) = p , 2 1 + x y2 4 2 ∇ f (2, −2) = i − j. 3 3 Tangent plane to z = f (x, y) at (2, −2, 3) has equation 4 2 (x − 2) − (y + 2) = z − 3, or 2x − 4y − 3z = 3. 3 3 Tangent line to f (x, y) = 3 at (2, −2) has equation 4 2 (x − 2) − (y + 2) = 0, or x − 2y = 6. 3 3 f (x, y) = ln(x 2 + y 2 ),

∇ f (x, y) =

7.

f (x, y, z) = x 2 y + y 2 z + z 2 x, f (1, −1, 1) = 1. ∇ f (x, y, z) = (2x y + z 2 )i + (x 2 + 2yz)j + (y 2 + 2zx)k, ∇ f (1, −1, 1) = −i − j + 3k. Tangent plane to f (x, y, z) = 1 at (1, −1, 1) has equation −(x − 1) − (y + 1) + 3(z − 1) = 0, or x + y − 3z = −3.

8.

f (x, y, z) = cos(x + 2y + 3z), π  11π f , π, π = cos = 0. 2 2 ∇ f (x, y, z) = − sin(x + 2y + 3z)(i + 2j + 3k), π  11π ∇f , π, π = − sin (i + 2j + 3k) = i + 2j + 3k. 2 2 π  Tangent plane to f (x, y, z) = 0 at , π, π has equa2 tion π x − + 2(y − π ) + 3(z − π ) = 0, 2

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SECTION 12.7 (PAGE 725)

11π . 2

or x + 2y + 3z =

9.

ADAMS and ESSEX: CALCULUS 8

16. Since x = r cos θ and y = r sin θ , we have ∂f ∂f ∂f = cos θ + sin θ ∂r ∂x ∂y ∂f ∂f ∂f = −r sin θ + r cos θ . ∂θ ∂x ∂y

√ 2 f (x, y, z) = ye−x sin z, f (0, 1, π/3) = 3/2. −x 2 sin zi+e −x 2 sin zj+ ye −x 2 cos zk, ∇ f (x, y, z) = −2x √ye 1 3 ∇ f (0, 1, π/3) = j + k. 2 2 √ 3 The tangent plane to f (x, y, z) = at 0, 1, π/3) has 2 equation √ 3 1 π (y − 1) + z− = 0, 2 2 3 √ √ π or 3y + z = 3 + . 3

10.

f (x, y) = 3x − 4y, ∇ f (0, 2) = ∇ f (x, y) = 3i − 4j, D−i f (0, 2) = −i • (3i − 4j) = −3.

11.

f (x, y) = x 2 y, ∇ f = 2x yi + x 2 j, ∇ f (−1, −1) = 2i + j.

Also, xi + yj = (cos θ )i + (sin θ )j r −yi + xj θˆ = = −(sin θ )i + (cos θ )j. r

rˆ =

Therefore, ∂f 1 ∂f ˆ rˆ + θ ∂r  r ∂θ  ∂f ∂f = cos2 θ + sin θ cos θ i ∂x ∂y   ∂f ∂f + cos θ sin θ + sin2 θ j ∂x ∂y   ∂f ∂f − sin θ cos θ i + sin2 θ ∂x ∂y   ∂f 2 ∂f + − cos θ sin θ + cos θ j ∂x ∂y ∂f ∂f i+ j = ∇ f. = ∂x ∂y

Rate of change of f at (−1, −1) in the direction of i + 2j is i + 2j 4 √ • (2i + j) = √ . 5 5

12.

f (x, y) =

x , 1+ y

∇ f (x, y) =

i−j ∇ f (0, 0) = i, u= √ ,   2 1 i−j Du f (0, 0) = i • √ = √ . 2 2

13.

f (x, y) = x 2 + y 2 ,

∇ f (1, −2) = 2i − 4j.

1 x i− j, 1+ y (1 + y)2

17.

−1 = Du f (2, 0)u • ∇ f (2, 0) = 2u 2 √ 1 3 if u 2 = − , and therefore u 1 = ± . At (2, 0), f has 2 2√ 1 3 rate of change −1 in the directions ± i − j. 2 2 3 If −3 = Du f (2, 0) = 2u 2 , then u 2 = − . This is 2 not possible for a unit vector u, so there is no direction at (2, 0) in which f changes at rate −3.

∇ f = 2xi + 2yj,

A unit vector in the direction making a 60◦ angle with √ 1 3 the positive x-axis is u = 2 i + 2 j. The rate of change of f at√(1, −2) in the direction of u is u • ∇ f (1, −2) = 1 − 2 3.

14.

f (x, y) = ln |r|, where r = xi+yj. Since |r| = we have

∇ f (x, y) =

15.

1 |r|



p

x 2 + y 2,

 x y r i + j = 2. |r| |r| |r|

|r|−n ,

f (x, y, where r = xi + yj + zk. Since pz) = |r| = x 2 + y 2 + z 2 , we have

∇ f (x, y, z) = −n|r| =−

−n−1

nr . |r|n+2



 x y z i+ j+ k |r| |r| |r|

f (x, y) = x y, ∇ f (x, y) = yi + xj, ∇ f (2, 0) = 2j. Let u = u 1 i + u 2 j be a unit vector. Thus u 21 + u 22 = 1. We have

If −2 = Du f (2, 0) = 2u 2 , then u 2 = −1 and u 1 = 0. At (2, 0), f has rate of change −2 in the direction −j.

18.

f (x, y, z) = x 2 + y 2 − z 2 .

∇ f (a, b, c) = 2ai + 2bj − 2ck. The maximum

rate of change of f at (a, b, c) is in the direction of ∇ f (a, b, c), and is equal to |∇ f (a, b, c)|. Let u be a unit vector making an angle θ with ∇ f (a, b, c). The rate of change of f at (a, b, c) in the direction of u will be half of the maximum rate of change of f at that point provided 1 |∇ f (a, b, c)| = u • ∇ f (a, b, c) = |∇ f (a, b, c)| cos θ, 2

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SECTION 12.7 (PAGE 725)

1 that is, if cos θ = , which means θ = 60◦ . At (a, b, c), 2 f increases at half its maximal rate in all directions mak◦ ing 60 angles with the direction ai + bj − ck.

19. Let ∇ f (a, b) = ui + vj. Then √ i+j u+v 3 2 = D(i+j)/√2 f (a, b) = √ • (ui + vj) = √ 2 2 3i − 4j 3u − 4v 5 = D(3i−4j)/5 f (a, b) = • (ui + vj) = . 5 5 Thus u + v = 6 and 3u − 4v = 25. This system has solution u = 7, v = −1. Thus ∇ f (a, b) = 7i − j.

d) If the ant moves at speed k in the direction −i − 2j, it experiences temperature changing at rate −i − 2j 12k √ • (4i + 4j)k = − √ , 5 5 √ that is, decreasing at rate 12k/ 5 degrees per unit time. e) To continue to experience maximum rate of cooling, the ant should crawl along the curve x = x(t), y = y(t), which is everywhere tangent to ∇ T (x, y). Thus we want dx dy i+ j = λ(2xi − 4yj). dt dt

20. Given the values Dφ1 f (a, b) and Dφ2 f (a, b), we can

1 dy 2 dx = − , from which we obtain, on y dt x dt integration,

solve the equations

Thus

f 1 (a, b) cos φ1 + f 2 (a, b) sin φ1 = Dφ1 f (a, b) f 1 (a, b) cos φ2 + f 2 (a, b) sin φ2 = Dφ2 f (a, b)

ln |y(t)| = −2 ln |x(t)| + ln |C|,

for unique values of f 1 (a, b) and f 2 (a, b) (and hence determine ∇ f (a, b) uniquely), provided the coefficients satisfy cos φ1 0 6= cos φ2

22. Let the curve  be y =  g(x). At (x, y) this curve has normal ∇ g(x) − y = g ′ (x)i − j.

sin φ1 = sin(φ2 − φ1 ). sin φ2

Thus φ1 and φ2 must not differ by an integer multiple of π.

21.

a) T (x, y) = x 2 − 2y 2 .

or yx 2 = C. Since the curve passes through (2, −1), we have yx 2 = −4. Thus, the ant should crawl along the path y = −4/x 2 .

A curve of the family x 4 + y 2 = C has normal ∇ (x 4 + y 2 ) = 4x 3 i + 2yj. These curves will intersect at right angles if their normals are perpendicular. Thus we require that 0 = 4x 3 g ′ (x) − 2y = 4x 3 g ′ (x) − 2g(x),

y

or, equivalently, T =−8 T =−2

T =2 T =0

x (2,−1) T =8

1 g ′ (x) = 3. g(x) 2x 1 Integration gives ln |g(x)| = − 2 + ln |C|, 4x 2 or g(x) = Ce−(1/4x ) . Since the curve passes through (1, 1), we must have 1 = g(1) = Ce−1/4 , so C = e1/4 . 2 The required curve is y = e(1/4)−(1/4x ) .

23. Let the curve be y = f (x). At (x, y) it has normal

dy i − j. dx The curve x 2 y 3 = K has normal 2x y 3 i + 3x 2 y 2 j. These curves will intersect at right angles if their normals are perpendicular, that is, if

T =−8

Fig. 12.7.21 b) ∇ T = 2xi − 4yj, ∇ T (2, −1) = 4i + 4j. An ant at (2, −1) should move in the direction of −∇ T (2, −1), that is, in the direction −i − j, in order to cool off as rapidly as possible. c) If the ant moves at speed k in the direction −i − j, it will experience √ temperature decreasing at rate |∇ T (2, −1)|k = 4 2k degrees per unit time.

dy − 3x 2 y 2 = 0 dx dy 3x = dx 2y 2y d y = 3x d x 3 y 2 = x 2 + C. 2

2x y 3

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SECTION 12.7 (PAGE 725)

ADAMS and ESSEX: CALCULUS 8

Since the curve must pass through (2, −1), we have 1 = 6 + C, so C = −5. The required curve is 3x 2 − 2y 2 = 10.

24. Let f (x, y) = e−(x

2 +y 2 )

. Then

∇ f (x, y) = −2e−(x The tion The tion

At (2, 3, 1) this second directional derivative has value −4/3.

26. At (1, −1, 1) the surface x 2 + y 2 = 2 has normal 2 +y 2 )

(xi + yj).

ai + bj vector u = √ is a unit vector in the direca 2 + b2 directly away from the origin at (a, b). first directional derivative of f at (x, y) in the direcof u is 2 2 2 u • ∇ f (x, y) = − √ (ax + by)e−(x +y ) . 2 2 a +b

The second directional derivative is   2 2 2 u • ∇ −√ (ax + by)e−(x +y ) a 2 + b2 2 2 2 (ai + bj) • e−(x +y ) =− 2 2 a + b    a − 2x(ax + by) i + b − 2y(ax + by) j .

n1 = ∇ (x 2 + y 2 )

(1,−1,1)

= 2i − 2j,

and y 2 + z 2 = 2 has normal n2 = ∇ (y + z ) 2

2

(1,−1,1)

= −2j + 2k.

A vector tangent to the curve of intersection of the two surfaces at (1, −1, 1) must be perpendicular to both these normals. Since (i − j) × (−j + k) = −(i + j + k), the vector i + j + k, or any scalar multiple of this vector, is tangent to the curve at the given point.

At (a, b) this second directional derivative is 2

2

 2e−(a +b )  2 a − 2a 4 − 2a 2 b2 + b2 − 2a 2 b2 − 2b4 − 2 2 a +b  2  2 2 2 2 2 −(a 2 +b2 ) = 2 2(a + b ) − a − b e a + b2  = 2 2(a 2 + b2 ) − 1 e−(a

2 +b2 )

.

2

25.

Remark: Since f (x, y) = e−r (expressed in terms of polar coordinates), the second directional derivative of f at (a, b) in the direction directly away from the origin (i.e., the direction of increasing r ) can be more easily calculated as d 2 −r 2 e 2 2 2. dr 2 r =a +b

f (x, y, z) = x yz, ∇ f (x, y, z) = yzi + x zj + x yk. The first directional derivative of f in the direction i − j − k is

i−j−k 1 √ • ∇ f (x, y, z) = √ (yz − x z − x y). 3 3 The second directional derivative in that direction is i−j−k 1 √ • √ ∇ (yz − x z − x y) 3 3 i i−j−k h = • −(y + z)i + (z − x)j + (y − x)k 3 i 2x − 2y − 2z 1h = −(y + z) − (z − x) − (y − x) = . 3 3

27. The vector n1 = i + j + k is normal to the plane

x + y + z = 6 at (1, 2, 3). A normal to the sphere x 2 + y 2 + z 2 = 14 at that point is n2 = ∇ (x + y + z ) 2

2

2

(1,2,3)

= 2i + 4j + 6k.

A vector tangent to the circle of intersection of the two surfaces at (1, 2, 3) is i n1 × n2 = 1 2

j 1 4

k 1 = 2i − 4j + 2k. 6

Any vector parallel to i − 2j + k is tangent to the circle at (1, 2, 3).

28. A vector tangent to the path of the fly at (1, 1, 2) is given by v = ∇ (3x 2 − y 2 − z) × ∇ (2x 2 + 2y 2 − z 2 ) (1,1,2) = (6xi − 2yj − k) × (4xi + 4yj − 2zk) (1,1,2)

= (6i − 2j − k) × (4i + 4j − 4k) j k i = 4 6 −2 −1 = 4(3i + 5j + 8k). 1 1 −1

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SECTION 12.7 (PAGE 725)

The temperature T = x 2 − y 2 + z 2 + x z 2 has gradient at (1, 1, 2) given by ∇ T (1, 1, 2) = (2x + z 2 )i − 2yj + 2z(1 + x)k

for some θ between 0 and 1. Since ∇ f is assumed to vanish throughout the disk x 2 + y 2 < r 2 , this implies that f (x, y) = f (0, 0) throughout the disk, that is, f is constant there. (Note that Theorem 3 of Section 3.6 can be used instead of Exercise 18 of Section 3.6 in this argument.)

(1,1,2)

= 6i − 2j + 8k.

Thus the fly, passing through (1, 1, 2) with speed 7, experiences temperature changing at rate 7×

32. Let f (x, y) = x 3 − y 2 . Then ∇ f (x, y) = 3x 2 i − 2yj

exists everywhere, but equals 0 at (0, 0). The level curve of f passing through (0, 0) is y 2 = x 3 , which has a cusp at (0, 0), so is not smooth there.

v 3i + 5j + 8k • ∇ T (1, 1, 2) = 7 √ • (6i − 2j + 8k) |v| 98 1 72 = √ (18 − 10 + 64) = √ . 2 2

y y 3 =x 2

We don’t know which direction the fly is moving along the curve, so all we can say √ is that it experiences temperature changing at rate 36 2 degrees per unit time.

29. If f (x, y, z) is differentiable at the point (a, b, c) and ∇ f (a, b, c) 6= 0, then ∇ f (a, b, c) is normal to the level surface of f which passes through (a, b, c).

x

Fig. 12.7.32

33. Let v = v 1 i + v 2 j + v 3 k. Thus ∂f ∂f ∂f + v2 + v3 Dv f = v 1 ∂x ∂y ∂z   ∂2 f ∂2 f ∂2 f i ∇(Dv f ) = v 1 2 + v 2 + v3 ∂x ∂ x∂ y ∂ x∂z   ∂2 f ∂2 f ∂2 f + v1 + v2 2 + v3 j ∂ y∂ x ∂y ∂ y∂z   ∂2 f ∂2 f ∂2 f + v2 + v3 2 k + v1 ∂ x∂z ∂ y∂z ∂z Dv (Dv f ) = v • ∇ (Dv f )

The proof is very similar to that of Theorem 6 of Section 3.7, modified to include the extra variable. The angle θ between ∇ f (a, b, c) and the secant vector from (a, b, c) to a neighbouring point (a + h, b + k, c + ℓ) on the level surface of f passing through (a, b, c) satisfies cos θ =

∇ f (a, b, c) • (hi + kj + ℓk) √ |∇ f (a, b, c)| h 2 + k 2 + ℓ2

h f 1 (a, b, c) + k f2 (a, b, c) + ℓ f 3 (a, b, c) √ |∇ f (a, b, c)| h 2 + k 2 + ℓ2 h −1 = √ f (a + h, b + k, c + ℓ) |∇ f (a, b, c)| h 2 + k 2 + ℓ2

=

→0

∂2 f ∂2 f ∂2 f + 2v 1 v 3 + 2v 1 v 2 2 ∂x ∂ x∂ y ∂ x∂z 2f 2f ∂ ∂ ∂2 f + v 22 2 + 2v 2 v 3 + v 32 2 ∂y ∂ y∂z ∂z

= v 12

i − f (a, b, c) − h f 1 (a, b, c) − k f2 (a, b, c) − ℓf 3 (a, b, c) as (h, k, ℓ) → (0, 0, 0)

π because f is differentiable at (a, b, c). Thus θ → , and 2 ∇ f (a, b, c) is normal to the level surface of f through (a, b, c).

30. The level surface of f (x, y, z) = cos(x + 2y + 3z) through (π, π, π ) has equation cos(x + 2y + 3z) = cos(6π ) = 1, which simplifies to x + 2y + 3z = 6π . This level surface is a plane, and is therefore its own tangent plane. We cannot determine this plane by the method used to find the tangent plane to the level surface of f through (π/2, π, π ) in Exercise 10, because ∇ f (π, π, π ) = 0, so the gradient does not provide a usable normal vector to define the tangent plane.

31. By the version of the Mean-Value Theorem in Exercise 18 of Section 3.6, f (x, y) = f (0, 0) + x f 1 (θ x, θ y) + y f 2 (θ x, θ y)

(assuming all second partials are continuous). Dv (Dv f ) gives the second time derivative of the quantity f as measured by an observer moving with constant velocity v.

34.

T = T (x, y, z). As measured by the observer, dT = Dv(t) T = v(t) • ∇ T dt d2T d = a(t) • ∇ T + v(t) • ∇ T 2 dt dt   d ∂T = Da(t) T + v 1 (t) +··· dt ∂ x   ∂T = Da(t) T + v 1 (t)v(t) • ∇ +··· ∂x   2 ∂ 2 T ∂2T = Da(t) T + v 1 (t) + v 1 (t)v 2 (t) + ··· ∂x2 ∂ y∂ x = Da(t) T + Dv(t) (Dv(t) T )

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SECTION 12.7 (PAGE 725)

ADAMS and ESSEX: CALCULUS 8

Section 12.8

(as in Exercise 37 above).

35.

T = T (x, y, z, t). The calculation is similar to that of Exercise 38, but produces a few more terms because of the dependence of T explicitly on time t. We continue to use ∇ to denote the gradient with respect to the spatial variables only. Using the result of Exercise 38, we have dT ∂T = + v(t) • ∇ T dt ∂t d2T d ∂T d = + v(t) • ∇ T dt 2 dt ∂t dt ∂2T ∂T = 2 + v(t) • ∂t ∂t ∂ + v(t) • ∇ T + Da(t) T + Dv(t) (Dv(t) T ) ∂t   2 ∂ T ∂T + Da(t) T + Dv(t) (Dv(t) T ). = 2 + 2Dv(t) ∂t ∂t

36.

  sin(x y) p 2 2 f (x, y) =  x +y 0

if (x, y) 6= (0, 0)

0−0 = 0 = f 2 (0, 0). Thus a) f 1 (0, 0) = lim h→0 h ∇ f (0, 0) = 0. √ b) If u = (i + j)/ 2, then

x y 3 + x 4 y = 2 defines x as a function of y. dx dx + 3x y 2 + 4x 3 y + x4 = 0 y3 dy dy x 4 + 3x y 2 dx =− 3 . dy y + 4x 3 y The given equation has a solution x = x(y) with this derivative near any point where y 3 + 4x 3 y 6= 0, i.e., y 6= 0 and y 2 + 4x 3 6= 0.

2.

x y 3 = y − z: x = x(y, z) ∂x y3 + 3x y 2 = 1 ∂y 1 − 3x y 2 ∂x = . ∂y y3 The given equation has a solution x = x(y, z) with this partial derivative near any point where y 6= 0.

3.

z2 + x y3 =

1 sin(h 2 /2) 1 √ = . h→0+ h 2 h2

37.

  2x 2 y f (x, y) = x 4 + y 2 if (x, y) 6= (0, 0) .  0 if (x, y) = (0, 0) Let u = ui + vj be a unit vector. If v 6= 0, then

y = y(x, z) ∂y x 2z ∂ y 2 e z + y − x ln y − =0 ∂z y ∂z ∂y x 2 ln y − ye yz x 2 y ln y − y 2 e yz = = . ∂z yze yz − x 2 z x 2z ze yz − y The given equation has a solution y = y(x, z) with this derivative near any point where y > 0, z 6= 0, and ye yz 6= x 2 . yz

1 2(h 2 u 2 )(hv) h→0+ h h 4 u 4 + h 2 v 2 2u 2 v 2u 2 = . = lim 2 4 2 h→0+ h u + v v

Du f (0, 0) = lim =

5.

h→0+

1 0 = 0. h h2

Thus f has a directional derivative in every direction at the origin even though it is not continuous there.

z = z(x, y)

2 4. e yz − = π:  x z ln y 

Du f (0, 0) = lim

If v = 0, then u = ±1 and

xz : y

∂z x ∂z xz 2z + 3x y 2 = − 2 ∂y y ∂y y xz 2 + 3x y x z + 3x y 4 ∂z y2 = x = . ∂y x y − 2y 2 z − 2z y The given equation has a solution z = z(x, y) with this derivative near any point where y 6= 0 and x 6= 2yz.

Du f (0, 0) = lim

c) f cannot be differentiable at (0, 0); if it were, then the directional derivative obtained in part (b) would have been u • ∇ f (0, 0) = 0.

(page 736)

1.

.

if (x, y) = (0, 0)

Implicit Functions

x 2 y 2 + y 2 z 2 + z 2 t 2 + t 2 w2 − xw = 0: ∂x ∂x 2x y 2 + 2t 2 w − w −x =0 ∂w ∂w 2 ∂x x − 2t w = . ∂w 2x y 2 − w

x = x(y, z, t, w)

The given equation has a solution with this derivative wherever w 6= 2x y 2 .

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INSTRUCTOR’S SOLUTIONS MANUAL

6.

SECTION 12.8 (PAGE 736)

F(x, y, x 2 − y 2 ) = y = y(x)  0: dy dy F1 + F2 + F3 2x − 2y =0 dx dx F1 (x, y, x 2 − y 2 ) + 2x F3 (x, y, x 2 − y 2 ) dy = . dx 2y F3 (x, y, x 2 − y 2 ) − F2 (x, y, x 2 − y 2 )

11.



x 2 + y 2 + z 2 + w2 = 1 x + 2y + 3z + 4w = 2 ∂x ∂y ∂x ∂y

2x

The given equation has a solution with this derivative near any point where F2 (x, y, x 2 − y 2 ) 6= 2y F3 (x, y, x 2 − y 2 ).

7.

G(x, y, z, u, v) = 0: u = u(x, y, z, v) ∂u G1 + G4 ∂x G 1 (x, y, z, u, v) ∂u =− . ∂x G 4 (x, y, z, u, v)

12.

F(x2 − z 2 , y 2 +x z) = 0: z = z(x, y)  ∂z ∂z F1 2x − 2z +z =0 + F2 x ∂x ∂x ∂z 2x F1 (x 2 − z 2 , y 2 + x z) + z F2 (x 2 − z 2 , y 2 + x z) = . ∂x 2z F1 (x 2 − z 2 , y 2 + x z) − x F2 (x 2 − z 2 , y 2 + x z)

10.

H (u 2 w, v 2 t, wt) = 0:  w = w(u,  v, t) ∂w 2 ∂w 2 H1 u + H2 v + H3 t +w = 0 ∂t ∂t ∂w H2 (u 2 w, v 2 t, wt)v 2 + H3 (u 2 w, v 2 t, wt)w =− . ∂t H1 (u 2 w, v 2 t, wt)u 2 + H3 (u 2 w, v 2 t, wt)t The given equation has a solution with this derivative near any point where t H3 (u 2 w, v 2 t, wt) 6= −u 2 H1 (u 2 w, v 2 t, wt).   x yuv = 1 y = y(x, u) ⇒ x +y+u+v =0 v = v(x, u)

2x

+

+

1

+

∂y ∂x ∂y ∂x

xuv

+ +

∂v ∂x ∂v ∂x

x yu

+

2

+

=

0

=

0

∂w ∂y ∂w 4 ∂y

2w

=

0

×2

=

0

×w

dy u dx

dy dx

+

(y 2 − 3u 2 ) du y dx

+

du dx

=

0

=

0

Multiply the first equation by u and the second by x 2 + 2yu and subtract: 2x(x 2 + yu) + (x 2 y + y 2 u + 3u 3 )

du =0 dx

du 2x(x 2 + yu) x =− 3 = − 3. dx 3u + x 2 y + y 2 u 2u

13.

The given equations have a solution with the indicated derivative near any point where u 6= 0.   u = u(x, y) x = u3 + v 3 ⇒ v = v(x, y) y = uv − v 2 Take partials with respect to x: 1

=

0

=

∂u ∂x ∂u v ∂x

3u 2

+ +

∂v ∂x ∂v (u − 2v) . ∂x 3v 2

At u = v = 1 we have

Multiply the last equation by x yu and subtract the two equations: ∂y yuv − x yu + (xuv − x yu) =0 ∂ x   ∂y y(x − v) = . ∂x u x(v − y)

(x 2 + 2yu)

+

Differentiate the given equations with respect to x: yuv

+

The given equations have a solution of the indicated form with this derivative near any point where w 6= 4x.   u = u(x) x 2 y + y 2u − u 3 = 0 ⇒ y = y(x) x 2 + yu = 1 2x y

The given equation has a solution with this derivative near any point where x F2 (x 2 − z 2 , y 2 + x z) 6= 2z F1 (x 2 − z 2 , y 2 + x z).

9.

2y

x = x(y, z) w = w(y, z)

∂x (4x − w) + 4y − 2w = 0 ∂y   ∂x 2w − 4y . = ∂y z 4x − w

The given equation has a solution with this derivative near any point where G 4 (x, y, z, u, v) 6= 0.

8.

+



⇒

1

=

0

=

∂u ∂x ∂u ∂x

3

+ −

∂v ∂x ∂v . ∂x

3

∂u ∂v 1 = = . ∂x ∂x 6 Similarly, differentiating the given equations with respect to y and putting u = v = 1, we get

Thus

The given equations have a solution of the indicated form with this derivative near any point where u 6= 0, x 6= 0 and y 6= v.

0 = 1 =

∂u ∂y ∂u ∂y

3

+ −

∂v ∂y ∂v . ∂y

3

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SECTION 12.8 (PAGE 736)

∂u ∂v 1 =− = . ∂y ∂y 2 Finally, ∂(u, v) = ∂(x, y)

ADAMS and ESSEX: CALCULUS 8

17. Let F(x, y, z, u, v) = x y 2 + zu + v 2 − 3

Thus

14.



1 6 1 2

x = r 2 + 2s y = s 2 − 2r

1 6

= −1. − 12 6

∂(x, y) 2r 2 = 4(r s + 1). = −2 2s ∂(r, s)

The given system can be solved for r and s as functions of x and y near any point (r, s) where r s 6= −1. We have ∂r ∂s 1 = 2r + 2 ∂x ∂x ∂r ∂s 0 = −2 + 2s ∂x ∂x ∂r ∂s 0 = 2r + 2 ∂y ∂y ∂s ∂r + 2s . 1 = −2 ∂y ∂y Thus ∂r s = ∂x 2(r s + 1) ∂s 1 = ∂x 2(r s + 1)

15.

x = r cos θ,

y = r sin θ ∂(x, y) cos θ = sin θ ∂(r, θ )

2 ∂(F, G, H ) y 2 = 3x z ∂(x, y, z) u − yz

−r sin θ = r. r cos θ

∂(F, G, H ) 1 = 3 ∂(x, y, z) 0

u 3 x . −x y

2 1 2 1 = 4. 0 −1

Since this Jacobian is not zero, the equations F = G = H = 0 can be solved for x, y, and z as functions of u and v near P0 . Also, 

∂y ∂u



v (1,1)

1 ∂(F, G, H ) 4 ∂(x, u, z) P0 2 z u 1 y 2 3 = − 3x z −v x 4 u − yz x −x y P0 1 1 1 1 3 = − 3 −1 1 = − . 4 0 1 −1 2

=−

18. Let F(x, y, z, u, v) = xe y + uz − cos v − 2

G(x, y, z, u, v) = u cos y + x 2 v − yz 2 − 1. If P0 is the point where (x, y, z) = (2, 0, 1) and (u, v) = (1, 0), then z ∂(F, G) sin v = x 2 P0 ∂(u, v) P0 cos y 1 0 = 4. = 1 4

x = R sin φ cos θ , y = R sin φ sin θ , z = R cos φ. sin φ cos θ R cos φ cos θ −R sin φ sin θ ∂(x, y, z) = sin φ sin θ R cos φ sin θ R sin φ cos θ ∂(R, φ, θ ) cos φ −R sin φ 0 R cos φ cos θ −R sin φ sin θ = cos φ R cos φ sin θ R sin φ cos θ sin φ cos θ −R sin φ sin θ + R sin φ sin φ sin θ R sin φ cos θ h i = R 2 cos φ cos φ sin φ cos2 θ + sin φ cos φ sin2 θ h i + R 2 sin φ sin2 φ cos2 θ + sin2 φ sin2 θ

2x y 2 v − xz

At point P0 where x = y = z = u = v = 1, we have

∂r 1 =− ∂y 2(r s + 1) ∂s r = . ∂y 2(r s + 1)

The transformation is one-to-one (and hence invertible) near any point where r 6= 0, that is, near any point except the origin.

16.

G(x, y, z, u, v) = x 3 z + 2y − uv − 2 H (x, y, z, u, v) = xu + yv − x yz − 1. Then

Since this Jacobian is not zero, the equations F = G = 0 can be solved for u, and v in terms of x, y and z near P0 . Also, 

= R 2 cos2 φ sin φ + R 2 sin3 φ = R 2 sin φ.

The transformation is one-to-one (and invertible) near any point where R 2 sin φ 6= 0, that is, near any point not on the z-axis.

482 Copyright © 2014 Pearson Canada Inc.

∂u ∂z





x,y (2,0,1)

1 ∂(F, G) 4 ∂(z, v) P0 1 u sin v = − 4 −2yz x 2 P0 1 1 0 = −1. = − 4 0 4 =−

INSTRUCTOR’S SOLUTIONS MANUAL

19.

(

F(x, y, z, w) G(x, y, z, w) H (x, y, x, w)

=0 =0 =0

⇒

(x

z w

SECTION 12.8 (PAGE 736)

=x(y) =z(y) =w(y)

dx dz dw + F2 + F3 + F4 =0 dy dy dy dx dz dw G1 + G2 + G3 + G4 =0 dy dy dy dx dz dw H1 + H2 + H3 + H4 =0 dy dy dy

There are 15 possible interpretations for We have

F1



x4 x6 x7 x8

⇒

∂(F, G, H ) ∂(x2 , x3 , x5 ) . =− ∂(F, G, H ) ∂(x1 , x3 , x5 )

z = z(x, y), then

∂z ∂z = 0, F2 + F3 =0 ∂x ∂y  2 ∂2z ∂z ∂z ∂z + F3 2 = 0. F11 + F13 + F31 + F33 ∂x ∂x ∂x ∂x F1 + F3

∂(F, G, H ) dx ∂(y, z, w) =− . ∂(F, G, H ) dy ∂(x, z, w)

21.



22. If F(x, y, z) = 0

By Cramer’s Rule,

20.

∂ x1 ∂ x2

∂ x1 . ∂ x2

Thus ∂2z 1 =− 2 ∂x F3



F1 − F3





F1 F11 + 2F13 + F33 − F3 i 1 h = − 3 F11 F32 − 2F1 F3 F13 + F12 F33 . F3

F(x, y, z, u, v) = 0 G(x, y, z, u, v) = 0 H (x, y, z, u, v) = 0 ∂x To calculate we require that x be one of three de∂y pendent variables, and y be one of two independent variables. The other independent variable can be z or u or v. The possible interpretations for this partial, and their values, are ∂(F, G, H )   ∂x ∂(y, u, v) =− ∂(F, G, H ) ∂y z ∂(x, u, v) ∂(F, G, H )   ∂x ∂(y, z, v) =− ∂(F, G, H ) ∂y u ∂(x, z, v) ∂(F, G, H )   ∂x ∂(y, z, u) =− . ∂(F, G, H ) ∂y v ∂(x, z, u)

F(x1 , x2 , . . . , x8 ) = 0 G(x1 , x2 , . . . , x8 ) = 0 H (x1 , x2 , . . . , x8 ) = 0 ∂ x1 To find we require that x1 be one of three depen∂ x2 dent variables, and that x2 be one of five independent variables. The other four independent variables must be chosen from among the six remaining variables. This can be done in   6! 6 = = 15 ways. 4 4!2!

"

2 #

Similarly, i 1 h ∂2z 2 2 = − F F − 2F F F + F F . 22 2 3 23 33 3 2 ∂ y2 F33 Also, F12 + F13

  ∂z ∂z ∂2z ∂z + F32 + F33 + F3 . ∂y ∂y ∂x ∂ y∂ x

Therefore      F2 F1 1 ∂2z =− F12 + F13 − + F23 − ∂ x∂ y F3 F3 F3 ! F1 F2 + F33 F32 h i 1 = − 2 F32 F12 − F2 F3 F13 − F1 F3 F23 + F1 F2 F33 . F3

23.

x = u + v, y = uv, z = u 2 + v 2 . The first two equations define u and v as functions of x and y, and therefore derivatives of z with respect to x and y can be determined by the Chain Rule. Differentiate the first two equations with respect to x: 1

=

0

=

∂u ∂x ∂u v ∂x

+ +

∂v ∂x ∂v u . ∂x

483 Copyright © 2014 Pearson Canada Inc.

SECTION 12.8 (PAGE 736)

Thus

ADAMS and ESSEX: CALCULUS 8

∂u u ∂v v = and = , and ∂x u−v ∂x v −u

∂z ∂u ∂z ∂v ∂z = + ∂x ∂u ∂ x ∂v ∂ x u v 2(u 2 − v 2 ) = 2u + 2v = = 2(u + v) = 2x. u−v v −u u−v Similarly, differentiating the first two of the given equations with respect to y, we get 0

=

1

=

∂u ∂y ∂u v ∂y

∂v ∂y ∂v + u . ∂y

4p , T = T ( p, V ) T2 ∂T 4 8p ∂T a) V = − 2 + 3 ∂p T T ∂p ∂T 8p ∂T p= + 3 . ∂V T ∂V Putting p = V = 1 and T = 2, we obtain

pV = T −

so

∂T = 2, ∂p

2

∂T = 1, ∂V

∂T ∂T 1 = 1 and = . ∂p ∂V 2

∂T ∂T dp + dV . ∂p ∂V If p = 1, |d p| ≤ 0.001, V = 1, and |d V | ≤ 0.002, then T = 2 and

b) d T =

1 |d T | ≤ (1)(0.001) + (0.002) = 0.002. 2 The approximate maximum error in T is 0.002.

25.

F(x, y, z) = 0   ∂x F1 + F2 = 0, ∂y z Similarly,





∂y ∂z

∂x ∂y



x

=−

  z

∂y ∂z

⇒

F3 , and F2

  x

∂z ∂x







y

∂z ∂x

∂x ∂y





=−

y

= (−1)5 = −1.

= (−1)n .

∂z 2u 2v 2(u − v) = + = = −2 ∂y v −u u−v v −u ∂2z = 0. ∂ x∂ y

2

For F(x, y, z, u, v) = 0 we have, similarly,           ∂y ∂z ∂u ∂v ∂x ∂ y z,u,v ∂z u,v,x ∂u v,x,y ∂v x,y,z ∂ x y,z,u In general, if F(x1 , x2 , . . . , xn ) = 0, then       ∂ x1 ∂ xn ∂ x2 ··· ∂ x2 x3 ,...,xn ∂ x3 x4 ,...,xn ,x1 ∂ x1 x2 ,...,xn−1

+

1 ∂v 1 ∂u = and = , and Thus ∂y v −u ∂y u−v

24.

For F(x, y, z, u) = 0 we have, similarly,         ∂x ∂y ∂z ∂u = (−1)4 = 1. ∂ y z,u ∂z u,x ∂u x,y ∂ x y,z

z

=−

F2 . F1

F1 . Hence F3

= (−1)3 = −1.

26. Given F(x, y, u, v) = 0, 1=

G(x, y, u, v) = 0, let

∂ F ∂G ∂ F ∂G ∂(F, G) = − . ∂(x, y) ∂x ∂y ∂y ∂x

Then, regarding the given equations as defining x and y as functions of u and v, we have ∂x 1 ∂(F, G) ∂y 1 ∂(F, G) =− =− ∂u 1 ∂(u, y) ∂u 1 ∂(x, u) ∂y 1 ∂(F, G) ∂x 1 ∂(F, G) =− . =− ∂v 1 ∂(x, v) ∂v 1 ∂(v, y) Therefore,   ∂(x, y) 1 ∂(F, G) ∂(F, G) ∂(F, G) ∂(F, G) = 2 − ∂(u, v) 1 ∂(u, y) ∂(x, v) ∂(v, y) ∂(x, u)    1 ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G = 2 − − 1 ∂u ∂ y ∂ y ∂u ∂ x ∂v ∂v ∂ x    ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G − − − ∂v ∂ y ∂ y ∂v ∂ x ∂u ∂u ∂ x  ∂ F ∂G ∂ F ∂G 1 ∂ F ∂G ∂ F ∂G − = 2 1 ∂u ∂ y ∂ x ∂v ∂ y ∂u ∂ x ∂v ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G − + ∂u ∂ y ∂v ∂ x ∂ y ∂u ∂v ∂ x ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G − + ∂v ∂ y ∂ x ∂u ∂v ∂ y ∂u ∂ x  ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G − + ∂ y ∂v ∂ x ∂u ∂ y ∂v ∂u ∂ x  1 ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G = 2 + 1 ∂u ∂ y ∂ x ∂v ∂ y ∂u ∂v ∂ x  ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G − − ∂v ∂ y ∂ x ∂u ∂ y ∂v ∂u ∂ x    1 ∂ F ∂G ∂ F ∂G ∂ F ∂G ∂ F ∂G = 2 − − 1 ∂x ∂y ∂y ∂x ∂u ∂v ∂v ∂u 1 ∂(F, G) ∂(F, G) = 2 1 ∂(x, y) ∂(u, v)  1 ∂(F, G) ∂(F, G) ∂(F, G) = = . 1 ∂(u, v) ∂(u, v) ∂(x, y)

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 12.8 (PAGE 736)

30. Let u = f (x, y) and v = g(x, y), and suppose that 27. By Exercise 26, with the roles of (x, y) and (u, v) re-

∂(u, v) ∂( f, g) = =0 ∂(x, y) ∂(x, y)

versed, we have ∂(u, v) ∂(F, G) = ∂(x, y) ∂(x, y)



for all (x, y). Thus

∂(F, G) . ∂(u, v)

∂ f ∂g ∂ f ∂g − = 0. ∂x ∂y ∂y ∂x

Apply this with

Now consider the equations u = f (x, y) and v = g(x, y) as defining u and y as functions of x and v. Holding v constant and differentiating with respect to x, we get

F(x, y, u, v) = f (u, v) − x = 0 G(x, y, u, v) = g(u, v) − y = 0 so that

∂g ∂g ∂ y + = 0, ∂x ∂y ∂x

∂(F, G) −1 0 =1 = 0 −1 ∂(x, y) and ∂(F, G) ∂( f, g) ∂(x, y) = = ∂(u, v) ∂(u, v) ∂(u, v)

and



and we obtain  ∂(x, y) ∂(u, v) =1 . ∂(x, y) ∂(u, v)

28. By the Chain Rule,  ∂x

∂x   ∂r ∂s  ∂y ∂y ∂r  ∂ x ∂s ∂ x ∂v ∂u + ∂u ∂r ∂v ∂r =  ∂ y ∂u ∂ y ∂v + ∂u ∂r ∂v  ∂ x ∂ x  ∂r∂u ∂v   ∂r =  ∂u ∂y ∂y ∂v ∂u ∂v ∂r

31. ∂ x ∂u + ∂u ∂s ∂ y ∂u + ∂u ∂s ∂u ∂s  ∂v . ∂s

∂x ∂v ∂y ∂v

∂v  ∂s  ∂v ∂s

Since the determinant of a product of matrices is the product of their determinants, we have



29. If f (x, y) = k g(x, y) , then   ∂g ∂f = k ′ g(x, y) , ∂x ∂x



v

∂f ∂ f ∂y + ∂x ∂y ∂x   ∂ f ∂g 1 ∂ f ∂g − = 0. = ∂g ∂ x ∂ y ∂y ∂x ∂y =

This says that u = u(x, v) is independent of x, and so depends only on v: u = k(v)  for some  function k of one variable. Thus f (x, y) = k g(x, y) , so f and g are functionally dependent.     2S 5 3h 2 N N 2/3 3N k − 3 Solving the equation, E = e , 4π m V for S we find " !   # 3N k 5 4π m E V 2/3 S= + ln = u(E, V , N ). 2 3 3h 2 N N Substituting for E using E = 32 N kT, we find that " !   # 3N k 5 2π mkT V 2/3 S= + ln = v(T, V , N ). 2 h2 N 3

32. The independent variables are T , V , and N . This suggests that we work with the Legendre transformation F = E − T S, which is called the Helmholtz free energy. Taking the differential of F gives,

∂(x, y) ∂(u, v) ∂(x, y) = . ∂(r, s) ∂(u, v) ∂(r, s) 

∂u ∂x

  ∂g ∂f = k ′ g(x, y) . ∂y ∂y

Therefore,   ∂(g, g) ∂( f, g) = k ′ g(x, y) = 0. ∂(x, y) ∂(r, s)

d F = d E − S dT − T dS = T dS − P dV + µ d N − S dT − T dS = −S d T − P d V + µ d N, which confirms that F does indeed depend on T , V , and N , and that   ∂F = F1 (T, V , N ) = −S ∂ T V,N   ∂F = F2 (T, V , N ) = −P. ∂ V T,N

485 Copyright © 2014 Pearson Canada Inc.

SECTION 12.8 (PAGE 736)

ADAMS and ESSEX: CALCULUS 8

Section 12.9 Taylor’s Formula, Taylor Series, and Approximations (page 742)

By equality of mixed partials, 

∂P ∂T



V,N

= −F21 (T, V , N ) = −F12 (T, V , N ) =



∂S ∂V



. T,N

1. Since the Maclaurin series for 33. The independent variables are S ,P, and N . This suggests that we work with the Legendre transformation W = E + PV , which is called the Enthalpy. Taking the differential of W gives,

1 is 1+t

1 − t + t2 − · · · =

∞ X (−1)n t n , n=0

the Taylor series for dW = d E + P dV + V d P = T dS − P dV + µ d N + P dV + V d P = T d S + V d P + µ d N,

f (x, y) =

which confirms that W does indeed depend on S, P, and N , and that about (0, 0) is 

 ∂W = W1 (S, P, N ) = T ∂ S P,N   ∂W = W2 (S, P, N ) = V . ∂ P S,N

∂V ∂S



P,N

1 x y2 1+ 2

∞ X x n y 2n (−1)n n+1 . 2 n=0

2. Since f (x, y) = ln(1 + x + y + x y)

  = ln (1 + x)(1 + y)

By equality of mixed partials, 

1 1 = 2 + x y2 2

= ln(1 + x) + ln(1 + y), the Taylor series for f about (0, 0) is

= W21 (S, P, N ) = W12 (S, P, N ) =



∂T ∂P



.

∞ X x n + yn . (−1)n−1 n n=1

S,N

34. The independent variables are T , P, and N . This suggests that we work with G = E − T S + PV , which is the

Gibbs free energy. Taking the differential of G gives,

3. Since f (x, y) = tan−1 (x + x y) = tan−1 (ux), where u = y + 1, the Taylor series for f about (0, −1) is

dG = d E − T d S − S d T + P d V + V d P = T dS − P dV + µ d N − T dS − S dT + P dV + V d P = −S d T + V d P + µ d N,

∞ ∞ X X (ux)2n+1 x 2n+1 (1 + y)2n+1 (−1)n = (−1)n . 2n + 1 2n + 1 n=0 n=0

which confirms that G does indeed depend on T , P, and N , and that

4. Let u = x − 1, v = y + 1. Thus



 ∂G = G 1 (T, P, N ) = −S ∂ T P,N   ∂G = G 2 (T, P, N ) = V . ∂ P T,N

f (x, y) = x 2 + x y + y 3

= (u + 1)2 + (u + 1)(v − 1) + (v − 1)3

= 1 + 2u + u 2 − 1 + v − u + uv + v 3 − 3v 2 + 3v − 1 = −1 + u + 4v + u 2 + uv − 3v 2 + v 3

By equality of mixed partials, 

∂S ∂P



T,N



∂V = −G 12 (T, P, N ) = −G 21 (T, P, N ) = − ∂T



.

= −1 + (x − 1) + 4(y + 1) + (x − 1)2

+ (x − 1)(y + 1) − 3(y + 1)2 + (y + 1)3 .

P,N

This is the Taylor series for f about (1, −1).

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INSTRUCTOR’S SOLUTIONS MANUAL

5.

2

SECTION 12.9 (PAGE 742)

2

f (x, y) = e x +y ∞ X (x 2 + y 2 )n = n! n=0 =

The Taylor polynomial of degree 3 for f near (1, 0) is 2(x − 1) − (x − 1)2 + y 2 − 2(x − 1)3 8 − 2(x − 1)y 2 + (x − 1)3 . 3

∞ n X 1 X n! x 2 j y 2n−2 j n! j !(n − j )! n=0 j =0

∞ X n X x 2 j y 2n−2 j = . j !(n − j )! n=0 j =0 This is the Taylor series for f about (0, 0).

6.

f (x, y) = sin(2x + 3y) = =

9.

=

∞ X (2x + 3y)2n+1 (−1)n (2n + 1)! n=0

x+y 2

Z

x+y 2

0

  1 − t 2 + · · · dt

 x+y 2 t3 + · · · = t− 3 0 1 2 = x + y − (x + y 2 )3 + · · · 3 x3 2 =x+y − +···. 3 The Taylor polynomial of degree 3 for f near (0, 0) is

∞ X X (−1)n 2n+1 (2n + 1)! (2x) j (3y)2n+1− j (2n + 1)! j !(2n + 1 − j )! n=0 j =0

∞ 2n+1 X X (−1)n 2 j 32n+1− j

x + y2 −

7. Let u = x − 2, v = y − 1. Then 1 1 = 2 + x − 2y 2 + (2 + u) − 2(v + 1) 1 1  = =  u − 2v 2 + u − 2v 2 1+ 2 # "     1 u − 2v u − 2v 2 u − 2v 3 = + − + ··· 1− 2 2 2 2

f (x, y) =

=

2

e−t dt

0

Z



x j y 2n+1− j . j !(2n + 1 − j )! n=0 j =0 This is the Taylor series for f about (0, 0). =

f (x, y) =

1 u v u2 uv − + + − 2 4 2 8 2 v2 u3 3u 2 v 3uv 2 v3 + − + − + +···. 2 16 8 4 2

10.

f (x, y) = cos(x + sin y)

(x + sin y)2 (x + sin y)4 + − ··· 2! 4!  2 y3 x+y− +··· (x + y − · · ·)4 6 =1− + −··· 2 4   3 4 xy y 1 x 2 + y 2 + 2x y − − + ··· =1− 2 3 3 1 + (x 4 + 4x 3 y + 6x 2 y 2 + 4x y 3 + y 4 + · · ·). 4 The Taylor polynomial of degree 4 for f near (0, 0) is =1−

The Taylor polynomial of degree 3 for f about (2, 1) is x −2 y − 1 (x − 1)2 1 − + + 2 4 2 8 (x − 2)(y − 1) (y − 1)2 (x − 2)3 − + − 2 2 16 3(x − 2)2 (y − 1) 3(x − 2)(y − 1)2 (y − 1)3 + − + . 8 4 2

x3 . 3

1−

11. Let u = x −

x2 y2 x4 − xy − + + x3y 2 2 4 7x y 3 5y 4 3x 2 y 2 + + . + 2 6 12

π , v = y − 1. Then 2 sin x sin(u + π/2) cos u = = y 1+v 1+v   u2 + · · · (1 − v + v 2 − · · ·) = 1− 2 u2 = 1−v − + v2 + · · · . 2

f (x, y) =

8. Let u = x − 1. Then f (x, y) = ln(x 2 + y 2 ) = ln(1 + 2u + u 2 + y 2 ) (2u + u 2 + y 2 )2 2 (2u + u 2 + y 2 )3 + − ··· 3

= (2u + u 2 + y 2 ) −

= 2u + u 2 + y 2 − 2u 2 − 2u 3 − 2uy 2 +

The Taylor polynomial of degree 2 for f near (π/2, 1) is 8u 3 + ···. 3

1 − (y − 1) −

1 π 2 x− + (y − 1)2 . 2 2 487

Copyright © 2014 Pearson Canada Inc.

SECTION 12.9 (PAGE 742)

12.

ADAMS and ESSEX: CALCULUS 8

1+x 1 + x 2 + y4   = (1 + x) 1 − (x 2 + y 4 ) + · · ·

Thus we must have

f (x, y) =

0 = 1 + a1 1 1 a1 = a2 − a12 2 2 1 1 a2 = a3 − a1 a2 + a13 2 3 1 1 1 a3 − a12 = a4 − a22 − a1 a3 + a12 a2 , 2 8 2

= 1 + x − x2 − · · · . The Taylor polynomial of degree 2 for f near (0, 0) is 1 + x − x 2.

13. The equation x sin y = y + sin x can be written

F(x, y) = 0 where F(x, y) = x sin y − y − sin x. Since F(0, 0) = 0, and F2 (0, 0) = −1 6= 0, the given equation has a solution of the form y = f (x) where f (0) = 0. Try y = a1 x + a2 x 2 + a3 x 3 + a4 x 4 + · · ·. Then sin y = y −

1 3 y + ··· 6

1 = a1 x + a2 x + a3 x + a4 x + · · · − (a1 x + · · ·)3 + · · · . 6 2

3

4

Substituting into the given equation we obtain   1 a1 x 2 + a2 x 3 + a3 − a13 x 4 + · · · 6 = a1 x + a2 x 2 + a3 x 3 + a4 x 4 + · · · + x −

and a1 = −1, a2 = 0, a3 = solution is

y = −x +

14.

a2 = a1 ,

a3 −

1 3 7 4 x − x +···. 3 24

15. The equation x + 2y + z + e2z = 1 can be written

F(x, y, z) = 0, where F(x, y, z) = x + 2y + z + e2z − 1. Since F(0, 0, 0) = 0 and F3 (0, 0, 0) = 3 6= 0, the given equation has a solution of the form z = f (x, y), where f (0, 0) = 0. Try z = Ax + By + C x 2 + Dx y + E y 2 + · · ·. Then

1 3 x + ···. 6

x + 2y + Ax + By + C x 2 + Dx y + E y 2 + · · ·

+ 1 + 2(Ax + By + C x 2 + Dx y + E y 2 + · · ·)

Comparing coefficients of various powers of x on both sides, we get a1 + 1 = 0,

1 7 , a4 = − . The required 3 24

1 = a2 . 6

Thus a1 = −1, a2 = −1, and a3 = −5/6. The required solution is 5 y = −x − x 2 − x 3 + · · · . 6 √ The equation 1 + x y = 1+ x +ln(1+ y) can be rewritten √ F(x, y) = 0, where F(x, y) = 1 + x y −1− x −ln(1+ y). Since F(0, 0) = 0 and F2 (0, 0) = −1 6= 0, the given equation has a solution of the form y = f (x) where f (0) = 0. Try y = a1 x + a2 x 2 + a3 x 3 + a4 x 4 + · · ·. We have p 1 + xy p = 1 + a1 x 2 + a2 x 3 + a3 x4 + · · · 1 = 1 + (a1 x 2 + a2 x 3 + a3 x 4 + · · ·) 2 1 − (a1 x 2 + · · ·)2 + · · · 8 1 + x + ln(1 + y)

= 1 + x + (a1 x + a2 x 2 + a3 x 3 + a4 x 4 + · · ·) 1 1 − (a1 x + a2 x 2 + a3 x 3 + · · ·)2 + (a1 x + a2 x 2 · · ·)3 − · · · 2 3

+ 2(Ax + By + · · ·)2 + · · · = 1.

Thus 1 + A + 2A = 0 2 + B + 2B = 0

⇒ ⇒

E + 2E + 2B 2 = 0

⇒

C + 2C + 2 A2 = 0 D + 2D + 4 AB = 0

⇒ ⇒

A = −1/3 B = −2/3

C = −2/27 D = −8/27

E = −8/27.

The Taylor polynomial of degree 2 for z is 1 2 2 2 8 8 2 − x− y− x − xy − y . 3 3 27 27 27

16. The coefficient of x 2 y in the Taylor series for f (x, y) = tan−1 (x + y) about (0, 0) is 1 1 f 112 (0, 0) = f 112 (0, 0). 2!1! 2 But 1 tan−1 (x + y) = x + y − (x + y)3 + · · · 3 1 3 = x + y − (x + 3x 2 y + 3x y 2 + y 3 ) + · · · 3

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INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 12 (PAGE 743)

y

so the coefficient of x 2 y is −1. Hence f 112 (0, 0) = −2.

T =24

1 . 1 + x 2 + y2 The coefficient of x 2n y 2n in the Taylor series for f (x, y) about (0, 0) is 1 ∂ 4n f (x, y) . 2n 2n (2n)!(2n)! ∂ x ∂ y (0,0)

17. Let f (x, y) =

-4

-3

-2

T =21 T =18 T =15.7

-1

1

2

T =27

3

4

5

x

However,

∞ X f (x, y) = (−1) j (x 2 + y 2 ) j j =0

j ∞ X X = (−1) j k=0

j =0

j! x 2k y 2 j −2k . k!( j − k)!

Fig. R-12.2

3. The graph is a saddle-like surface with downward slopes for legs and a tail, thus monkey saddle. y C=−16

C=16

The coefficient of x 2n y 2n is (−1)2n

(2n)! (2n)! = . n!n! (n!)2

C=0 C=8 C=16

[(2n)!]3 ∂ 4n = Thus f (x, y) . ∂ x 2n ∂ y 2n (n!)2 (0,0)

Review Exercises 12

1.

x

(page 743)

4y 2 =C x x 2 + 4y 2 = C x  2 x − (C/2) y2 + =1 (C/2)2 (C/4)2 Ellipse: centre ((C/2), 0), semi-axes: C/2, C/4, with the origin deleted. y C = −2 C =1 C = −3 C = −1 C =2 C=3 C =4 C = −4

Fig. R-12.3

x+

4.

f (x, y) =



x 3 /(x 2 + y 2 ) 0

if (x, y) 6= (0, 0) . if (x, y) = (0, 0)

(h 3 − 0)/ h 2 =1 h→0 h 0−0 f 2 (0, 0) = lim = 0. k→0 k f 1 (0, 0) = lim

For (x, y) 6= (0, 0), we have

x

x 4 + 3x 2 y 2 (x 2 + y 2 )2 2x 3 y f 2 (x, y) = − 2 (x + y 2 )2 0−1 f 1 (0, k) − f 1 (0, 0) = lim does not exist f 12 (0, 0) = lim k→0 k→0 k k f 2 (h, 0) − f 2 (0, 0) 0−0 f 21 (0, 0) = lim = lim = 0. h→0 h→0 h h f 1 (x, y) =

Fig. R-12.1

2.

140 + 30x 2 − 60x + 120y 2 8 + x 2 − 2x + 4y 2 100 = 30 − (x − 1)2 + 4y 2 + 7 Ellipses: centre (1, 0), values of T between 30 − (100/7) (minimum) at (1, 0) and 30 (at infinite distance from (1, 0)). T =

5.

f (x, y) =

x 3 − y3 (x − y)(x 2 + x y + y 2 ) = . 2 2 x −y (x − y)(x + y)

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REVIEW EXERCISES 12 (PAGE 743)

ADAMS and ESSEX: CALCULUS 8

f is continuous except on the lines x = y and x = −y where it is not defined. It has a continuous extension, x 2 + x y + y2 namely , to all points of x = y except the x+y origin. It cannot be extended so as to be continuous at the origin. For example, if (x, y) → (0, 0) along the curve y = −x + x 4 , then f (x, y) =

c) The tangent √ plane of part (a) is parallel to the plane x + y + 2 2z = 97 if √ ai + bj + 4ck = t (i + j + 2 2k),

x 2 − x 2 + x 5 + (x 4 − x)2 x6 − x3 + 1 = , x4 x2

which → ∞ as x → 0.

8.

If we define f (0, 0) = 0, then f (h, 0) − f (0, 0) h = lim = 1 h→0 h h f (0, k) − f (0, 0) k f 2 (0, 0) = lim = lim = 1. k→0 k→0 k k

f 1 (0, 0) = lim

h→0

6.

f (x, y) = e x

2 −2x−4y 2 +5

f 1 (x, y) = 2(x − 1)e

f 2 (x, y) = −8ye x

1 d R 5 1 2 13 1 · + · = . ≤ 20 R 100 100 25 100 1002

f (1, −1) = 1 f 1 (1, −1) = 0 f 2 (1, −1) = 8.

x 2 −2x−4y 2 +5

2 −2x−4y 2 +5

Thus |d R/R| ≤ 13/500; R can be in error by about 2.6%.

a) The tangent plane to z = f (x, y) at (1, −1, 1) has equation z = 1 + 8(y + 1), or z = 8y + 9.

9. The measured sides of the field are x = 150 m and

y = 200 m with |d x| = 1 and |d y| = 1, and the contained angle between them is θ = 30◦ with |dθ | = 2◦ = π/90 rad. The area A of the field satisfies

b) f (x, y) = C ⇒ (x − 1)2 − 4y 2 + 4 = ln C

⇒ (x − 1)2 − 4y 2 = ln C − 4. These are hyperbolas with centre (1, 0) and asymptotes x = 1 ± 2y. y C=1 C=10

-4

-3

-2

-1

1 x y sin θ ≈ 7, 500 2 y x xy d A = sin θ d x + sin θ d y + cos θ dθ 2 2√ 2 175 3 π + 15, 000 · ≈ 541. = 2 2 90 A=

C=100

1

2

3

4

5

x

The area is 7,500 m2 , accurate to within about 540 m2 for a percentage error of about 7.2%.

10.

C=1,000 C=10,000

Fig. R-12.6

7. Let f (x, y, z) = x 2 + y 2 + 4z 2 . Then S has equation f (x, y, z) = 16.

a) ∇ f (a, b, c) = 2ai + 2bj + 8ck. The tangent plane to S at (a, b, c) has equation 2a(x − a) + 2b(y − b) + 4c(z − c) = 0,

√ that is, a = t, b = t, c = t/ 2. Then 16 = a 2 + b2 + 4c2 = 4t 2 , so t = ±2. The two points on S where √ the tangent plane√is parallel to x +√y + 2 2z = 97 are (2, 2, 2) and (−2, −2, − 2). 1 1 1 = + R R1 R2 1 1 1 − 2 d R = − 2 d R1 − 2 d R2 R R1 R2 If R1 = 100 and R2 = 25, so that R = 20, and if |d R1 /R1 | = 5/100 and |d R2 /R2 | = 2/100, then

T = x 3 y + y 3 z + z 3 x.

a) ∇ T = (3x 2 y + z 3 )i + (3y 2 z + x 3 )j + (3z 2 x + y 3 )k ∇ T (2, −1, 0) = −12i + 8j − k. A unit vector in the direction from (2, −1, 0) towards (1, 1, 2) is u = (−i + 2j + 2k)/3. The directional derivative of T at (2, −1, 0) in the direction of u is u • ∇ T (2, −1, 0) =

or

ax + by + 4cz = a 2 + b2 + 4c2 = 16.

b) The tangent plane ax + by + 4cz = 16 passes through (0, 0, 4) if 16c = 16, that is, if c = 1. In this case a 2 +b2 = 16−4c2 = 12. These √ points (a, b, c) lie on a horizontal circle of radius 12 centred at (0, 0, 1) in the plane z = 1.

12 + 16 − 2 26 = . 3 3

b) Since ∇(2x 2 + 3y 2 + z 2 ) = 4xi + 6yj + 2zk, at t = 0 the fly is at (2, −1, 0) and is moving in the direction ±(8i − 6j), so its velocity is

490 Copyright © 2014 Pearson Canada Inc.

±5

8i − 6j = ±(4i − 3j). 10

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 12 (PAGE 743)

Since the fly is moving in the direction of increasing T , the rate at which it experiences T increasing is

c) The tangent plane at P0 to the level surface of f through P0 has equation   ∇ f (P0 ) • (x − 1)i + (y − 1)j + (z + 1)k = 0

dT = |(4i − 3j) • (−12i + 8j − k)| = 48 + 24 = 72. dt

11.

2(x − 1) + (y − 1) − 4(z + 1) = 0 2x + y − 4z = 7.

d) The bird is flying in direction

f (x, y, z) = x 2 y + yz + z 2 . a)

∇ f (x, y, z) = 2x yi + (x 2 + z)j + (y + 2z)k ∇ f (1, −1.1) = −2i + 2j + k.

(2 − 1)i + (−1 − 1)j + (1 + 1)k = i − 2j + 2k, a vector of length 3. Since the bird’s speed is 5, its velocity is 5 v = (i − 2j + 2k). 3 The rate of change of f as experienced by the bird is

The directional derivative of f in the direction i + k at (1, −1, 1) is i+k 1 √ • (−2i + 2j + k) = − √ . 2 2

df 5 40 = v • ∇ f (P0 ) = (2 − 2 − 8) = − . dt 3 3

b) The plane x + y +z = 1 intersects the level surface of f through (1, −1, 1) in a curve whose tangent vector at (1, −1, 1) is perpendicular to both ∇ f (1, −1, 1) and the normal vector i+j+k to the plane. Thus the ant is crawling in the direction of the cross product of these vectors:

13. j k i ± −2 2 1 = ±(i + 3j − 4k). 1 1 1

c) The second ant is crawling in the direction of the vector projection of ∇ f (1, −1, 1) onto the plane x + y + z = 1, which is ∇ f (1, −1, 1) minus its vector projection onto the normal to that plane:

∇ f (1, −1, 1) −

∇ f (1, −1, 1) • (i + j + k)

(i + j + k) |i + j + k|2 −7i + 5j + 2k 1 , = −2i + 2j + k − (i + j + k) = 3 3

that is, in the direction −7i + 5j + 2k.

12.

πxy + yz 2, P0 = (1, 1, −1). 2  πy 2 πxy  πxy + (x + z 2 ) cos i a) ∇ f = 2x sin 2  2 πx 2 π x y + (x 2 + z 2 ) cos + z2 j 2 2  πxy + 2z sin +y k 2 ∇ f (P0 ) = 2i + j − 4k.

14.

e) To experience the greatest rate of increase of f while flying through P0 at speed 5, the bird should fly in the direction of ∇ f (P0 ), that is, 2i + j − 4k.  x y u = k ln cos − ln cos k  k  x x 1 = − tan u x = k − tan k k k   y y 1 tan = tan uy = k k k k 1 2 x u xx = − sec k k 1 2 y u yy = sec k k u xy = 0

(1+u 2x )u yy − uu x u y u xy + (1 + u 2y )u xx 1 x y y x 1 = sec2 sec2 − 0 − sec2 sec2 = 0. k k k k k k If F(x, y, z) = 0, G(x, y, z) = 0 are solved for x = x(y), z = z(y), then dz dx + F2 + F3 =0 dy dy dx dz G1 + G2 + G3 = 0. dy dy

f (x, y, z) = (x 2 + z 2 ) sin

b) Since f (P0 ) = 2 + 1 = 3, the linearization of f at P0 is

F1

Eliminating dz/d y from these equations, we obtain dx F2 G 3 − F3 G 2 =− . dy F1 G 3 − F3 G 1 Similarly, if the equations are solved for x = x(z), y = y(z), then

L(x, y, z) = 3 + 2(x − 1) + (y − 1) − 4(z + 1).

dy F3 G 1 − F1 G 3 =− , dz F2 G 1 − F1 G 2

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REVIEW EXERCISES 12 (PAGE 743)

ADAMS and ESSEX: CALCULUS 8

and if the equations are solved for y = y(x), z = z(x), then dz F1 G 2 − F2 G 1 =− . dx F3 G 2 − F2 G 3

At x = 1, y = 2, ∂y ∂x +4 =1 ∂u ∂u ∂y ∂x −8 = 0, −6 ∂u ∂u 2

Hence

d x d y dz · · d y dz d x F2 G 3 − F3 G 2 F3 G 1 − F1 G 3 F1 G 2 − F2 G 1 =− · · = 1. F1 G 3 − F3 G 1 F2 G 1 − F1 G 2 F3 G 2 − F2 G 3

15.

from which we obtain ∂ x/∂u = −1 and ∂ y/∂u = 3/4 at (5, −7).

b) If z = ln(y 2 − x 2 ), then

  ∂z 1 ∂y ∂x = 2 + 2y . −2x ∂u ∂u ∂u y − x2

x = u 3 − uv

y = 3uv + 2v 2 Assume these equations define u = u(x, y) and v = v(x, y) near the point P where (u, v, x, y) = (−1, 2, 1, 2). a) Differentiating both equations with respect to x, we get ∂u ∂u ∂v 1 = 3u 2 −v −u ∂x ∂x ∂x ∂v ∂v ∂u + 3u + 4v . 0 = 3v ∂x ∂x ∂x At P, these equations become 1=

∂u ∂v + , ∂x ∂x

0=6

∂u ∂v +5 , ∂x ∂x

from which we obtain ∂u/∂ x = −5. P Similarly, differentiating the given equations with respect to y leads to 0=

∂u ∂v + , ∂y ∂y

1=6

∂u ∂v +5 , ∂y ∂y

At (u, v) = (5, −7), we have (x, y) = (1, 2), and so    ∂z 1 3 5 = −2 (−1) + 4 = . ∂u 3 4 3

Challenging Problems 12 1.

a) If f is differentiable at (a, b), then its graph has a nonvertical tangent plane at (a, b, f (a, b)). Any line through that point, part of which lies on the surface z = f (x, y) near (a, b), must be tangent to that surface at (a, b), so must lie in the tangent plane. b) The surface S with equation z = y g(x/y) has the property that if P = (x0 , y0 , z 0 ) is any point on it, then all points other than the origin on the line joining P0 to the origin also lie on S. Specifically, if t 6= 0, then (t x0 , t y0 , t z 0 ) lies on S, because t z 0 = t y0 g

from which we obtain ∂u/∂ y = 1. P

b) Since u(1, 2) = −1, we have ∂u ∂u u(1.02, 1.97) ≈ −1 + (0.02) + (−0.03) ∂ x P ∂ y P = −1 − 5(0.02) + 1(−0.03) = −1.13.

16.

2

u=x +y

2

v = x 2 − 2x y 2 Assume these equations define x = x(u, v) and y = y(u, v) near the point (u, v) = (5, −7), with x = 1 and y = 2 at that point. a) Differentiate the given equations with respect to u to obtain ∂x ∂y 1 = 2x + 2y ∂u ∂u ∂y 2 ∂x 0 = 2(x − y ) − 4x y . ∂u ∂u

(page 744)



t x0 t y0



⇔ z 0 = y0 g



x0 y0



.

Thus S consists entirely of lines through the origin; it is some kind of “cone” with vertex at the origin. By part (a), all tangent planes to S contain the lines on S through the points of contact, so all tangent planes must pass through the origin.

2. Let the position vector of the particle at time t be r = x(t)i + y(t)j + z(t)k. Then the velocity of the particle is dy dz dx i+ j+ k. v= dt dt dt This velocity must be parallel to

∇ f (x, y, z) = −2xi − 4yj + 6zk at every point of the path, that is, dx = −2t x, dt

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dy = −4t y, dt

dz = 6t z, dt

INSTRUCTOR’S SOLUTIONS MANUAL

so that we get

CHALLENGING PROBLEMS 12 (PAGE 744)

dx dy dz = = . Integrating these equations, −2x −4y 6z

ln |y| = 2 ln |x| + C1 ,

ln |z| = −3 ln |x| + C2 .

with all three terms evaluated at (r sin( p) cos(t), r sin( p) sin(t), r cos( p)), thus confirming the identity.

4. If u(x, y, z, t) = v(R, t) =

Since the path passes through (1, 1, 8), C1 and C2 are determined by ln 1 = 2 ln 1 + C1 ,

ln 8 = −3 ln 1 + C2 .

Thus C1 = 0 and C2 = ln 8. The path therefore has equations y = x 2 , z = 8/x 3 . Evidently (2, 4, 1) lies on the path, and (3, 7, 0) does not.

3. We used Maple V to verify the stated identity. Using r, p, and t to represent R, φ, and θ , respectively, we defined > >

v := (r,p,t) -> u(r*sin(p)*cos(t), r*sin(p)*sin(t), > r*cos(p)); and then asked Maple to calculate and simplify the left side of the identity: > > > > >

simplify(diff(v(r,,p,t),r$2) +(2/r)*diff(v(r,p,t),r) +(cot(p)/rˆ2)*diff(v(r,p,t),p) +(1/rˆ2)*diff(v(r,p,t),p$2) +(1/(r*sin(p))ˆ2)*diff(v(r,p,t),t$2));

Maple responded with D1,1 (u) + D3,3 (u) + D2,2 (u),

θ and φ, then

f (R − ct) is independent of R

∂ 2u ∂ 2u ∂ 2u ∂2v 2 ∂v + + = + ∂x2 ∂ y2 ∂z 2 ∂ R2 R ∂R by Problem 3. We have ∂v ∂R ∂ 2v ∂ R2 ∂v ∂t ∂ 2v ∂t 2 ∂ 2v ∂ R2

f ′ (R − ct) f (R − ct) − R R2 f ′′ (R − ct) 2 f ′ (R − ct) 2 f (R − ct) − = + R R2 R3 ′ c f (R − ct) =− R c2 f ′′ (R − ct) = R 2 ∂v + R ∂R f ′′ (R − ct) 2 f ′ (R − ct) 2 f (R − ct) = − + R R2 R3 2 f ′ (R − ct) 2 f (R − ct) + − R2 R3 ′′ f (R − ct) = R 1 ∂ 2v 1 ∂ 2u = 2 2 = 2 2. c ∂t c ∂t =

The function f (R − ct)/R represents the shape of a symmetrical wave travelling uniformly away from the origin at speed c. Its amplitude at distance R from the origin decreases as R increases; it is proportional to the reciprocal of R.

493 Copyright © 2014 Pearson Canada Inc.

SECTION 13.1 (PAGE 752)

ADAMS and ESSEX: CALCULUS 8

1 1 Thus B 2 − AC = − < 0, and (−4, 2) is a local 16 4 maximum.

CHAPTER 13. APPLICATIONS OF PARTIAL DERIVATIVES Section 13.1 Extreme Values 1.

2.

3.

6.

f (x, y) = cos(x + y), f 1 = − sin(x + y) = f 2 . All points on the lines x + y = nπ (n is an integer) are critical points. If n is even, f = 1 at such points; if n is odd, f = −1 there. Since −1 ≤ f (x, y) ≤ 1 at all points in R2 , f must have local and absolute maximum values at points x + y = nπ with n even, and local and absolute minimum values at such points with n odd.

7.

f (x, y) = x sin y. For critical points we have

(page 752)

f (x, y) = x 2 + 2y 2 − 4x + 4y f 1 (x, y) = 2x − 4 = 0 if x = 2 f 2 (x, y) = 4y + 4 = 0 if y = −1. Critical point is (2, −1). Since f (x, y) → ∞ as x 2 + y 2 → ∞, f has a local (and absolute) minimum value at that critical point. f (x, y) = x y − x + y, f 1 = y − 1, f 2 = x + 1 A = f 11 = 0, B = f 12 = 1, C = f 22 = 0. Critical point (−1, 1) is a saddle point since B 2 − AC > 0.

f 1 = sin y = 0,

Since sin y and cos y cannot vanish at the same point, the only critical points correspond to x = 0 and sin y = 0. They are (0, nπ ), for all integers n. All are saddle points.

f (x, y) = x 3 + y 3 − 3x y

f 1 (x, y) = 3(x 2 − y), f 2 (x, y) = 3(y 2 − x). 2 For critical points: x = y and y 2 = x. Thus x 4 − x = 0, that is, x(x − 1)(x 2 + x + 1) = 0. Thus x = 0 or x = 1. The critical points are (0, 0) and (1, 1). We have A = f 11 (x, y) = 6x, C = f 22 (x, y) = 6y.

8.

B = f 12 (x, y) = −3,

At (0, 0): A = C = 0, B = −3. Thus AC < B 2 , and (0, 0) is a saddle point of f . At (1, 1): A = C = 6, B = −3, so AC > B 2 . Thus f has a local minimum value at (1, 1).

4.

f (x, y) = x 4 +y 4 −4x y,

f 1 = 4(x 3 −y),

f 2 = 4(y 3 −x)

A = f 11 = 12x 2 , B = f 12 = −4, C = f 22 = 12y 2 . For critical points: x 3 = y and y 3 = x. Thus x 9 = x, or x(x 8 − 1) = 0, and x = 0, 1, or −1. The critical points are (0, 0), (1, 1) and (−1, −1). At (0, 0), B 2 − AC = 16 − 0 > 0, so (0, 0) is a saddle point. At (1, 1) and (−1, −1), B 2 − AC = 16 − 144 < 0, A > 0, so f has local minima at these points.

5.

x 8 + −y y x 1 8 f 1 (x, y) = − 2 = 0 if 8y = x 2 y x x f 2 (x, y) = − 2 − 1 = 0 if x = −y 2 . y For critical points: 8y = x 2 = y 4 , so y = 0 or y = 2. f (x, y) is not defined when y = 0, so the only critical point is (−4, 2). At (−4, 2) we have

9.

f (x, y) = cos x + cos y, f 1 = − sin x, f 2 = − sin y A = f 11 = − cos x, B = f 12 = 0, C = f 22 = − cos y. The critical points are points (mπ, nπ ), where m and n are integers. Here B 2 − AC = − cos(mπ ) cos(nπ ) = (−1)m+n+1 which is negative if m + n is even, and positive if m + n is odd. If m + n is odd then f has a saddle point at (mπ, nπ ). If m + n is even and m is odd then f has a local (and absolute) minimum value, −2, at (mπ, nπ ). If m + n is even and m is even then f has a local (and absolute) maximum value, 2, at (mπ, nπ ). f (x, y) = x 2 ye−(x

16 1 =− , 4 x3 2x = 3 = −1. y

C = f 22

B = f 12 = −

1 1 =− , 4 y2

2 +y 2 )

f 1 (x, y) = 2x y(1 − x 2 )e−(x f 2 (x, y) = x 2 (1 − 2y 2 )e−(x

2 +y 2 )

2 +y 2 )

A = f 11 (x, y) = 2y(1 − 5x 2 + 2x 4 )e−(x

2 +y 2 )

B = f 12 (x, y) = 2x(1 − x 2 )(1 − 2y 2 )e−(x

C = f 22 (x, y) = 2x 2 y(2y 2 − 3)e−(x

2 +y 2 )

2 +y 2 )

.

For critical points:

f (x, y) =

A = f 11 =

f 2 = x cos y = 0.

x y(1 − x 2 ) = 0

x 2 (1 − 2y 2 ) = 0. √ The critical √ points are (0, y) for all y, (±1, 1/ 2), and (±1, −1/ 2). Evidently, f (0, y) = 0. Also f (x, y) > 0 if y > 0 and x 6= 0, and f (x, y) < 0 if y < 0 and x 6= 0. Thus f has a local minimum at (0, y) if y > 0, and a local maximum if y < 0. The origin is a saddle point. √ √ At (±1, 1/ 2): A = C = −2 2e−3/2 , B = 0, and so AC > B 2 . Thus f has local maximum values at these two points.

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.1 (PAGE 752)

√ √ At (±1, −1/ 2): A = C = 2 2e−3/2 , B = 0, and so AC > B 2 . Thus f has local minimum values at these two points.

10.

11.

Since f (x, 0 as √ x 2 + y 2 → ∞, the value √ y) → −3/2 f (±1, 1/ 2) = e / 2 is the absolute √ maximum √ value for f , and the value f (±1, −1/ 2) = −e−3/2 / 2 is the absolute minimum value. xy f (x, y) = 2 + x 4 + y4 (2 + x 4 + y 4 )y − x y4x 3 y(2 + y 4 − 3x 4 ) f1 = = (2 + x 4 + y 4 )2 (2 + x 4 + y 4 )2 x(2 + x 4 − 3y 4 ) f2 = . (2 + x 4 + y 4 )2 For critical points, y(2 + y 4 − 3x 4 ) = 0 and x(2 + x 4 − 3y 4 ) = 0. One critical point is (0, 0). Since f (0, 0) = 0 but f (x, y) > 0 in the first quadrant and f (x, y) < 0 in the second quadrant, (0, 0) must be a saddle point of f . Any other critical points must satisfy 2 + y 4 − 3x 4 = 0 and 2 + x 4 − 3y 4 = 0, that is, y 4 = x 4 , or y = ±x. Thus 2 − 2x 4 = 0 and x = ±1. Therefore there are four other critical points: (1, 1), (−1, −1), (1, −1) and (−1, 1). f is positive at the first two of these, and negative at the other two. Since f (x, y) → 0 as x 2 + y 2 → ∞, f must have maximum values at (1, 1) and (−1, −1), and minimum values at (1, −1) and (−1, 1). f (x, y) = xe−x

f 2 (x, y) = 3x y 2 e−x

xy + y2 2 (x + y 2 )y − 2x 2 y f 1 (x, y) = (x 2 + y 2 )2 y(y 2 − x 2 ) = 2 (x + y 2 )2 x(x 2 − y 2 ) (by symmetry). f 2 (x, y) = 2 (x + y 2 )2 Both partial derivatives are zero at all points of the lines y = ±x for x 6= 0. Also f (x, x) = 21 , and f (x, −x) = − 21 for x 6= 0. Since x 2 ± 2x y + y 2 = (x ± y)2 ≥ 0, we have |x y| ≤ 21 (x 2 + y 2) for all (x, y) 6= (0, 0), so | f (x, y)| ≤ on its domain. Thus, f has absolute maximum value 12 at all points (x, x) for x 6= 0, and absolute minimum value − 12 at all points (x, −x) for all x 6= 0. f (x, y) =

1 1 − x + y + x 2 + y2 1 . =  2   1 1 2 1 x− + y+ + 2 2 2

1 2

f (x, y) =

Since



 1 1 ,− . 2 2

1 − 2x (1 − x + y + x 2 + y 2 )2 1 + 2y f 2 (x, y) = − , (1 − x + y + x 2 + y 2 )2

3 +y 3

3

x2

Evidently f has absolute maximum value 2 at

3 +y 3

B = f 12 (x, y) = −3y 2 (3x 3 − 1)e−x

f 1 (x, y) =

3 +y 3 3

C = f 22 (x, y) = 3x y(3y 3 + 2)e−x +y For critical points: 3x 3 = 1 and 3x y 2 = 0. The only critical point is (3−1/3 , 0). At that point we have B = C = 0 so the second derivative test is inconclusive. 3 3 However, note that f (x, y) = f (x, 0)e y , and e y has an inflection point at y = 0. Therefore f (x, y) has neither a maximum nor a minimum value at (3−1/3 , 0), so has a saddle point there. x2 + y2 2 (x + y 2 )2x − 2x 3 2x y 2 f 1 (x, y) = = 2 2 2 2 (x + y ) (x + y 2 )2 2 2x y f 2 (x, y) = − 2 . (x + y 2 )2 Both partial derivatives are zero at all points of the coordinate axes. Also f (x, 0) = 1 for x 6= 0, and f (0, y) = 0 for y 6= 0. Evidently 0 ≤ f (x, y) ≤ 1 for all (x, y) 6= (0, 0). Thus, f has absolute maximum value 1 at all points x2

14.

3 +y 3

A = f 11 (x, y) = 3x 2 (3x 3 − 4)e−x

f (x, y) =

13.

3 +y 3

f 1 (x, y) = (1 − 3x 3 )e−x

12.

(x, 0) for x 6= 0, and absolute minimum value 0 at all points (0, y) for all y 6= 0.



15.

1 1 ,− 2 2



is the only critical point of f .

    1 1 1 1 f (x, y) = 1 + 1+ + x y x y (x + 1)(y + 1)(x + y) = x 2 y2 (y + 1)(x y + x + 2y) f 1 (x, y) = − x 3 y2 (x + 1)(x y + y + 2x) f 2 (x, y) = − x 2 y3 2(y + 1)(x y + x + 3y) A = f 11 (x, y) = x 4 y2 2(x y + x + y) B = f 12 (x, y) = x 3 y3 2(x + 1)(x y + y + 3x) C = f 22 (x, y) = . x 2 y4

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Thus (2, −2, −2) is a saddle point. By symmetry, so are the remaining two critical points.

For critical points:

and

y = −1 or x = −1 or

x y + x + 2y = 0, x y + y + 2x = 0.

17.

If y = −1, then x = −1 or x − 1 = 0. If x = −1, then y = −1 or y − 1 = 0. If x 6= −1 and y 6= −1, then x − y = 0, so x 2 + 3x = 0. Thus x = 0 or x = −3. However, the definition of f excludes x = 0. Thus, the only critical points are (1, −1),

(−1, 1),

(−1, −1),

f 3 (x, y, z) = x 2 − 2z.


The only critical point is 1, 1, 12 . We have

(−3, −3).

and

D = f 1 + h, 1 + k,

At (1, −1), (−1, 1), and (−1, −1) we have AC = 0 and B 6= 0. Therefore these three points are saddle points of f. At (−3, −3), A = C = 4/243 and B = 2/243, so AC > B 2 . Therefore f has a local minimum value at (−3, −3).

16.

0 = f 2 = x z−2y,

Let u = ui + vj + wk, where

u2

+ v2

+ w2

h 2 (1 + 2h) If m = h and k = 0, then D = > 0 for small 2 |h|. If h = k = 0, then D = −m 2 < 0 for
m 6= 0. Thus f has a saddle point at 1, 1, 12 .

18.

= 1. Then

− (1 + m)4 − 1 = 4(1 + h + k + m + hk + hm + km + hkm)

+ (zu − 2v + xw)v + (yu + xv − 2w)w.   At (0, 0, 0), Du Du f (0, 0, 0) = −2u 2 − 2v 2 − 2w2 < 0 for u 6= 0, so f has a local maximum value at (0, 0, 0).

− (1 + 4h + 6h 2 + 4h 3 + h 4 ) − (1 + 4k + 6k 2 + 4k 3 + k 4 )

− (1 + 4m + 6m 2 + 4m 3 + m 4 ) − 1

= 4(hk + hm + km) − 6(h 2 + k 2 + m 2 ) + · · · ,

At (2, 2, 2), we have   Du Du f (2, 2, 2) = (−2u + 2v + 2w)u + (2u − 2v + 2w)v

where · · · stands for terms of degree 3 and 4 in the variables h, k, and m. Completing some squares among the quadratic terms we obtain

+ (2u + 2v − 2w)w

= −2(u + v 2 + w2 ) + 4(uv + vw + wu)

h i D = −2 (h −k)2 +(k −m)2 +(h −m)2 +h 2 +k 2 +m 2 +· · ·

= −2[(u − v − w)2 − 4vw]  < 0 if v = w = 0, u 6= 0 > 0 if v = w 6= 0, u − v − w = 0.

which is negative if |h|, |k| and |m| are small and not all 0. (This is because the terms of degree 3 and 4 are smaller in size than the quadratic terms for small values of the variables.) Hence f has a local maximum value at (1, 1, 1).

Thus (2, 2, 2) is a saddle point. At (2, −2, −2), we have   Du Du f = −2(u 2 + v 2 + w2 + 2uv + 2uw − 2vw) = −2[(u + v + w)2 − 4vw]  < 0 if v = w = 0, u 6= 0 > 0 if v = w 6= 0, u + v + w = 0.

f (x, y, z) = 4x yz − x 4 − y 4 − z 4 D = f (1 + h, 1 + k, 1 + m) − f (1, 1, 1)

= 4(1 + h)(1 + k)(1 + m) − (1 + h)4 − (1 + k)4

Du f (x, y, z) = (yz − 2x)u + (x z − 2y)v + (x y − 2z)w   Du Du f (x, y, z) = (−2u + zv + yw)u

2



+ m − f 1, 1, 12

1 + 2h + h 2 + (1 + 2h + h 2 )m 2   1 3 − 1 − 2h − h 2 − 1 − k − − m − m 2 − − 4 4 2 2 h (2m − 1) + 2h(k + 2m) − 2m = . 2

0 = f 3 = x y−2z.

Thus x yz = 2x 2 = 2y 2 = 2z 2 , so x 2 = y 2 = z 2 . Hence x 3 = ±2x 2 , and x = ±2 or 0. Similarly for y and z. The only critical points are (0, 0, 0), (2, 2, 2), (−2, −2, 2), (−2, 2, −2), and (2, −2, −2).

1 2

= 1 + h + k + hk +

f (x, y, z) = x yz − x 2 − y 2 − z 2 . For critical points we have 0 = f 1 = yz−2x,

f (x, y, z) = x y + x 2 z − x 2 − y − z 2 f 1 (x, y, z) = y + 2x(z − 1) f 2 (x, y, z) = x − 1

19.

f (x, y) = x ye−(x

2 +y 4 )

f 1 (x, y) = y(1 − 2x 2 )e−(x

f 2 (x, y) = x(1 − 4y 4 )e−(x

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SECTION 13.1 (PAGE 752)

For critical points y(1 − 2x 2 ) = 0 and x(1 − 4y 4 ) = 0. The critical points are     1 1 1 1 (0, 0), ±√ , √ , ±√ ,−√ . 2 2 2 2 We have

Since f (x, y) → 0 as x 2 + y 2 → ∞, the maximum and 1 1 minimum values of f are e−3/4 and − e−3/4 respec2 2 tively. x f (x, y) = 1 + x 2 + y2 1 + y2 − x 2 f 1 (x, y) = (1 + x 2 + y 2 )2 −2x y f 2 (x, y) = . (1 + x 2 + y 2 )2

For critical points, x 2 − y 2 = 1, and x y = 0. The critical points are (±1, 0). f (±1, 0) = ± 21 . Since f (x, y) → 0 as x 2 + y 2 → ∞, the maximum and minimum values of f are 1/2 and −1/2 respectively.

21.

f (x, y, z) = x yze−(x

f (x, y) = x + 8y + f 1 (x, y) = 1 −

1 , xy

1

=0 x2y 1 f 2 (x, y) = 8 − 2 = 0 xy

(x > 0,

y > 0)

⇒

x2y = 1

⇒

8x y 2 = 1.

The critical points must satisfy

f (0, 0) = 0     1 1 1 1 1 f √ ,√ = f −√ ,−√ = e−3/4 > 0 2  2 2  2  2 1 1 1 1 1 f −√ , √ = f √ ,−√ = − e−3/4 < 0 2 2 2 2 2

20.

22.

2 +y 2 +z 2 )

f 1 (x, y, z) = yz(1 − 2x 2 )e−(x f 2 (x, y, z) = x z(1 − 2y 2 )e−(x

f 3 (x, y, z) = x y(1 − 2z 2 )e−(x

x2y x = = 8, y x y2 that is, x = 8y. Also, x 2 y = 1, so 64y 3 = 1.
Thus y = 1/4, and x = 2; the critical point is 2, 14 . Since f (x, y) → ∞ if x → 0+, y → 0+, or x 2 + y 2 → ∞, the critical point must give a minimum value for f . The minimum value is
f 2, 41 = 2 + 2 + 2 = 6.

23. Let the length, width, and height of the box be x, y, and z, respectively. Then V = x yz. The total surface area of the bottom and sides is S = x y + 2x z + 2yz = x y + 2(x + y) = xy +

2V 2V + , x y

where x > 0 and y > 0. Since S → ∞ as x → 0+ or y → 0+ or x 2 + y 2 → ∞, S must have a minimum value at a critical point in the first quadrant. For CP: ∂S =y− ∂x ∂S =x− 0= ∂y

2 +y 2 +z 2 )

0=

2 +y 2 +z 2 ) 2 +y 2 +z 2 )

.

Any critical point must satisfy 1 yz(1 − 2x 2 ) = 0 i.e., y = 0 or z = 0 or x = ± √ 2 1 x z(1 − 2y 2 ) = 0 i.e., x = 0 or z = 0 or y = ± √ 2 1 x y(1 − 2z 2 ) = 0 i.e., x = 0 or y = 0 or z = ± √ . 2 Since f (x, y, z) is positive at some points, negative at others, and approaches 0 as (x, y, z) recedes to infinity, f must have maximum and minimum values at critical points. Since f (x, y, z) = 0 if x = 0 or y = 0 or z = 0, the maximum and minimum values must√occur among√ the eight critical √ points where x = ±1/ 2, y = ±1/ 2, and z = ±1/ 2. At four of these points, f has the value 1 √ e−3/2 , the maximum value. At the other four f has 2 2 1 the value − √ e−3/2 , the minimum value. 2 2

V xy

2V x2 2V . y2

Thus x 2 y = 2V = x y 2 , so that x = y = (2V )1/3 and z = V /(2V )2/3 = 2−2/3 V 1/3 .

24. Let the length, width, and height of the box be x, y, and z, respectively. Then V = x yz. If the top and side walls cost $k per unit area, then the total cost of materials for the box is C = 2kx y + kx y + 2kx z + 2kyz     2V 2V V = k 3x y + + , = k 3x y + 2(x + y) xy x y where x > 0 and y > 0. Since C → ∞ as x → 0+ or y → 0+ or x 2 + y 2 → ∞, C must have a minimum value at a critical point in the first quadrant. For CP:  ∂C = k 3y − ∂x  ∂C 0= = k 3x − ∂y

0=

 2V x2  2V . y2

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ADAMS and ESSEX: CALCULUS 8

Thus 3x 2 y = 2V = 3x y 2 , so that x = y = (2V /3)1/3 and z = V /(2V /3)2/3 = (9V /4)1/3 .

25. Let (x, y, z) be the coordinates of the corner of the box

Hence 9b + 6c = 180 = 6b + 8c, from which we obtain 3b = 2c = 30. The three numbers are b = 10, c = 15, and a = 30 − 10 − 15 = 5.

that is in the first octant of space. Thus x, y, z ≥ 0, and x2 y2 z2 + + = 1. a2 b2 c2

27. Differentiate the given equation 2

s

V = (2x)(2y)(2z) = 8cx y 1 −

x2 y2 − 2 2 a b

for x ≥ 0, y ≥ 0, and (x 2 /a 2 )+(y 2 /b2 ) ≤ 1. For analysis it is easier to deal with V 2 than with V :   x 4 y2 x 2 y4 V 2 = 64c2 x 2 y 2 − 2 − 2 . a b Since V = 0 if x = 0 or y = 0 or (x 2 /a 2 ) + (y 2 /b2 ) = 1, the maximum value of V 2 , and hence of V , will occur at a critical point of V 2 where x > 0 and y > 0. For CP:   4x 3 y 2 2x y 4 ∂V 2 = 64c2 2x y 2 − − ∂x a2 b2   2 2 2x y = 128c2 x y 2 1 − 2 − 2 a b   2 2 ∂V x 2y 2 0= = 128c2 x 2 y 1 − 2 − 2 . ∂y a b

with respect to x and y, regarding z as a function of x and y: e

2zx−x 2

e2zx−x

2

    ∂z ∂z 2zy+y 2 + 2z − 2x − 3e 2y 2x =0 ∂x ∂x     ∂z ∂z 2 − 3e2zy+y 2y 2x + 2z + 2y = 0 ∂y ∂y

∂z ∂z = 0 and = 0, and it ∂x ∂y follows from the equations above that z = x and z = −y. Substituting these into the given equation, we get

For a critical point we have

0=

Hence we must have 2x 2 y2 x2 2y 2 + 2 =1= 2 + 2 , 2 a b a b √ √ so that x 2 /a 2 = y 2 /b2 = 1/3, and x = a/ 3, y = b/ 3. The largest box has volume r 8abc 1 1 8abc V = 1 − − = √ cubic units. 3 3 3 3 3

26. Given that a > 0, b > 0, c > 0, and a + b + c = 30, we want to maximize

P = ab2 c3 = (30 − b − c)b2 c3 = 30b2 c3 − b3 c3 − b2 c4 . Since P = 0 if b = 0 or c = 0 or b + c = 30 (i.e., a = 0), the maximum value of P will occur at a critical point (b, c) satisfying b > 0, c > 0, and b + c < 30. For CP: ∂P = 60bc3 − 3b2 c3 − 2bc4 = bc3 (60 − 3b − 2c) ∂b ∂P 0= = 90b2 c2 − 3b3 c2 − 4b2 c3 = b2 c2 (90 − 3b − 4c). ∂c

0=

2

e2zx−x − 3e2zy+y = 2

The volume of the box is

2

2

e z − 3e−z = 2 2

2

2

2

(e z )2 − 2e z − 3 = 0

(e z − 3)(e z + 1) = 0.

2

2

2

Thus e z = 3 or e z = −1. Since e z = −1 is not possi√ 2 ble, we have e z = 3, so√z = ± √ln 3. The√critical √ points are ( ln 3, − ln 3), and (− ln 3, ln 3).

28. We will use the second derivative test to classify the two critical points calculated in Exercise 25. To calculate the second partials

A=

∂2z , ∂x2

B=

∂2z , ∂ x∂ y

C=

∂2z , ∂ y2

we differentiate the expressions (∗), and (∗∗) obtained in Exercise 25. Differentiating (∗) with respect to x, we obtain 2  ∂z + 2z − 2x e 2x ∂x  ∂z ∂2z +4 + 2x 2 − 2 ∂x ∂x  2  2 ∂z ∂2z − 3e2zy+y 2y + 2y 2 = 0. ∂x ∂x 2zx−x 2

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(∗) (∗∗)

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At a critical point, z 2 = ln 3, so

SECTION 13.1 (PAGE 752)

∂z = 0, z = x, z = −y, and ∂x

   3 ∂2z ∂2z 3 2x 2 − 2 − 2y 2 = 0, 3 ∂x ∂x 2 6 ∂ z = . A= ∂x2 6x − 2y 

Differentiating (∗∗) with respect to y gives 2  ∂z ∂2z + 2x 2 ∂y ∂y  2  ∂z 2 ∂z ∂2z 2y − 3e2zy+y + 2z + 2y + 4 + 2y 2 + 2 = 0, ∂y ∂y ∂y

e2zx−x

2



so B 2 − AC < 0, and f has a local maximum at that critical point.

29.

f (x, y) = (y − x 2 )(y − 3x 2 ) = y 2 − 4x 2 y + 3x 4

f 1 (x, y) = −8x y + 12x 3 = 4x(3x 2 − 2y)

f 2 (x, y) = 2y − 4x 2 .

Since f 1 (0, 0) = f 2 (0, 0) = 0, therefore (0, 0) is a critical point of f . Let g(x) = f (x, kx) = k 2 x 2 − 4kx 3 + 3x 4 . Then g ′ (x) = 2k 2 x − 12kx 2 + 12x 3

2x

and evaluation at a critical point gives     ∂2z ∂2z 3 2y 2 + 2 = 0, 3 2x 2 − ∂y 3 ∂y ∂2z 2 C= 2 = . ∂y 6x − 2y Finally, differentiating (∗) with respect to y gives    ∂z ∂z 2zx−x 2 + 2z − 2x 2x 2x e ∂x ∂y  ∂2z ∂z + 2x +2 ∂ x∂ y ∂y    ∂z ∂z 2 2zy+y + 2z + 2y 2y − 3e 2y ∂y ∂x  2 ∂z ∂ z +2 + 2y = 0, ∂x ∂ x∂ y

g ′′ (x) = 2k 2 − 24kx + 36x 2 .

Since g ′ (0) = 0 and g ′′ (0) = 2k 2 > 0 for k 6= 0, g has a local minimum value at x = 0. Thus f (x, kx) has a local minimum at x = 0 if k 6= 0. Since f (x, 0) = 3x 4 and f (0, y) = y 2 both have local minimum values at (0, 0), f has a local minimum at (0, 0) when restricted to any straight line through the origin. However, on the curve y = 2x 2 we have f (x, 2x 2 ) = x 2 (−x 2 ) = −x 4 , which has a local maximum value at the origin. Therefore f does not have an (unrestricted) local minimum value at (0, 0). =0 Note that A = f 11 (0, 0) = (−8y + 36x 2 ) (0,0) B = f 12 (0, 0) = −8x = 0. (0,0)

Thus AC = B 2 , and the second derivative test is indeterminate at the origin.

30. We have

and, evaluating at a critical point, ∂2z (6x − 2y) = 0, ∂ x∂ y so that

∂2z = 0. ∂ x∂ y √ √ At the critical point ( ln 3, − ln 3) we have B=

A=

6 , 8 ln 3

B = 0,

C=

2 , 8 ln 3

so B 2 − AC < 0, and f has a local minimum at that critical point. √ √ At the critical point (− ln 3, ln 3) we have 6 A=− , 8 ln 3

B = 0,

2 C=− , 8 ln 3

Q(u, v) = Au 2 + 2Buv + Cv 2     2B B2 B2 = A u2 + uv + 2 v 2 + C − v2 A A A   Bv 2 AC − B 2 2 = A u+ + v . A A A B = AC − B 2 > 0, both terms above have If B C the same sign, positive if A > 0 and negative if A < 0, ensuring that Q is positive definite or negative definite respectively, since the two terms cannot both vanish if (u, v) 6= (0, 0). If AC − B 2 < 0, Q(u, v) is a difference of squares, and must be indefinite.

31. Let Q(u, v, w) = Au 2 + Bv 2 + Cw2 + 2Duv + 2Euw + 2Fvw

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Section 13.2 Extreme Values of Functions Defined on Restricted Domains (page 758)

and let K 1 = A, A K3 = D E

A D = AB − D 2 K2 = D B D E B F = ABC + 2D E F − B E 2 − C D 2 − AF 2 . F C

1.

0 = f 1 (x, y) = 1 − 2x,

Suppose that K 1 6= 0, K 2 6= 0, and K 3 6= 0. We have Q(u, v, w) "

= A u 2 + 2u

Dv + Ew + A



Dv + Ew A

2 #

AB − D 2 2 AC − E 2 2 2(AF − D E) v + w + vw A A A  2 Dv + Ew = A u+ A !   2 AB − D 2(AF − D E) AF − D E 2 2 2 + v + vw + w A AB − D 2 AB − D 2   (AF − D E)2 AC − E 2 − w2 + A A(AB − D 2 )    2 AB − D 2 Dv + Ew 2 AF − D E + = A u+ v+ w A A AB − D 2 A(ABC − B E 2 − AF 2 − C D 2 + 2D E F) 2 + w A(AB − D 2 )  2  2 Dv + Ew K2 AF − D E = K1 u + + v+ w A K1 AB − D 2 K3 2 w . + 2. K2 +

If K 1 > 0, K 2 > 0, and K 3 > 0, then all three squares the last expression above have positive coefficients, and so Q is positive definite. If K 1 < 0, K 2 > 0, and K 3 < 0, then all three squares the last expression above have negative coefficients, and so Q is negative definite. In all other cases where none of the K i = 0, the coefficients of the squares are not all of the same sign so choices of (u, v, w) can be made which make the expression either positive or negative, and Q is indefinite.

f (x, y) = x y − 2x on R = {(x, y) : −1 ≤ x ≤ 1, 0 ≤ y ≤ 1}. For critical points: 0 = f 2 (x, y) = x.

The only CP is (0, 2), which lies outside R. Therefore the maximum and minimum values of f on R lie on one of the four boundary segments of R. On x = −1 we have f (−1, y) = 2 − y for 0 ≤ y ≤ 1, which has maximum value 2 and minimum value 1. On x = 1 we have f (1, y) = y − 2 for 0 ≤ y ≤ 1, which has maximum value −1 and minimum value −2. On y = 0 we have f (x, 0) = −2x for −1 ≤ x ≤ 1, which has maximum value 2 and minimum value −2. On y = 1 we have f (x, 1) = −x for −1 ≤ x ≤ 1, which has maximum value 1 and minimum value −1. Thus the maximum and minimum values of f on the rectangle R are 2 and −2 respectively.

D = f 12 (a, b, c), E = f 23 (a, b, c), F = f 23 (a, b, c).

Then f has a local minimum value at (a, b, c) if K 1 > 0, K 2 > 0, and K 3 > 0, a local maximum value at (a, b, c) if K 1 < 0, K 2 > 0, and K 3 < 0, and a saddle point at (a, b, c) if K 1 , K 2 , K 3 are all nonzero but satisfy neither of the above conditions.

0 = f 2 (x, y) = 2y.

The only CP is (1/2, 0), which lies on the boundary of R. The boundary consists of four segments; we investigate each. On x = 0 we have f (x, y) = f (0, y) = y 2 for 0 ≤ y ≤ 1, which has minimum value 0 and maximum value 1. On y = 0 we have f (x, y) = f (x, 0) = x − x 2 = g(x) for 0 ≤ x ≤ 2. Since g ′ (x) = 1 − 2x = 0 at x = 1/2, g(1/2) = 1/4, g(0) = 0, and g(2) = −2, the maximum and minimum values of f on the boundary segment y = 0 are 1/4 and −2 respectively. On x = 2 we have f (x, y) = f (2, y) = −2 + y 2 for 0 ≤ y ≤ 1, which has minimum value −2 and maximum value −1. On y = 1, f (x, y) = f (x, 1) = x − x 2 + 1 = g(x) + 1 for 0 ≤ x ≤ 2. Thus the maximum and minimum values of f on the boundary segment y = 1 are 5/4 and −1 respectively. Overall, f has maximum value 5/4 and minimum value −2 on the rectangle R.

0 = f 1 (x, y) = y − 2,

If f has continuous partial derivatives of order two and (a, b, c) is a critical point of f (x, y, z), let A = f 11 (a, b, c), B = f 22 (a, b, c), C = f 33 (a, b, c),

f (x, y) = x − x 2 + y 2 on R = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1}. For critical points:

3.

f (x, y) = x y − y 2 on D = {(x, y) : x 2 + y 2 ≤ 1}. For critical points: 0 = f 1 (x, y) = y,

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0 = f 2 (x, y) = x − 2y.

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SECTION 13.2 (PAGE 758)

On the side x = 1 we have f (1, y) = y − y 2 = g(y), (0 ≤ y ≤ 1). g has maximum value 1/4 at its critical point y = 1/2. On the side y = 1 we have f (x, 1) = x − x 3 = h(x), (0 ≤ x ≤ 1).√ h has critical point given by 1 − 3x 2 = 0; only  x=  1/ 3 is on the side of S. 1 2 1 h √ = √ > . 4 3 3 3 On the square S, f (x, y) has minimum value 0 (on the sides x = 0 and y = 0 and at the corner (1, 1) of √ the √ square), and maximum value 2/(3 3) at the point (1/ 3, 1). There is a smaller local maximum value at (1, 1/2).

The only CP is (0, 0), which lies inside D. We have f (0, 0) = 0. The boundary of D is the circle x = cos t, y = sin t, −π ≤ t ≤ π . On this circle we have g(t) = f (cos t, sin t) = cos t sin t − sin2 t i 1h = sin 2t + cos 2t − 1 , (−π ≤ t ≤ π ). 2 g(0) = g(2π ) = 0 g ′ (t) = cos 2t − sin 2t. The critical points of g satisfy cos 2t = sin 2t, that is, 5π π 5π π tan 2t = 1, so 2t = ± or ± , and t = ± or ± . 4 4 8 8 We have

6.

π 

1 1 1 1 1 = √ − + √ = √ − >0 8 2 2 2 2 2 2 2  π 1 1 1 1 g − =− √ − + √ =− 8 2 2 2 2 2 2   5π 1 1 1 1 1 g = − √ − − √ = −√ − 8 2 2 2 2 2 2 2   1 1 5π 1 1 = √ − − √ =− . g − 8 2 2 2 2 2 2 g

0 = f 1 (x, y) = y(1−2x−y),

4.

0 = f 2 (x, y) = x(1−x−2y).

The only critical points are (0, 0), (1, 0) and (0, 1), which are on the boundary of T , and (1/3, 1/3), which is inside T . The maximum value of f over T is f (1/3, 1/3) = 1/27.

Thus the maximum and minimum values of f on the 1 1 1 1 disk D are √ − and − √ − respectively. 2 2 2 2 x2

f (x, y) = x y(1 − x − y) on the triangle T shown in the figure. Evidently f (x, y) = 0 on all three boundary segements of T , and f (x, y) > 0 inside T . Thus the minimum value of f on T is 0, and the maximum value must occur at an interior critical point. For critical points:

y

1

y2

f (x, y) = x + 2y on the closed disk + ≤ 1. Since f 1 = 1 and f 2 = 2, f has no critical points, and the maximum and minimum values of f , which must exist because f is continuous on a closed, bounded set in the plane, must occur at boundary points of the domain, that is, points of the circle x 2 + y 2 = 1. This circle can be parametrized x = cos t, y = sin t, so that

x+y=1 T 1 x

Fig. 13.2.6

f (x, y) = f (cos t, sin t) = cos t + 2 sin t = g(t), say. ′ (t) = − sin t + 2 cos t. For critical points of g: 0 = g√ √ Thus tan t = 2, and x √ = ±1/ √5, y = ±2/ 5. The critical points √ are (−1/ √ √ 5, −2/ 5), where f √has value − 5, and (1/ 5, 2/ 5), where f has value 5. Thus the maximum and minimum values of f (x, y) on the √ √ disk are 5 and − 5 respectively.

5.

f (x, y) = x y − x 3 y 2 on the square S: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. f 1 = y − 3x 2 y 2 = y(1 − 3x 2 y), f 2 = x − 2x 3 y = x(1 − 2x 2 y). (0, 0) is a critical point. Any other critical points must satisfy 3x 2 y = 1 and 2x 2 y = 1, that is, x 2 y = 0. Therefore (0, 0) is the only critical point, and it is on the boundary of S. We need therefore only consider the values of f on the boundary of S. On the sides x = 0 and y = 0 of S, f (x, y) = 0.

7. Since −1 ≤ f (x, y) = sin x cos y ≤ 1 everywhere, and

since f (π/2, 0) = 1, f (3π/2, 0) = −1, and both (π/2, 0) and (3π/2, 0) belong to the triangle bounded by x = 0, y = 0 and x + y = 2π , therefore the maximum and minimum values of f over that triangle must be 1 and −1 respectively.

8.

f (x, y) = sin x sin y sin(x + y) on the triangle T shown in the figure. Evidently f (x, y) = 0 on the boundary of T , and f (x, y) > 0 at all points inside T . Thus the minimum value of f on T is zero, and the maximum value must occur at an interior critical point. For critical points inside T we must have 0 = f 1 (x, y) = cos x sin y sin(x + y) + sin x sin y cos(x + y) 0 = f 2 (x, y) = sin x cos y sin(x + y) + sin x sin y cos(x + y).

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SECTION 13.2 (PAGE 758)

ADAMS and ESSEX: CALCULUS 8

Therefore cos x sin y = cos y sin x, which implies x = y for points inside T , and cos x sin x sin 2x + sin2 x cos 2x = 0

2 sin2 x cos2 x + 2 sin2 x cos2 x − sin2 x = 0 4 cos2 x = 1.

Thus cos x = ±1/2, and x = ±π/3. The interior √ critical point is (π/3, π/3), where f has the value 3 3/8. This is the maximum value of f on T . y

Any critical points must satisfy 1 − x 2 + y 2 + 2x y = 0 2 2 and −1 − x 2 + y 2 − 2x y = 0, and hence √ x = y and 2x y = −1. Therefore y = −x = ±1/ √ 2. The√only critical point in the region y√≥ 0 is (−1/ 2, 1/ 2), where f has the value −1/ 2. On the boundary y = 0 we have f (x, 0) =

π

x = g(x), 1 + x2

(−∞ < x < ∞).

Evidently, g(x) → 0 as x → ±∞. 1 − x2 Since g ′ (x) = , the critical points of g are (1 + x 2 )2 1 x = ±1. We have g(±1) = ± . 2 The maximum and minimum values √ of f on the upper half-plane y ≥ 0 are 1/2 and −1/ 2 respectively.

z+y=π T π x

Fig. 13.2.8

9.

2

2

T = (x + y)e−x −y on D = {(x, y) : x 2 + y 2 ≤ 1}. For critical points:   2 2 ∂T 0= = 1 − 2x(x + y) e−x −y ∂x   2 2 ∂T 0= = 1 − 2y(x + y) e−x −y . ∂y

The critical points are given by 2x(x + y) = 1 = 2y(x + y), which forces x = y and 1 4x 2 = 1, so x = y = ± . 2     1 1 1 1 and − , − , The two critical points are , 2 2 2 2 both of which lie inside D. T takes the values ±e−1/2 at these points. On the boundary of D, x = cos t, y = sin t, 0 ≤ t ≤ 2π , so that T = (cos t + sin t)e−1 = g(t),

(0 ≤ t ≤ 2π ).

We have g(0) = g(2π ) = e−1 . For critical points of g: ′

0 = g (t) = (cos t − sin t)e

10.

−1

,

so tan t =√1 and t = π/4 or t = 5π/4. √ −1Observe that g(π/4) = 2e−1 , and g(5π/4) = − 2e . √ Since e−1/2 > 2e−1 (because e > 2), the maximum and minimum values of T on the disk are ±e−1/2 , the values at the interior critical points. x−y f (x, y) = on the half-plane y ≥ 0. 1 + x 2 + y2 For critical points: 1 − x 2 + y 2 + 2x y (1 + x 2 + y 2 )2 −1 − x 2 + y 2 − 2x y 0 = f 2 (x, y) = . (1 + x 2 + y 2)2 0 = f 1 (x, y) =

11. Let f (x, y, z) = x y 2 +yz 2 on the ball B: x 2 +y 2 +z 2 ≤ 1. First look for interior critical points: 0 = f1 = y 2,

0 = f 2 = 2x y + z 2 ,

0 = f 3 = 2yz.

All points on the x-axis are CPs, and f = 0 at all such points. Now consider the boundary sphere z 2 = 1 − x 2 − y 2 . On it f (x, y, z) = x y 2 +y(1−x 2−y 2 ) = x y 2 +y−x 2 y−y 3 = g(x, y), where g is defined for x 2 + y 2 ≤ 1. Look for interior CPs of g: 0 = g1 = y 2 − 2x y = y(y − 2x) 0 = g2 = 2x y + 1 − x 2 − 3y 2 .

Case I: y = 0. Then g = 0 and f = 0. Case II: y = 2x. Then 4x 2 + 1 − x 2 − 12x 2 = 0, so 9x 2 = 1 and x = ±1/3. This case produces critical points  2 4 1 2 , ,± , where f = , and 3 3 3 9   1 2 2 4 − ,− ,± , where f = − . 3 3 3 9



Now we must consider the boundary x 2 + y 2 = 1 of the domain of g. Here g(x, y) = x y 2 = x(1 − x 2 ) = x − x 3 = h(x)

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SECTION 13.2 (PAGE 758)

for −1 ≤ x ≤ 1. At the endpoints x = ±1, h = 0, so g = 0 and f = 0. For CPs of h:

13.

0 = h ′ (x) = 1 − 3x 2 , √ √ so x = ±1/ 3 √ and y = ± 2/3. The √ value of h at such points is ±2/(3 3). However 2/(3 3) < 4/9, so the maximum value of f is 4/9, and the minimum value is −4/9.

12. Let f (x, y, z) = x z + yz on the ball x 2 + y 2 + z 2 ≤ 1. First look for interior critical points: 0 = f 1 = z,

0 = f 2 = z,

0 = f 3 = x + y.

All points on the line z = 0, x + y = 0 are CPs, and f = 0 at all such points. Now consider the boundary sphere x 2 + y 2 + z 2 = 1. On it q f (x, y, z) = (x + y)z = ±(x + y) 1 − x 2 − y 2 = g(x, y), where g has domain x 2 + y 2 ≤ 1. On the boundary of its domain, g is identically 0, although g takes both positive and negative values at some points inside its domain. Therefore, we need consider only critical points of g in x 2 + y 2 < 1. For such CPs: q (x + y)(−2x) 1 − x 2 − y2 + p 2 1 − x 2 − y2 2 2 2 1 − x − y − x − xy p = 1 − x 2 − y2 2 1 − x − y2 − x y − y2 p 0 = g2 = . 1 − x 2 − y2 0 = g1 =

2x 2 + y 2 + x y

x 2 + 2y 2 + x y,

Therefore =1= from which x 2 = y 2. Case I: x = −y. Then g = 0, so f = 0. Case II: x = y. Then 2x 2 + x 2 + x 2 = 1, so x 2 = 1/4 and x = ±1/2. g (which is really two functions depending on our choice of the “+” or “−” sign) has four CPs, two corresponding to x = y = 1/2 and two to x = √ y = −1/2. The values of g at these four points are ±1/ 2. Since we have considered all points where f can have extreme values, we conclude that the maximum value √ of f on the ball is 1/ 2 (which occurs at the√boundary points ±( 12 , 12 , √1 )) and minimum value −1/ 2 (which

f (x, y) = x ye−xy on Q = {(x, y) : x ≥ 0, y ≥ 0}. 2 Since f (x, kx) = kx 2 e−kx → 0 as x → ∞ if k > 0, and f (x, 0) = f (0, y) = 0, we have f (x, y) → 0 as (x, y) recedes to infinity along any straight line from the origin lying in the  first quadrant Q.  1 However, f x, = 1 and f (x, 0) = 0 for all x > 0, x   1 even though the points x, and (x, 0) become arx bitrarily close together as x increases. Thus f does not have a limit as x 2 + y 2 → ∞. Observe that f (x, y) = r e−r = g(r ) on the hyperbola x y = r > 0. Since g(r ) → 0 as r approaches 0 or ∞, and g ′ (r ) = (1 − r )e−r = 0 ⇒ r = 1, f (x, y) is everywhere on Q less than g(1) = 1/e. Thus f does have a maximum value on Q.

14.

f (x, y) = x y 2 e−xy on Q = {(x, y) : x ≥ 0, y ≥ 0}. As in Exercise 13, f (x, 0) = f (0, y) = 0 and 2 lim x→∞ f (x, kx) = k 2 x 3 e−x = 0. y 1 Also, f (0, y) = 0 while f ,y = → ∞ as y e 2 2 y → ∞, so that f has no limit as x + y → ∞ in Q, and f has no maximum value on Q.

15. If brewery A produces x litres per month and brewery B produces y litres per month, then the monthly profits of the two breweries are given by P = 2x −

2x 2 + y 2 , 106

Q = 2y −

4y 2 + x 2 . 2 × 106

STRATEGY I. Each brewery selects its production level to maximize its own profit, and assumes its competitor does the same. Then A chooses x to satisfy 0=

4x ∂P =2− 6 ∂x 10

⇒

x = 5 × 105 .

B chooses y to satisfy 0=

∂Q 8y =2− ∂y 2 × 106

⇒

y = 5 × 105 .

The total profit of the two breweries under this strategy is 3 × 25 × 1010 5 × 25 × 1010 + 106 − 6 10 2 × 106 = $625, 000.

P + Q = 106 −

STRATEGY II. The two breweries cooperate to maximize the total profit

2

occurs at the boundary points ±( 12 , 12 , − √1 )). 2

T = P + Q = 2x + 2y −

5x 2 + 6y 2 2 × 106

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SECTION 13.2 (PAGE 758)

ADAMS and ESSEX: CALCULUS 8

by choosing x and y to satisfy

Substituting the first equation into the second we obtain  50  1 − 2 sin2 θ + 2 sin θ − 100 sin θ = 0 2 − sin θ 50(1 − 2 sin2 θ + 2 sin θ ) = 100(2 sin θ − sin2 θ ) 50 = 100 sin θ.

∂T 10x =2− , ∂x 2 × 106 ∂T 12y 0= =2− . ∂y 2 × 106 0=

1 Thus x = 4 × 105 and y = × 106 . 3 In this case the total monthly profit is 2 80 × 1010 + × 1012 2 6 3 P + Q = 8 × 10 + × 10 − 3 2 × 106 ≈ $733, 333. 5

Observe that the total profit is larger if the two breweries cooperate and fix prices to maximize it.

16. Let the dimensions be as shown in the figure. Then 2x + y = 100, the length of the fence. For maximum area A of the enclosure we will have x > 0 and 0 < θ < π/2. Since h = x cos θ , the area A is 1 A = x y cos θ + 2 × (x sin θ )(x cos θ ) 2 = x(100 − 2x) cos θ + x 2 sin θ cos θ 1 = (100x − 2x 2 ) cos θ + x 2 sin 2θ. 2

Thus sin θ = 1/2, and θ = π/6. 50 = Therefore x = 2 − (1/2) 100 y = 100 − 2x = . 3

100 , and 3

The maximum area for the enclosure is   √   √ 100 2 3 100 2 1 3 2500 A= + = √ 3 2 3 2 2 3 square units. All three segments of the fence will be the same length, and the bend angles will be 120◦ .

17. To maximize Q(x, y) = 2x + 3y subject to x ≥ 0,

y ≥ 0,

y ≤ 5,

x + 2y ≤ 12,

4x + y ≤ 12.

The constraint region is shown in the figure. y

4x+y=12

We look for a critical point of A satisfying x > 0 and 0 < θ < π/2. y=5

wall

x+2y=12 h

h

x



x

θ

θ y

∂A = (100 − 4x) cos θ + x sin 2θ ∂x ⇒ cos θ (100 − 4x + 2x sin θ ) = 0 ⇒ 4x − 2x sin θ = 100 ⇒ x =

x

Fig. 13.2.17

Fig. 13.2.16

0=



7 4 ,5

50 2 − sin θ

∂A 0= = −(100x − 2x 2 ) sin θ + x 2 cos 2θ ∂θ ⇒ x(1 − 2 sin2 θ ) + 2x sin θ − 100 sin θ = 0.

Observe that any point satisfying y ≤ 5 and 4x + y ≤ 12 automatically satisfies x +2y ≤  12. Since y = 5 and 7 4x + y = 12 intersect at , 5 , the maximum value of 4 Q(x, y) subject to the given constraints is   7 37 7 Q , 5 = + 15 = . 4 2 2

18. Minimize F(x, y, z) = 2x + 3y + 4z subject to x ≥ 0, x + y ≥ 2,

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y ≥ 0, y + z ≥ 2,

z ≥ 0, x + z ≥ 2.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.3 (PAGE 765)

Here the constraint region has vertices (1, 1, 1), (2, 2, 0), (2, 0, 2), and (0, 2, 2). Since F(1, 1, 1) = 9, F(2, 2, 0) = 10, F(2, 0, 2) = 12, and F(0, 2, 2) = 14, the minimum value of F subject to the constraints is 9.

To produce the maximum revenue, the manufacturer should produce 20, 000/3 ≈ 6, 667 kg of each grade of fabric.

20. If the developer builds x houses, y duplex units, and z

z

apartments, his profit will be P = 40, 000x + 20, 000y + 16, 000z.

x+y=2

x=0

The legal constraints imposed require that

y=0

x y z + + ≤ 10, 6 8 12

(0,2,2) (2,0,2)

that is 4x + 3y + 2z ≤ 240,

and also (1,1,1)

z ≥ x + y.

x+z=2

y+z=2 (2,2,0)

Evidently we must also have x ≥ 0, y ≥ 0, and z ≥ 0. The planes 4x + 3y + 2z = 240 and z = x + y intersect where 6x + 5y = 240. Thus the constraint region has vertices (0, 0, 0), (40, 0, 40), (0, 48, 48), and (0, 0, 120), which yield revenues of $0, $2,240,000, $1,728,000, and $1,920,000 respectively. For maximum profit, the developer should build 40 houses, no duplex units, and 40 apartments.

y

x z=0

Fig. 13.2.18

19. Suppose that x kg of deluxe fabric and y kg of standard fabric are produced. Then the total revenue is R = 3x + 2y. The constraints imposed by raw material availability are

Section 13.3 (page 765)

1. First we observe that f (x, y) = x 3 y 5 must have a max-

imum value on the line x + y = 8 because if x → −∞ then y → ∞ and if x → ∞ then y → −∞. In either case f (x, y) → −∞. Let L = x 3 y 5 + λ(x + y − 8). For CPs of L:

20 10 x+ y ≤ 2, 000, ⇔ 2x + y ≤ 20, 000 100 100 50 40 x+ y ≤ 6, 000, ⇔ 5x + 4y ≤ 60, 000 100 100 30 50 x+ y ≤ 6, 000, ⇔ 3x + 5y ≤ 60, 000. 100 100

∂L = 3x 2 y 5 + λ ∂x ∂L 0= = 5x 3 y 4 + λ ∂y ∂L 0= = x + y − 8. ∂λ

0=

The lines 2x + y =  20, 000 and 5x +  4y = 60, 000 20, 000 20, 000 intersect at the point , , which satisfies 3 3 3x + 5y ≤ 60, 000, so lies in the constraint region. We have   20, 000 20, 000 f , ≈ 33, 333. 3 3 The lines 2x + y =20, 000 and 3x +5y = 60, 000 in40, 000 60, 000 tersect at the point , , which does not 7 7 satisfy 5x + 4y ≤ 60, 000 and so does not lie in the constraint region. The lines 5x + 4y =  60, 000 and 3x +5y = 60, 000 in60, 000 120, 000 tersect at the point , , which satisfies 13 13 2x + y ≤ 20, 000 and so lies in the constraint region. We have   60, 000 120, 000 f , ≈ 32, 307. 13 13

Lagrange Multipliers

The first two equations give 3x 2 y 5 = 5x 3 y 4 , so that either x = 0 or y = 0 or 3y = 5x. If x = 0 or y = 0 then f (x, y) = 0. If 3y = 5x, then x + 53 x = 8, so 8x = 24 and x = 3. Then y = 5, and f (x, y) = 33 55 = 84, 375. This is the maximum value of f on the line.

2.

a) Let D be the distance from (3, 0) to the point (x, y) on the curve y = x 2 . Then D 2 = (x − 3)2 + y 2 = (x − 3)2 + x 4 . d D2 = 2(x − 3) + 4x 3 . Thus dx 2x 3 + x − 3 = 0. Clearly x = 1 is a root of this cubic equation. Since

For a minimum, 0 =

2x 3 + x − 3 = 2x 2 + 2x + 3, x −1

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SECTION 13.3 (PAGE 765)

ADAMS and ESSEX: CALCULUS 8

and 2x 2 + 2x + 3 has negative discriminant, x = 1 is the only critical point. Thus thepminimum distance √ from (3, 0) to y = x 2 is D = (−2)2 + 14 = 5 units. b) We want to minimize D 2 = (x − 3)2 + y 2 subject to the constraint y = x 2 . Let L = (x − 3)2 + y 2 + λ(x 2 − y). For critical points of L we want ∂L = 2(x − 3) + 2λx ∂x ⇒ (1 + λ)x − 3 = 0 ∂L = 2y − λ 0= ∂y ∂L 0= = x 2 − y. ∂λ

Therefore Y = Z = 32 and X = 13 , and the distance is 1 unit as in part (a). c) The point (X, Y, Z) must be a critical point of the Lagrange function L = x 2 + y 2 + z 2 + λ(x + 2y + 2z − 3). To find these critical points we have ∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ

0=

0=

(A) (B) (C)

Eliminating λ from (A) and (B), we get x + 2x y − 3 = 0. Substituting (C) then leads to 2x 3 + x − 3 = 0, or (x − 1)(2x 2 + 2x + 3) = 0. The only real solution is x = 1, so the point on y = x 2 closest to (3, 0) is (1, 1). 2 Thus p the minimum distance √ from (3, 0) to y = x is 2 2 D = (1 − 3) + 1 = 5 units.

3. Let (X, Y, Z) be the point on the plane x + 2y + 2z = 3

= 2x + λ = 2y + 2λ = 2z + 2λ = x + 2y + 2z − 3.

The first three equations yield y = z = −λ, x = −λ/2. Substituting these into the fourth equation we get λ = − 32 , so that the critical point is once
again 31 , 23 , 23 , whose distance from the origin is 1 unit.

4. Let f (x, y, z) = x + y − z, and define the Lagrange function

closest to (0, 0, 0).

a) The vector ∇ (x + 2y + 2z) = i + 2j + 2k is perpendicular to the plane, so must be parallel to the vector Xi + Y j + Zk from the origin to (X, Y, Z). Thus

L = x + y − z + λ(x 2 + y 2 + z 2 − 1). Solutions to the constrained problem will be found among the critical points of L. To find these we have

Xi + Y j + Zk = t (i + 2j + 2k),

∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ 0=

for some scalar t. Thus X = t, Y = 2t, Z = 2t, and, since (X, Y, Z) lies on the plane, 3 = X + 2Y + 2Z = t + 4t + 4t = 9t. 1 3,

1 3

2 3.

Thus t = and we have X = and Y = Z = The minimum√distance from the origin to the plane is therefore 13 1 + 4 + 4 = 1 unit.

b) (X, Y, Z) must minimize the square of the distance from the origin to (x, y, z) on the plane. Thus it is a critical point of S = x 2 + y 2 + z 2 . Since x + 2y + 2z = 3, we have x = 3 − 2(y + z), and  2 S = S(y, z) = 3 − 2(y + z) + y 2 + z 2 . The critical points of this function are given by   ∂S = −4 3 − 2(y + z) + 2y = −12 + 10y + 8z ∂y   ∂S 0= = −4 3 − 2(y + z) + 2z = −12 + 8y + 10z. ∂z

0=

= 1 + 2λx, = 1 + 2λy, = −1 + 2λz, = x 2 + y 2 + z 2 − 1.

Therefore 2λx = 2λy = −2λz. Either λ = 0 or x = y = −z. λ = 0 is not possible. (It implies 0 = 1 from the first equation.) From x = y = −z we obtain 1 1 = x 2 + y 2 + z 2 = 3x 2 , so x = ± √ . L has critical 3     1 1 1 1 1 1 points at √ , √ , − √ and √ ,−√ , √ . 3 √ 3 3 − 3 3 3 At the first f = 3, which is the maximum value of √ f on the sphere; at the second f = − 3, which is the minimum value.

5. The distance D from (2, 1, −2) to (x, y, z) is given by D 2 = (x − 2)2 + (y − 1)2 + (z + 2)2 .

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.3 (PAGE 765)

We can extremize D by extremizing D 2 . If (x, y, z) lies on the sphere x 2 + y 2 +z 2 = 1, we should look for critical points of the Lagrange function L = (x − 2)2 + (y − 1)2 + (z + 2)2 + λ(x 2 + y 2 + z 2 − 1). Thus ∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ

0=

= 2(x − 2) + 2λx

⇔

= 2(y − 1) + 2λy

⇔

= 2(z + 2) + 2λz

⇔

= x 2 + y 2 + z 2 − 1.

2 1+λ 1 y= 1+λ −2 z= 1+λ x=

Substituting the solutions of the first three equations into the fourth, we obtain 1 (4 + 1 + 4) = 1 (1 + λ)2 (1 + λ)2 = 9 1 + λ = ±3.
Thus we must consider the two points P = 32 , 13 , − 32 ,
and Q = − 32 , − 31 , 23 for giving extreme values for D. At P, D = 2. At Q, D = 4. Thus the greatest and least distances from (2, 1, −2) to the sphere x 2 + y 2 + z 2 = 1 are 4 units and 2 units respectively.

6. Let L = x 2 + y 2 + z 2 + λ(x yz 2 − 2). For critical points: ∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ

0=

= 2x + λyz 2

⇔

−λx yz 2 = 2x 2

= 2y + λx z 2

⇔

−λx yz 2 = 2y 2

= 2z + 2λx yz

⇔

−λx yz 2 = z 2

For critical points of L: ∂L ∂a ∂L 0= ∂b ∂L 0= ∂c ∂L 0= ∂λ 0=

4π abc subject to the con3 1 4 1 straint 2 + 2 + 2 = 1. Note that abc cannot be zero. a b c Let   4π abc 1 4 1 L= +λ + + − 1 . 3 a2 b2 c2

⇔ ⇔

2π abc λ = 2 3 a 2π abc 4λ = 2 3 b λ 2π abc = 2 3 c

− 1.

4 1 1 1 = 2 = 2 = , a2 b c 3 √ √ √ so a = ± 3, b = ±2 3, and c = ± 3.

8. Let L = x 2 + y 2 + λ(3x 2 + 2x y + 3y 2 − 16). We have ∂L = 2x + 6λx + 2λy ∂x ∂L = 2y + 6λy + 2λx. 0= ∂y

0=

(A) (B)

Multiplying (A) by y and (B) by x and subtracting we get 2λ(y 2 − x 2 ) = 0. Thus, either λ = 0, or y = x, or y = −x. λ = 0 is not possible, since it implies x = 0 and y = 0, and the point (0, 0) does not lie on the √ given ellipse. If y = x, then 8x 2 = 16, so x = y = ± 2. If y = −x, then 4x 2 = 16, so x = −y = ±2. √ √ The points √ on√the ellipse nearest the origin are ( 2, 2) and (− 2, − 2). The points farthest from the origin are (2, −2) and (−2, 2). The major√axis of the ellipse lies along y = −x and has length 4 2. The minor axis lies along y = x and has length 4.

9. Let L = x yz + λ(x 2 + y 2 + z 2 − 12). For CPs of L: ∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ 0=

= x yz − 2.

7. We want to minimize V =

⇔

abc 6= 0 implies λ 6= 0, and so we must have

2

From the first three equations, x 2 = y 2 and z 2 = 2x 2 . The fourth equation then gives x 2 y 2 4z 4 = 4, or x 8 = 1. Thus x 2 = y 2 = 1 and z 2 = 2. The shortest distance from the origin to the surface x yz 2 = 2 is √ 1 + 1 + 2 = 2 units.

4π bc 2λ − 3 3 a 4π ac 8λ = − 3 3 b 4π ab 2λ = − 3 3 c 1 4 1 = 2 + 2 + 2 a b c =

= yz + 2λx

(A)

= x z + 2λy

(B)

= x y + 2λz

(C)

= x 2 + y 2 + z 2 − 12.

(D)

Multiplying equations (A), (B), and (C) by x, y, and z, respectively, and subtracting in pairs, we conclude that λx 2 = λy 2 = λz 2 , so that either λ = 0 or x 2 = y 2 = z 2 . If λ = 0, then (A) implies that yz = 0, so x yz = 0. If x 2 = y 2 = z 2 , then (D) gives 3x 2 = 12, so x 2 = 4. We obtain eight points (x, y, z) where each coordinate is either 2 or −2. At four of these points x yz =8, which is the maximum value of x yz on the sphere. At the other four x yz = −8, which is the minimum value.

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SECTION 13.3 (PAGE 765)

ADAMS and ESSEX: CALCULUS 8

10. Let L = x + 2y − 3z + λ(x 2 + 4y 2 + 9z 2 − 108). For CPs of L:

∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ

0=

12. The maximum will occur at a critical point of the Lagrange function

(A)

= 1 + 2λx = 2 + 8λy

(B)

= −3 + 18λz

(C)

= x 2 + 4y 2 + 9z 2 − 108.

(D)

L=

1 2 3 =− = , 2x 8y 18z

so x = 2y = −3z. From (D): x2 + 4



x2 4



+9



x2 9



= 108,

L = x 2 + y 2 + z 2 + λx y 2 z 4 − 32,

0=

−1 .

The first n equations show that xi = −1/(2λ) for each i , so xi = x1 for 1 ≤ i ≤ n. The constraint equation now gives n X 1 1= xi2 = nx12 , so x1 = ± √ . n i=1 √

n.

critical points of L: 0= 0= 0= 0= 0=

2 4

= x y 2 z 4 − 1.

xi = nx1 =

13. Let L = x + λ(x + y − z) + µ(x 2 + 2y 2 + 2z 2 − 8). For

x 6= 0, y 6= 0, z 6= 0.

2x = 2x + λy z ⇐⇒ = −λ y2z4 1 = 2y + 2λx yz 4 ⇐⇒ = −λ x z4 1 = 2z + 4λx y 2 z 3 ⇐⇒ = −λ 2x y 2 z 2

n X i=1

Thus we calculate

0=

i=1

!

∂L = 1 + 2xi λ, −1 ≤ i ≤ n ∂ xi n X ∂L = xi2 − 1. 0= ∂λ i−1

and is not bounded, so it has no farthest point from the origin. To find the closest point we look at critical points of

0=

xi2

0=

11. The surface has no points where any coordinate is zero,

0=

xi + λ

The maximum value is thus

so x 2 = 36, and x = ±6. There are two CPs: (6, 3, −2) and (−6, −3, 2). At the first, x + 2y − 3z = 18, the maximum value, and at the second, x + 2y − 3z = −18, the minimum value.

∂L ∂x ∂L ∂y ∂L ∂z ∂L ∂λ

i=1

n X

For a critical point we have:

From (A), (B), and (C), λ=−

n X

∂L ∂x ∂L ∂y ∂L ∂z ∂L ∂λ ∂L ∂µ

= 1 + λ + 2µx

(A)

= λ + 4µy

(B)

= −λ + 4µz

(C)

=x +y−z

(D)

= x 2 + 2y 2 + 2z 2 − 8.

(E)

From (B) and (C) we have µ(y + z) = 0. Thus µ = 0 or y + z = 0. CASE I. µ = 0. Then λ = 0 by (B), and 1 = 0 by (A), so this case is not possible. CASE II. y + z = 0. Then z = −y and, by (D), x = −2y. Therefore, by (E), 4y 2 + 2y 2 + 2y 2 = 8, and so y = ±1. From this case we obtain two points: (2, −1, 1) and (−2, 1, −1). The function f (x, y, z) = x has maximum value 2 and minimum value −2 when restricted to the curve x + y = z, x 2 + 2y 2 + 2z 2 = 8.

The first three equations imply that 1 1 2x = 4 = , y2z4 xz 2x y 2 z 2 from which it follows that y 2 = 2x 2 and z 2 = 2y 2 = 4x 2 . The constraint equation then gives x(2x 2 )(16x 4 )√= 32, or, simply, x 7 = 1. This implies x = 1, so y = ± 2 and z = ±2. The distance from the surface to the origin is √ √ 1 + 2 + 4 = 7 units.

14. Let L = x 2 + y 2 + z 2 + λ(x 2 + y 2 − z 2 ) + µ(x − 2z − 3). For critical points of L:

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0=

∂L = 2x(1 + λ) + µ ∂x

(A)

INSTRUCTOR’S SOLUTIONS MANUAL

∂L ∂y ∂L 0= ∂z ∂L 0= ∂λ ∂L 0= ∂µ 0=

SECTION 13.3 (PAGE 765)

= 2y(1 + λ)

(B)

= 2z(1 − λ) − 2µ

(C)

= x 2 + y2 − z2

(D)

= x − 2z − 3.

(E)

From (B), either y = 0 or λ = −1. CASE I. y = 0. Then (D) implies x = ±z. If x = z then (E) implies z = −3, so we get the point (−3, 0, −3). If x = −z then (E) implies z = −1, so we get the point (1, 0, −1). CASE II. λ = −1. Then (A) implies µ = 0 and (C) implies z = 0. By (D), x = y = 0, and this contradicts (E), so this case is not possible. If f (x, y, z) = x 2 + y 2 + z 2 , then f (−3, 0, −3) = 18 is the maximum value of f on the ellipse x 2 + y 2 = z 2 , x − 2z = 3, and f (1, 0, −1) = 2 is the minimum value.

15. Let L = 4 − z + λ(x 2 + y 2 − 8) + µ(x + y + z − 1). For critical points of L:

0= 0= 0= 0= 0=

∂L ∂x ∂L ∂y ∂L ∂z ∂L ∂λ ∂L ∂µ

= 2λx + µ

(A)

= 2λy + µ

(B)

= −1 + µ

(C)

= x 2 + y2 − 8

(D)

= x + y + z − 1.

(E)

From (C), µ = 1. From (A) and (B), λ(x − y) = 0, so either λ = 0 or x = y. CASE I. λ = 0. Then µ = 0 by (A), and this contradicts (C), so this case is not possible. CASE II. x = y. Then x = y = ±2 by (D). If x = y = 2, then z = −3 by (E). If x = y = −2, then z = 5 by (E). Thus we have two points, (2, 2, −3) and (−2, −2, 5), where f (x, y, z) = 4 − z takes the values 7 (maximum), and −1 (minimum) respectively.

16. The max and min values of f (x, y, z) = x + y 2 z subject to the constraints y 2 + z 2 = 2 and z = x will be found among the critical points of L = x + y 2 z + λ(y 2 + z 2 − 2) + µ(z − x).

Thus

∂L = 1 − µ = 0, ∂x ∂L = 2yz + 2λy = 0, 0= ∂y ∂L 0= = y 2 + 2λz + µ = 0, ∂z ∂L 0= = y 2 + z 2 − 2, ∂λ ∂L 0= = z − x. ∂µ From the first equation µ = 1. From the second, either y = 0 or z = −λ. If√y = √ 0 then z 2 =√ 2, z =√x, so critical points are √ ( 2, 0, 2) and (− 2, 0, − 2). f has the values ± 2 at these points. If z = −λ then y 2 − 2z 2 + 1 = 0. Thus 2z 2 − 1 = 2 − z 2 , or z 2 = 1, z = ±1. This leads to critical points (1, ±1, 1) and (−1, ±1, −1) where f has values ±2. The maximum value of f subject to the constraints is 2; the minimum value is −2. 0=

17. Let L = (x − a)2 + (y − b)2 + (z − c)2 + λ(x − y) + µ(y − z) + σ (a + b) + τ (c − 2). For critical points of L, we have ∂L = 2(x − a) + λ (A) ∂x ∂L 0= = 2(y − b) − λ + µ (B) ∂y ∂L 0= = 2(z − c) − µ (C) ∂z ∂L 0= = −2(x − a) + σ (D) ∂a ∂L 0= = −2(y − b) + σ (E) ∂b ∂L 0= = −2(z − c) + τ (F) ∂c ∂L 0= =x−y (G) ∂λ ∂L 0= = y−z (H ) ∂µ ∂L 0= = a+b (I ) ∂σ ∂L 0= = c − 2. (J ) ∂τ Subtracting (D) and (E) we get x − y = a − b. From (G), x = y, and therefore a = b. From (I), a = b = 0, and from (J), c = 2. Adding (A), (B) and (C), we get x + y +z = a +b+c = 2. From (G) and (H), x = y = z = 2/3. The minimum distance between the two lines is s √  2  2  2 r 2 2 2 24 2 6 −0 + −0 + −2 = = units. 3 3 3 9 3 0=

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SECTION 13.3 (PAGE 765)

ADAMS and ESSEX: CALCULUS 8

For critical points of L,

18. Let the width, depth, and height of the box be x, y and z respectively. We want to minimize the surface area S = x y + 2x z + 2yz subject to the constraint that x yz = V , where V is a given positive volume. Let L = x y + 2x z + 2yz + λ(x yz − V ). For critical points of L, ∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ 0=

= y + 2z + λyz

⇔

−λx yz = x y + 2x z

= x + 2z + λx z

⇔

−λx yz = x y + 2yz

= 2x + 2y + λx y

⇔

−λx yz = 2x z + 2yz

∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ

0=

= yz + λ(y + 3z)

⇔

−x yz = λ(x y + 3x z)

= x z + λ(x + 2z)

⇔

−x yz = λ(x y + 2yz)

= x y + λ(2y + 3x)

From the first three equations x y = 2yz = 3x z. From the fourth equation, the sum of these expressions is 18. Thus x y = 2yz = 3x z = 6. Thus the maximum volume of the box is V = x yz =

z as shown in the figure. Let the cost per unit area of the back and sides be $k. Then the cost per unit area of the front and bottom is $5k. We want to minimize C = 5k(x z + x y) + k(2yz + x z)

Let

For critical points of L, 0= 0= 0= 0=

∂L ∂x ∂L ∂y ∂L ∂z ∂L ∂λ

= yz + 4λ

⇔

x yz + 4λx = 0

= x z + 2λ

⇔

x yz + 2λy = 0

= xy + λ

⇔

x yz + λz = 0

= 4x + 2y + z − 2 = 0.

The first three equations imply that z = 2y = 4x (since we cannot have λ = 0 if V is positive). The fourth equation then implies that 12x = 2. Hence x = 1/6, y = 1/3, and z = 2/3. The largest box has volume V =

1 1 1 2 × × = cubic units. 6 3 3 27

20. We want to maximize x yz subject to x y +2yz +3x z = 18. Let

L = x yz + λ(x y + 2yz + 3x z − 18).

p √ (x y)(yz)(x z) = 6 × 3 × 2 = 6 cubic units.

21. Let the width, depth, and height of the box be x, y, and

19. We want to maximize V = x yz subject to 4x +2y+z = 2. L = x yz + λ(4x + 2y + z − 2).

−x yz = λ(2yz + 3x z)

= x y + 2yz + 3x z − 18.

= x yz − V .

From the first three equations, x y = 2x z = 2yz. Since x, y, and z are all necessarily positive, we must therefore have x = y = 2z. Thus the most economical box with no top has width and depth equal to twice the height.

⇔

subject to the constraint x yz = V (constant). Let L = k(5x y + 6x z + 2yz) + λ(x yz − V ). For critical points of L, ∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ 0=

= 5ky + 6kz + λyz

⇔

−λx yz = 5kx y + 6kx z

= 5kx + 2kz + λx z

⇔

−λx yz = 5kx y + 2kyz

= 6kx + 2ky + λx y

⇔

−λx yz = 6kx z + 2kyz

= x yz − V .

From the first three of these equations we obtain 5x 5x y = 6x z = 2yz. Thus y = 3x and z = . From the 2 15 3 fourth equation, V = x yz = x .  2 1/3   2V 2V 1/3 The largest box has width , depth 3 , 15 15  1/3 5 2V and height . 2 15

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SECTION 13.3 (PAGE 765)

23. In this problem we do the boundary analysis for Exercise 22 using the suggested parametrization of the sphere x 2 + y 2 + z 2 = 1. We have

back

f (x, y, z) = x y + z 2

side

= sin2 φ sin θ cos θ + cos2 φ 1 = sin2 φ sin 2θ + cos2 φ 2 = g(φ, θ )

z side

front bottom

y

x

for 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π . For critical points of g, Fig. 13.3.21

22.

0 = g1 (φ, θ ) = sin φ cos φ sin 2θ − 2 sin φ cos φ = sin φ cos φ(sin 2θ − 2)

f (x, y, z) = x y + z 2 on B = {(x, y, z) : x 2 + y 2 + z 2 ≤ 1}. For critical points of f , 0 = f 1 (x, y, z) = y, 0 = f 3 (x, y, z) = 2z.

0 = g2 (φ, θ ) = sin2 φ cos 2θ.

0 = f 2 (x, y, z) = x,

Thus the only critical point is the interior point (0, 0, 0), where f has the value 0, evidently neither a maximum nor a minimum. The maximum and minimum must therefore occur on the boundary of B, that is, on the sphere x 2 + y 2 + z 2 = 1. Let L = x y + z 2 + λ(x 2 + y 2 + z 2 − 1). For critical points of L, 0= 0= 0= 0=

∂L ∂x ∂L ∂y ∂L ∂z ∂L ∂λ

24.

= y + 2λx

(A)

= x + 2λy

(B)

= 2z(1 + λ)

(C)

= x 2 + y 2 + z 2 − 1.

(D)

From (C) either z = 0 or λ = −1.

The first of these equations implies that either sin φ = 0 or cos φ = 0. If sin φ = 0, then both equations are satisfied. Since cos φ = ±1 in this case, we have g(φ, θ ) = 1. If cos φ = 0, then sin φ = ±1, and the second equation π 3π requires cos 2θ = 0. Thus θ = ± or ± . In this case 4 4 1 g(φ, θ ) = ± . 2 Again we find that f (x, y, z) = x y + z 2 has maximum 1 value 1 and minimum value − when restricted to the 2 surface of the ball B. These are the maximum and minimum values for the whole ball as noted in Exercise 22. x y z Let L = sin sin sin + λ(x + y + z − π ). Then 2 2 2 ∂L 1 x y z = cos sin sin + λ ∂x 2 2 2 2 ∂L 1 x y z 0= = sin cos sin + λ ∂y 2 2 2 2 ∂L 1 x y z 0= = sin sin cos + λ. ∂z 2 2 2 2 0=

 1 1 √ ,±√ ,0 2 2

and



 1 1 −√ ,±√ ,0 2 2

1 1 f takes the values and − . 2 2 CASE II. λ = −1. (A) and (B) imply that x = y = 0, and so by (D), z = ±1. f has the value 1 at the points (0, 0, ±1). Thus the maximum and minimum values of f on B are 1 and −1/2 respectively.

(B) (C)

For any triangle we must have 0 ≤ x ≤ π , 0 ≤ y ≤ π and 0 ≤ z ≤ π . Also

CASE I. z = 0. (A) and (B) imply that y 2 = x 2 and (D) then implies that x 2 = y 2 = 1/2. At the four points 

(A)

P = sin

x y z sin sin 2 2 2

is 0 if any of x, y or z is 0 or π . Subtracting equations (A) and (B) gives 1 z x−y sin sin = 0. 2 2 2 It follows that we must have x = y; all other possibilities lead to a zero value for P. Similarly, y = z. Thus the triangle for which P is maximum must be equilateral: x = y = z = π/3. Since sin(π/3) = 1/2, the maximum value of P is 1/8.

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SECTION 13.3 (PAGE 765)

ADAMS and ESSEX: CALCULUS 8

25. We are given that g2 (a, b) 6= 0, and therefore that the

equation g(x, y) = C has a solution  of the form y = h(x) valid near (a, b). Since g x, h(x) = C holds identically for x near a, we must have 0=



 d  = g1 (a, b) + g2 (a, b)h ′ (a). g x, h(x) dx x=a

If f (x, y), subject to the constraint g(x,y) = C, has an extreme value at (a, b), then F(x) = f x, h(x) has an extreme value at x = a, so 0 = F ′ (a) = f 1 (a, b) + f 2 (a, b)h ′ (a).

(Since g2 (a, b) 6= 0, therefore, if g1 (a, b) = 0, then f 1 (a, b) = 0 also.) It follows that 0 = f 2 (a, b) + λg2 (a, b),

so (a, b) is a critical point of L = f (x, y) + λg(x, y).

26. As can be seen in the figure, the minimum distance √

2 from √ (0, −1) to points of the semicircle y = 1 − x is 2, the closest points to (0, −1) on the semicircle being (±1, 0). These points will not be found by the method of Lagrange multipliers because the level curve f (x, y) = 2 of the function f giving the square of the distance from (x, y) to (0, −1) is not tangent to the semicircle at (±1, 0). This could only have happened because (±1, 0) are endpoints of the semicircle. y √

(−1,0)

then we will have λ = 0.

Section 13.4 (page 775)

1. Let L = x1 + x2 + · · · + xn + λ(x12 + x22 + · · · + xn2 − 1). For critical points of L we have

∂L ∂L = 1 + 2λx1 , . . . 0 = = 1 + 2λxn ∂ x1 ∂ xn ∂L = x12 + x22 + · · · + xn2 − 1. 0= ∂λ

x1 = x2 = · · · = xn = −

1 , 2λ

and the final equation gives 1 1 1 + 2 + · · · + 2 = 1, 4λ2 4λ 4λ √ so that 4λ2 = n, and λ = ± n/2. The maximum and minimum values of x1 + x2 + · · · + xn √ n subject to x12 + · · · + xn2 = 1 are ± , that is, n and 2λ √ − n respectively.

2. Let L = x1 + 2x2 + · · · + nxn + λ(x12 + x22 + · · · + xn2 − 1). For critical points of L we have

1 ∂L = 1 + 2λx1 ⇔ x1 = − ∂ x1 2λ ∂L 2 0= = 2 + 2λx2 ⇔ x2 = − ∂ x2 2λ ∂L 3 0= = 3 + 2λx3 ⇔ x3 = − ∂ x3 2λ .. . ∂L n 0= = n + 2λxn ⇔ xn = − ∂ xn 2λ ∂L 0= = x12 + x22 + · · · + xn2 − 1. ∂λ

0=

1−x 2

(1,0)

Lagrange Multipliers in n-Space

The first n equations give

f 1 (a, b) f 2 (a, b) = = −λ (say). g1 (a, b) g2 (a, b)

y=

∇ f (x0 , y0 ) = 0

0=

Together these equations imply that g1 (a, b) f 2 (a, b) = g2 (a, b) f 1 (a, b), and therefore that

0 = f 1 (a, b) + λg1 (a, b),

as shown in the text. The argument given there holds whether or not ∇ f (x0 , y0 ) is 0. However, if

x

(0,−1)

Thus Fig. 13.3.26

27. If f (x, y) has an extreme value on g(x, y) = 0 at a point (x0 , y0 ) where ∇ g 6= 0, and if ∇ f exists at that point, then ∇ f (x0 , y0 ) must be parallel to ∇ g(x0 , y0 ); ∇ f (x0 , y0 ) + λ∇ g(x0 , y0 ) = 0

4 9 n2 1 + + + · · · + =1 4λ2 4λ2 4λ2 4λ2 n(n + 1)(2n + 1) 4λ2 = 1 + 4 + 9 + · · · + n 2 = 6 r 1 n(n + 1)(2n + 1) λ=± . 2 6

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.4 (PAGE 775)

Thus the maximum and minimum values of x1 + 2x2 + · · · + nxn over the hypersphere x12 + x22 + · · · + xn2 = 1 are

For a critical point we need ∂L = 2xi + λ + i µ, ∂ xi 10 X ∂L 0= = xi − 10 ∂λ i=1

0=

s

6 (12 + 22 + 32 + · · · + n 2 ) n(n + 1)(2n + 1) r n(n + 1)(2n + 1) =± . 6

±

0=

L=

xi2

i=1

+λ

10 X

xi

i=1

!

!

− 10 + µ

10 X

i xi

i=1

!

!

− 55 .

For a critical point we need ∂L = 2xi + λ + i µ, ∂ xi 10 X ∂L 0= = xi − 10 ∂λ i=1 0=

(1 ≤ i ≤ 10)

   10  1 X 2i 2 X 4 8i 4i 2 340 1 + + + . = = S= 0 3 33 9 99 1089 33 i=1 i=1 Since

110 + 55λ + 385µ.

Together these latter equations imply that λ = −2 and µ = 0, so that each xi = 2 and S = 10. Since

L=

i=1

xi2

+λ

i=1

xi

!

if i = j , if i 6= j

Its critical points must satisfy 0= 0= 0= 0=

− 10 + µ

1 0

L(x, y, u, v, λ, µ) = (x−u)2 +(y−v)2 +λ(y−x 2)+µ(v−2u 2 −1).

if i = j , if i 6= j

!



5. The Lagrange function is

4. The Lagrange function here is 10 X

∂2 L = ∂ xi ∂ x j

the Hessian matrix for L at any point is diagonal, with all diagonal elements equal to 2, and so all eigenvalues equal 2 and the matrix is positive definite. Thus,the above value of S is a local minimum value. Since there are no other critical points, and S ≥ 0 for all choices of the 10 numbers xi satisfying the constraints, it is an absolute minimum.

the Hessian matrix for L at any point is diagonal, with all diagonal elements equal to 2, and so all eigenvalues equal 2 and the matrix is positive definite. Thus, S = 10 is a local minimum value. Since there are no other critical points, and S ≥ 0 for all choices of the 10 numbers xi satisfying the constraints, it is an absolute minimum.

10 X

2 2i + . 3 33

The corresponding value of S is

Multiplying first equation by i and summing leads to P the 2 (since 10 i=1 i = 385),

1 0

120 + 55λ + 385µ.

xi =

Summing the first equation and using the other two we get 20 + 10λ + 55µ = 0.



Multiplying first equation by i and summing leads to P the 2 (since 10 i=1 i = 385), Together these latter equations imply that λ = −4/3 and µ = −4/33, so that

10 X ∂L = i xi − 55. 0= ∂µ i=1

∂2 L = ∂ xi ∂ x j

10 X ∂L = i xi − 60. ∂µ i=1

Summing the first equation and using the other two we get 20 + 10λ + 55µ = 0.

3. The Lagrange function here is 10 X

(1 ≤ i /le10)

10 X i=1

i xi

!

!

− 60 .

0= 0=

∂L ∂x ∂L ∂y ∂L ∂u ∂L ∂v ∂L ∂λ ∂L ∂µ

= 2(x − u) − 2λx

(A)

= 2(y − v) + λ

(B)

= −2(x − u) − 4µu

(C)

= −2(y − v) + µ

(D)

= y − x2

(E)

= v − 2u 2 − 1.

(F)

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SECTION 13.4 (PAGE 775)

ADAMS and ESSEX: CALCULUS 8

From (B) and (D), λ = −µ. Then, from (A) and (C), either λ = 0 or x = 2u. If λ = 0, then µ = 0 and (A) and (C) imply x = u and y = v. This is not possible because then (E) and (F) would imply x 2 + 1 = 0. Thus λ 6= 0 and x = 2u. Then A implies u = 0 or λ = 1/2. If u = 0, then x = 0, y = 0, v = 1, and λ = 2. One critical point is P = (0, 0, 0, 1, 2, −2). If λ = 1/2, then µ = −1/2, x = 2u, and (E) and (F) then force u 2 = 6/16. This leads to two more critical points ! √ √ 6 3 6 7 1 1 Q= , , , , ,− and 2 2 4 4 2 2 ! √ √ 1 6 3 6 7 1 , ,− , , ,− . R= − 2 2 4 4 2 2 Now the Hessian matirx for L is  2 − 2λ 0 −2 2 0  0 H= −2 0 2 − 4µ 0 −2 0

 0 −2   0 2

√ which √ has eigenvalues (29 + 793)/14 ≈ 4.08 and (29 − 793)/14 ≈ 0.6, both positive. So S has a local minimum at Q, and by symmetry, also at R. Proceeding similarly with P, we evaluate 6  0 HP =  −2 0 

√ which has eigenvalues 0, 4, 6 ± 2 37, again inconclusive. At P, the normal vectors to the constraint manifold are U1 = e y and U2 = ev , so an orthonormal basis for the tangent space is given by vectors V1 = ex and V2 = eu . Thus, if   1 0 0 0 E=  0 1 0 0 , then the restriction of H P to the tangent space to the constraint manifold at P is T

H P|T g = E H P E =

At Q this Hessian becomes 1  0 HQ =  −2 0 

 0 −2 0 2 0 −2  , 0 4 0 −2 0 2

and its eigenvalues r satisfy 1−r 0 0= −2 0

0 2−r 0 −2

−2 0 4−r 0

0 −2 0 2−r

which, on simplification by minors turns out to be r 2 (r − 4)(r − 5) = 0. The eigenvalues are 0, 0, 4, 5, so H Q is not positive or negative definite. We calculate the restriction of H Q to the tangent space T g to the constraint manifold at Q. Observe that the vec√ tors U1 = −2xex + e y = − 6ex + e y and √ U2 = −4ueu + ev = − 6eu + ev are normal to T g. We need two perpendicular unit vectors each perpendicu√ √ lar to both v 1 and√v 2 The√vectors V 1 = (ex + 6e y )/ 7 and V 2 = (eu + 6ev )/ 7 will clearly do. Accordingly, √ let   0 √1/ 7 0√   (6/7) E= . 0 1/ √ 7 0 (6/7) Then the Hessian restricted to the tangent space at Q is   13/7 −2 H Q|T g = ET H E = −2 16/7

 0 −2 0 2 0 −2  , 0 18 0 −2 0 2



−6 −2 −2 18



,

√ which has eigenvalues 6 ± 2 37, of opposite sign. Thus P has saddle behaviour. The √ minimum distance between √ the two curves is the value of S at Q (or R) and is 7/4 units.

Section 13.5 (page 781)

The Method of Least Squares

1. If the power plant is located at (x, y), then x and y should minimize (and hence be a critical point of) S=

n h X i=1

i (x − xi )2 + (y − y2 )2 .

Thus we must have n n X X ∂S =2 (x − xi ) = 2 nx − xi 0= ∂x i=1 i=1

!

! n n X X ∂S yi . (y − yi ) = 2 ny − 0= =2 ∂y i=1 i=1

n n 1X 1X xi = x, ¯ and y = yi = y¯ . n i=1 n i=1 Place the power plant at the position whose coordinates are the averages of the coordinates of the machines.

Thus x =

514 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

2. We want to minimize S = 0=

and a =

2 i=1 x i yi

3. We minimize S =

2 i=1 (ax i

− yi )2 . Thus

n X dS = 2(axi2 − yi )xi2 da i=1

=2 Pn

Pn

SECTION 13.5 (PAGE 781)






Pn

n X (axi4 − xi2 yi ), i=1

Pn

4 i=1 x i

i=1 (ae

xi

and a =

Pn

i=1 yi e

xi




choose p and q to minimize

S=

− yi )2 . Thus

Pn

i=1 e

2xi

4. We choose a, b, and c to minimize S=

6. The relationship y = p + qx 2 is linear in p and q, so we


.

n X dS i =2 (ae xi − yi )e x , 0= da i=1



When written in terms of the components of the vectors involved, these three equations are the same as the equations for a, b, and c encountered in Exercise 4, and so they have the same solution as given for that exercise.




0= 0=

i=1

0= 0=

n X ∂S (axi + byi + c − z i )xi =2 ∂a i=1

n X ∂S =2 (axi + byi + c − z i )yi ∂b i=1

n X ∂S =2 (axi + byi + c − z i ). ∂c i=1

P 2 P P P Let AP= xi , B x i yi , P C = xi , D = P yi2 , P= E = yi , F = xi z i , G = yi z i , and H = z i . In terms of these quantities the above equations become Aa Ba Ca

+ + +

Bb Db Eb

+ + +

Cc Ec nc

= = =

F G H.

By Cramer’s Rule (Theorem 5 of Section 1.6) the solution is 1 A F C 1 F B C b= a= G D E , B G E , 1 H E n 1 C H n A B C 1 A B F c= B D G , where 1 = B D E . 1 C E H C E n

5. If x = (x1 , . . . , xn ), y = (y1 , . . . , yn ), z = (z 1 , . . . , z n ),

w = (1, . . . , 1), and p = ax + by + cw, we want to choose a, b, and c so that p is the vector projection of z onto the subspace of R3 spanned by x, y and w. Thus p − z must be perpendicular to each of x, y, and w: (p − z) • x = 0,

(p − z) • y = 0,

n X ∂S =2 ( p + qxi2 − yi ) ∂p i=1

n X ∂S =2 ( p + qxi2 − yi )xi2 , ∂q i=1

that is,

Thus 0=

i=1

Thus

.

n  2 X axi + byi + c − z i .

n X ( p + qxi2 − yi )2 .

so

np P 2
xi p

P 2
x q P i4
xi q

+ +

= =

P y P 2i x i yi ,

P
P 4
P 2
P 2
yi xi − x i yi xi p= P 4
P 2
2 n x − x P 2
i P
Pi 2
n x i yi − yi xi q= . P 4
P 2
2 n xi − xi

This is the result obtained by direct linear regression. (No transformation of variables was necessary.)

7. We transform y = peq x into the form ln y = ln p + qx, which is linear in ln p and q. We let ηi = ln yi and use the regression line η = a + bx obtained from the data (xi , ηi ), with b = q and a = ln p. Using the formulas for a and b obtained in the text, we have

P
P xi ln yi − xi ln yi P 2
P
2 n xi − xi


P 2 P P
P xi ln yi − xi xi ln yi q=b= P 2
P
2 n xi − xi

ln p = a =

n

P

p = ea .

These values of p and q are not the same values that minimize the expression

(p − z) • w = 0.

S=

n X (yi − peq xi )2 . i=1

515 Copyright © 2014 Pearson Canada Inc.

SECTION 13.5 (PAGE 781)

ADAMS and ESSEX: CALCULUS 8

8. We transform y = ln( p + qx) into the form e y = p + qx,

which is linear in p and q. We let ηi = e yi and use the regression line η = ax + b obtained from the data (xi , ηi ), with a = q and b = p. Using the formulas for a and b obtained in the text, we have
P P
P y
n x i e yi − xi e i q=a= P 2
P
2 n x − xi
P 2
P i yi
P
P xi e − xi x i e yi . p=b= P 2
P
2 n xi − xi

These values of p and q are not the same values that minimize the expression

These values of p and q are not the same values that minimize the expression S=

S=

that is,

so

= =

P x y P 2i i x i yi ,


P 4
P 2
P 3
x i yi xi − x i yi xi P 2
P 4
P 3
2 x x − xi
P 3
P 2
Pi 2
i P xi x i yi − x i yi xi q= . P 2
P 4
P 3
2 xi xi − xi p=

10.

P 3
x q P i4
xi q

P

2

.

that is,

n X ∂S =2 ( pxi + qxi2 − yi )xi2 , 0= ∂q i=1

+ +

i=1

pe xi + qe−xi − yi

n   X ∂S pe xi + qe−xi − yi e xi =2 ∂p i=1 n   X ∂S pe xi + qe−xi − yi e−xi . =2 0= ∂q i=1

i=1

n X ∂S =2 ( pxi + qxi2 − yi )xi ∂p i=1

P 2
x p P i3
xi p

n  X

0=

we choose p and q to minimize

0=

.

Thus

9. The relationship y = px + qx 2 is linear in p and q, so

Thus

2

we choose p and q to minimize

i=1

n X ( pxi + qxi2 − yi )2 .

i=1

pxi + q − yi

11. The relationship y = pe x + qe−x is linear in p and q, so

n  2 X S= ln( p + qxi ) − yi .

S=

n  X √

This is the result obtained by direct linear regression. (No transformation of variables was necessary.) √ We transform y = ( px + q) into the form y 2 = px + q, which is linear in p and q. We let ηi = yi2 and use the regression line η = ax + b obtained from the data (xi , ηi ), with a = p and b = q. Using the formulas for a and b obtained in the text, we have
P P
P 2
n xi yi2 − xi yi p=a= P 2
P
2 n x − xi
P 2
Pi 2
P
P yi − xi xi yi2 xi q =b= . P 2
P
2 n xi − xi

P so


e2xi p np

+ +

P

nq
e−2xi q

= =

P xi e y P −xi i e yi ,


P x
P −xi
e yi e i yi − n e−2xi p= P 2x
P −2x
e i e i − n2 P x
P 2x
P −x
e i yi e i yi − n e i q= . P 2x
P −2x
e i e i − n2 P

This is the result obtained by direct linear regression. (No transformation of variables was necessary.)

12. We use the result of Exercise 6. We have n = 6 and X X

Therefore

xi2 = 115, yi = 55.18,

X

X

xi4 = 4051,

xi2 yi = 1984.50.

P
P 4
P 2
P 2
yi xi − x i yi xi P 4
P 2
2 n xi − xi 55.18 × 4051 − 1984.50 × 115 = ≈ −0.42 2 6 × 4051 − 115 P 2
P
P 2
n x i yi − xi yi q= P 2
2 P 4
n xi − xi 6 × 1984.50 − 55.18 × 115 = ≈ 0.50. 6 × 4051 − 1152 p=

We have (approximately) y = −0.42 + 0.50x 2 . The predicted value of y at x = 5 is −0.42 + 0.50 × 25 ≈ 12.1.

516 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.5 (PAGE 781)

13. Choose a, b, and c to minimize S=

n  X

axi2

i=1

+ bxi + c − yi

2

.

16. To maximize I = a so that

Thus 0= 0= 0=

Aa Ba Ca

+ + +

Bb Cb Db

+ + +

Cc Dc nc

= = =

H I J.

By Cramer’s Rule (Theorem 5 of Section 1.6) the solution is 1 A H C 1 H B C b= a= I C D , B I D , 1 J D n 1 C J n A B C 1 A B H c= B C I , where 1 = B C D . 1 C D J C D n

14. Since y = pe x + q + r e−x is equivalent to x

x 2

0

Z

π 0



17. To minimize I = b so that

dI = da

1 0

dx =

π 480 − 5 ≈ 0.00227. 2 π

(ax 2 + b − x 3 )2 d x, we choose a and

Z 1 2a 2b 1 ∂I = 2(ax 2 + b − x 3 )x 2 d x = + − ∂a 5 3 3 0 Z 1 ∂I 2a 1 0= = 2(ax 2 + b − x 3 ) d x = + 2b − . ∂b 3 2 0

Solving these two equations, we get a = 15/16 and b = −1/16. The minimum value of I is Z

18. To minimize

1  15x 2

16

Z

0

c so that

0=2

1

Thus a = 5/6, and the minimum value of I is ! Z 1 25x 4 5x 5 6 − +x dx 36 3 0

Z

2

0=

0

2(ax 2 − x 3 )x 2 d x 0 ! 1 x5 2x 6 2a 1 = 2a − = − . 5 6 0 5 3

0=

0

120 x(π − x) − sin x π5

0

Z

d x, we choose

(We have omited the details of evaluation of these integrals.) Hence a = 120/π 5 . The minimum value of I is

e y = p(e ) + qe + r,

15.

2

π 5a = − 8. 15

x

we let ξi = e xi and ηi = e xi yi for i = 1, 2, . . . , n. We then have p = a, q = b, and r = c, where a, b, and c are the values calculated by the formulas in Exercise 13, but for the data (ξi , ηi ) instead of (xi , yi ). Z 1 (ax 2 − x 3 )2 d x, we choose a so that To minimize I =

ax(π − x) − sin x

Z π   dI = 2 ax(π − x) − sin x x(π − x) d x da Z π 0 Z π = 2a x 2 (π − x)2 d x − 2 x(π − x) sin x d x

n X ∂S =2 (axi2 + bxi + c − yi )xi ∂b i=1

P 4 P 3 P 2 P Let A = x , B = xi , C = x , D = xi , P 2 i P Pi H = x i yi , I = xi yi , and J = yi . In terms of these quantities the above equations become

0

π

0=

n X ∂S =2 (axi2 + bxi + c − yi )xi2 ∂a i=1

n X ∂S =2 (axi2 + bxi + c − yi ). ∂c i=1

Z

0=2 0=2 that is,

5 5 1 1 = − + = . 36 18 7 252

1

−

1 − x3 16

2

dx =

1 . 448

(x 3 − ax 2 − bx − c)2 d x, choose a, b and

Z

1

0

Z

1

0

Z

0

a 5 a 4 a 3

1

(x 3 − ax 2 − bx − c)(−x 2 ) d x (x 3 − ax 2 − bx − c)(−x) d x (x 3 − ax 2 − bx − c)(−1) d x,

+ + +

b 4 b 3 b 2

+ + +

c 3 c 2 c

= = =

1 6 1 5 1 4

517 Copyright © 2014 Pearson Canada Inc.

SECTION 13.5 (PAGE 781)

19.

ADAMS and ESSEX: CALCULUS 8

3 3 1 for which the solution is a = , b = − , and c = . 2 5 20 Z π To minimize (sin x − ax 2 − bx)2 d x we choose a and b so that

0

0=2 0=2

π

Z

Z0 π 0

a

5 π4 a 4

21. To minimize

+ +

π4

b

4 π3 b 3

=

π2 − 4

=

π,

= −2

(x − a sin π x − b sin 2π x − c sin 3π x) d x.

(x − a sin π x − b sin 2π x − c sin 3π x) sin π x d x

2 (π a − 2) π ∂J 0= ∂b Z 1 = −2 (x − a sin π x − b sin 2π x − c sin 3π x) sin 2π x d x =

−1

2 = (π b + 1) π ∂J 0= ∂c Z 1 = −2 (x − a sin π x − b sin 2π x − c sin 3π x) sin 3π x d x −1

2 = (3π c − 2). 3π

0

1

−1

!

−

1 2



Z

d x,

! n a0 X ak cos kx (− cos nx) d x f (x) − − 2 k=1

π

0

sin mπ x sin nπ x d x = 0

2 π

Z

π

f (x) d x,

0

and, since π 0

cos kx cos nx d x =

(

0 π 2

if k 6= n if k = n = 1, 2, . . .

we also have an =

2 π

Z

π

f (x) cos nx d x

0

(n = 1, 2, . . .).

22. The Fourier sine series coefficients for f (x) = x on (0, π ) are

bn =

2 π

Z

0

π

x sin(nx) d x = (−1)n−1

2 n

for n = 1, 2, . . .. Thus the series is ∞ X 2 (−1)n−1 sin nx. n n=0

We have omitted the details of evaluation of these integrals, but note that Z

n a0 X − ak cos kx f (x) − 2 k=1

π

a0 =

Z

1

−1

Z

2

∂J ∂a Z

dx

for n = 1, 2, . . .. Thus

To minimize J , choose a, b, and c to satisfy 0=

0

!2

we require

∂I 0= =2 ∂an

1 −1

n a0 X f (x) − − ak cos kx 2 k=1

π

and

20 2 (π − 16) π5 12 b = 4 (20 − π 2 ). π

J=

Z

∂I 0= =2 ∂a0

a=

20.

I =

(sin x − ax 2 − bx)(−x) d x.

for which the solution is

Z

The equations above imply that a = 2/π , b = −1/π , and c = 2/(3π ). These are the values that minimize J .

(sin x − ax 2 − bx)(−x 2 ) d x

We omit the details of the evaluation of the integrals. The result of the evaluation is that a and b satisfy π5

if m and n are different integers.

Since x and the functions sin nx are all odd functions, we would also expect the series to converge to x on (−π, 0).

518 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.6 (PAGE 790)

23. The Fourier cosine series coefficients for f (x) = x on

and the sum will increase if x is outside that interval. In this case the value of x which minimizes the sum is not unique unless it happens that xn/2 = x(n/2)+1 .

(0, π ) are

2 π

Z

π

2 an = π (

Z

π

a0 =

=

x dx = π

0

0

x cos(nx) d x = −

0

−

n2π

if n ≥ 2 is even

4 2 n π

Section 13.6 (page 790)

  2 1 − (−1)n

1.

if n ≥ 1 is odd.

Thus the Fourier cosine series is π−

F(x) =

Since the terms of this series are all even functions, and the series converges to x if 0 < x < π , it will converge to −x = |x| if −π < x < 0. Remark: since |x| is continuous at x = 0, the series also converges at x = 0 to 0. It follows that ∞ X 1 1 π2 1 1 + 2 + 2 + ··· = = . 2 4 3 5 (2n + 1) n=0

2.

1

0

Z

1

24. We are given that x1 ≤ x2 ≤ x3 ≤ . . . ≤ xn .

x2

x3

x4

Z

|x − xi |

= (x3 − x1 ) + (x3 − x2 ) + 0 + (x4 − x3 ) + (x5 − x3 ) = (x5 − x1 ) + (x4 − x2 ).

i=1

−2xt 2 e−t

∞

Z

∞ −∞

2 x2

(∗)

2 −t 2

√

t e

dt =

−2xt 4 e−x

2t2

dt = −

√ 3 π . 2x 4

Divide by −2 and let x = 1: Z

If x moves away from x3 in either direction, then 5 X

∞

π . 2 −∞ Differentiate (∗) with respect to x again:

If x = 1 we get

If x = x3 , then

i=1

(x > −1)

t x ln t dt = −

Z

x5

Fig. 13.5.24

5 X

1 x +1

√ π dt = − 2 x −∞ √ Z ∞ π 2 2 t 2 e−x t dt = 3 . 2x −∞

To motivate the method, look at a special case, n = 5 say.

x1

t x dt =

1 (x + 1)2 0 Z 1 2 F ′′ (x) = t x (ln t)2 dt = (x + 1)3 0 .. . Z 1 (−1)n n! F (n) (x) = t x (ln t)n dt = . (x + 1)n+1 0 Z ∞ √ 2 e−u du = π Let u = xt −∞ du = x dt √ Z ∞ π −x 2 t 2 e dt = . x −∞ Differentiate with respect to x: F ′ (x) =

∞ 4X cos((2n + 1)x) . π n=0 (2n + 1)2

Z

Parametric Problems

∞ −∞

2

t 4 e−t dt =

√ 3 π . 4

|x − xi | = (x5 − x1 ) + (x4 − x2 ) + |x − x3 |.

Thus the minimum sum occurs P if x = x3 . In general, if n is odd, then ni=1 |x − xi | is minimum if x = x(n+1)/2 , the middle point of the set of points {x1 , x2 , . . . , xn }. The value of x is unique in this case. If n is even and x satisfies xn/2 ≤ x ≤ x(n/2)+1 , then n X i=1

|x − xi | =

n/2 X i=1

3. Let I (x, y) = y > 0. Then

Z

∞ −∞

∂I =− ∂x

|xn+1−i − xi |,

Z

2

2

e−xt − e−yt dt, where x > 0 and t2 ∞

√ Let xt = s √ −∞ x dt = ds √ Z ∞ 1 2 π = −√ e−s ds = − √ . x −∞ x 2

e−xt dt

519 Copyright © 2014 Pearson Canada Inc.

SECTION 13.6 (PAGE 790)

ADAMS and ESSEX: CALCULUS 8

√ ∂I π = √ . Now ∂y y

Similarly,

Z √ √ dx I (x, y) = − π √ = −2 π x + C1 (y) x √ π ∂I ∂C1 √ = ⇒ C1 (y) = 2 π y + C2 √ = y ∂y ∂y √  √ √ y − x + C2 . I (x, y) = 2 π

F(x) =

2

∞

Z

4. Let I (x, y) = Then

Z

1 tx 0

∂I ∂x ∂I ∂y

− ty dt, where x > −1 and y > −1. ln t Z 1 1 = t x dt = x +1 0 1 =− . y+1

7.

dx = ln(x + 1) + C1 (y) x +1 −1 ∂I ∂C1 = = ⇒ C1 (y) = − ln(y + 1) + C2 y+1 ∂y ∂y   x +1 I (x, y) = ln + C2 . y+1

5.

1 tx

0

− ty dt = ln ln t



x +1 y+1



for x > −1 and y > −1. Z ∞ 1 if x > 0. e−xt sin t dt = 1 + x2 0 Multiply by −1 and differentiate with respect to x twice: Z

∞

2x (1 + x 2 )2 Z ∞ 2(3x 2 − 1) . t 2 e−xt sin t dt = (1 + x 2 )3 0

6.

F(x) = ′

F (x) =

∞

Z

Z0 ∞ 0

0

e−xt

te−xt sin t dt =

sin t dt t

−e−xt sin t dt = −

1 1 + x2

(x > 0).

Z

0

∞

e−s s sin ds. s x

R ∞ sin t π dt = lim F(x) = . 0 x→0 t 2 ∞ Z ∞ dt 1 t π = tan−1 = for x > 0. 2 + t2 x x x 2x 0 0 Differentiate with respect to x: Z ∞ −2x dt π =− 2 (x 2 + t 2 )2 2x 0 Z ∞ π dt = 3. (x 2 + t 2 )2 4x 0

Differentiate with respect to x again: Z ∞ −4x dt 3π =− 4 (x 2 + t 2 )3 4x 0 Z ∞ dt 3π = . (x 2 + t 2 )3 16x 5 0

But I (x, x) = 0, so C2 = 0. Thus Z

sin(s/x) ds = s/x x

In particular,

Z

I (x, y) =

e−s

Hence −

Thus I (x, y) =

0

∞

π π + C = 0, and C = . Therefore 2 2 Z ∞ π sin t dt = − tan−1 x. F(x) = e−xt t 2 0

2

√ √ √  e−xt − e−yt dt = 2 π y− x . 2 t

−∞

Z

Since | sin(s/x)| ≤ s/x if s > 0, x > 0, we have Z ∞ 1 1 |F(x)| ≤ e−s ds = → 0 as x → ∞. |x| 0 |x|

But I (x, x) = 0. Therefore C2 = 0, and I (x, y) =

Z

dx = − tan−1 x + C. 1 + x2 Now, make the change of variable xt = s in the integral defining F(x), and obtain Therefore F(x) = −

8.

x π dt 1 −1 t = tan for x > 0. = 2 2 x x 0 4x 0 x +t Differentiate with respect to x: Z x 1 −2x dt π + =− 2 2 + t 2 )2 2x 2 (x 4x 0   Z x π dt 1 1 =− − 2− 2 2 2 2 2x 4x 2x 0 (x + t ) π 1 = 3 + 3. 8x 4x

Z

x

Differentiate with respect to x again:   Z x −4x dt 1 3 π 1 + = − + 2 2 3 4x 4 x4 8 4 0 (x + t )   Z x dt 1 3π 3 1 = − − − − 2 2 3 4x 8x 4 4x 4 4x 4 0 (x + t ) 3π 1 = + 5. 32x 5 4x

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.6 (PAGE 790)

Z x f (x) = 1 + (x − t)n f (t) dt ⇒ f (a) = 1 a Z x f ′ (x) = n (x − t)n−1 f (t) dt a Z x f ′′ (x) = n(n − 1) (x − t)n−2 f (t) dt

9.

Therefore C A −1− +x 2 3



 C − A = 0. 2

This can hold for all x only if

a

A C −1− =0 2 3

.. .

f (n) (x) = n! f

10.

(n+1)

Z

x

f (t) dt

a

(x) = n! f (x)

(n+1)

f

⇒

(a) = n! f (a) = n!.

Z x f (x) = C x + D + (x − t) f (t) dt ⇒ 0 Z x f ′ (x) = C + f (t) dt ⇒ f ′ (0) = C

f (0) = D

11.

f (x) = x +

Z

x

(x − 2t) f (t) dt ⇒ 0 Z x f ′ (x) = 1 − x f (x) + f (t) dt ⇒

f (0) = 0 f ′ (0) = 1

0

f ′′ (x) = − f (x) − x f ′ (x) + f (x) = −x f ′ (x). du x2 If u = f ′ (x), then = −x d x, so ln u = − + ln C1 . u 2 Therefore ′ −x 2 /2 f (x) = u = C1 e . We have 1 = f ′ (0) = C1 , so f ′ (x) = e−x f (x) =

x

Z

e−t

2 /2

0

2 /2

and

dt + C2 .

A 2A − = 1, so that A = −6 and 2 3 C = −12. Therefore f (x) = −6 − 12x.

13. We eliminate c from the pair of equations f (x, y, c) = 2cx − c2 − y = 0 ∂ f (x, y, c) = 2x − 2c = 0. ∂c

12.

Thus c = x and 2x 2 − x 2 − y = 0. The envelope is y = x 2.

14. We eliminate c from the pair of equations f (x, y, c) = y − (x − c) cos c − sin c = 0 ∂ f (x, y, c) = cos c + (x − c) sin c − cos c = 0. ∂c Thus c = x and y − 0 − sin x = 0. The envelope is y = sin x.

15. We eliminate c from the pair of equations f (x, y, c) = x cos c + y sin c − 1 = 0 ∂ f (x, y, c) = −x sin c + y cos c = 0. ∂c Squaring and adding these equations yields x 2 + y 2 = 1, which is the equation of the envelope.

But 0 = f (0) = C2 , and so f (x) =

16. We eliminate c from the pair of equations x

Z

e−t

2 /2

x y + −1 = 0 cos c sin c ∂ x sin c y cos c f (x, y, c) = − = 0. ∂c cos2 c sin2 c

dt.

f (x, y, c) =

0

Z 1 f (x) = 1 + (x + t) f (t) dt 0 Z 1 f ′ (x) = f (t) dt = C, say,

From the second equation, y = x tan3 c. Thus x (1 + tan2 c) = 1 cos c

0

since the integral giving f ′ (x) does not depend on x. Thus f (x) = A + C x, where A = f (0). Substituting this expression into the given equation, we obtain A + Cx = 1 +

Z

0

C − A = 0. 2

Thus C = 2 A and

0

f ′′ (x) = f (x) ⇒ f (x) = A cosh x + B sinh x D = f (0) = A, C = f ′ (0) = B ⇒ f (x) = D cosh x + C sinh x.

and

which implies that x = cos3 c, and hence y = sin3 c. The envelope is the astroid x 2/3 + y 2/3 = 1.

17. We eliminate c from the pair of equations

1

(x + t)(A + Ct) dt

= 1 + Ax +

A Cx C + + . 2 2 3

f (x, y, c) = c + (x − c)2 − y = 0

∂ f (x, y, c) = 1 + 2(c − x) = 0. ∂c

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SECTION 13.6 (PAGE 790)

18.

ADAMS and ESSEX: CALCULUS 8

y

1 1 Thus c = x − . The envelope is the line y = x − . 2 4 We eliminate c from the pair of equations f (x, y, c) = (x − c)2 + (y − c)2 − 1 = 0

circles x 2 +(y−c)2 =kc2

∂ f (x, y, c) = 2(c − x) + 2(c − y) = 0. ∂c

envelope (1−k)x 2 =ky 2

Thus c = (x + y)/2, and 

x−y 2

2

+



y−x 2

2

x

=1

√ or x − y = ± 2. These two parallel lines constitute the envelope of the given family which consists of circles of radius 1 with centres along the line y = x.

Fig. 13.6.20

21. We eliminate c from the equations

19. Not every one-parameter family of curves in the plane

f (x, y, c) = y 3 − (x + c)2 = 0

has an envelope. The family of parabolas y = x 2 + c evidently does not. (See the figure.) If we try to calculate the envelope by eliminating c from the equations f (x, y, c) = y − x 2 − c = 0

∂ f (x, y, c) = −2(x + c) = 0. ∂c

Thus x = −c, and we obtain the equation y = 0 for the envelope. However, this is not really an envelope at all. The curves y 3 = (x + c)2 all have cusps along the x-axis; none of them is tangent to the axis.

∂ f (x, y, c) = −1 = 0, ∂c

we fail because the second equation is contradictory.

y

y

x f (x,y,c)=y 3 −(x+c)2 =0

Fig. 13.6.21

y=x 2 +c

22. If the family of surfaces f (x, y, z, λ, µ) = 0 has an

Fig. 13.6.19

envelope, that envelope will have parametric equations

20. The curve x√2 +(y −c)2 = kc2 is a circle with centre (0, c) and radius

x

kc, provided k > 0. Consider the system:

f (x, y, c) = x 2 + (y − c)2 − kc2 = 0

∂ f (x, y, c) = −2(y − c) − 2kc = 0. ∂c

The second equation implies that y − c = −kc, and the first equation then says that x 2 = k(1 − k)c2 . This is only possible if 0 ≤ k ≤ 1. The cases k = 0 and k = 1 are degenerate. If k = 0 the “curves” are just points on the y-axis. If k = 1 the curves are circles, all of which are tangent to the x-axis at the origin. There is no reasonable envelope in either case. If 0 < k < 1, the envelope is the pair of lines given √ √ k by x 2 = y 2 , that is, the lines 1 − kx = ± k y. 1−k √ These lines make angle sin−1 k with the y-axis.

x = x(λ, µ),

y = y(λ, µ),

z = z(λ, µ),

giving the point on the envelope where the envelope is tangent to the particular surface in the family having parameter values λ and µ. Thus   f x(λ, µ), y(λ, µ), z(λ, µ), λ, µ = 0. Differentiating with respect to λ, we obtain f1

∂x ∂y ∂z + f2 + f3 + f 4 = 0. ∂λ ∂λ ∂λ

However, since for fixed µ, the parametric curve x = x(t, µ),

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y = y(t, µ),

z = z(t, µ)

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.6 (PAGE 790)

is tangent to the surface f (x, y, z, λ, µ) = 0 at t = λ, its tangent vector there, T=

λ2 + µ2 . 2 Differentiate with respect to λ and µ:

24. (x − λ)2 + (y − µ)2 + z 2 =

∂x ∂y ∂z i+ j+ k, ∂λ ∂λ ∂λ

−2(x − λ) = λ, Thus λ = 2x, µ = 2y, and

is perpendicular to the normal

x 2 + y 2 + z 2 = 2x 2 + 2y 2 .

N = ∇ f = f 1 i + f 2 j + f 3 k, so

−2(y − µ) = µ.

The envelope is the cone z 2 = x 2 + y 2 .

∂x ∂y ∂z f1 + f2 + f3 = 0. ∂λ ∂λ ∂λ

25.

∂f Hence we must also have = f 4 (x, y, z, λ, µ) = 0. ∂λ ∂f Similarly, = 0. ∂µ The parametric equations of the envelope must therefore satisfy the three equations f (x, y, z, λ, µ) = 0 ∂ f (x, y, z, λ, µ) = 0 ∂λ ∂ f (x, y, z, λ, µ) = 0. ∂µ

y + ǫ sin(π y) = x ⇒ y = y(ǫ, x) ∂y ∂y + sin(π y) + π ǫ cos(π y) =0 ∂ǫ ∂ǫ  2 ∂2 y ∂y ∂y 2 + 2π cos(π y) − π ǫ sin(π y) ∂ǫ ∂ǫ ∂ǫ 2 ∂2 y + π ǫ cos(π y) 2 = 0. ∂ǫ If ǫ = 0 then y = x, so y(x, 0) = x. Also, at ǫ = 0, yǫ (x, 0)(1 + 0) = − sin(π y(x, 0)) = − sin(π x), that is, yǫ (x, 0) = − sin(π x). Also, yǫǫ (x, 0)(1 + 0) = −2π cos(π x)yǫ (x, 0) + 0 = 2π cos(π x) sin(π x) = π sin(2π x).

The envelope can be found by eliminating λ and µ from these three equations.

Thus

23. To find the envelope we eliminate λ and µ from the equations x sin λ cos µ + y sin λ sin µ + z cos λ = 1 x cos λ cos µ + y cos λ sin µ − z sin λ = 0 − x sin λ sin µ + y sin λ cos µ = 0.

y = y(x, ǫ) = y(x, 0) + ǫyǫ (x, 0) +

(1) (2) (3)

Multiplying (1) by cos λ and (2) by sin λ and subtracting the two gives z = cos λ. Therefore (2) and (3) can be rewritten x cos µ + y sin µ = sin λ x sin µ − y cos µ = 0. Squaring and adding these equations gives

= x − ǫ sin(π x) +

26.

ǫ2 yǫǫ (x, 0) + · · · 2!

ǫ2 π sin(2π x) + · · · 2

2

y 2 + ǫe−y = 1 + x 2 2

2

2yyǫ + e−y − 2yǫe−y yǫ = 0   2 2 2y 1 − ǫe−y yǫ + e−y = 0     2 2 2 2yǫ 1 − ǫe−y yǫ − 2ye−y yǫ + 2y 2yǫe−y yǫ yǫ   2 2 + 2y 1 − ǫe−y yǫǫ − 2ye−y yǫ = 0. √ At ǫ = 0 we have y(x, 0) = 1 + x 2 , and p 2 2 1 + x 2 yǫ (x, 0) + e−(1+x ) = 0 1 2 yǫ (x, 0) = − √ e−(1+x ) 2 2 1+x

x 2 + y 2 = sin2 λ. Therefore

2

x 2 + y 2 + z 2 = sin2 λ + cos2 λ = 1; the envelope is the sphere of radius 1 centred at the origin.

2yǫ2 − 4ye−y yǫ + 2yyǫǫ = 0 2

yyǫǫ = 2yyǫ e−y − yǫ2   1 1 2 e−2(1+x ) . yǫǫ (x, 0) = − √ + 2 3/2 2 4(1 + x ) 1+x

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SECTION 13.6 (PAGE 790)

ADAMS and ESSEX: CALCULUS 8

Thus ǫ2 y = y(x, ǫ) = y(x, 0) + ǫyǫ (x, 0) + yǫǫ (x, 0) + · · · 2! p ǫ 2 = 1 + x2 − √ e−(1+x ) 2 1 + x2   ǫ2 1 1 2 e−2(1+x ) + · · · . − √ + 2 3/2 2 2 4(1 + x ) 1+x

27.

ǫx =1 1 + y2 x 2ǫx yyǫ 2yǫ + − =0 1 + y2 (1 + y 2 )2   4x yyǫ ∂ 2x yyǫ 2yǫǫ − − ǫ = 0. (1 + y 2 )2 ∂ǫ (1 + y 2 )2 2y +

At ǫ = 0 we have y(x, 0) =

1 1 5 1 1 − × + × 2 32 100 256 2 × 1002 1 105 × + ··· − 4096 6 × 1003 ≈ 0.49968847

y=

with error less than 10−8 in magnitude.

29. Let x(ǫ) and y(ǫ) be the solution of x x

x

=

x x

ǫ2 yǫǫ (x, 0) + · · · 2!

28. Let y(x, ǫ) be the solution of y + ǫy 5 = have

1 . Then we 2

yǫ 1 + 5ǫy 4 + y 5 = 0   yǫǫ 1 + 5ǫy 4 + 20ǫy 3 yǫ2 + 10y 4 yǫ = 0     yǫǫǫ 1 + 5ǫy 4 + yǫǫ 60ǫy 3 yǫ + 15y 4 + 60ǫyǫ3 y 2 + 60y 3 yǫ2 = 0.

At ǫ = 0 we have

+

2y ′

x′

−

y′

+

x ′′

3 0.

ǫe−x x ′′ ǫe−y y ′′

−

2y y

2y ′′ y ′′

= =

3 0



⇒x =y=1

1  1 −  ′ e ⇒ x = − e 1 y′ = 0 = −  e 2 ) −2 = − 2 ⇒ x ′′ = y ′′ = 2 . e 3e = 0 =

Thus x =1−

For ǫ =

1 2

+ −

x′

x ′′



y(x, 0) =

= =

At ǫ = 0 we have

1 2ǫx 16ǫ 2 x 2 − − +···. 2 5 125



ǫe−x ǫe−y

+ +

x ′ + 2y ′ + e−x − ǫe−x x ′ = 0 x ′ − y ′ + e−y − ǫe−y y ′ = 0 x ′′ + 2y ′′ − 2e−x x ′ + ǫe−x (x ′ )2 − x ′′ − y ′′ − 2e−y y ′ + ǫe−y (y ′ )2 −

Thus y = y(x, ǫ) = y(x, 0) + ǫyǫ (x, 0) +

2y y

+ −

Thus

2x =− 1 5 1+  4   1 2x 4x − 1 32x 2 2 5 = =− .  2 2 125 1 1+ 4

1 yǫ (x, 0) = − 2

yǫǫ

1 , and 2

1 we have 100

For ǫ =

ǫ2 ǫ − 2 +···, e 3e

y =1−

ǫ2 +···. 3e2

1 we have 100

1 32   10 1 5 yǫǫ (x, 0) = − − = 2 16 32 16     5 15 60 1 2 105 yǫǫǫ (x, 0) = − 2 − − =− . 16 16 8 32 4096 yǫ (x, 0) = −

524 Copyright © 2014 Pearson Canada Inc.

1 1 + +··· 100e 30, 000e2 1 y =1− + ···. 30, 000e2

x =1−

= =

0 0.

INSTRUCTOR’S SOLUTIONS MANUAL

Section 13.7

Newton’s Method

SECTION 13.7 (PAGE 794)

(page 794)

For each of Exercises 1–6, and 9, we sketch the graphs of the two given equations, f (x, y) = 0 and g(x, y) = 0, and use their intersections to make initial guesses x 0 and y0 for the solutions. These guesses are then refined using the formulas x n+1

f g2 − g f 2 = xn − , f1 g2 − g1 f 2 (xn ,yn )

yn+1

f 1 g − g1 f = yn − . f 1 g2 − g1 f 2 (xn ,yn )

NOTE: The numerical values in the tables below were obtained by programming a microcomputer to calculate the iterations of the above formulas. In most cases the computer was using more significant digits than appear in the tables, and did not truncate the values obtained at one step before using them to calculate the next step. If you use a calculator, and use the numbers as quoted on one line of a table to calculate the numbers on the next line, your results may differ slightly (in the last one or two decimal places).

1. y y=e x

x=sin y x

Fig. 13.7.1 f (x, y) = y − e x g(x, y) = x − sin y

f 1 (x, y) = −e x f 2 (x, y) = 1

We start with x 0 = 0.9, y0 = 2.0. n 0 1 2 3 4

xn

g1 (x, y) = 1 . g2 (x, y) = − cos y yn

0.9000000 0.8100766 0.7972153 0.7971049 0.7971049

2.0000000 2.2384273 2.2191669 2.2191071 2.2191071

f (x n , yn ) −0.4596031 −0.0096529 −0.0001851 0.0000000 0.0000000

g(x n , yn ) −0.0092974 0.0247861 0.0001464 0.0000000 0.0000000

Thus x = 0.7971049, y = 2.2191071.

2.

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SECTION 13.7 (PAGE 794)

ADAMS and ESSEX: CALCULUS 8

y x 2 +y 2 =1 y=e x x

Fig. 13.7.2 f (x, y) = x 2 + y 2 − 1 g(x, y) = y − e x

f1 (x, y) = 2x f2 (x, y) = 2y

g1(x, y) = −e x . g2(x, y) = 1

Evidently one solution is x = 0, y = 1. The second solution is near (−1, 0). We try x 0 = −0.9, y0 = 0.2. n xn yn f (x n , yn ) g(x n , yn ) 0 1 2 3 4

−0.9000000 −0.9411465 −0.9170683 −0.9165628 −0.9165626

0.2000000 0.3898407 0.3995751 0.3998911 0.3998913

−0.1500000 0.0377325 0.0006745 0.0000004 0.0000000

−0.2065697 −0.0003395 −0.0001140 −0.0000001 0.0000000

The second solution is x = −0.9165626, y = 0.3998913. 3. y 4

x 4 +y 2 =16

xy=1 x

2

Fig. 13.7.3 f (x, y) = x 4 + y 2 − 16 g(x, y) = x y − 1

f 1 (x, y) = 4x 3 f 2 (x, y) = 2y

g1 (x, y) = y . g2 (x, y) = x

There are four solutions as shown in the figure. We will find the two in the first quadrant; the other two are the negatives of these by symmetry. The first quadrant solutions appear to be near (1.9, 0.5) and (0.25, 3.9). n xn yn f (x n , yn ) g(x n , yn ) 0 1 2

1.9000000 1.9990542 1.9921153

0.5000000 0.5002489 0.5019730

−2.7179000 0.2200049 0.0011548

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−0.0500000 0.0000247 −0.0000120

INSTRUCTOR’S SOLUTIONS MANUAL

3 4 n 0 1 2 3

SECTION 13.7 (PAGE 794)

1.9920783 1.9920783

0.5019883 0.5019883

xn

0.0000000 0.0000000 f (x n , yn )

yn

0.2500000 0.2499499 0.2500305 0.2500305

3.9000000 4.0007817 3.9995117 3.9995115

−0.7860937 0.0101569 0.0000016 0.0000000

0.0000000 0.0000000 g(x n , yn ) −0.0250000 −0.0000050 −0.0000001 0.0000000

The four solutions are x = ±1.9920783, ±y = 0.5019883, and x = ±0.2500305, y = ±3.9995115. 4. y x(1+y 2 )=1 2

y(1+x 2 )=2

1 x

Fig. 13.7.4 f (x, y) = x(1 + y 2 ) − 1

f 1 (x, y) = 1 + y 2 f 2 (x, y) = 2x y

2

g(x, y) = y(1 + x ) − 2

g1(x, y) = 2x y

g2(x, y) = 1 + x

The solution appears to be near x = 0.2, y = 1.8. n xn 0 1 2 3

.

f (x n , yn )

yn

0.2000000 0.2169408 0.2148268 0.2148292

2

1.8000000 1.9113487 1.9117785 1.9117688

−0.1520000 0.0094806 −0.0000034 0.0000000

g(x n , yn ) −0.1280000 0.0013031 0.0000081 0.0000000

The solution is x = 0.2148292, y = 1.9117688. 5. y y=sin x

−1

1 x −1

x 2 +(y+1)2 =2

Fig. 13.7.5

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SECTION 13.7 (PAGE 794)

f (x, y) = y − sin x 2

2

g(x, y) = x + (y + 1) − 2

ADAMS and ESSEX: CALCULUS 8

f 1 (x, y) = − cos x f 2 (x, y) = 1

Solutions appear to be near (0.5, 0.3) and (−1.5, −1). n xn 0 1 2 3 4

0.5000000 0.3761299 0.3727877 0.3727731 0.3727731

g1 (x, y) = 2x . g2 (x, y) = 2(y + 1) f (x n , yn )

yn

0.3000000 0.3707193 0.3642151 0.3641995 0.3641995

n

xn

yn

0 1 2 3 4

−1.5000000 −1.4166667 −1.4141680 −1.4141606 −1.4141606

−1.0000000 −0.9916002 −0.9877619 −0.9877577 −0.9877577

−0.1794255 0.0033956 0.0000020 0.0000000 0.0000000

f (x n , yn ) −0.0025050 −0.0034547 −0.0000031 0.0000000 0.0000000

g(x n , yn ) −0.0600000 0.0203450 0.0000535 0.0000000 0.0000000

g(x n , yn ) 0.2500000 0.0070150 0.0000210 0.0000000 0.0000000

The solutions are x = 0.3727731, y = 0.3641995, and x = −1.4141606, y = −0.9877577. 6. y

(π/2,π )

y 2 =x 3

sin x + sin y = 1 (π,π/2)

π/2

x

π/2

Fig. 13.7.6 f (x, y) = sin x + sin y − 1

g(x, y) = y 2 − x 3

f 1 (x, y) = cos x f 2 (x, y) = cos y

g1(x, y) = −3x 2 . g2(x, y) = 2y

There are infinitely many solutions for the given pair of equations, since the level curve of f (x, y) = 0 is repeated periodically throughout the plane. We will find the two solutions closest to the origin in the first quadrant. From the figure, it appears that these solutions are near (0.6, 0.4) and (2, 3). n xn yn f (x n , yn ) g(x n , yn ) 0 1 2 3 4 n

0.6000000 0.5910405 0.5931130 0.5931105 0.5931105 xn

0.4000000 0.4579047 0.4567721 0.4567761 0.4567761 yn

−0.0459392 −0.0007050 −0.0000015 0.0000000 0.0000000 f (x n , yn )

−0.0560000 0.0032092 −0.0000063 0.0000000 0.0000000 g(x n , yn )

0

2.0000000

3.0000000

0.0504174

1.0000000

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INSTRUCTOR’S SOLUTIONS MANUAL

1 2 3 4

SECTION 13.7 (PAGE 794)

2.0899016 2.0854887 2.0854779 2.0854779

3.0131366 3.0116804 3.0116770 3.0116770

−0.0036336 −0.0000086 0.0000000 0.0000000

−0.0490479 −0.0001199 0.0000000 0.0000000

The solutions are x = 0.5931105, y = 0.4567761, and x = 2.0854779, y = 3.0116770. 7. By analogy with the two-dimensional case, the Newton’s Method iteration formulas are

x n+1

z n+1

8.

1 f f 2 f 3 = x n − g g 2 g 3 1 h h h (xn ,yn ,zn ) 2 3 f f f 1 2 1 = z n − g1 g2 g 1 h 1 h 2 h (xn ,yn ,zn )

f (x, y, z) = y 2 + z 2 − 3 f 1 (x, y, z) = 0 f 2 (x, y, z) = 2y f 3 (x, y, z) = 2z

1 f 1 f f3 yn+1 = yn − g1 g g3 1 h 1 h h 3 (x n ,yn ,z n ) f 1 f 2 f 3 where 1 = g1 g2 g3 h 1 h 2 h 3 (xn ,yn ,zn ) h(x, y, z) = x 2 − z h 1 (x, y, z) = 2x h 2 (x, y, z) = 0 h 3 (x, y, z) = −1

g(x, y, z) = x 2 + z 2 − 2 g1 (x, y, z) = 2x g2 (x, y, z) = 0 g3 (x, y, z) = 2z

It is easily seen that the system f (x, y, z) = 0, has first-quadrant solution x = z = 1, y = n xn yn 0 1 2 3 4 5 9.

f (x, y) = y − x 2

g(x, y) = y − x 3

2.0000000 1.3000000 1.0391403 1.0007592 1.0000003 1.0000000

g(x, y, z) = 0,

√ 2. Let us start at the “guess” x 0 = y0 = z 0 = 2. zn f (x n , yn , z n ) g(x n , yn , z n )

2.0000000 1.5500000 1.4239564 1.4142630 1.4142136 1.4142136

f 1 (x, y) = −2x f 2 (x, y) = 1

h(x, y, z) = 0

2.0000000 1.2000000 1.0117647 1.0000458 1.0000000 1.0000000

5.0000000 0.8425000 0.0513195 0.0002313 0.0000000 0.0000000

6.0000000 1.1300000 0.1034803 0.0016104 0.0000006 0.0000000

h(x n , yn , z n ) 2.0000000 0.4900000 0.0680478 0.0014731 0.0000006 0.0000000

g1 (x, y) = −3x 2 g2 (x, y) = 1 y

(1,1) y=x 2 y=x 3 x

Fig. 13.7.9

529 Copyright © 2014 Pearson Canada Inc.

SECTION 13.7 (PAGE 794)

n 0 1 2 3 4 5 .. .

ADAMS and ESSEX: CALCULUS 8

xn 0.1000000 0.0470588 0.0229337 0.0113307 0.0056327 0.0028083

yn

n

0.1000000 −0.0005882 −0.0000561 −0.0000062 −0.0000007 −0.0000001

0 1 2 3 4

xn

yn

0.9000000 1.0285714 1.0015038 1.0000045 1.0000000

0.9000000 1.0414286 1.0022771 1.0000068 1.0000000

15 0.0000027 0.0000000 16 0.0000014 0.0000000 17 0.0000007 0.0000000 18 0.0000003 0.0000000 Eighteen iterations were needed to obtain the solution x = y = 0 correct to six decimal places, starting from x = y = 0.1. This slow convergence is due to the fact that the curves y = x 2 and y = x 3 are tangent at (0, 0). Only four iterations were needed to obtain the solution x = y = 1 starting from x = y = 0, because, although the angle between the curves is small at (1, 1), it is not 0. The curves are not tangent there. Section 13.8 Calculations with Maple (page 799)

>

vars := {x=0.9, y=0.2}:

>

xy := fsolve(eqns,vars);

> z=evalf(subs(xy,x*y)); 1. The equation z = x y can be used to reduce the given system of three equations in three variables to a system of 2 equations in two variables: 2

2

>

vars := {x=0.5, y=0.8}:

>

xy := fsolve(eqns,vars);

2 2

x +y +x y =1

6x 2 y = 1.

> z=evalf(subs(xy,x*y));

The first equation can only be satisfied by points (x, y) satisfying |x| ≤ 1 and |y| ≤ 1. >

x y := {x = 0.968971, y = 0.177512} z = 0.172004

Digits := 6:

> eqns := {xˆ2+yˆ2+(x*y)ˆ2=1, 6*xˆ2*y=1}:

x y := {y = 0.812044, x = 0.453038} z = 0.367887 The four solutions are (x, y, z) = (±0.96897, 0.17751, ±0.17200) and (x, y, z) = (±0.45304, 81204, ±0.36789), rounded to five figures.

We use plots[implicitplot] to locate suitable starting points for fsolve. > plots[implicitplot](eqns,x=-1..1, y=-1..1); The resulting plot (omitted here) shows four roots; two in the first quadrant near (.9, .2) and (.5, .8), and two more that are reflections of these in the yaxis. We use fsolve to find the two first-quadrant roots and calculate the corresponding values for z by substitution.

2. The equation y = sin z can be used to reduce the given system of three equations in three variables to a system of 2 equations in two variables: x 4 + sin2 z + z 2 = 1

z + z 3 + z 4 = x + sin z. The first equation can only be satisfied by points (x, z) satisfying |x| ≤ 1 and |z| ≤ 1.

530 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.8 (PAGE 799)

>

Digits := 6:

>

eqns := {xˆ4+(sin(z))ˆ2+zˆ2=1,

> z+zˆ3+zˆ 4=x+sin(z)}: We use plots[implicitplot] to locate suitable starting points for fsolve. > plots[implicitplot](eqns,x=-1..1, z=-1..1); The resulting plot shows two roots in the x z-plane, one near (0.6, 0.7) and the other near (−0.2, −0.7). We use fsolve to find them more precisely, and we then calculate the corresponding values for y by substitution. >

vars := {x=0.6, z=0.7}:

>

xz := fsolve(eqns,vars);

x y := {z = 0.686259, x = 0.597601} y = 0.633648

>

vars := {x=-0.2, z=-0.7}:

>

xy := fsolve(eqns,vars);

> y=evalf(subs(xz,sin(z))); x y := {z = −0.738742, x = −0.170713} y = −0.673358

The two solutions are (x, y, z) = (0.59760, 0.63365, 0.68626) and (x, y, z) = (−0.17071, −0.67336, −0.73874), each rounded to five figures. 3. First define the expression f :

Digits := 6:

>

eqns := {diff(f,x), diff(f,y)}:

>

vars := {x=-0.3, y=-0.6}:

>

cp := fsolve(eqns,vars);

> val=evalf(subs(cp,f)); cp := {x = −.338532, y = −.520621} val = 0.810414 >

vars := {x=0.2, y=0.6}:

>

cp := fsolve(eqns,vars);

cp := {x = 0.133192, y = 0.536823} val = − .665721

There are only two critical points and the values of f at them have opposite sign. Since f → 0 as x 2 + y 2 → ∞, f has absolute maximum value 0.81041 at (−0.33853, −0.52062) and absolute minimum value −0.66572 at (0.13319, 0.53682), all numerical values rounded to five figures. 4. We begin with >

Digits := 6:

>

f := 1 - 10*xˆ4 - 8*yˆ4 - 7*zˆ4:

>

g := y*z - x*y*z - x - 2*y + z:

>

h := f + g:

f := (x*y-x-2*y)/(1+xˆ2+yˆ2)ˆ2:

Because the numerator grows much more slowly than the denominator for large x 2 + y 2 , global max and min values will be near the origin. We plot contours of f on, say, the square |x| ≤ 2, |y| ≤ 2. > >

>

> val=evalf(subs(cp,f));

> y=evalf(subs(xz,sin(z)));

>

The resulting plot (which we omit here) indicates the only likely critical points are near (−0.3, −0.6) and (0.2, 0.6). We determine them using fsolve and use substitution to evaluate f .

contourplot(f(x,y), x=-2..2, y=-2..2, contours=16);

Since h = 1 at (0, 0, 0) and h → −∞ as x 2 + y 2 + z 2 increases, the maximum value of g will be near (0, 0, 0). We can try various choices of starting points including (0, 0, 0) itself. It turns out they all lead to the same critical point:

531 Copyright © 2014 Pearson Canada Inc.

SECTION 13.8 (PAGE 799)

ADAMS and ESSEX: CALCULUS 8

> eqns := {diff(h,x),diff(h,y),diff(h,z)}: > vars := x=0,y=0,z=0: > cp := fsolve(eqns,vars); val = evalf(subs(cp,h));

>

> f := (x+1.1*y-0.9*z+1)/(1+xˆ2+yˆ2+zˆ2):

cp := {x = −.28429, y = −.372953, z = 0.265109} val = 1.91367

> eqns := {diff(f,x),diff(f,y),diff(f,z)}: > vars := x=1,y=1,z=-1: > cp := fsolve(eqns,vars); val = evalf(subs(cp,f));

The absolute maximum value of h is 1.91367 (to five decimal places). 5. Because of the small coefficients on the x y and x z terms and the fact that without them f would certainly have a minimum value near the origin, we can use fsolve starting with various points near the origin. It turns out they all lead to only one critical point. >

Digits := 6:

>

f := xˆ2 + yˆ2 + zˆ2

Digits := 6:

This attempt fails; fsolve cannot locate a solution. We try a guess closer to the origin. > vars := x=0.5,y=0.5,z=-0.5: > cp := fsolve(eqns,vars); val = evalf(subs(cp,f)); cp := {y = 0.366057, z = −.299501, x = 0.332779} val = 1.50250 > vars := x=-0.5,y=-0.5,z=+0.5: > cp := fsolve(eqns,vars); val = evalf(subs(cp,f));

> +0.2*x*y-0.3*x*z+4*x-y: > eqns := {diff(f,x),diff(f,y),diff(f,z)}: > vars := x=0,y=0,z=0: > cp := fsolve(eqns,vars); val = evalf(subs(cp,f));

cp := {x = −.995031, z = 0.895528, y = −1.09453} val = − .502494

The eigenvalues of the Hessian matrix of f at each of these critical points confirms that the first is a local maximum and gives f its absolute maximum value 1.50250 and the second is a local minimum so the absolute minimum value of f is −0.502494.

cp := {x = −2.11886, y = 0.711886, z = −.317829} val = − 4.59368

To confirm that this CP does give a local minimum, you can calculate VectorCalculus[Hessian](f,[x,y,z]) and then evalf the result of LinearAlgebra[Eigenvalues](subs(cp,%)) and observe that all three eigenvalues are positive. The minimum value of f is −4.59368.

Section 13.9 Entropy in Statistical Mechanics and Information Theory (page 804) 1.

If the expression is universal, it must apply to each of n > 1 independent subsystems of a larger combined system, each denoted by index i , Si = k ln Wi + C.

6. First define the function: >

f := (x+1.1y-0.9z+1)/(1+xˆ2+yˆ2);

Since f (x, y, z) → 0 as x 2 + y 2 + z 2 → ∞ we expect f to have maximum and minimum values in some neighbourhood of the origin. If the numerator were instead x + y − z, we would expect the extreme values to occur along the line x = y = −z by symmetry. Accordingly, we use starting points along this line.

By additivity the entropy of the combined system is S=

n X i=1

Si = k ln

n Y i=1

Wi + nC = k ln W + nC.

We must require nC = C to have the same constant in the composite system. Thus C(n − 1) = 0. Since n > 1, C = 0.

532 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 13.9 (PAGE 804)

Pn 2. To Pnextrenize S = −k i=1 pi ln pi subject to i=1 pi = 1 we look for critical points of L = −k

n X i=1

n X

pi ln pi + λ

pi

i=1

!

4.

−1 .

1 2

At any such critical point we will have

=− =−

i=1

J X j =1

pi log2 pi − qj

I X i=1

J X I X

pi log2 pi −

I X

pi

i=1

i=1

= k N ln N −

J X

M X i=1

q j log2 q j

and

i=1

K X k=1

pi = 1 and

πk =

I X J X i=1 j =1

j =1 q j

pi q j =

n i ln n i .

i=1

and

M X i=1

n i ǫi = E

M X i=1

n i ln n i − α

M X i=1

ni − β

M X

n i ǫi .

i=1

For a critical point we require

j =1

0=

∂L = − ln n i − 1 − α − βǫi , ∂n i


so that we have n i = e− (1+α)+βǫi = Ae−Bǫi , where A = e−(1+α) and B = β. The values of the Lagrange multipliers α and β, (or, equivalently, the two constants A and B) are determined by the two constraint equations.

πk log2 πk , 6.

k=1

PJ

ni = N

L = N ln N −

The probabilities of the combination of independent sub systems are given as the product pi q j = πk . Each term can be uniquely accounted for with a single index k = i + I ( j − 1) between 1 and K = I J . Thus

PI

!

we look for a critical point of the Lagrange function

pi q j log2 pi q j .

K X

M X

For maximum S (or equivalently S/k subject to the constraints

j =1 i=1

= H1 + H2 = −

= 0.875



q j log2 q j

j =1



 N! n 1 !n 2 ! . . . n M ! ! M X ≈ k N ln N − N − (n i ln n i − n i )

S = k ln

The Hessian matrix for L is diagonal, has (equal) negative eigenvalues, and is negative definite on Rn , so the critical point must provide a local maximum value for PS by Theorem 5 in Section 13.4. Since S = k ni=1 ((ln n)/n) > 0 and S → 0 as points ( p1 , p2 , . . . , pn ) approach the P boundary of the region of Rn defined by pi ≥ 0, ni=1 = 1, the local minimum must, in fact be absolute. PI P 3. Since i=1 pi = 1 and jJ=1 q j = 1, we have H1 + H2 = −

1 2 3 3 + + + 2 4 8 8

5. Using the Modified Stirling approximation, we have

It follows that ln pi = 1 + (λ/k) for each i and so all the pi are equal. The constraint equation then gives pi = 1/n for each i ; there is only one critical point. Now  ∂2 L 0 if ⊂6= j = −kn < 0 if i = j . ∂ pi ∂ p j

J X



An optimal encoding scheme that achieves this is a → 0, b → 11, c → 101, d → 100.

∂L = −k (ln pi − 1) + λ. 0= ∂ pi

I X

The best compression is

!

= 1 also imply that

(a) Z Since n(u) ≥ 0 for all real u and ∞ n(u) du = N, p(u) = n(u)/N satisfies −∞

I X i=1

pi

J X j =1

q j = 1.

Z

∞ −∞

p(u) du = 1

533 Copyright © 2014 Pearson Canada Inc.

SECTION 13.9 (PAGE 804)

ADAMS and ESSEX: CALCULUS 8

and p(u) is a probability density function. By Definition 7 in Section 7.8, f (u) =

The probability density for the other two components of the velocity are the same, and since the three components are independent, the probability density for the three components of the velocity v = v 1 i + v 2 j + v 3 k is

1 2 2 √ e−(u−µ) /2σ σ 2π

is the density function of a normally distributed random variable with mean µ and variance σ 2 . Evidently, p(u) =

p(v) = p(v 1 , v 2 , v 3 ) = p(v 1 ) p(v 2 ) p(v 3 )  m 3 m(v 2 +v 2 +v 2 ) 1 2 3 2 − 2kT = e 2πkT  m 3 2 −m|v|2 /2kT = e . 2πkT

A −Bmu 2 1 2 2 e /2 = √ e−(u−µ) /2σ = f (u) N σ 2π

This is the classical Maxwell-Boltzmann distribution for an ideal gas.

provided µ = 0 and σ2 =

1 Bm

and

1 A √ = . N σ 2π

Review Exercises 13 (page 805)

It follows that A=

1.

√

f 1 (x, y) = (y − x y)e−x+y = y(1 − x)e−x+y

N N Bm = √ . √ σ 2π 2π

f 2 (x, y) = (x + x y)e−x+y = x(1 + y)e−x+y

A = f 11 = (−2y + x y)e−x+y

U2

B = f 12 = (1 − x + y − x y)e−x+y

(b) As shown in Section 7.8, the expectation of for such a normally distributed random variable with mean µ = 0 and variance σ 2 is Z

∞ −∞

C = f 22 = (2x + x y)e−x+y . For CP: either y = 0 or x = 1, and either x = 0 or y = −1. The CPs are (0, 0) and (1, −1).

u 2 p(u) du = σ 2 .

Accordingly, the expected value of the part of the kinetic energy of a particle due to motion in 1 1 the x-direction is mσ 2 = . By symme2 2B try the same result will obtain for the expected value of the part of the kinetic energy due to motion in the other two coordinate directions, so the expected value of the total kinetic energy of 3 , and that of all N particles is the particle is 2B 3N E= . 2B (c) Comparing the result of part (b) with the known 1 3 . Thus formula E = NkT , we see that B = 2 kT r m A = N and the probability density for 2πkT one component of velocity is p(u) =

 m 1 2 2 − mu e 2kT . 2πkT

f (x, y) = x ye−x+y

CP (0, 0) (1, −1) 2.

A 0 e−2

B 1 0

C 0 e−2

AC − B 2 −1 e−4

class saddle loc. min

f (x, y) = x 2 y − 2x y 2 + 2x y

f 1 (x, y) = 2x y − 2y 2 + 2y = 2y(x − y + 1)

f 2 (x, y) = x 2 − 4x y + 2x = x(x − 4y + 2) A = f 11 = 2y B = f 12 = 2x − 4y + 2 C = f 22 = −4x. For CP: either y = 0 or x − y + 1 = 0, and either x = 0 or x − 4y + 2 = 0. The CPs are (0, 0), (0, 1), (−2, 0), and (−2/3, 1/3). CP (0, 0) (0, 1) (−2, 0) (− 32 , 13 )

534 Copyright © 2014 Pearson Canada Inc.

A 0 2 0 2 3

B 2 −2 −2 − 23

C 0 0 8 8 3

AC − B 2 −4 −4 −4 4 3

class saddle saddle saddle loc. min

INSTRUCTOR’S SOLUTIONS MANUAL

3.

4.

REVIEW EXERCISES 13 (PAGE 805)

4 9 1 + + x y 4−x −y 1 9 f 1 (x, y) = − 2 + x (4 − x − y)2 9 4 f 2 (x, y) = − 2 + y (4 − x − y)2 2 18 A = f 11 = 3 + x (4 − x − y)3 18 B = f 12 = (4 − x − y)3 8 18 C = f 22 = 3 + . y (4 − x − y)3 For CP: y 2 = 4x 2 so that y = ±2x. If y = 2x, then 9x 2 = (4 − 3x)2 , from which x = 2/3, y = 4/3. If y = −2x, then 9x 2 = (4 + x)2 , from which x = −1 or x = 2. The CPs are (2/3, 4/3), (−1, 2), and (2, −4). f (x, y) =

CP

A

B

C

(−1, 2) (2, −4) ( 23 , 43 )

− 34 1 3

2 3 1 12 9 4

5 3

9

AC − B 2 − 83 1 − 48 729 16

1 − 24 45 8

5.

6.

x 2 + y 2 + z 2 − x y − x z − yz 1 = (x 2 − 2x y + y 2 ) + (x 2 − 2x z + z 2 ) 2  + (y 2 − 2yz + z 2 )  1 = (x − y)2 + (x − z)2 + (y − z)2 ≥ 0. 2 The minimum value, 0, is assumed at the origin and at all points of the line x = y = z.

7.

f (x, y) = x ye−x

class saddle saddle loc. min

0 = f 1 = e−x 0 = f 2 = e−x

B = f 12 = 4x − 3x 2 − 4x y

C = f 22 = −2x 2. (0, y) is a CP for any y. If x 6= 0 but y = 0, then x = 2 from the second equation. Thus (2, 0) is a CP. If neither x nor y is 0, then x + 2y = 2 and 3x + 2y = 4, so that x = 1 and y = 1/2. The third CP is (1, 1/2). CP (0, y) (2, 0) (1, 21 )

A 4y

− 2y 2

0 − 32

B

C

0 0 −4 −8 −1 −2

AC − B 2

class

0 −16 2

? saddle loc. max

The second derivative test is unable to classify the line of critical points along the y-axis. However, direct inspection of f (x, y) shows that these are local minima if y(2 − y) > 0 (that is, if 0 < y < 2) and local maxima if y(2 − y) < 0 (that is, if y < 0 or y > 2). The points (0, 0) and (0, 2) are neither maxima nor minima, so they are saddle points.

lim

x 2 +y 2 →∞

f (x, y) = 0.

2 −4y 2 2 −4y 2

(y − 2x 2 y) = e−x (x − 8x y 2) = e−x

2 −4y 2 2 −4y 2

y(1 − 2x 2 )

x(1 − 8y 2).
1 , √

√1 2 2 2

1 (where f = 1/4e), and ± √1 , − √ (where 2 2 2 f = −1/4e). Thus f has maximum value 1/4e and minimum value −1/4e.

f 1 (x, y) = 4x y − 3x y − 2x y = x y(4 − 3x − 2y)

A = f 11 = 4y − 6x y − 2y 2

satisfies

The CPs are (0, 0) (where f = 0), ±

2

f 2 (x, y) = 2x 2 − x 3 − 2x 2 y = x 2 (2 − x − 2y)

2 −4y 2

Since f (1, 1) > 0 and f (−1, 1) < 0, f must have maximum and minimum values and these must occur at critical points. For CP:

f (x, y) = x 2 y(2 − x − y) = 2x 2 y − x 3 y − x 2 y 2 2

f (x, y, z) = g(s) = s +(1/s), where s = x 2 + y 2 +z 2 . Since g(s) → ∞ as s → ∞ or s → 0+, g must have a minimum value at a critical point in (0, ∞). For CP: 0 = g ′ (s) = 1 − (1/s 2 ), that is, s = 1. g(1) = 2. The minimum value of f is 2, and is assumed at every point of the sphere x 2 + y 2 +z 2 = 1.

8.

f (x, y) = (4x 2 − y 2 )e−x

f 1 (x, y) = e−x

2 +y 2 2

2




2 +y 2

2x(4 − 4x 2 + y 2 )

f 2 (x, y) = e−x +y (−2y)(1 − 4x 2 + y 2 ). f has CPs (0, 0), (±1, 0). f (0, 0) = 0. f (±1, 0) = 4/e. 2

a) Since f (0, y) = −y 2 e y → −∞ as y → ±∞, and since f (x, x) = 3x 2 e0 = 3x 2 → ∞ as x → ±∞, f does not have a minimum or a maximum value on the x y-plane. 2

b) On y = 3x, f (x, 3x) = −5x 2e8x → −∞ as x → ∞. Thus f can have no minimum value on the wedge 0 ≤ y ≤ 3x. However, as noted in (a), f (x, x) → ∞ as x → ∞. Since (x, x) is in the wedge for x > 0, f cannot have a maximum value on the wedge either. 9. Let the three pieces of wire have lengths x, y, and L − x − y cm, respectively. The sum of areas of the squares is S=


1 2 x + y 2 + (L − x − y)2 , 16 535

Copyright © 2014 Pearson Canada Inc.

REVIEW EXERCISES 13 (PAGE 805)

ADAMS and ESSEX: CALCULUS 8

for which we must find extreme values over the triangle x ≥ 0, y ≥ 0, x + y ≤ L. For critical points:
1 ∂S = x − (L − x − y) ∂x 8
∂S 1 0= = y − (L − x − y) , ∂y 8

0=

from which we obtain x = y = L/3. This CP is inside the triangle, and S = L 2 /48 at it.

or, about 9.26 cubic feet. 11. The ellipse (x/a)2 +(y/b)2 = 1 contains the rectangle −1 ≤ x ≤ 1, −2 ≤ y ≤ 2, if (1/a 2 ) + (4/b2) = 1. The area of the ellipse is A = πab. We minimize A by looking for critical points of   4 1 + − 1 . L = πab + λ a 2 b2 For CPs:

On the boundary segment x = 0, we have
1 2 y + (L − y)2 , S= 16

∂L 2λ = πb − 3 ∂a a 8λ ∂L = πa − 3 0= ∂b b ∂L 1 4 0= = 2 + 2 − 1. ∂λ a b 0=

(0 ≤ y ≤ L).

At y = 0 or y = L, we have S = L 2 /16. For critical points
1 dS 0= = y − (L − y) , dy 8 so y = L/2 and S = L 2 /32. By symmetry the extreme values of S on the other two boundary segments are the same. Thus the minimum value of S is L 2 /48, and corresponds to three equal squares. The maximum value of S is L 2 /16, and corresponds to using the whole wire for one square.

10. Let the length, width, and height of the box be x, y, and z in, respectively. Then the girth is g = 2x + 2y. We require g + z ≤ 120 in. The volume V = x yz of the box will be maximized under the constraint 2x + 2y + z = 120, so we look for CPs of

12. The ellipsoid (x/a)2 + (y/b)2 + (z/c)2 = 1 contains the rectangle −1 ≤ x ≤ 1, −2 ≤ y ≤ 2, −3 ≤ z ≤ 3, provided (1/a 2 ) + (4/b2 ) + (9/c)2 = 1. The volume of the ellipsoid is V = 4πabc/3. We minimize V by looking for critical points of   4π 1 4 9 L= abc + λ + + − 1 . 3 a 2 b2 c2 ∂L ∂a ∂L 0= ∂b ∂L 0= ∂c ∂L 0= ∂λ

0= = yz + 2λ

(A)

= x z + 2λ

(B)

= xy + λ

(C)

= 2x + 2y + z − 120.

(D)

Comparing (A), (B), and (C), we see that x = y = z/2. Then (D) implies that 3z = 120, so z = 40 and x = y = 20 in. The largest box has volume V = (20)(20)(40) = 16, 000

in3 ,

(C)

Multiplying (A) by a and (B) by b, we obtain 2λ/a 2 = 8λ/b2 , so that either λ = 0 or b = 2a. Now λ = 0 implies b = 0, which is inconsistent with 2 (C). If √ b = 2a, then (C) implies that 2/a = 1, so a = 2. The smallest area of the ellipse is V = 4π square units.

For CPs: ∂L ∂x ∂L 0= ∂y ∂L 0= ∂z ∂L 0= ∂λ

(B)

For CPs:

L = x yz + λ(2x + 2y + z − 120).

0=

(A)

4π 2λ bc − 3 3 a 4π 8λ = ac − 3 3 b 4π 18λ = ab − 3 3 c 1 4 9 = 2 + 2 + 2 − 1. a b c =

(A) (B) (C) (D)

Multiplying (A) by a, (B) by b, and (C) by c, we obtain 2λ/a 2 = 8λ/b2 = 18λ/c2 , so that either λ = 0 or b = 2a, c = 3a. Now λ = 0 implies bc = 0, which is inconsistent with (D). If b = 2a √ and c = 3a, then (D) implies that 3/a 2 = 1, so a = 3. The smallest volume of the ellipsoid is V =

√ √ √ 4π √ ( 3)(2 3)(3 3) = 24 3π cubic units. 3

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INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 13 (PAGE 805)

13. The box −1 ≤ x ≤ 1, −2 ≤ y ≤ 2, 0 ≤ z ≤ 2 is contained in the region z

  y2 x2 0≤z ≤a 1− 2 − 2 b c

z

provided that (2/a) + (1/b2) + (4/c2) = 1. The volume of the region would normally be calculated via a “double integral” which we have not yet encountered. (See Chapter 5.) It can also be done directly by slicing. A horizontal plane at height z (where 0 ≤ z ≤ a) intersects the region in an elliptic disk bounded by the ellipse

y

y

x

Fig. R-13.14 x2 b2

+

y2 c2

=1−

z . a

The area of the window is

The area of this disk is  r  r   z z z A(z) = π b 1 − c 1− = πbc 1 − . a a a Thus the region has volume V = πbc

Z

a 0



1−

For maximum A, we look for critical points:

Thus we look for critical points of L=

  s 2 1 x ∂A = L − x − 2z + z 2 −  0= ∂x 2 4  



2 1 4 πabc +λ + 2 + 2 −1 . 2 a b c

For critical points: ∂L ∂a ∂L 0= ∂b ∂L 0= ∂c ∂L 0= ∂λ

0=

π 2λ bc − 2 2 a π 2λ = ac − 3 2 b π 8λ = ab − 3 2 c 2 1 4 = + 2 + 2 − 1. a b c =

or, since x + 2y + 2z = L,   s 2 x x A= L − x − 2z + z 2 −  . 2 4

z πabc dz = . a 2



s x 2 x2 A = xy + z − , 2 4

(A) (B) (C) (D)

Multiplying (A) by a, (B) by b, and (C) by c, we obtain 2λ/a = 2λ/b2 = 8λ/c2 , so that either λ = 0 or b2 = a, c2 = 4a. Now λ = 0 implies bc = 0, which is inconsistent with (D). If b2 = a and c2 = 4a, then (D) implies that 4/a = 1, so a = 4. The smallest volume of the region is V = π(4)(2)(4)/2 = 16π cubic units.

 x −1 − r x   2 2 x 4 z2 − 4 L 2z 2 − x 2 = −x−z+ r 2 x2 4 z2 − 4 ∂A xz 0= = −x + r . ∂z x2 2 2 z − 4 +

(A)

(B)

Now (B) p implies that either x = 0 or z = 2 z 2 − (x 2 /4). But x = 0 gives zero area rather than maximum area, so the second √ alternative must hold, and it implies that z = x/ 3. Then (A) gives

14.

  L 1 x = 1+ √ x + √ , 2 3 2 3

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REVIEW EXERCISES 13 (PAGE 805)

ADAMS and ESSEX: CALCULUS 8

√ from which we obtain x = L/(2 + 3). The maximum area of the window is, therefore, 1 L2 √ = A √ L/ 3 L 4 2+ 3 x= √ , z= √ 2+ 3

16. The envelope of y = (x − c)3 + 3c is found by eliminating c from that equation and 0=

∂ [(x − c)3 + 3c] = −3(x − c)2 + 3. ∂c

2+ 3

≈ 0.0670L 2 sq. units.

15. If $1, 000x widgets per month are manufactured and sold for $y per widget, then the monthly profit is $1, 000P, where x 2 y3 − x. P = xy − 27

∂P 2x y 3 =y− −1 ∂x 27 ∂P x 2 y2 0= =x− . ∂y 9

(B)

dP = 3 − y 2, dy

√ √ so y = 3 and P = 2 3 − 3 ≈ 0.4641. On segment y = 2, 0 ≤ x ≤ 3, we have P = x − (8x 2 /27), which has values P = 0 at x = 0 and P = 1/3 at x = 3. It also has a critical point given by dP 16x 0= =1− , dx 27 so x = 27/16 and P = 27/32 ≈ 0.84375. It appears that the greatest monthly profit corresponds to manufacturing 27, 000/16 ≈ 1, 688 widgets/month and selling them for $2 each.

ǫ2 yǫǫ (x, 0) + · · · . 2!

Putting ǫ = 0 in the given equation, we get y(x, 0) = −2x. Now differentiate the given equation with respect to ǫ twice: yǫ + xe y + ǫxe y yǫ = 0

(A)

(B) implies that x = 0, which yields zero profit, or x y 2 = 9, which, when substituted into (A), gives y = 3 and x = 1. Unfortunately, the critical point (1, 3) lies outside of R. Therefore the maximum P must occur on the boundary of R. We consider all four boundary segments of R. On segment x = 0, we have P = 0. On segment y = 0, we have P = −x ≤ 0. On segment x = 3, 0 ≤ y ≤ 2, we have P = 3y − (y 3 /3) − 3, which has values P = −3 at y = 0 and P = 1/3 at y = 2. It also has a critical point given by 0=

17. Look for a solution of y + ǫxe y = −2x in the form of a Maclaurin series y = y(x, ǫ) = y(x, 0) + ǫyǫ (x, 0) +

We are required to maximize P over the rectangular region R satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 2. First look for critical points: 0=

This later equation implies that (x − c)2 = 1, so x − c = ±1. The envelope is y = (±1)3 + 3(x ∓ 1), or y = 3x ± 2.

yǫǫ + 2xe y yǫ + ǫxe y yǫ2 + ǫxe y yǫǫ = 0. The first of these equations gives yǫ (x, 0) = −xe y(x,0) = −xe−2x . The second gives yǫǫ (x, 0) = −2xe y(x,0) yǫ (x, 0) = 2x 2 e−4x . Thus y = −2x − 2ǫxe−2x + ǫ 2 x 2 e−4x + · · · . Z ∞ tan−1 (x y) 18. a) G(y) = dx x Z0 ∞ x 1 G ′ (y) = d x Let u = x y x 1 + x 2 y2 0 du = y d x Z 1 ∞ du π = = for y > 0. y 0 1 + u2 2y Z ∞ tan−1 (π x) − tan−1 x b) dx x 0 Z π Z π π dy π ln π = G(π) − G(1) = G ′ (y) d y = = . 2 1 y 2 1 Challenging Problems 13

(page 805)

1. To minimize In =

Z

π −π

538 Copyright © 2014 Pearson Canada Inc.

"

n a0 X − (ak cos kx + bk sin kx) f (x) − 2 k=1

#2

dx

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 13 (PAGE 805)

we choose ak and bk to satisfy

2. If f (x) =

∂ In 0= ∂a0 # Z π " n a0 X =− f (x) − − (ak cos kx + bk sin kx) d x 2 −π k=1   Z π = πa0 − f (x) d x −π

∂ In 0= ∂am # Z π " n a0 X − (ak cos kx + bk sin kx) = −2 f (x) − 2 −π k=1 cos mx d x

= 2am

Z

π −π

cos2 mx d x −

Z

π

f (x) cos mx d x −π

∂ In ∂bm # Z π " n a0 X − (ak cos kx + bk sin kx) = −2 f (x) − 2 −π k=1

0=

sin mx d x

= 2bm

Z

π −π

sin2 mx d x −

Z

π

f (x) sin mx d x. −π

The simplifications in the integrals above resulted from the facts that for any integers k and m, Z

cos kx cos mx d x = 0 unless k = m

Z −π π −π

sin kx sin mx d x = 0 unless k = m, and

Z

π −π

cos2 mx d x =

Z

π

= Z

−π

−π

Z 1 π f (x) cos mx d x for 0 ≤ m ≤ n, and π −π Z π 1 bm = f (x) sin mx d x for 1 ≤ m ≤ n. π −π

!

dx

n X πa02 +π (ak2 + bk2 ). 2 k=0

Therefore Z π " In = f (x) − −π

=

In is minimized when am =

n X πa02 +π (ak2 + bk2 ) 2 k=0

n a0 X f (x) + (ak cos kx + bk sin kx) 2 k=1

π

=

sin2 mx d x = π,

for −π ≤ x < 0 , then for 0 ≤ x ≤ π

Z 1 π π x dx = π 0 2 Z 1 π x cos kx d x ak = π 0 U=x d V = cos kx d x 1 dU = d x V = sin kx k π Z π   1 = x sin kx − sin kx d x πk 0 0( 0 if k is even cos kπ − 1 = = − 2 2 if k is odd πk πk 2 Z π 1 bk = x sin kx d x π 0 U=x d V = sin kx d x 1 dU = d x V = − cos kx π Z kπ   1 =− x cos kx − cos kx d x πk 0 0 k+1 (−1) . = k Because of the properties of trigonometric integrals listed in the solution to Problem 1, !2 Z π n a0 X + (ak cos kx + bk sin kx) d x 2 −π k=1

cos kx sin mx d x = 0.

Since

0 x

a0 =

π

Z−ππ



Z

π

−π

+ =

Z

π

−π

n a0 X + (ak cos kx + bk sin kx) 2 k=1 ! n X
2 πa02 2 2 f (x) d x − 2 +π (ak + bk ) 2 k=0

!#2

n X πa02 +π (ak2 + bk2 ) 2 k=0


2

f (x) d x −

! n X πa02 +π (ak2 + bk2) . 2 k=0

In fact, it can be shown that In → 0 as n → ∞.

539 Copyright © 2014 Pearson Canada Inc.

dx

CHALLENGING PROBLEMS 13 (PAGE 805)

3. Let I (x) = I ′ (x) =

Z

x 0

ADAMS and ESSEX: CALCULUS 8

y

ln(1 + t x) dt. Then 1 + t2

ln(1 + x 2 ) + 1 + x2

Z

x 0

t (1 +

t 2 )(1 + t x)

P1

D1

dt.

θ1

If we expand the latter integrand in partial fractions with respect to t, we obtain

D2

Z

0

x 0

(x + t) dt = (1 + x 2 )(1 + t 2 )

x

+ ln(1 + t 2 ) x

If Di = |P Pi | for i = 1, 2, 3, then Di2 = (x − x i )2 + (y − yi )2 ∂ Di = 2(x − x i ) 2Di ∂x ∂ Di x − xi = = cos θi ∂x Di −−→ where θi is the angle between P Pi and i. Similarly ∂ Di /∂y = sin θi . To minimize S = D1 + D2 + D3 we look for critical points: ∂S = cos θ1 + cos θ2 + cos θ3 ∂x ∂S 0= = sin θ1 + sin θ2 + sin θ3 . ∂y 0=

Thus ln(1 + x 2 ) 1 d ln(1 + x 2 ) + tan−1 x ln(1 + x 2 ) − 2 2 dx 1+x 1 + x2 1 d = tan−1 x ln(1 + x 2 ). 2 dx

I ′ (x) =

1 Therefore, I (x) = tan−1 x ln(1 + x 2 ) + C. Since 2 I (0) = 0, we have C = 0, and

4.

x 0

D3

Fig. C-13.4 2xtan−1 t

2(1 + x 2 ) 0 −1 2xtan x + ln(1 + x 2 ) = 2(1 + x 2 ) 1 d tan−1 x ln(1 + x 2 ) = 2 dx Z x dt x dt x = (1 + x 2 )(1 + t x) 1 + x2 0 1 + tx Let u = 1 + t x du = x dt Z 1+x 2 1 du ln(1 + x 2 ) = = . 2 1+x 1 u 1 + x2

Z

θ3

P3

Now we have x

P

P2

t x +t x = − . 2 2 2 2 (1 + t )(1 + t x) (1 + x )(1 + t ) (1 + x )(1 + t x)

Z

θ2

ln(1 + t x) 1 d x = tan−1 x ln(1 + x 2 ). 1 + t2 2

Thus cos θ1 + cos θ2 = − cos θ3 and sin θ1 + sin θ2 = − sin θ3 . Squaring and adding these two equations we get 2 + 2(cos θ1 cos θ2 + sin θ1 sin θ2 ) = 1, or cos(θ1 − θ2 ) = −1/2. Thus θ1 − θ2 = ±2π/3. Similarly θ1 − θ3 = θ2 − θ3 = ±2π/3. Thus P −−→ −−→ −−→ should be chosen so that P P1 , P P2 ,and P P3 make 120◦ angles with each other. This is possible only if all three angles of the triangle are less than 120◦. If the triangle has an angle of 120◦ or more (say at P1 ), then P should be that point on the side P2 P3 such that P P1 ⊥ P2 P3 .

540 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.1 (PAGE 812)

CHAPTER 14. MULTIPLE INTEGRATION

Section 14.1 Double Integrals 1.

The solid is split by the vertical plane through the zaxis and the point (3, 2, 0) into two pyramids, each with a trapezoidal base; one pyramid’s base is in the plane y = 0 and the other’s is in the plane z = 0. I is the sum of the volumes of these pyramids:

(page 812)

f (x, y) = 5 − x − y h R = 1 × f (0, 1) + f (0, 2) + f (1, 1) + f (1, 2) i + f (2, 1) + f (2, 2)

I =

7.

= 4 + 3 + 3 + 2 + 2 + 1 = 15

2.

h R = 1 × f (1, 1) + f (1, 2) + f (2, 1) + f (2, 2) i + f (3, 1) + f (3, 2) =3+2+2+1+1+0= 9

3.

h R = 1 × f (0, 0) + f (0, 1) + f (1, 0) + f (1, 1) i + f (2, 0) + f (2, 1) = 5 + 4 + 4 + 3 + 3 + 2 = 21

4.

8. 9.

J=

h R = 4 × 1 × 5 + 5 + 4 + 4 + 2] = 80 J = area of disk = π(52 ) ≈ 78.54

11.

R = 1 × (e1/2 + e1/2 + e3/2 + e3/2 + e5/2 + e5/2 ) ≈ 32.63

12.

f (x, y) = x 2 + y 2

h R = 4 × 1 × f ( 21 , 12 ) + f ( 32 , 12 ) + f ( 52 , 12 ) + f ( 72 , 12 )

R = 1 × f (1, 0) + f (1, 1) + f (2, 0) + f (2, 1) i + f (3, 0) + f (3, 1)

+ f ( 29 , 12 ) + f ( 12 , 23 ) + f ( 32 , 32 ) + f ( 52 , 32 ) + f ( 72 , 32 ) + f ( 92 , 23 )

+ f ( 12 , 52 ) + f ( 32 , 25 ) + f ( 52 , 52 ) + f ( 72 , 52 )

i + f ( 12 , 72 ) + f ( 32 , 27 ) + f ( 52 , 72 ) + f ( 12 , 92 ) + f ( 32 , 92 )

h

I =

ZZ

figure.

D

10.

= 918

13.

= 4 + 3 + 3 + 2 + 2 + 1 = 15

6.

   5+2 1 5+3 (3)(2) + (2)(3) = 15. 2 3 2

h R = 4 × 1 × 4 + 4 + 4 + 3 + 0] = 60

h

R = 1 × f ( 12 , 12 ) + f ( 12 , 32 ) + f ( 32 , 12 ) + f ( 32 , 32 ) i + f ( 52 , 12 ) + f ( 25 , 32 )



1dA h R = 4 × 1 × 5 + 5 + 5 + 5 + 4] = 96

= 4 + 3 + 3 + 2 + 2 + 1 = 15

5.

ZZ

1 3

RR

R

d A = area of R = 4 × 5 = 20. y 1

−1 D

3 x

(5 − x − y) d A is the volume of the solid in the R

z

z =5−x − y

−4

Fig. 14.1.13 5

14.

3

ZZ

D

(x + 3) d A =

2 x

2

3 Fig. 14.1.6

y

ZZ

D

x dA+3

ZZ

dA D

= 0 + 3(area of D)

π 22 =3× = 6π. 2 The integral of x over D is zero because D is symmetrical about x = 0.

541 Copyright © 2014 Pearson Canada Inc.

SECTION 14.1 (PAGE 812)

ADAMS and ESSEX: CALCULUS 8

y

y

√

2

y=

x 2 +y 2 =1

4−x 2

1 D

x 2 x

−2

Fig. 14.1.14

Fig. 14.1.17

15. ZZ T is symmetric about the line x + y = 0. Therefore, T

18.

(x + y) d A = 0.

y

ZZ

x 2 +y 2 ≤a 2

q a2 − x 2 − y 2 d A

= volume of hemisphere shown in the figure   2 1 4 3 π a = π a3 . = 2 3 3

(2,2)

(−1,1)

z T a

x

√

z=

a 2 −x 2 −y 2

(1,−1) (−2,−2) a

Fig. 14.1.15

y

x 2 +y 2 =a 2

x

Fig. 14.1.18

16.

ZZ





x 3 cos(y 2 ) + 3 sin y − π d A  = 0 + 0 − π area bounded by |x| + |y| = 1 |x|+|y|≤1

1 = −π × 4 × (1)(1) = −2π. 2 (Each of the first two terms in the integrand is an odd function of one of the variables, and the square is symmetrical about each coordinate axis.)

19.

ZZ

x 2 +y 2 ≤a 2



a−

q

 x 2 + y2 d A

= volume of cone shown in the figure 1 = π a3 . 3 z a

y

√

y=a−

x 2 +y 2

1

1 −1

x

a

−1

Fig. 14.1.19

Fig. 14.1.16

17.

ZZ

x 2 +y 2 ≤1

20. By the symmetry of S with respect to x and y we have ZZ

(4x 2 y 3 − x + 5) d A

= 0 − 0 + 5(area of disk) = 5π.

y

x 2 +y 2 =a 2

x

(by symmetry)

S

(x + y) d A = 2

ZZ

x dA S

= 2 × (volume of wedge shown in the figure) 1 = 2 × (a 2 )a = a 3 . 2

542 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.2 (PAGE 819)

z

2. z=x

S x

y

3.

ZZ

T

Z

= volume of the tetrahedron shown in the figure  11 1 = (1)(1) (1) = . 3 2 6

0 1  x2y

π 0

=

(1 − x − y) d A

1Z y

(x y + y 2 ) d x d y  x=y Z dy = + x y 2 2 0 x=0 Z 1 3 3 = y3 d y = . 2 0 8 0

(a,a,0)

Fig. 14.1.20

21.

Z

Z

x

Z

0

y=−x

Z

=2

cos y d y d x y=x dx sin y

−x π

π sin x d x = −2 cos x = 4.

π

0

0

z

(0,0,1)

4.

z=1−x−y

2

dy 0

y

(1,0,0)

Z

y

y 2 e xy d x

0

2

=

Z Z

2

=

(0,1,0)

T x

Z

2

y dy

0

0

2

x=y ! 1 xy e y x=0

2 2 e y − y 2 e4 − 5 = . 2 2 0

y(e y − 1) d y =

Fig. 14.1.21

22.

5.

ZZ q b2 − y 2 d A R

ZZ

R

(x 2 + y 2 ) d A =

z=

a

dx 0

b

Z

0

(x 2 + y 2 ) d y

 y=b y 3 3 y=0 0  Z a 1 bx 2 + b3 d x = 3 0  a 1 1 3 = bx + b3 x = (a 3 b + ab3 ). 3 3 0

= volume of the quarter cylinder shown in the figure 1 1 = (π b2 )a = π ab2 . 4 4 z √ b

Z

=

b2 −y 2

Z

a



dx

x2y +

y b y x

b

a

Fig. 14.1.22

R

Section 14.2 Iteration of Double Integrals in Cartesian Coordinates (page 819) 1.

Z

1

Z

x

a

x

Fig. 14.2.5 2

(x y + y ) d y   y=x 1 x y2 y 3 = dx + 2 3 y=0 0 Z 1 5 5 = x3 dx = . 6 0 24 dx

0

x

0

Z

6.

ZZ

R

x 2 y2 d A = =

Z

a

0 a3

x2 dx

b3 = 3 3

Z

b

y2 d y

0 a 3 b3

9

.

543 Copyright © 2014 Pearson Canada Inc.

SECTION 14.2 (PAGE 819)

7.

ADAMS and ESSEX: CALCULUS 8

ZZ

(sin x + cos y) d A S Z π/2 Z π/2 = dx (sin x + cos y) d y 0 0 Z π/2   y=π/2 = d x y sin x + sin y 0

Z

=

9.

ZZ

xy d A =

Z

=

Z

2

R

=

y=0

π/2  π

 sin x + 1 d x

2 0  π  π/2 π π = + = π. = − cos x + x 2 2 2 0

= =

y

1

Z

x dx 0

√

x

y2 d y

x2

 y=√x 1 3 x dx y 3 0 y=x 2 Z  1 1  5/2 x − x7 dx 3 0   1 1 2 7/2 x 8 x − 3 7 8 0   1 2 1 3 − = . 3 7 8 56 1



y

π 2

x=y 2

y=x 2

R

S

(1,1)

x x

π 2

Fig. 14.2.7

8.

ZZ

T

(x − 3y) d A =

Z

a

dx

Z

Fig. 14.2.9

(x − 3y) d y

0

0

10.

b(1−(x/a))

 y=b(1−(x/a)) 3 2 dx xy − y = 2 0 y=0     Z a  2 x 2x 3 2 x2 b x− = − b 1− + 2 dx a 2 a a 0  2  a x b x3 3 3 b2 x 2 1 b2 x 3 = b − − b2 x + − 2 a 3 2 2 a 2 a 2 0 a2 b ab2 = − . 6 2 Z

a



y

b

x

x

ZZ

x cos y d A D

=

Z

=

Z

1

x dx

0

1−x 2

cos y d y

0

1

0

Z

Z

y=1−x 2 x d x (sin y) y=0

1

x sin(1 − x 2 ) d x

Let u = 1 − x 2 du = −2x d x 0 Z 0 1 1 1 − cos(1) =− sin u du = cos u = . 2 1 2 2 1 =

0

y

1

x y a + b =1

y=1−x 2 D 1

T

x x

a

x

Fig. 14.2.8

Fig. 14.2.10

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.2 (PAGE 819)

y

11. For intersection: x y = 1, 2x + 2y = 5. 2x 2

Thus − 5x + 2 = 0, or (2x − 1)(x − 2) = 0. The intersections are at x = 1/2 and x = 2. We have ZZ

D

ln x d A =

Z

2



Z

2

Z

ln x d x 1/2

y

(5/2)−x

ln x

y=x

dy

T

1/x



5 1 −x− dx 2 x 1/2   Z 2 5 1  2 2 = ln x − x dx − ln x 2 2 1/2 1/2

=

(a,a)



 5 U = ln x d V = − x dx 2 dx dU = 5 x2 x V = x− 2 2 2     1 1 5 x2 =− (ln 2)2 − (ln )2 + x− ln x 2 2 2 2 1/2  Z 2  5 x − − dx 2 1/2 2   5 1 1 15 15 = (5 − 2) ln 2 − − ln − + 4 8 2 4 16 45 33 ln 2 − . = 8 16

a

x

Fig. 14.2.12

13.

Z 1 y Z √y x y e e dA = dy x dx R y 0 y y Z 1 1 = (1 − y)e y d y 2 0 U = 1 − y dV = ey dy dU = −d y V = ey # " 1 Z 1 1 y y e dy (1 − y)e + = 2

ZZ

0

0

1 1 e = − + (e − 1) = − 1. 2 2 2 y

(1,1) y



y=x



1 2 ,2

D xy=1

R

y=x 2 x

  2x+2y=5 1 2, 2 Fig. 14.2.13 x

14. Fig. 14.2.11

12.

ZZ q

a2

T

−

y2 d A

=

Z

a 0

q

a2

−

y2 d y

Z

ZZ

T

a

dx y

0

Z

1

0

q a (a − y) a 2 − y 2 d y = 0 Z aq Z a q y a2 − y 2 d y =a a2 − y 2 d y − Z

Z x x d x y dy 4 0 1+x 0 Z 1 x3 1 = dx 2 0 1 + x4 1 ln 2 1 . = ln(1 + x 4 ) = 8 8

xy dA = 1 + x4

y

(1,1)

0

Let u = a 2 − y 2 du = −2y d y Z π a2 1 0 1/2 =a + u du 4 2 a2 a 2   π a3 1 π 1 = − u 3/2 = − a3 . 4 3 4 3 0

y=x T 1

x

Fig. 14.2.14

545 Copyright © 2014 Pearson Canada Inc.

SECTION 14.2 (PAGE 819)

15.

Z

1

dy 0

Z

=

1

Z e

1 y

−x 2

2

e−x d x = dx

0

Z

1

Z

x

Z

ADAMS and ESSEX: CALCULUS 8

y

2

e−x d x

(R as shown)

R

dy

R

0

y=x

2

xe−x d x

Let u = x 2 0 du = 2x d x 1   Z 1 1 −u 1 1 −u 1 = e du = − e = 1− . 2 0 2 2 e 0

=

x

y

Fig. 14.2.17

(1,1)

1

y=x

18.

R 1

x

Fig. 14.2.15

16.

Z

π/2

dy 0

Z

=

0

(1,1)

1

π/2

Z

Z x 1/3 q 1 − y4 d y dx 0 ZZ qx = 1 − y 4 d A (R as shown) R Z 1 q Z 1 q 4 = y 1 − y dy − y3 1 − y4 d y

Z

1

0

0

Let u = y 2 Let v = 1 − y 4 du = 2y d y dv = −4y 3 d y Z 1p Z 1 0 1/2 1 1 − u 2 du + v dv = 2 0 4 1 0  1 1 π π 1 = × 12 + v 3/2 = − . 2 4 6 8 6 1

π/2

ZZ sin x sin x dx = d A (R as shown) x y R x Z x Z π/2 sin x dx dy = sin x d x = 1. x 0 0

y

y

(π/2,π/2)

1

y=x 1/3

(1,1)

R y=x

y=x R π/2

x

x

Fig. 14.2.18

Fig. 14.2.16

17.

Z 1 yλ dx dy (λ > 0) 2 2 0 x x +y ZZ yλ = d A (R as shown) 2 2 R x +y Z 1 Z y dx = yλ d y 2 2 0 0 x +y   x=y Z 1 1 x = yλ d y tan−1 y y x=0 0 1 Z 1 π π y λ π y λ−1 d y = = . = 4 0 4λ 0 4λ

Z

1

19.

V = =

20.

Z

1

dx

0

Z

Z

0

1

0

x

(1 − x 2 ) d y

(1 − x 2 )x d x =

Z y dy (1 − x 2 ) d x 0 0  Z 1 y3 1 1 5 = y− dy = − = cu. units. 3 2 12 12 0

V =

Z

1

1 1 1 − = cu. units. 2 4 4

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INSTRUCTOR’S SOLUTIONS MANUAL

21.

V =

Z

=

Z

1

0

0

1−x

Z

dx

0

1

SECTION 14.2 (PAGE 819)

25.

(1 − x 2 − y 2 ) d y

(1 − x 2 )y −

22.

Z

z = 1 − y 2 and z = x 2 intersect on the cylinder x 2 + y 2 = 1. The volume lying below z = 1 − y 2 and above z = x 2 is ZZ

V =

(1 − y 2 − x 2 ) d A Z √

Z

23.

Z

2

dx

x

Z

dx

1

(1 − x 2 − 2y 2 ) d A Z √ (1−x 2 )/2

1−x 2

1

1 dy x + y 1 0 y=x ! Z 2 d x ln(x + y) = 1 y=0 Z 2 Z = (ln 2x − ln x) d x = ln 2

V =

E

y √ 1/ 2

x 2 +y 2 ≤1

(1 − x 2 − y 2 ) d y 0 0 √   y= 1−x 2 Z 1 3 y =4 d x (1 − x 2 )y − 3 y=0 0 Z 1 8 = (1 − x 2 )3/2 d x Let x = sin u 3 0 d x = cos u du Z Z π/2 8 π/2 2 = cos4 u du = (1 + cos 2u)2 du 3 0 3 0   Z 1 + cos 4u 2 π/2 1 + 2 cos 2u + du = 3 0 2 23π π = = cu. units. 32 2 2 =4

Z

ZZ

dx (1 − x 2 − 2y 2 ) d y 0 0  Z 1 1 2 (1 − x 2 )3/2 =4 √ (1 − x 2 )3/2 − √ dx 3 2 2 2 0 √ Z 4 2 1 (1 − x 2 )3/2 d x Let x = sin θ = 3 0 d x = cos θ dθ √ √ Z   Z 4 4 2 π/2 2 π/2 1 + cos 2θ 2 dθ cos4 θ dθ = = 3 3 2 0 0 √ Z π/2   2 1 + cos 4θ = 1 + 2 cos 2θ + dθ 3 0 2 √   π/2 2 3θ 1 π = = √ cu. units. + sin 2θ + sin 4θ 3 2 8 2 2 0 =4

 (1 − x)3 (1 − x 2 )(1 − x) − dx 3 0  Z 1 2 4x 3 2 2 1 1 = − 2x 2 + d x = − + = cu. units. 3 3 3 3 3 3 0

=

1

 y=1−x y 3 dx 3 y=0

Vol =

2

1

1

d x = ln 2 cu. units.

x 2 +2y 2 =1

1 x

Fig. 14.2.25

26.

ZZ 

x y 2− − dA a b T Z a Z b(1−(x/a))  x y 2− − = dx dy a b  Z 0 a  0   x x 1 2 x 2 2− b 1− − b 1− dx = a a 2b a 0  Z a b 4x x2 = 3− + 2 dx 2 0 a a  a  b x 3 2x 2 2 = + 2 = ab cu. units. 3x − 2 a 3a 3 0

Vol =

y

b

x y a + b =1

24.

V =

Z

π 1/4

dy

y

x 2 sin(y 4 ) d x

0

0

1 = 3

Z

Z

1 = 12

π 1/4

T 4

y sin(y ) d y

0

Z

3

π 0

y4

Let u = du = 4y 3 d y

a

1 sin u du = cu. units. 6

x

Fig. 14.2.26

547 Copyright © 2014 Pearson Canada Inc.

SECTION 14.2 (PAGE 819)

27.

ADAMS and ESSEX: CALCULUS 8

Vol = 8 × part in the first octant Z a Z √a 2 −x 2 p =8 dx a2 − x 2 d y 0 Z0 a =8 (a 2 − x 2 ) d x 0   a x 3 16 3 = 8 a2 x − = a cu. units. 3 0 3

30. Since F ′ (x) = f (x) and G ′ (x) = g(x) on a ≤ x ≤ b, we have

ZZ

f (x)g(x) d A =

T

=

z

=

a

x 2 +z 2 =a 2

ZZ

f (x)g(x) d A =

T

y

Z

=

b a

a

  f (x) G(x) − G(a) d x

f (x)G(x) d x − G(a)F(b) + G(a)F(a) Z b b g(y) d y F ′ (x) d x y

b

a

f (x)G(x) d x = F(b)G(b)−F(a)G(a)−

F(y)g(y) d x.

Z

b

g(y)F(y) d y. a

(b,b)

 8 − x − (y − 4) d A

12 d A

b

y

cylinder x 2 = 2y 2 = 8 lies above z = 0. The part of the plane z = y − 4 inside the cylinder lies below z = 0. Thus the required volume is

2 +2y 2 ≤8

a

G ′ (y) d y

a

b

28. The part of the plane z = 8 − x lying inside the elliptic

ZZ x

b

x

Thus

Fig. 14.2.27



Z

a

Z

a

x 2 +y 2 =a 2

ZZ

Z

f (x) d x

  = g(y) F(b) − F(y) d y a Z = F(b)G(b) − F(b)G(a) −

a

Vol =

Z

b

Z

a x

Z

T

(by symmetry)

x 2 +2y 2 ≤8

(a,a)

y2 x2 + =1 = 12 × area of ellipse 8√ 4 √ = 12 × π(2 2)(2) = 48 2π cu. units.

(b,a)

x

Fig. 14.2.30

29. With g(x) and G(x) defined as in the statement of the problem, we have Z

a

x

G(u) du = = =

where C =

Z

c

Z

x

du

a

Z

d

dt

c

Z d c

Z

Z

Section 14.3 Improper Integrals and a MeanValue Theorem (page 824)

d

f 1 (u, t) dt c x

1.

f 1 (u, t) du

a

 f (x, t) − f (a, t) dt = g(x) − C,

d

f (a, t) dt is independent of x. Applying

ZZ

Q

e−x−y d A = =

2.

ZZ

Q

Z

∞

e−x d x

0

lim (−e

R→∞

dA = (1 + x 2 )(1 + y 2 )

the Fundamental Theorem of Calculus we obtain Z x d G(u) du = G(x). g ′ (x) = dx a

548 Copyright © 2014 Pearson Canada Inc.

=

Z

−x

∞

e−y d y

0

R !2 = 1 (converges) ) 0

Z

∞

Z

∞

dy 1 + y2 R !2 π2 −1 lim (tan x) = R→∞ 4 0

0

dx 1 + x2

(converges)

0

INSTRUCTOR’S SOLUTIONS MANUAL

3.

4.

ZZ

ZZ

S

y dA = 1 + x2

Z

=

1 2

1 √ dA = x y

1

dx 1 + x2 R! π = lim tan−1 x) (converges) S→−∞ 2 S R→∞ Z

2x

dx dy √ y 0 x x √ Z 1 √ 2( 2x − x) = dx x 0 Z 1 dx √ √ = 2( 2 − 1) √ = 4( 2 − 1) (converges) x 0 y

T

which diverges to infinity. Thus the given double integral diverges to infinity by comparison.

−∞

1

Z

∞

Z

y dy

0

SECTION 14.3 (PAGE 824)

7.

ZZ

e−(|x|+|y|) d A = 4 R2 =4

ZZ Z

x ≥0 y≥0

∞

e−(x+y) d A

e−x d x

0

=4

Z

∞

e−y d y

0

lim −e

R→∞

R !2 =4

−x

0

(The integral converges.)

(1, 2)

y = 2x

8. On the strip S between the parallel lines x + y = 0 and

T

x + y = 1 we have e−|x+y| = e−(x+y) ≥ 1/e. Since S has infinite area, ZZ e−|x+y| d A = ∞.

(1, 1)

y=x

S

Since e−|x+y| > 0 for all (x, y) in R2 , we have

x Fig. 14.3.4

5.

6.

ZZ

x2

ZZ

y2

+ dA (1 + x 2 )(1 + y 2 ) ZZ x2 d A =2 (by symmetry) 2 2 Q (1 + x )(1 + y ) Z ∞ 2 Z ∞ Z ∞ 2 x dx dy x dx =2 = π , 2 2 1 + x 1 + y 1 + x2 0 0 0 2 2 which diverges to infinity, since x /(1 + x ) ≥ 1/2 on [1, ∞). ZZ

Q

H

=

Z

=

Z

dA = 1+x + y ∞

0

Z

∞

Z

dx

0

1 0

e−|x+y| d A > R2

y x+y=1

S

ln

0



2+x 1+x



dx =

Z

∞

0

Fig. 14.3.8

 ln 1 +

1 1+x



d x.

ln(1 + u) = 1, we have ln(1 + u) ≥ u/2 on u→0+ u some interval (0, u 0 ). Therefore

Since lim



1 ln 1 + 1+x



≥

1 2(1 + x)

on some interval (x0 , ∞), and Z

0

∞

 ln 1 +

1 1+x



dx ≥

x

x+y=0

y=0

∞

e−|x+y| d A, S

and the given integral diverges to infinity.

1 dy 1+x +y

y=1 ! ln(1 + x + y) dx

ZZ

Z

∞ x0

1 d x, 2(1 + x)

9.

Z x dx e−y/x d y x3 0 T 1 y=x ! Z ∞ dx −y/x = −xe 3 x 1 y=0  Z ∞ 1 dx = 1− e x2 1 R !   1 1 1 = 1− lim − =1− R→∞ e x 1 e

ZZ

1 −y/x e dA = x3

Z

∞

(The integral converges.)

549 Copyright © 2014 Pearson Canada Inc.

SECTION 14.3 (PAGE 824)

ADAMS and ESSEX: CALCULUS 8

y

13.

a) I =

y=x

=

T

(1,1)

Z

S 1

0

1

Z

dA = x+y

dx

0

Z

1 0

y=1 ! d x ln(x + y)

dy x+y

y=0

i 1 = lim (x + 1) ln(x + 1) − x ln x c→0+ h

x

1

ZZ

c

= lim 2 ln 2 − 0 − (c + 1) ln(c + 1) + c ln c = 2 ln 2. c→0+

y

y

Fig. 14.3.9

10.

ZZ

T

dA = x 2 + y2

Z

∞

Z

(1,1)

x

dy 2 + y2 x 1 0 y=x ! Z ∞ 1 −1 y tan = dx x x 1 y=0 Z ∞ π dx = =∞ 4 1 x (The integral diverges to infinity.) dx

S T

x

e−xy d A > Q

ZZ

c

e−xy d A, R

e−xy d A > Q

Z

∞

dx

1

Z

1/x

e−xy d y >

0

1 e

Z

∞ 1

dx = ∞. x

c→0+ c

Vol =

15.

ZZ

y 1 y= x Q

R

12.

Z ∞ Z 1/x 1 1 1 1 sin d A = sin d x dy x x R x 2/π x 0 Z ∞ 1 1 = sin d x Let u = 1/x 2 x 2/π x du = −1/x 2 d x 0 Z 0 =− sin u du = cos u =1

ZZ

π/2

(The integral converges.)

π/2

Z

1

0

d x = 2 ln 2.

ZZ

0

x

Fig. 14.3.11

(ln 2x − ln x) d x = 2 ln 2

2x y dA 2 + y2 x S ZZ 2x y =4 d A (T as in #9(b)) 2 + y2 x T Z 1 Z x y dy =4 x dx Let u = x 2 + y 2 2 2 0 0 x +y du = 2y d y Z 2x 2 Z 1 du x dx =2 u x2 0 Z 1 = 2 ln 2 x d x = ln 2 cu. units.

14.

The given integral diverges to infinity.

1

y=0

1

Z

= 2 lim

where R satisfies 1 ≤ x < ∞, 0 ≤ y ≤ 1/x. Thus ZZ

x

Fig. 14.3.13a Fig. 14.3.13b Z 1 Z x ZZ dy dA = 2 lim c → 0+ dx b) I = 2 x + y x +y c 0 T y=x ! Z 1 = 2 lim c → 0+ d x ln(x + y)

11. Since e−xy > 0 on Q we have ZZ

(1,1)

16.

17.

Z 1 1 d x Z xk dA dy = x k−a d x, which con= a a Dk x 0 x 0 0 verges if k − a > −1, that is, if k > a − 1. Z

Z

1

Z

∞

xk

1

x k(b+1) d x if b+1 Dk 0 0 0 b > −1. This latter integral converges if k(b + 1) > −1. Thus, the given integral converges if b > −1 and k > −1/(b + 1). ZZ

ZZ

yb d A =

a

Rk

x dA =

dx

1

a

Z

x dx

yb d y =

Z

0

Z

xk

dy =

Z

∞

x k+a d x, which

1

converges if k + a < −1, that is, if k < −(a + 1).

550 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

18.

19.

20.

SECTION 14.3 (PAGE 824)

Z ∞ Z xk Z ∞ k(1−b) dA dy x = d x = d x if b < 1. b b y y 1−b Rk 1 0 1 This latter integral converges if k(1 − b) < −1. Thus, the given integral converges if b < 1 and k < −1/(1 − b). ZZ

Z

1

Z

∞

xk

1

x a+(b+1)k x a yb d A = xa dx yb d y = d x, b+1 Dk 0 0 0 if b > −1. This latter integral converges if a + (b + 1)k > −1. Thus, the given integral converges if b > −1 and k > −(a + 1)/(b + 1). ZZ

ZZ

Rk

x a yb d A =

Z

xa dx

1

Z

Z

xk 0

yb d y =

Z

1

∞

These seemingly contradictory results are explained by the fact that the given double integral is improper and does not, in fact, exist, that is, it does not converge. To see this, we calculate the integral over a certain subset of the square S, namely the triangle T defined by 0 < x < 1, 0 < y < x. ZZ

T

S

ZZ

S

1

dx 0

1

Z

x+1

0

1

Z

Z

x−y dy (x + y)3

x−y dA = (x + y)3

Z Z

dx

1

dy 0

Z

1

Z

y+1

0

1

x−y dx (x + y)3

dy

Z

x

0

0

ZZ 1 x2 d A (b − a)(d − c) R Z b Z d 1 = x2 dx dy (b − a)(d − c) a c a 2 + ab + b2 1 b3 − a 3 = . = b−a 3 3

Let u = x + y du = d y

y d R c

a

b

x

Fig. 14.3.22

23. The average value of x 2 + y 2 over the triangle T is Let u = x + y du = d x

u − 2y du u3 y 0  u=y+1 Z 1  y 1 = dy − u u=y u2 0  Z 1 y 1 1 1 = − − + dy (y + 1)2 y+1 y y 0 1 Z 1 dx 1 1 =− = =− . 2 y+1 0 2 0 (y + 1) =

dx

22. The average value of x 2 over the rectangle R is

2x − u du u3 0 x   Z 1 u=x+1 1 x dx = − 2 u u 0 u=x  Z 1 1 x 1 1 = − + dx − x + 1 (x + 1)2 x x 0 1 Z 1 dx 1 1 = = . = − 2 x + 1 0 2 0 (x + 1) =

Other iteration:

Z

1

which diverges to infinity.

21. One iteration: x−y dA = (x + y)3

Z

x−y dy (x + y)3 Let u = x + y du = d y Z 1 Z 2x 2x − u du = dx u3 0 x   u=2x Z 1 1 x = dx − 2 u u 0 u=x Z 1 1 dx = 4 0 x

x a+(b+1)k d x, b+1

if b > −1. This latter integral converges if a + (b + 1)k < −1. Thus, the given integral converges if b > −1 and k < −(a + 1)/(b + 1).

ZZ

x−y dA = (x + y)3

2 a2

ZZ

(x 2 + y 2 ) d A Z a Z a−x 2 = 2 dx (x 2 + y 2 ) d y a 0 0  y=a−x  Z a y 3 2 = 2 dx x2y + a 0 3 y=0 Z ah i 2 = 2 3x 2 (a − x) + (a − x)3 d x 3a 0 Z ah i 2 a2 = 2 a 3 − 3a 2 x + 6ax 2 − 4x 3 d x = . 3 3a 0 T

551 Copyright © 2014 Pearson Canada Inc.

SECTION 14.3 (PAGE 824)

ADAMS and ESSEX: CALCULUS 8

y

y a

√

y=

a

a 2 −x 2

y=a−x Q

T a x

a

x

x+y=0

Fig. 14.3.23

Fig. 14.3.25

26. Let R be the region 0 ≤ x < ∞, 0 ≤ y ≤ 1/(1 + x 2 ). If

24. The area of region R is Z

0

1

f (x, y) = x, then Z Z f (x, y) d A =

The average value of 1/x over R is

3

ZZ

R

1

√

x dx dy 0 x x2 Z 1  9 =3 x −1/2 − x d x = . 2 0

dA =3 x

Z

Z

then ZZ

=

(1,1) y=x 2

x

28.

x

Fig. 14.3.24

25. The distance √ from (x, y) to the line x + y = 0 is

(x + y)/ 2. The average value of this distance over the quarter-disk Q is √ ZZ 4 2 x+y √ dA = x dA π a2 2 Q Q √ √ Z Z a 2 −x 2 4 2 a dy = x d x π a2 0 0 √ Z a p 4 2 = x a 2 − x 2 d x Let u = a 2 − x 2 2 πa 0 du = −2x d x √ √ Z a2 4 2a 2 2 u 1/2 du = = . 3π π a2 0

4 π a2

ZZ

R

f (x, y) d A =

Z

1/(1+x 2 )

dy =

Z

∞

Z

∞

x dx

0

Z

1/(1+x 2 )

y dy

0

1 2

Z

1 4 Z

Z

∞

0 ∞

x dx (1 + x 2 )2

Let u = 1 + x 2 du = 2x d x

du 1 = 2 u 4 1 ∞ dx π Area = = . 2 1 + x 2 0 2 1 1 Thus f (x, y) has average value × = on R. π 4 2π The integral in Example 2 reduced to  Z ∞  1 ln 1 + 2 d x x 1   1 U = ln 1 + 2 dV = dx x V =x 2 dx dU = − x(x 2 + 1) " #   R Z R dx 1 = lim x ln 1 + 2 + 2 2 R→∞ x 1 1+x 1   π ln 1 + (1/R 2 ) π − − ln 2 + lim =2 R→∞ 2 4 1/R 3 π −(2/R )  = − ln 2 + lim  R→∞ 1 + (1/R 2 ) (−1/R 2 ) 2 =

R

x dx

27. If f (x, y) = x y on the region R of the previous exercise,

y

x=y 2

∞

x dx 1 + x2 R 0 0 0 which diverges to infinity. Thus f has no average value on R.

√ 1 ( x − x 2 ) d x = sq. units. 3

=

π − ln 2. 2

29. By the Mean-Value Theorem (Theorem 3),

552 Copyright © 2014 Pearson Canada Inc.

ZZ

Rhk

f (x, y) d A = f (x0 , y0 )hk

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.4 (PAGE 834)

for some point (x0 , y0 ) in Rhk . Since (x0 , y0 ) → (a, b) as (h, k) → (0, 0), and since f is continuous at (a, b), we have ZZ 1 lim f (x, y) d A (h,k)→(0,0) hk Rhk =

lim

(h,k)→(0,0)

=

R

f 12 (x, y) d A =

Z

a+h h

a

Z

a+h

Z

dx

a

7.

R

ZZ

D

(x 2 + y 2 ) d A =

Z

Q

ydA =

2π

dθ

0

Z

a

3. 4.

5.

0

a

r 4 cos2 θ sin2 θr dr

0

dθ

Z

a

r sin θ r dr

0 π/2

0

a3 a3 = 3 3

9.

ZZ

(x + y) d A =

e

x 2 +y 2

Q

dA =

π/2

Z

dθ

0

π = 2

Z

a

2

er r dr

0



 a 2 π(ea − 1) 1 r 2 e = 2 4 0

Z a 2 Z π/2 2x y 2r sin θ cos θ dθ d A = r dr 2 2 r2 Q x +y 0 0 π/2 Z a2 a 2 π/2 a 2 cos(2θ ) = = sin(2θ ) dθ = − 2 0 4 2 0

10.

ZZ

11.

ZZ

S

r 2r dr

0

Z

π/3

Z

a

(r cos θ + r sin θ )r dr Z a Z π/3 r 2 dr (cos θ + sin θ ) dθ = 0 0 π/3 a3 = (sin θ − cos θ ) 3 0 # " √ ! √ 3 1 a3 ( 3 + 1)a 3 = − = − (−1) 2 2 3 6

(x + y) d A =

dθ

0

0

y √ y= 3x

x 2 +y 2 =a 2

0

π/2 3 4a 3 a = 4 sin θ = 3 3 0

S π/3

π a4 x2 d A = ; by symmetry the value of this inte4 D gral is half of that in Exercise 1.

ZZ

Z

2a 3 ; by symmetry, the value is twice 3 Q that obtained in the previous exercise.

8.

ZZ

π a4 a4 = 4 2 ZZ q Z 2π Z a 2π a 3 x 2 + y2 d A = dθ r r dr = 3 D 0 0 ZZ Z 2π Z a dA r dr p = dθ = 2π a 2 2 r D x +y 0 0 ZZ Z π/2 Z a |x| d A = 4 dθ r cos θ r dr D

π/2

0

= 2π

2.

Z

f 21 (x, y) d A.

Section 14.4 Double Integrals in Polar Coordinates (page 834) ZZ

ZZ

R

Divide both sides of this identity by hk and let (h, k) → (0, 0) to obtain, using the result of Exercise 31, f 12 (a, b) = f 21 (a, b).

1.

0

= (− cos θ )

= f (a + h, b + k) − f (a + h, b) − f (a, b + k) + f (a, b). f 12 (x, y) d A =

dθ

Z a 6 π/2 2 sin (2θ ) dθ 6 0 Z  a 6 π/2  π a6 = 1 − cos(4θ ) dθ = 12 0 24

b

b

π/2

=

f 12 (x, y) d y

i f 1 (x, b + k) − f 1 (x, b) d x

ZZ

D

Z

x 2 y2 d A = 4

b+k

= f (a + h, b + k) − f (a, b + k) − f (a + h, b) + f (a, b) ZZ Z b+k Z a+h f 21 (x, y) d A = dy f 21 (x, y) d x R b a Z b+k h i = f 2 (a + h, y) − f 2 (a, y) d y Thus

ZZ

f (x0 , y0 ) = f (a, b).

30. If R = {(x, y) : a ≤ x ≤ a + h, b ≤ y ≤ b + k}, then ZZ

6.

a

x

Fig. 14.4.11

553 Copyright © 2014 Pearson Canada Inc.

SECTION 14.4 (PAGE 834)

12.

ZZ

S

x dA = 2

Z

π/4

dθ

Z

ADAMS and ESSEX: CALCULUS 8

√ 2

14.

r cos θ r dr

sec θ

0

Z  √  2 π/4 cos θ 2 2 − sec3 θ dθ = 3 0 √ π/4 π/4 2 4 2 − tan θ = sin θ 3 3 0 0 4 2 2 = − = 3 3 3

S 1

x 2 +y 2 ≤1 Z 1

= 4π

Z

ln(x 2 + y 2 ) d A =

2π

dθ 0

Z

1

(ln r 2 )r dr

0

r ln r dr

0

U = ln r d V = r dr dr r2 dU = V = r 2 # " 1 Z 1 1 r2 ln r − r dr = 4π 2 2 0 0   1 = 4π 0 − 0 − = −π 4

y

π/4

ZZ

(Note that the integral is improper, but converges since limr→0+ r 2 ln r = 0.)

√ 2 x

15. The average distance from the origin to points in the disk D: x 2 + y 2 ≤ a 2 is 1 π a2

Fig. 14.4.12

ZZ q D

x 2 + y2 d A =

1 π a2

Z

area region is

13.

Z π/4 Z sec θ (x 2 + y 2 ) d A = dθ r 3 dr T 0 0 Z 1 π/4 4 sec θ dθ = 4 0 Z π/4 1 = (1 + tan2 θ ) sec2 θ dθ Let u = tan θ 4 0 du = sec2 θ dθ Z 1 1 = (1 + u 2 ) du 4 0   1 1 u 3 1 = u+ = 4 3 3

− a 2 ).

ZZ

D

dA = 2 (x + y 2 )k

Z

0

(1,1)

1

Fig. 14.4.13

x

2a . 3

Let u = r 2 du = 2r dr

2π

dθ

Z

0

1

r −2k r dr = 2π

Z

1

r 1−2k dr

0

which converges if 1 − 2k > −1, that is, if k < 1. In this case the value of the integral is 2π

π/4

0

r 2 dr =

x 2 + y 2 ≤ b has 2 2 The average value of e−(x +y ) over the

y

T

a

17. If D is the disk x 2 + y 2 ≤ 1, then

0

x=1 r=sec θ

Z

p

ZZ 1 2 2 e−(x +y ) d A π(b2 − a 2 ) R Z 2π Z b 1 2 = dθ e−r r dr 2 2 π(b − a ) 0 a Z b2 1 1 (2π ) e−u du = 2 2 π(b − a ) 2 a2  1  −a 2 −b2 = 2 e − e . b − a2

ZZ

y=x

dθ

0

16. The annular region R: 0 < a ≤ π(b2

2π

18.

ZZ

1 r 2−2k π = . 2 − 2k 0 1−k

dA 2 + y 2 )k (1 + x R Z 2π Z ∞ r dr = dθ Let u = 1 + r 2 (1 + r 2 )k 0 0 du = 2r dr Z ∞ −π −k =π u du = if k > 1. 1−k 1 2

554 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

The integral converges to

19.

ZZ

D

π/4

Z

Z

SECTION 14.4 (PAGE 834)

π if k > 1. k −1

22. One quarter of the required volume lies in the first octant.

a

r cos θr sin θr dr Z Z a 1 π/4 r 3 dr = sin 2θ dθ 2 0 0   π/4 a4 a4 cos 2θ = = − . 8 2 16 0

xy d A =

dθ

0

0

y

x 2 +y 2 =a 2 y=x D π/4 a

x

Fig. 14.4.19

20.

ZZ

C

ydA =

Z

= =

π

dθ

0Z

1 3 1 3

π

0

Z

0

2

Z

1+cos θ

r sin θr dr 0

sin θ (1 + cos θ )3 dθ 2 4 u 4 = 12 0 3

u 3 du = y

Let u = 1 + cos θ du = − sin θ dθ

(See the figure.) In polar coordinates the cylinder x 2 + y 2 = ax becomes r = a cos θ . Thus, the required volume is ZZ q V =4 a2 − x 2 − y 2 d A D Z π/2 Z a cos θ p =4 dθ a 2 − r 2r dr Let u = a 2 − r 2 0 0 du = −2r dr Z π/2 Z a2 =2 dθ u 1/2 du 2 sin2 θ 0 a   a 2 Z 4 π/2 3/2  = dθ u 3 0 a 2 sin2 θ Z π/2 4 = a3 (1 − sin3 θ ) dθ 3 0   Z π/2 4 3 π 2 = a − sin θ (1 − cos θ ) dθ 3 2 0 Let v = cos θ dv = − sin θ dθ Z 2π a 3 4a 3 1 = − (1 − v 2 ) dv 3 3 0   1 2π a 3 4a 3 v 3 − v− = 3 3 3 0

2π a 3 8a 3 2 = − = a 3 (3π − 4) cu. units. 3 9 9 z

r=1+cos θ a x 2 +y 2 +z 2 =a 2

C 2 x

x 2 +y 2 =ax

Fig. 14.4.20

21. The paraboloids z = x 2 + y 2 and 3z = 4 − x 2 − y 2

intersect where 3(x 2 + y 2 ) = 4 − (x 2 + y 2 ), i.e., on the cylinder x 2 + y 2 = 1. The volume they bound is given by  4 − x 2 − y2 2 2 V = − (x + y ) d A 3 x 2 +y 2 ≤1  Z 2π Z 1 4 − r2 = dθ − r 2 r dr 3 0 0 Z 8π 1 (r − r 3 ) dr = 3 0   1 8π r 2 r 4 2π = − = cu. units. 3 2 4 3 ZZ



0

D x

a y

a

Fig. 14.4.22

23. The volume inside the sphere x 2 + y 2 + z 2 = 2a 2 and the cylinder x 2 + y 2 = a 2 is Z π/2 Z ap 2a 2 − r 2r dr V =8 dθ

Let u = 2a 2 − r 2 du = −2r dr Z 2a 2   3 √ 4π a = 2π u 1/2 du = 2 2 − 1 cu. units. 3 a2 0

0

555 Copyright © 2014 Pearson Canada Inc.

SECTION 14.4 (PAGE 834)

ADAMS and ESSEX: CALCULUS 8

z

z √ 2a x 2 +y 2 +z 2 =2a 2

x 2 +y 2 =a 2 a x 2 +z 2 =a 2 x=y

x 2 +y 2 =a 2

a

√ 2a

a y

y

x

x

y 2 +z 2 =a 2

Fig. 14.4.25

Fig. 14.4.23

24.

D

a

Z

2π

Z

2

(r cos θ + r sin θ + 4)r dr Z 2π Z 2 Z = (cos θ + sin θ ) dθ r 2 dr + 8π

Volume =

dθ

0

26. One quarter of the required volume V is shown in the

0

0

figure. We have

2

r dr

0

0

2

= 0 + 4π(2 ) = 16π cu. units.

25. One eighth of the required volume lies in the first octant. This eighth is divided into two equal parts by the plane x = y. One of these parts lies above the circular sector D in the x y-plane specified in polar coordinate by 0 ≤√ r ≤ a, 0 ≤ θ ≤ π/4, and beneath the cylinder z = a 2 − x 2 . Thus, the total volume lying inside all three cylinders is

ZZ

√

ydA Z Z 2 sin θ √ r sin θ r dr =4 dθ 0 0 2 sin θ ! Z π/2 √ 2 5/2 =4 sin θ dθ r 5 0 0 √ Z π/2 √ 32 2 2 64 3 = cu. units. sin dθ = 5 15 0

V =4

ZZ p a2 − x 2 d A D Z π/4 Z ap = 16 dθ a 2 − r 2 cos2 θr dr

D π/2

V = 16

0

z

0

Let u = a 2 − r 2 cos2 θ du = −2r cos2 θ dr Z π/4 Z a2 dθ u 1/2 du =8 cos2 θ a 2 sin2 θ 0 Z 16a 3 π/4 1 − sin3 θ = dθ 3 cos2 θ 0   Z 16a 3 π/4 1 − cos2 θ = sin θ dθ sec2 θ − 3 cos2 θ 0   π/4 16a 3 1 = tan θ − − cos θ 3 cos θ 0   √ 16a 3 1 = 1−0− 2+1− √ +1 3 2   1 3 = 16 1 − √ a cu. units. 2

y=z 2

x 2 +y 2 =2y 2

1

y

D

x

Fig. 14.4.26

27. By symmetry, we need only calculate the average distance from points in the sector S: 0 ≤ θ ≤ π/4,

556 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.4 (PAGE 834)

0 ≤ r ≤ 1 to the line x = 1. This average value is 8 π

ZZ

S

Z Z 1 8 π/4 dθ (1 − r cos θ )r dr π 0 0   Z π/4 Z 1 8 π 2 = − cos θ dθ r dr π 8 0 √ 0 8 4 2 units. =1− √ =1− 3π 3 2π

(1 − x) d A =

y

S

1 x

Let x = au, y = bv. Then ∂(x, y) du dv = ab du dv. d x d y = ∂(u, v)

The region E corresponds to the quarter disk Q: u 2 + v 2 ≤ 1, u, v ≥ 0 in the uv-plane. Thus ZZ p V = 8abc 1 − u 2 − v 2 du dv Q   1 = 8abc × × volume of ball of radius 1 8 4 = π abc cu. units. 3

30. We use the same regions and change of variables as in the previous exercise. The required volume is

28. The area of x¯ =

4π

=

4π

=

4π

=

4π

Fig. 14.4.27 √ S is (4π − 3 3)/3 sq. units. Thus ZZ 3 √ x dA −3 3 S Z 2 Z π/3 6 r cos θ r dr dθ √ −3 3 0 sec θ Z π/3 2 cos θ (8 − sec3 θ ) dθ √ −3 3 0 √ π/3 ! √ 6 3 2 = √ 4 3 − tan θ √ . −3 3 4π − 3 3 0

The segment has centroid

! √ 6 3 √ ,0 . 4π − 3 3

y

√ 3 π/3 1 S

2

2 x

Fig. 14.4.28

29. Let E be the region in the first quadrant of the x y-plane bounded by the coordinate axes and the ellipse x2 y2 + = 1. The volume of the ellipsoid is a2 b2 s ZZ x2 y2 V = 8c 1 − 2 − 2 d x d y. a b E

 ZZ  y2 x2 1 − 2 − 2 dx dy a b EZZ = ab (1 − u 2 − v 2 ) du dv.

V =

Q

Now transform to polar coordinates in the uv-plane: u = r cos θ , v = r sin θ . V = ab =

Z

π/2

dθ 0

π ab 2



Z

1

(1 − r 2 )r dr

0  1 r4

r2 − 2 4

= π ab cu. units. 8 0

u+v u−v , y = , so that x + y = u and 2 2 x − y = v. We have 1 1 2 | du dv = 1 du dv. d x d y = | 12 1 2 2 −2

31. Let x =

Under the above transformation the square |x| + |y| ≤ a corresponds to the square S: −a ≤ u ≤ a, −a ≤ v ≤ a. Thus ZZ ZZ 1 e x+y d A = eu du dv 2 S |x|+|y|≤a Z a Z 1 a u dv e du = 2 −a −a = a(ea − e−a ) = 2a sinh a.

32. The parallelogram P bounded by x + y = 1, x + y = 2,

3x + 4y = 5, and 3x + 4y = 6 corresponds to the square S bounded by u = 1, u = 2,v = 5, and v = 6 under the transformation u = x + y,

v = 3x + 4y,

557 Copyright © 2014 Pearson Canada Inc.

SECTION 14.4 (PAGE 834)

ADAMS and ESSEX: CALCULUS 8

v

or, equivalently, x = 4u − v,

y = v − 3u. v

y

v=2

v=6 u=1 R u=2 v=5

x+y=1 3x+4y=5

3x+4y=6 P

u

Fig. 14.4.33 u

Fig. 14.4.32a

34. Under the transformation u = x 2 − y 2 , v = x y, the

region R in the first quadrant of the x y-plane bounded by y = 0, y = x, x y = 1, and x 2 − y 2 = 1 corresponds to the square S in the uv-plane bounded by u = 0, u = 1, v = 0, and v = 1. Since

Fig. 14.4.32b

∂(x, y) 4 −1 = 1, = −3 1 ∂(u, v)

∂(u, v) 2x = y ∂(x, y)

so d x d y = du dv. Also

x 2 + y 2 = (4u − v)2 + (v − 3u)2 = 25u 2 − 14uv + 2v 2 .

we therefore have

Thus we have ZZ ZZ (x 2 + y 2 ) d x d y = (25u 2 − 14uv + 2v 2 ) du dv P S Z 2 Z 6 7 = du (25u 2 − 14uv + 2v 2 ) dv = . 2 1 5

∂(u, v) y = −y/x 2 ∂(x, y)

y x = 2 = 2v, 1/x x

∂(x, y) 1 = . The region D in the first quadrant ∂(u, v) 2v of the x y-plane bounded by x y = 1, x y = 4, y = x, and y = 2x corresponds to the rectangle R in the uv-plane bounded by u = 1, u = 4, v = 1, and v = 2. Thus the area of D is given by ZZ ZZ 1 du dv dx dy = 2v R D Z 4 Z 2 1 3 dv = = ln 2 sq. units. du 2 1 2 1 v

−2y = 2(x 2 + y 2 ), x 1 du dv. 2

(x 2 + y 2 ) d x d y = Hence, ZZ

33. Let u = x y, v = y/x. Then

so that

u=4

v=1

x+y=2 x

We have

R

u=1

35.

I =

ZZ

R

(x 2 + y 2 ) d x d y =

ZZ

S

1 1 du dv = . 2 2

e(y−x)/(y+x) d A. T

a) I =

Z

π/2

dθ

0

1 = 2

Z

Z

1/(cos θ +sin θ )

0 cos θ −sin θ e sin θ +cos θ

π/2

cos θ −sin θ

e sin θ +cos θ r dr

dθ (cos θ + sin θ )2 0 cos θ − sin θ Let u = sin θ + cos θ 2 dθ du = − (sin θ + cos θ )2 Z 1 1 u e − e−1 = e du = . 4 −1 4

y v

y y=2x

1 y=x

(−1,1) x+y=1

D

xy=4

1

(1,1)

T′

T 1

xy=1

x x

Fig. 14.4.35

Fig. 14.4.33

558 Copyright © 2014 Pearson Canada Inc.

u

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.4 (PAGE 834)

b) If u = y − x, v = y + x then

where S is the square 0 ≤ s ≤ x, 0 ≤ t ≤ x. By symmetry,

∂(u, v) −1 1 = −2, = 1 1 ∂(x, y)



1 du dv. Also, T corresponds 2 ′ to the triangle T bounded by u = −v, u = v, and v = 1. Thus so that d A = d x d y =

where T is the triangle 0 ≤ s ≤ x, 0 ≤ t ≤ s. t

(x,x)

ZZ 1 eu/v du dv 2 T′ Z Z v 1 1 dv eu/v du 2 0 −v Z
v 1 1 dv veu/v 2 0 −v Z 1 1 e − e−1 . e − e−1 v dv = 2 4 0

I = = = =

ZZ 2 8 2 2 Erf(x) = e−(s +t ) ds dt, π T

s=t

T x

s

Fig. 14.4.37 Now transform to polar coordinates in the st-plane. We have

36. The region R whose area we must find is shown in part (a) of the figure. The change of variables x = 3u, y = 2v maps the ellipse 4x 2 + 9y 2 = 36 to the circle u 2 + v 2 = 1, and the line 2x + 3y = 1 to the line u + v = 1. Thus it maps R to the region S in part (b) of the figure. Since 3 d x d y = | 0

the area of R is

A=

ZZ

R



0 | du dv = 6 du dv, 2

dx dy = 6

ZZ

Z Z x sec θ 2 8 π/4 2 Erf(x) = dθ e−r r dr π 0 0 Z   x sec θ 4 π/4 −r 2 dθ −e = π 0 0 Z π/4   4 −x 2 / cos2 θ dθ. = 1−e π 0

Since cos2 θ ≤ 1, we have e−x

du dv.



S

But the area of S is (π/4) − (1/2), so A = (3π/2) − 3 square units. v y (a) (b) 1

38.

S R 3

a) Ŵ(x) =

1

x

u b) Ŵ

2 π 

Z

0

x

2 2 e−t dt = √ π

2 4 Erf(x) = π

ZZ

Z

x

2

e−s ds. Thus

0

e−(s

2 +t 2 )

t x−1 e−t dt

=2

2

≤ e−x , so

∞

2

Let t = s 2 dt = 2s ds

s 2x−1 e−s ds.

0

Z

∞ 0

2

e−s ds = 2

√ √ π = π 2


1√ 1 Ŵ = Ŵ 12 = π. 2 2 Z 1 c) B(x, y) = t x−1 (1 − t) y−1 dt 3 2

Fig. 14.4.36

37. Erf(x) = √

2

Z

θ

2 2 Erf(x) ≥ 1 − e−x q 2 Erf(x) ≥ 1 − e−x .

0

=2
1

∞

Z

2 / cos2




ds dt,

S

(x > 0, y > 0)

0

let t = cos2 θ , dt = −2 sin θ cos θ dθ Z π/2 =2 cos2x−1 θ sin2y−1 θ dθ. 0

559 Copyright © 2014 Pearson Canada Inc.

SECTION 14.4 (PAGE 834)

ADAMS and ESSEX: CALCULUS 8

z

d) If Q is the first quadrant of the st-plane,

c

 Z ∞  Z ∞  2 2 Ŵ(x)Ŵ(y) = 2 s 2x−1 e−s ds 2 t 2y−1 e−t dt 0 0 ZZ 2x−1 2y−1 −(s 2 +t 2 ) =4 s t e ds dt

x y z a + b + c =1

Q

R

(change to polar coordinates) Z π/2 Z ∞ 2 =4 dθ r 2x−1 cos2x−1 θr 2y−1 sin2y−1 θ e−r r dr 0 0  Z π/2  = 2 cos2x−1 θ sin2y−1 θ dθ 0  Z ∞  2 × 2 r 2(x+y)−1 e−r dr

b x

Thus B(x, y) =

by (a) and (c).

Fig. 14.5.4

Section 14.5 Triple Integrals 1.

5.

Ŵ(x)Ŵ(y) . Ŵ(x + y)

(page 840)

R is symmetric about the coordinate planes and has volume 8abc. Thus ZZZ (1 + 2x − 3y) d V = volume of R + 0 − 0 = 8abc. R

2.

ZZZ

B

x yz d V =

Z

1

Z

0

Z

4

x dx y dy z dz −2  1 4 15 1 16 − 1 − =− . = 2 2 2 2 0



3. The hemispherical dome x 2 + y 2 + z 2 ≤ 4, z ≥ 0, is

symmetric about the planes x = 0 and y = 0. Therefore ZZZ

4.

ZZZ

R

D

(3 + 2x y) d V = 3

ZZZ

D

dV + 2

ZZZ

xy dV D

2 = 3 × π(23 ) + 0 = 16π. 3 Z

a

Z

x
b 1− a

Z

x y
c 1− a − b

dy dz 0
Z a Z b 1− x  a x y =c x dx 1− − dy a b 0 0  Z a   x 2 b 2  x 2 =c x b 1− − 1− dx a 2b a 0 Z a bc x 2 = 1− x d x Let u = 1 − (x/a) 2 0 a du = −(1/a) d x Z 1 2 a bc a 2 bc = u 2 (1 − u) du = . 2 24 0

x dV =

x dx

0

0

y

a

x

0

= B(x, y)Ŵ(x + y)

y

R is the cube 0 ≤ x, y, z ≤ 1. By symmetry, ZZZ ZZZ (x 2 + y 2 ) d V = 2 x2 dV R R Z 1 Z 1 Z =2 x2 dx dy 0

0

1

0

dz =

2 . 3

6. As in Exercise 5, ZZZ

R

(x 2 + y 2 + z 2 ) d V = 3

ZZZ

R

x2 dV =

3 = 1. 3

7. The set R: 0 ≤ z ≤ 1 − |x| − |y| is a pyramid, one

quarter of which lies in the first octant and is bounded by the coordinate planes and the plane x + y + z = 1. (See the figure.) By symmetry, the integral of x y over R is 0. Therefore, ZZZ ZZZ (x y + z 2 ) d V = z2 d V R R Z 1 Z 1−z Z 1−z−y =4 z 2 dz dy dx 0 0 0 Z 1 Z 1−z =4 z 2 dz (1 − z − y) d y 0 0  Z 1  1 =4 z 2 (1 − z)2 − (1 − z)2 dz 2 0 Z 1 1 =2 (z 2 − 2z 3 + z 4 ) dz = . 15 0

560 Copyright © 2014 Pearson Canada Inc.

z 1 z=1−x−y y+z=1 R 1 x

1

x+y=1

Fig. 14.5.7

y

INSTRUCTOR’S SOLUTIONS MANUAL

8.

SECTION 14.5 (PAGE 840)

z

R is the cube 0 ≤ x, y, z ≤ 1. We have ZZZ = =

1 (0,1,1)

yz 2 e−xyz d V R 1

Z

0

Z

1

Z

z dz

0

1

z dz

0

x=0

1

Z

(1,0,1)


x=1 d y −e−xyz

0

(1 − e

−yz

) dy !

y=1 1 z 1 + e−yz dz z 0 y=0 Z 1 1 = + (e−z − 1) dz 2 0 1 1 1 1 −z = −1−e = − . 2 2 e 0

=

9.

ZZZ

Z

1

sin(π y ) d V =

Z

=

Z

3

R

1

3

sin(π y ) d y

0

Z

1 y x (1,1,0)

Fig. 14.5.10

11.

y

dz

0

1

0

x+y+z=2

R y+z=1

Z

R is bounded by z = 1, z = 2, y = 0, y = z, x = 0, and x = y + z. These bounds provide an iteration of the triple integral without our having to draw a diagram. ZZZ

y

dx 0

dV + y + z)3 Z z Z dz dy

R (x Z 2

y+z

dx (x + y + z)3 1 0 0   x=y+z Z 2 Z z −1 = dz dy 2 2(x + y + z) 1 0 x=0 Z Z z 3 2 dy = dz 2 8 1 0 (y + z)  y=z Z 2 3 −1 = dz 8 1 y + z y=0 Z 2 3 dz 3 = ln 2. = 16 1 z 16

1 cos(π y 3 ) y 2 sin(π y 3 ) d y = − 3π 0

=

2 . = 3π z

(0,1,1) z=y (1,1,1)

R

12. We have

1 y x

x=y (1,1,0)

Fig. 14.5.9

10.

ZZZ

R

y dV = =

Z

Z

1

y dy 0

1

1−y 1

1

y dy 0

Z

Z

1−y

dz

Z

1



cos x cos y cos z d V ZRπ Z π −x Z π −x−y = cos x d x cos y d y cos z dz 0 0 0 z=π −x−y Z π Z π −x = cos x d x cos y d y (sin z) 0

2−y−z

dx 0

(2 − y − z) dz

 z=1 z 2 = y d y (2 − y)z − 2 z=1−y 0  Z 1  1 2 = y (2 − y)y − 1 − (1 − y) dy 2 0 Z 1   5 1 2y 2 − y 3 d y = . = 24 0 2 Z

ZZZ

=

Z

0

π

cos x d x

0

Z

z=0

π −x

0

cos y sin(x + y) d y

 1 recall that sin a cos b = sin(a + b) + sin(a − b) Z π Z π −x h 2 i 1 cos x d x = sin(x + 2y) + sin x d y 2 0 0   y=π −x Z 1 π cos(x + 2y) cos x d x − = + y sin x 2 0 2 y=0 Z  1 π cos x cos(2π − x) cos2 x = + − 2 0 2 2

561 Copyright © 2014 Pearson Canada Inc.

SECTION 14.5 (PAGE 840)

π

ADAMS and ESSEX: CALCULUS 8

z

 + (π − x) cos x sin x d x

Z

∞

Z

2

e−u du = −∞ √ √ k > 0, let u = kt, so that du = k dt. Thus

13. By Example 4 of Section 5.4,

Z

∞

−∞

2

e−kt dt =

r

(1,1,1)

y x (1,1,0)

Fig. 14.5.15

16.

√

z

z = y2

π. If

π . k x

3

ZR∞

e−x

2 −2y 2 −3y 2

x+y=1

dV

ZZZ

R

f (x, y, z) d V = =

14. Let E be the elliptic disk bounded by x 2 + 4y 2 = 4.

=

Then E has area π(2)(1) = 2π square units. The volume of the region of 3-space lying above E and beneath the plane z = 2 + x is V = since

15.

ZZZ

T

RR

E

ZZ

E

(2 + x) d A = 2

ZZ

d A = 4π cu. units,

E

x d A = 0 by symmetry.

x dV =

Z

1

x dx

0

Z

1

Z

1

1−x Z 1

dy

Z

Z

1

dx

0

Z

1

dy

0

Z

1

dy

0

=

Z

=

Z

=

Z

1

Z

Z

1

dz

1−y

0

dx

y2

Z

f (x, y, z) dz

0

0

1

√

z

y2

Z

f (x, y, z) dz

0 1−y

y2

Z

0

dy

0

dz 1

1−x

dx

Z

0

0

Z

Z dz f (x, y, z) d x 0 Z 1−y dy f (x, y, z) d x 0

(1−x)2

0 Z 1−√z 0

dz dx

Z

Z

1−x

√

f (x, y, z) d y

z 1−x

√

f (x, y, z) d y. z

1

dz

2−x−y

x dx (x + y − 1) d y 0 1−x   Z 1 1 (x − 1)2 +x − dx = x 2 2 0 Z 1 3 x 1 = dx = . 8 0 2 =

y

Fig. 14.5.16

Z ∞ Z ∞ 2 2 2 e−x d x e−2y d y e−3z dz −∞ −∞ −∞ r r √ π π π 3/2 = √ . = π 2 3 6 =

(0, 1, 1)

x = (1 − x)2

Thus ZZZ

(0,1,1)

(1,0,1)

π−x sin 2x d x 2 0 U = π − x d V = sin 2x d x cos 2x dU = −d x V =− 2 " # π Z π 1 1 π −x = − cos 2x − cos 2x d x 4 2 2 0 0 " π # π 1 sin 2x = . = π− 8 2 0 8 1 = 2

Z

17. = =

1

dz 0 ZZZ

Z

1−z

dy 0

Z

0

f (x, y, z) d x 0

f (x, y, z) d V (R is the prism in the figure) Z Z 1 1 1−y dx dy f (x, y, z) dz. R

Z

1

562 Copyright © 2014 Pearson Canada Inc.

0

0

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.5 (PAGE 840)

z 1

20.

(1,0,1)

y+z=1

=

R 1 x

y

=

1 (1,1,0)

1

Z

dy 0 ZZZ R

Z

1

dx

0

Z √1−y 2 0

18.

1

dz 0 ZZZ

Z

1

dy z

0

f (x, y, z) d x y 2 +z 2

x−y 2

x

f (x, y, z) dz.

dy

0

0

z

x=y 2 +z 2

y

Z

f (x, y, z) d x R

0

f (x, y, z) d V (R is the pyramid in the figure) R Z 1 Z 1 Z y = dx dy f (x, y, z) dz. =

1

f (x, y, z) d V (R is the paraboloid in the figure) Z √ Z √

Fig. 14.5.17

Z

Z

dz

x

1

0

z

y

x

Fig. 14.5.20 (0,1,1) z=y

21.

I =

Z

1

dz

0

Z

1−z

dy 0

Z

1

f (x, y, z) d x.

0

The given iteration corresponds to (1,1,1)

0 ≤ z ≤ 1, x=0

y

I =

y=1

y=x (1,1,0)

z=0

Fig. 14.5.18

0 ≤ x ≤ 1.

Thus 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − 0 = 1, 0 ≤ z ≤ 1 − y, and

R

x

0 ≤ y ≤ 1 − z,

22.

I =

Z

1

dz

0

Z

Z

1

dx 0

1

Z

dy

0

1

dy

Z

1−y

f (x, y, z) dz.

0

y

Z

f (x, y, z) d x.

0

z

The given iteration corresponds to Z

19.

1

dz 0 ZZZ

Z

1

dx

Z

x−z

0 ≤ z ≤ 1,

f (x, y, z) d y

0

z

0

0

0 ≤ x ≤ y.

Thus 0 ≤ x ≤ 1, x ≤ y ≤ 1, 0 ≤ z ≤ y, and

f (x, y, z) d V (R is the tetrahedron in the figure) R Z 1 Z x Z x−y = dx dy f (x, y, z) dz. =

z ≤ y ≤ 1,

Z

I =

1

dx

0

0

Z

1

dy x

Z

y

f (x, y, z) dz.

0

z

23.

(1,0,1)

I =

Z

1

dz

0

Z

1

dx z

Z

x−z

f (x, y, z) d y.

0

The given iteration corresponds to

z=x−y

0 ≤ z ≤ 1,

z ≤ x ≤ 1,

0 ≤ y ≤ x − z.

Thus 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ x − y, and

1 x y (1,1,0)

I =

Z

0

1

dx

Z

0

x

dy

Z

x−y

f (x, y, z) dz.

0

Fig. 14.5.19

563 Copyright © 2014 Pearson Canada Inc.

SECTION 14.5 (PAGE 840)

24.

I =

1

Z

dy

0

Z √1−y 2

Z

dz

0

ADAMS and ESSEX: CALCULUS 8

1

27.

f (x, y, z) d x. y 2 +z 2

The given iteration corresponds to 0 ≤ y ≤ 1,

q

0≤z≤

Thus 0 ≤ x ≤ 1, 0 ≤ y ≤ I =

25.

I =

1

Z

dy

0

1

Z

Z

dx

0

1

dz y

Z

√

√

x

dy 0

y 2 + z 2 ≤ x ≤ 1.

1 − y 2,

x, 0 ≤ z ≤

Z √x−y 2

p

x − y 2 , and

Z 1 Z x 3 dz dx ex d y 0 z 0 ZZZ 3 = ex d V (R is the pyramid in the figure) R Z 1 Z x Z x 3 = ex d x dy dz 0 0 0 Z 1 e−1 3 = x 2ex d x = . 3 0 Z

1

z

f (x, y, z) dz.

0

(1,0,1)

(1,1,1)

z

Z

z=x

y=0

f (x, y, z) d x.

0

The given iteration corresponds to 0 ≤ y ≤ 1,

y ≤ z ≤ 1,

(1,0,0) x

0 ≤ x ≤ z.

y z=0

Thus 0 ≤ x ≤ 1, x ≤ z ≤ 1, 0 ≤ y ≤ z, and 1

Z

I =

dx

0

Z

1

dz x

Z

(1,1,0)

Fig. 14.5.27

z

f (x, y, z) d y.

0

26.

y=x

R

x=1

28. z

(0, 1, 1) (1, 1, 1)

(0, 1, 0)

x

Z

1

dx 0 ZZZ

Z

1−x

dy 0

Z

1 y

sin(π z) dz z(2 − z)

sin(π z) dV (R is the pyramid in the figure) = R z(2 − z) Z 1 Z z Z 1−y sin(π z) = dz dy dx 0 z(2 − z) 0 0 Z z Z 1 sin(π z) dz (1 − y) d y = 0 0 z(2 − z)   Z 1 sin(π z) z2 = z− dz 2 0 z(2 − z) Z 1 1 1 = sin(π z) dz = . 2 0 π z

(0,0,1)

y

z=1 x=0 (0,1,1)

(1,0,1)

Fig. 14.5.26

y=0

I =

Z

0

1

dx

Z

1

dy x

Z

y

f (x, y, z) dz =

x

ZZZ

y

where P is the triangular pyramid (see the figure) with vertices at (0, 0, 0), (0, 1, 0), (0, 1, 1), and (1, 1, 1). If we we reiterate I to correspond to the horizontal slice shown then Z

1

dz 0

Z

1

dy z

Z

0

z=y

f (x, y, z) d V , P x

y=1−x

(1,0,0)

Fig. 14.5.28

29. The average value of f (x, y, z) over R is

z

f (x, y, z) d x. f¯ =

564 Copyright © 2014 Pearson Canada Inc.

1 volume of R

ZZZ

f (x, y, z) d V . R

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.6 (PAGE 847)

If f (x, y, z) = x 2 + y 2 + z 2 and R is the cube 0 ≤ x, y, z ≤ 1, then, by Exercise 6, 1 f¯ = 1

ZZZ

R

V =

bounded, connected set D in 3-space, then there exists a point (x0 , y0 , z 0 ) in D such that

D

f (x, y, z) d V = f (x0 , y0 , z 0 ) × (volume of D).

Apply this with D = Bǫ (a, b, c), which has volume 4 3 π ǫ , to get 3 ZZZ

Bǫ (a,b,c)

r2

(x 2 + y 2 + z 2 ) d V = 1.

30. If the function f (x, y, z) is continuous on a closed,

ZZZ

√ r intersects the sphere r 2 + z 2 = 2 where + r − 2 = 0. This equation has positive root r = 1. The required volume is

2. The surface z =

Z

2π

dθ

0

Z √2−r 2

1

Z

r dr

√ r

0

Z 1 p √  dθ 2 − r 2 − r r dr 0 0 Z 1 p  2 = 2π r 2 − r 2 dr − Let u = 2 − r 2 5 0 du = −2r dr Z 2 4π 1/2 =π u du − 5 1 √  4π 2π  √ 4 2π 22π = = − cu. units. 2 2−1 − 3 5 3 15 =

Z

2π

dz

z r 2 +z 2 =2

4 f (x, y, z) d V = f (x0 , y0 , z 0 ) π ǫ 3 3

√ z= r

for some (x0 , y0 , z 0 ) in Bǫ (a, b, c). Thus y

3 lim ǫ→0 4π ǫ 3

ZZZ

f (x, y, z) d V

x

Bǫ (a,b,c)

= lim f (x0 , y0 , z 0 ) = f (a, b, c) ǫ→0

Fig. 14.6.2 since f is continuous at (a, b, c).

3. The paraboloids z = 10 − r 2 and z = 2(r 2 − 1) intersect Section 14.6 Change of Variables in Triple Integrals (page 847)

where r 2 = 4, that is, where r = 2. The volume lying between these surfaces is V =

1.

V = =

Z

2π

0

2π a 3

dθ 3 

Z

π/4

Z

a

2π

2

dθ

0

= 2π

sin φ dφ R dR 0  1 1− √ cu. units. 2 0

Z

Z

0

Z

0

2

2

[10 − r 2 − 2(r 2 − 1)]r dr

(12r − 3r 3 ) dr = 24π cu. units. z z=10−r 2

z

2 R=a π/4

y z=2(r 2 −1)

x

y

Fig. 14.6.1

Fig. 14.6.3

565 Copyright © 2014 Pearson Canada Inc.

SECTION 14.6 (PAGE 847)

ADAMS and ESSEX: CALCULUS 8

4. The paraboloid z = r 2 intersects the sphere r 2√+ z 2 = 12 4 2 where r + r − 12 = 0, that is, where r = required volume is V =

Z

2π

dθ

0

= 2π =π =

Z

√

0

Z

9

12

3

Z

0

3. The

√ 3 p

 12 − r 2 − r 2 r dr

p 9π r 12 − r 2 dr − 2

Let u = 12 − r 2 du = −2r dr

9π 2

u 1/2 du −

6. The required volume V lies above z = 0, below

z = 1 − r 2 , and between θ = −π/4 and θ = π/3. Thus Z π/3 Z 1 V = dθ (1 − r 2 )r dr −π/4 0   7π 1 1 7π = − = cu. units. 12 2 4 48

7. Let R be the region in the first octant, inside the ellipsoid

 9π √ 2π  3/2 45π = 16 3π − cu. units. 12 − 27 − 3 2 2

x2 y2 z2 + 2 + 2 = 1, 2 a b c and between the planes y = 0 and y = x. Under the transformation

2

x = au, z=x 2 +y 2

y = bv,

z = cw,

R corresponds to the region S in the first octant of uvwspace, inside the sphere u 2 + v 2 + w2 = 1,

x 2 +y 2 +z 2 =12

Fig. 14.6.4

and between the planes v = 0 and bv = au. Therefore, the volume of R is

5. One half of the required volume V lies in the first octant, inside the cylinder with polar equation r = 2a sin θ . Thus Z π/2 Z 2a sin θ V =2 dθ (2a − r )r dr 0 0 Z Z π/2 2 π/2 3 3 8a sin θ dθ = 2a 4a 2 sin2 θ dθ − 3 0 0 Z Z π/2 16a 3 π/2 3 = 4a 3 sin θ dθ (1 − cos 2θ ) dθ − 3 0 0 32a 3 cu. units. = 2π a 3 − 9 z

V =

ZZZ

R

d x d y dz = abc

ZZZ

du dv dw. S

Using spherical coordinates in uvw-space, S corresponds to 0≤φ≤

0 ≤ R ≤ 1,

π , 2

0 ≤ θ ≤ tan−1

a . b

Thus V = abc =

Z

tan−1 (a/b)

0

dθ

Z

π/2

0

1 a abc tan−1 cu. units. 3 b

sin φ dφ

Z

1

R2 d R

0

z=2a−r

8. One eighth of the required volume V lies in the first octant. Call this region R. Under the transformation

r=2a sin θ

x = au,

2a y

x

Fig. 14.6.5

y = bv,

z = cw,

R corresponds to the region S in the first octant of uvwspace bounded by w = 0, w = 1, and u 2 + v 2 − w2 = 1. Thus V = 8abc × (volume of S).

566 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.6 (PAGE 847)

The volume of S can be determined by using horizontal slices: V = 8abc

Z

1

13.

z

14.

(x 2 + y 2 + z 2 ) d V

R Z 2π

Z tan−1 (1/c) Z a dθ sin φ dφ R4 d R 0 0 0      2π a 5 1 2π a 5 c = 1 − cos tan−1 = 1− √ . 5 c 5 c2 + 1 ZZZ (x 2 + y 2 ) d V =

π 8 (1 + w2 ) dw = π abc cu. units. 4 3

0

ZZZ

R 2π

=

b

a x

y

x 2 y 2 z2 + + =1 a 2 b2 c2

9. Let x = au, y = bv, z = w. The indicated region R cor-

responds to the region S above the uv-plane and below the surface w = 1 − u 2 − v 2 . We use polar coordinates in the uv-plane to calculate the volume V of R: V =

ZZZ

15.

0

2π a 5 5

Z

=

2π a 5 5

Z

10.

11.

12.

ZZZ

0 tan−1 (1/c)

0 1

√

c/

c2 +1

Z

sin3 φ dφ

a

R4 d R

0

sin φ(1 − cos2 φ) dφ

Let u = cos φ du = − sin φ dφ

(1 − u 2 ) du

  1 u 3 u− √ 3 c/ c2 +1   c 2π a 5 2 c3 −√ . = + 5 3 c2 + 1 3(c2 + 1)3/2 √ z = r 2 and z = 2 − r 2 intersect where r 4 + r 2 − 2 = 0, that is, on the cylinder r = 1. Thus 2π a 5 5

ZZZ

ZZZ

d V = ab du dv dw S Z 2π Z 1 π ab = ab dθ (1 − r 2 )r dr = cu. units. 2 0 0 R

tan−1 (1/c)

Z

dθ

=

=

Fig. 14.6.8

Z

R

z dV =

Z

2π

dθ

0

=π

Z

1

r dr

1 0

z dz

r2

0

Z

Z √2−r 2

(2 − r 2 − r 4 )r dr =

7π . 12

16. By symmetry, both integrals have the same value:

(x 2 + y 2 + z 2 ) d V R Z 2π Z a Z h = dθ r dr (r 2 + z 2 ) dz 0 0 0  Z a 1 r 3 h + r h 3 dr = 2π 3 0   4 a h a2 h 3 π a4 h π a2 h 3 = 2π + = + . 4 6 2 3 ZZZ (x 2 + y 2 ) d V B Z 2π Z π Z a = dθ sin φ dφ R 2 sin2 φ R 2 d R 0 Z 0 0 Z a π = 2π sin3 φ dφ R4 d R 0 0   5 4 a 8π a 5 = 2π = . 3 5 15 ZZZ (x 2 + y 2 + z 2 ) d V B Z 2π Z π Z a 4π a 5 . = dθ sin φ dφ R4 d R = 5 0 0 0

ZZZ

17.

ZZZ

R

x dV =

ZZZ

z dV Z a Z π/2 R3 d R cos φ sin φ dφ dθ = 0 0 0   π 1 a4 π a4 = = . 2 2 4 16 Z

Z

R π/2

π/2

Z

a

Z

h(1−(r/a))

r dr r cos θ dz R 0 0 0 Z π/2 Z a  ha 3 r =h dr = , cos θ dθ r2 1 − a 12 0 0 ZZZ Z π/2 Z a Z h(1−(r/a)) z dV = dθ r dr z dz R 0 0 0 Z   π h2 a r 2 = 1− r dr 4 0 a   a π h2 r 2 2r 3 r 4 π a2 h 2 = − + 2 = . 4 2 3a 4a 48 0 x dV =

dθ

18. If

x = au,

y = bv,

z = cw,

567 Copyright © 2014 Pearson Canada Inc.

SECTION 14.6 (PAGE 847)

ADAMS and ESSEX: CALCULUS 8

then the volume of a region R in x yz-space is abc times the volume of the corresponding region S in uvw-space. If R is the region inside the ellipsoid x2 y2 z2 + + =1 a2 b2 c2 and above the plane y + z = b, then the corresponding region S lies inside the sphere 2

2

2

u +v +w =1 and above the plane bv + cw = b. The distance from the origin to this plane is b D= √ b2 + c2

(assuming b > 0)

by Example 7 of Section 1.4. By symmetry, the volume of S is equal to the volume lying inside the sphere u 2 + v 2 + w2 = 1 and above the plane w = D. We calculate this latter volume by slicing; it is

π

Z

  1 w3 (1 − w2 ) dw = π w − 3 D D   2 D3 =π −D+ . 3 3 1

Hence, the volume of R is

π abc



2 b b3 −√ + 2 3 3(b + c2 )3/2 b2 + c2



(The θ coordinates are identical in the two systems.) Observe that z, r, R, and φ play, respectively, the same roles that x, y, r , and θ play in the transformation from Cartesian to polar coordinates in the plane. We can exploit this correspondence to avoid repeating the calculations of partial derivatives of a function u, since the results correspond to calculations made (for a function z) in Example 10 of Section 3.5. Comparing with the calculations in that Example, we have ∂u ∂u ∂u = cos φ + sin φ ∂R ∂z ∂r ∂u ∂u ∂u = −R sin φ + R cos φ ∂φ ∂z ∂r 2u 2 ∂ 2u ∂ ∂ 2u 2 2 ∂ u = cos φ + 2 cos φ sin φ + sin φ ∂ R2 ∂z 2 ∂z∂r ∂r 2  2u ∂u ∂ ∂2u = −R + R 2 sin2 φ 2 ∂φ 2 ∂R ∂z  2 ∂ u ∂2u − 2 cos φ sin φ + cos2 φ 2 . ∂z∂r ∂r Substituting these expressions into the expression for 1u given in the statement of this exercise in terms of spherical coordinates, we obtain the expression in terms of cylindrical coordinates established in the previous exercise: ∂ 2u + ∂ R2 ∂ 2u = 2 ∂r ∂ 2u = ∂x2

cot φ ∂u 1 ∂ 2u 2 ∂u 1 ∂ 2u + + + R ∂R R 2 ∂φ R 2 ∂φ 2 R 2 sin2 φ ∂θ 2 2 2 1 ∂u 1 ∂ u ∂ u + + 2 2 + 2 r ∂r r ∂θ ∂z ∂ 2u + 2 = 1u ∂y

cu. units. by Exercise 33.

21. Consider the transformation 19. By Example 10 of Section 3.5, we know that 1 ∂ 2u ∂ 2u ∂ 2u ∂2u 1 ∂u + 2 = 2 + + 2 2. 2 ∂x ∂y ∂r r ∂r r ∂θ

The required result follows if we add

∂ 2u to both sides. ∂z 2

x = x(u, v, w),

y = y(u, v, w),

z = z(u, v, w),

and let P be the point in x yz-space corresponding to u = a, v = b, w = c. Fixing v = b, w = c, results in a parametric curve (with parameter u) through P. The vector ∂y ∂z −→ ∂ x i+ j+ k PQ = ∂u ∂u ∂u and corresponding vectors

20. Cylindrical and spherical coordinates are related by z = R cos φ,

r = R sin φ.

568 Copyright © 2014 Pearson Canada Inc.

∂y ∂z −→ ∂ x PR = i+ j+ k ∂v ∂v ∂v ∂x ∂y ∂z − → PS = i+ j+ k ∂w ∂w ∂w

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.7 (PAGE 855)

span a parallelepiped in x yz-space corresponding to a rectangular box with volume du dv dw in uvw-space. The parallelepiped has volume

Thus

4.

∂(x, y, z) −→ −→ − → du dv dw. |( P Q × P R) • P S| = ∂(u, v, w) ∂(x, y, z) du dv dw. d V = d x d y dz = ∂(u, v, w)

q z = 2 1 − x 2 − y2 ∂z 2x 2y ∂z , = −p = −p 2 2 ∂x ∂ y 1−x − y 1 − x 2 − y2 s s 4(x 2 + y 2 ) 1 + 3(x 2 + y 2 ) dS = 1 + d A = dA 1 − x 2 − y2 1 − x 2 − y2 ZZ S= dS x 2 +y 2 ≤1 s Z 2π Z 1 1 + 3r 2 = dθ r dr Let u 2 = 1 − r 2 1 − r2 0 0 u du = −r dr Z 1p √ 4 − 3u 2 du Let 3u = 2 sin v = 2π √ 0 3 du = 2 cos v dv

Section 14.7 Applications of Multiple Integrals (page 855)

1.

Z

= 2π

4π = √ 3

π/3 0

Z

2 dv (2 cos2 v) √ 3

π/3

(1 + cos 2v) dv

0

 π/3  4π 2 sin 2v 4π = √ + π sq. units. = √ v+ 2 3 3 3 0

∂z ∂z z = 2x + 2y, =2= ∂x ∂y p d S = 1 + 22 + 22 d A = 3 d A ZZ S= 3 d A = 3π(12 ) = 3π sq. units. x 2 +y 2 ≤1

5. 2.

3 ∂z 4 ∂z = , = z = (3x − 4y)/5, ∂x 5 ∂y 5 s √ 32 + 42 dS = 1 + dA = 2dA 52 ZZ √ √ √ S= 2 d A = 2π(2)(1) = 2 2π sq. units.

∂z x ∂z y ∂z = 2x, = , = 3z 2 = x 2 + y 2 , 6z ∂x ∂x 3z ∂y 3z s s x 2 + y2 9z 2 + 3z 2 2 dS = 1 + d A = dA = √ dA 9z 2 9z 2 3 ZZ 2 24π 2 S= √ d A = √ π(12) = √ sq. units. 3 3 3 x 2 +y 2 ≤12

(x/2)2 +y 2 ≤1

3.

q z = a2 − x 2 − y 2 ∂z x y ∂z = −p = −p , 2 2 2 2 ∂x ∂ y a −x −y a − x 2 − y2 s x 2 + y2 a dS = 1 + 2 dA = p dA 2 a − x 2 − y2 a − x 2 − y2 ZZ adA p S= (use polars) 2 x 2 +y 2 ≤a 2 a − x 2 − y2 Z 2π Z a r dr =a dθ √ Let u = a 2 − r 2 a2 − r 2 0 0 du = −2r dr Z a2 = πa u −1/2 du = 2π a 2 sq. units. 0

6.

∂z ∂z z = 1 − x 2 − y 2, = −2x, = −2y ∂x ∂y q d S = 1 + 4x 2 + 4y 2 d A ZZ q S= 1 + 4(x 2 + y 2 ) d A x 2 +y 2 ≤1, x≥0, y≥0 Z 1p π/2

=

Z

=

π 16

0

π = 16

dθ

0

Z

5

1 + 4r 2 r dr

Let u = 1 + 4r 2 du = 8r dr

u 1/2 du

1



√  5 2 3/2 π(5 5 − 1) u sq. units. = 3 24 1 569

Copyright © 2014 Pearson Canada Inc.

SECTION 14.7 (PAGE 855)

ADAMS and ESSEX: CALCULUS 8

7. The triangle is defined by 0 ≤qy ≤ 1, 0 ≤ x ≤ y.

∂z z = y 2, = 2y, d S = 1 + 4y 2 d A ∂y Z 1 Z yq S= dy 1 + 4y 2 d x 0 0 Z 1 q = y 1 + 4y 2 d y Let u = 1 + 4y 2 0 du = 8y d y √  5  Z 5 5 5−1 1 2 3/2 1 1/2 u du = u sq. units. = = 8 1 8 3 12 1

8.

9.

r ∂z 1 1 = √ , dS = 1 + dA ∂x 4x 2 x √ r r Z 1 Z x Z 1 1 4x + 1 √ S= dx 1+ dy = x dx 4x 4x 0 0 0 Z 1 √ 1 = 4x + 1 d x Let u = 4x + 1 2 0 du = 4 d x √   5 Z 5 5−1 1 2 3/2 1 5 1/2 u du = u sq. units. = = 8 1 8 3 12 1 z=

√

x,

∂z ∂z x z 2 = 4 − x 2 , 2z = −2x, =− ∂ x ∂ x z s 2 x 2 2 dS = 1 + 2 d A = d A = √ dA z z 4 − x2 (since z ≥ 0 on the part of the surface whose area we want to find) Z 2 Z x 2 S= dx dy √ 4 − x2 0 0 Z 2 2x √ = d x Let u = 4 − x 2 4 − x2 0 du = −2x d x 4 Z 4 √ −1/2 = u du = 2 u = 4 sq. units. 0

0

p 1 + x 2 + y 2 d A. Oneeighth of the part of the surface above −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, lies above the triangle T : given by 0 ≤ x ≤ 1, 0 ≤ y ≤ x, or, in polar coordinates, by 0 ≤ θ ≤ π/4, 0 ≤ r ≤ 1/ cos θ = sec θ . Thus ZZ q S=8 1 + x 2 + y2 d A T Z π/4 Z sec θ p =8 dθ 1 + r 2 r dr Let u = 1 + r 2 0 0 du = 2r dr Z π/4 Z 1+sec2 θ √ =4 dθ u du 0 0 Z π/4 h i 8 (1 + sec2 θ )3/2 − 1 dθ = 3 0 Z 8 π/4 2π = (1 + sec2 θ )3/2 dθ − . 3 0 3

11. If z = 12 (x 2 + y 2 ), then d S =

Using a TI-85 numerical integration routine, we obtain the numerical value S ≈ 5.123 sq. units.

12. As the figure suggests, the area√of the canopy is the

area of a hemisphere of radius 2 minus four times the area of half of a spherical cap cut off from the sphere x 2 + y 2 + z 2 = 2 by a plane at distance 1 from the origin,psay the plane z = 1. Such a spherical cap, z = 2 − x 2 − y 2 , lies above the disk ∂z ∂z 2 2 x + y ≤ 2 − 1 = 1. Since = −x/z and = −y/z ∂x ∂y on it, the area of the spherical cap is s ZZ x 2 + y2 1+ dA z2 x 2 +y 2 ≤1 √ Z 1 r dr = 2 2π √ Let u = 2 − r 2 2 − r2 0 du = −2r dr √ √ √ √ Z 2 −1/2 u du = 2 2( 2 − 1) = 4 − 2 2. = 2π 1

Thus the area of the canopy is √ √ √ 1 S = 2π( 2)2 −4× ×(4−2 2) = 4(π + 2)−8 sq. units. 2 z

10. The area elements on z = 2x y and z = x 2 + y 2 , respectively, are

q q 1 + (2y)2 + (2x)2 d A = 1 + 4x 2 + 4y 2 d x d y, q q d S2 = 1 + (2x)2 + (2y)2 d A = 1 + 4x 2 + 4y 2 d x d y.

d S1 =

Since these elements are equal, the area of the parts of both surfaces defined over any region of the x y-plane will be equal.

570 Copyright © 2014 Pearson Canada Inc.

y

x Fig. 14.7.12

INSTRUCTOR’S SOLUTIONS MANUAL

13.

SECTION 14.7 (PAGE 855)

Z π Z a AR 2 d R dθ sin φ dφ 2 0 0 B + R Z a 0 B = 4π A 1− 2 dR R +B  0 √ a units. = 4π A a − B tan−1 √ B

Mass =

Z

2π

15. The force is

F = 2π kmρ

14. A slice of the ball at height z, having thickness dz, is a √ circular disk of radius a 2 − z 2 and areal density ρ dz. As calculated in the text, this disk attracts mass m at (0, 0, b) with vertical force b−z

d F = 2π kmρdz 1 − p a 2 − z 2 + (b − z)2

!

Z

h 0

b−z

1− p a 2 + (b − z)2

Z

a



b−z

z (0,0,b)

. h z



dz a 2 + b2 − 2bz let v = a 2 + b2 − 2bz, dv = −2b dz −a

1− √

dz

Let u = a 2 + (b − z)2 du = −2(b − z) dz ! Z 2 2 1 a +b du = 2π kmρ h − √ 2 a 2 +(b−h)2 u   p p = 2π kmρ h − a 2 + b2 + a 2 + (b − h)2 .

Thus the ball attracts m with vertical force F = 2π kmρ

!

a 2 + b2 − v b2 − a 2 + v then b − z = b − = 2b " # 2b Z (b+a)2 2 1 b − a2 + v = 2π kmρ 2a − 2 dv √ 4b (b−a)2 v   b2 − a 2  = 2π kmρ 2a − b + a − (b − a) 2 2b  1  3 3 − 2 (b + a) − (b − a) 6b 3 4π kmρa km M = = , 3b2 b2 where M = (4/3)π a 3 ρ is the mass of the ball. Thus the ball attracts the external mass m as though the ball were a point mass M located at its centre.

a x

y

Fig. 14.7.15

16. The force is Z

b

b−z 1− p 2 0 a (b − z)2 + (b − z)2  Z b 1 1− √ dz = 2π kmρ a 2 +1 0  1 = 2π kmρb 1 − √ . 2 a +1

F = 2π kmρ

z

!

dz

z (0,0,b) b r z=b− a

z

a

y

x ab x

y

Fig. 14.7.16

Fig. 14.7.14

571 Copyright © 2014 Pearson Canada Inc.

SECTION 14.7 (PAGE 855)

ADAMS and ESSEX: CALCULUS 8

z

17. The force is

1

Z

a



b−z 1− √ dz 2 a + b2 − 2bz 0 use the same substitution as in Exercise 2) ! Z a 2 +b2 2 b − a2 + v 1 √ dv = 2π kmρ a − 2 4b (b−a)2 v   b2 − a 2 p 2 = 2π kmρ a − a + b2 − (b − a) 2 2b  1  2 − 2 (a + b2 )3/2 − (b − a)3 6b  p 2π kmρ  3 3 2 2 2 + b2 . a 2b + a − (2b − a ) = 3b2

F = 2π kmρ

P

x

21.

a 2 −x 2 −y 2

y

x

Fig. 14.7.17

18.

Z a (x 2 + y 2 + z 2 ) dz dy 0 0 0 Z a Z a Z a =3 x2 dx dy dz = a 5 0 0 0 Z a Z a Z a = x dx dy (x 2 + y 2 + z 2 ) dz 0 0 0  Z a Z a a3 2 2 a(x + y ) + = x dx dy 3 0 0  Z a 4 2a 7a 6 = + a2 x 2 x d x = . 3 12 0

m=

M x=0

Z

a

dx

Z

a

Thus x¯ = M x=0 /m =

7a . 12

By symmetry, the centre of mass is



 7a 7a 7a , , . 12 12 12

19. Since the base triangle has centroid



 1 1 , , 0 , the cen3 3

troid of the prism is



 1 1 1 , , . 3 3 2

Fig. 14.7.19 Z ∞ 2π 2 dθ e−r r dr = π . By

0

0

symmetry, the moments about x = 0 and y = 0 are both zero. We have Z 2π Z ∞ Z e−r 2 Mz=0 = dθ r dr z dz 0 0 Z ∞ 0 π 2 =π r e−2r dr = . 4 0

(0,0,b)

√

Z

20. Volume of region =

z

z=

y

1

1

The centroid is (0, 0, 1/4).   π a3 1 4 3 πa = . By symmetry, the The volume is 8 3 6 moments about all three coordinate planes are equal. We have Z π/2 Z π/2 Z a Mz=0 = dθ sin φ dφ R cos φ R 2 d R 0 0 0 Z π a 4 π/2 π a4 = . sin φ cos φ dφ = 8 0 16 Thus z¯ = Mz=0 /volume = 3a/8.   3a 3a 3a The centroid is , , . 8 8 8 z 1

r=a 2 z

y x

Fig. 14.7.21

22. The cube has centroid (1/2, 1/2, 1/2). The tetrahedron lying above the plane x + y + x = 2 has centroid (3/4, 3/4, 3/4) and volume 1/6. Therefore the part of the cube lying below the plane has centroid (c, c, c) and volume 5/6, where 5 3 1 1 c + × = × 1. 6 4 6 2   9 9 9 Thus c = 9/20; the centroid is , , . 20 20 20

572 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.7 (PAGE 855)

z

27.

1

1 x

y

1

Fig. 14.7.22

23. The model still involves angular acceleration to spin the

24.

25.

26.

ball — it doesn’t just fall. Part of the gravitational potential energy goes to producing this spin as the ball falls, even in the limiting case where the fall is vertical. Z 2π Z a Z h I =ρ dθ r 3 dr dz 0 0 0  4 a πρha 4 = 2πρh . = 4 2 p a m = πρa 2 h, D¯ = I /m = √ . 2 Z 2π Z a Z h dθ r dr (x 2 + z 2 ) dz I =ρ 0 0 0  Z a Z 2π h3 hr 2 cos2 θ + dθ r dr =ρ 3 0 0  Z 2π  4 ha h 3 a2 dθ =ρ cos2 θ + 4 6 0    2  π ha 4 π h 3a2 h2 a 2 =ρ + = πρa h + 4 3 4 3 s p h2 a2 + . m = πρa 2 h, D¯ = I /m = 4 3 Z 2π Z a Z h(1−(r/a)) I =ρ dθ r 3 dr dz 0 0 0 Z a   r πρa 4 h r3 1 − = 2πρh dr = , a 10 0 r p πρa 2 h 3 m= , D¯ = I /m = a. 3 10

Z h(1−(r/a)) r dr (x 2 + z 2 ) dz 0 0 0 Z 2π Z a  r 2 =ρ dθ h 1− r cos2 θ a 0 0  h3  r 3 + 1− r dr 3 a  Z Z a 2πρh 3 a  r 3 r4 dr + r 1− dr = πρh r3 − a 3 h 0 0 in the second integral put u = 1 − (r/a) Z πρa 4 h 2πρa 2 h 3 1 = + (1 − u)u 3 du 20 3 0 πρa 4 h 2πρa 2 h 3 πρa 2 h = + = (3a 2 + 2h 2 ), 20 60 60 s

I =ρ

m=

28.

Z

2π

dθ

Z

πρa 2 h , 3

I =ρ

ZZZ

= 2ρ

Z

Q a

D¯ =

m = ρa 3 ,

p

I /m =

3a 2 + 2h 2 . 20

(x 2 + y 2 ) d V a

a

2ρa 5 , dz = 3 0 0 r p 2 D¯ = I /m = a. 3

x2 dx

0

a

Z

dy

Z

z

a a a

y x

Fig. 14.7.28

z

h z r h + a =1

29. The distance s from (x, y, z) to the line x = y, z = 0

satisfies s 2 = u 2 + z 2 , where u is the distance from (x, y, 0) to the line x = y in √ the x y-plane. By Example 7 of Section 1.4 u = |x − y|/ 2, so

x

a

a

y

s2 =

Fig. 14.7.26

(x − y)2 + z2. 2

573 Copyright © 2014 Pearson Canada Inc.

SECTION 14.7 (PAGE 855)

ADAMS and ESSEX: CALCULUS 8

The moment of inertia of the cube about this line is  Z a Z a Z a (x − y)2 I =ρ dx dy + z 2 dz 2 0 0 0  Z a Z a a3 a =ρ dx (x − y)2 + d y Let u = x − y 2 3 0 0 du = −d y Z Z x ρa a ρa 5 2 + dx u du = 3 2 0 x−a Z ρa 5 ρa a = + (3ax 2 − 3a 2 x + a 3 ) d x 3 6 0   ρa 5 ρa 5ρa 5 3a 4 = + + a4 = , a4 − 3 6 2 12 r p 5 m = ρa 3 , D¯ = I /m = a. 12

32.

I =ρ

Z

2π

Z

dθ 0

c

dz

0

m = πρc(b2 − a 2 ),

b

πρc(b4 − a 4 ) r 3 dr = , 2 a s b2 + a 2 D¯ = . 2

Z

z

c

r=a r=b

y

x

Fig. 14.7.32

30. The line L through the origin parallel to the vector v = i + j + k is a diagonal of the cube Q. By Example 8 of Section 1.4, the distance from the point with position vector r = xi + yj + zk to L is s = |v × r|/|v|. Thus, the square of the distance from (x, y, z) to L is (x − y)2 + (y − z)2 + (z − x)2 3  2 2 2 2 x + y + z − x y − x z − yz . = 3

s2 =

We have ZZZ

Q

ZZZ

Q

33.

m = 2ρ

Z

2π

dθ 0

Z

a

x2 dV = xy dV =

Q

ZZZ

Q

y2 d V = yz d V =

ZZZ

Q

ZZZ

z2 d V =

3

a5 xz dV = . 4 Q

Therefore, the moment of inertia of Q about L is ! 2ρ a5 a5 ρa 5 I = 3× −3× = . 3 3 4 6

b

I = 2ρ

0

31.

p

Z

a

b

a

= 4πρ = 2πρ

Z

a 2 −b2

b

a 2 −r 2

r 3 dr

dz

0

p r 3 a 2 − r 2 dr

Let u = a 2 − r 2 du = −2r dr

√ (a 2 − u) u du

 0 2 2 2 2 = 2πρ a (a − b2 )3/2 − (a 2 − b2 )5/2 3 5 1 2 2 3/2 1 2 = 4πρ(a − b ) (2a + 3b2 ) = m(2a 2 + 3b2 ). 15 5 z

a I /m = √ . 6

Z b Z c dx dy (x 2 + y 2 ) dz −c −a −b  Z a 2b3 = 2ρc 2bx 2 + dx 3 −a 8ρabc 2 = (a + b2 ), 3 s

I =ρ

dz

0

p r a 2 − r 2 dr

Z

The mass of Q is m = ρa 3 , so the radius of gyration is D¯ =

r dr

Z √a 2 −r 2

Let u = a 2 − r 2 du = −2r dr Z a 2 −b2 √ 4πρ 2 (a − b2 )3/2 , = 2πρ u du = 3 0 Z Z Z √ = 4πρ

a5

a

b

2π

ZZZ

Z

a

m = 8ρabc,

D¯ =

p

I /m =

b

a y

x

a 2 + b2 . 3

574 Copyright © 2014 Pearson Canada Inc.

Fig. 14.7.33

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 14.7 (PAGE 855)

34. By Exercise 26, the cylinder has moment of inertia

36. The kinetic energy of the oscillating pendulum is

πρa 4 h ma 2 = , 2 2

I =

KE =

where m is its mass. Following the method of Example 4(b), the kinetic energy of the cylinder rolling down the inclined plane with speed v is 1 2 mv + 2 1 = mv 2 + 2

KE =

1 2 I 2 1 2 v2 3 ma 2 = mv 2 . 4 a 4

The potential energy of the cylinder when it is at height h is mgh, so, by conservation of energy,

−

1 I 2

I

2

.



dθ dt

2

− mga cos θ = constant.



dθ dt



d 2θ + mga sin θ dt 2



dθ dt



= 0,

or d 2θ mga + sin θ = 0. dt 2 I For small oscillations we have sin θ ≈ θ , and the above equation is approximated by d 2θ + ω2 θ = 0, dt 2

35. By Exercise 35, the ball with hole has moment of inertia

where ω2 = mga/I . The period of oscillation is

m (2a 2 + 3b2 ) 5

2π = 2π T = ω

about the axis of the hole. The kinetic energy of the rolling ball is 1 2 m v2 mv + (2a 2 + 3b2 ) 2 2 10 a   2a 2 + 3b2 7a 2 + 3b2 2 1 = mv + = mv 2 . 2 2 10a 10a 2

KE =

s

I . mga

A θ

a C

By conservation of energy, mv 2

dθ dt

Differentiating with respect to time t, we obtain

dv 2 = g sin α. dt 3

I =



The potential energy is mgh, where h is the distance of C above A. In this case, h = −a cos θ . By conservation of energy,

3 2 mv + mgh = constant. 4 Differentiating this equation with respect to time t, we obtain 3 dv dh 0 = mv + mg 2 dt dt 3 dv = mv + mgv sin α. 2 dt Thus the cylinder rolls down the plane with acceleration

1 I 2

7a 2 + 3b2 + mgh = constant. 10a 2

Differentiating with respect to time, we obtain Fig. 14.7.36

7a 2 + 3b2 dv mv + mgv sin α = 0. dt 5a 2 Thus the ball rolls down the plane (with its hole remaining horizontal) with acceleration −

37. If the centre of mass of B is at the origin, then

dv 5a 2 = 2 g sin α. dt 7a + 3b2

M x=0 =

ZZZ

B

xρ d V = 0.

575 Copyright © 2014 Pearson Canada Inc.

SECTION 14.7 (PAGE 855)

ADAMS and ESSEX: CALCULUS 8

If line L 0 is the z-axis, and L k is the line x = k, y = 0, then the moment of inertia Ik of B about L k is ZZZ   Ik = (x − k)2 + y 2 ρ d V ZZZ B = (x 2 + y 2 + k 2 − 2kx) ρ d V

Review Exercises 14 (page 857)

1. By symmetry, ZZ

B

= I0 + k 2 m − 2k M x=0 = I0 + k 2 m,

R

(x + y) d A = 2 =2

where m is the mass of B and I0 is the moment about L 0. z

=2

L0 Lk

ZZ

R 1

Z

0



y

x dA = 2

Z

1

x dx 0

Z

√

dy x2

(x 3/2 − x 3 ) d x

  1  2 1 2 5/2 x 4 3 = 2 = x − − 5 4 0 5 4 10

B

x

(1, 1)

√ y= x x = y2

k y

R

y = x2

Fig. 14.7.37

x

38. The moment of inertia of the ball about the point where it contacts the plane is, by Example 4(b) and Exercise 39,   4 8 πρa 5 + πρa 3 a 2 I = 15 3   2 7 2 = + 1 ma = ma 2 . 5 5

Fig. R-14.1

2.

=

The kinetic energy of the ball, regarded as rotating about the point of contact with the plane, is therefore KE =

ZZ

(x 2 + y 2 ) d A =

P

Z

0

Z

dy

+ x y 2 3 x=y

39. By Example 7 of Section 1.4, the distance from the point

Z

1  (2

Z

2+y y

(x 2 + y 2 ) d x

dy

 + y)3 y3 + y 2 (2 + y) − − y3 d y 3 3 0  Z 1 8 8 4 = + 4y + 4y 2 d y = + 2 + = 6 3 3 3 0 y

=

1 2 7 v2 7 I = ma 2 2 = mv 2 . 2 10 a 10

with position vector r = xi + yj + zk to the straight line L through the origin parallel to the vector a = Ai + Bj + Ck is |a × r| s= . |a|

The moment of inertia of the body occupying region R about L is, therefore, ZZZ 1 |a × r|2 ρ d V I = 2 |a| R ZZZ h 1 = 2 (Bz − C y)2 + (C x − Az)2 A + B2 + C 2 R i + (Ay − Bx)2 ρ d V h 1 = 2 (B 2 + C 2 )Pxx + (A2 + C 2 )Pyy A + B2 + C 2 i + (A2 + B 2 )Pzz − 2 AB Pxy − 2 AC Pxz − 2BC Pyz .

1

0  x=2+y

1  x3

(1, 1) y=x

P 2 Fig. R-14.2

3.

ZZ

D

y dA = x

576 Copyright © 2014 Pearson Canada Inc.

Z

0

π/4

dθ

Z

0

(3, 1) x =2+y x

2

tan θ r dr

0 π/4 2 2 r

= ln sec θ

x

= 2 ln 2 0

√ 2 = ln 2

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 14 (PAGE 857)

y

we need to ensure that   k 2 1− √ = 1. k2 + 1

√ Thus k 2 + 1 = (2k)2 , and so 3k 2 = 1, and k = 1/ 3.

y=x S

4.

a) I = =

√

Z

Z √4−y 2

3

dy

0 ZZ

e−x

√ y/ 3

2 −y 2

e−x

2 −y 2

a

x

2 Fig. R-14.3

dA

Fig. R-14.5

R

6.

I = =

√

y=

2

Z

dy

0 ZZ

y

Z

dx

0

+ c) I =

Z

√ 3x

e−x Z √

2 −y 2

I =

dx

0

e

2

Z

−x 2 −y 2

dy

(2, 2)

x Fig. R-14.6

7.

φ0

Z

a

R2 d R   2π a 3 2π a 3 k = (1 − cos φ0 ) = 1− √ . 3 3 k2 + 1 sin φ dφ

0

0

To have 1 4



4 3 πa 3



=

J=

Z

1

dz

0

Z

0

z

dy

Z

y

f (x, y, z) d x

0

corresponds to the region

p

V =

f (x, y) d y. x

2

e−r r dr

φ0 = Thus the volume inside the cone and inside the sphere x 2 + y 2 + z 2 = a 2 is dθ

6−x 2

y=x

! 2 −4 = π(1 − e ) − 2 6 0

Z

f (x, y) d x

0

2

y = 6 − x2

0

2π

Z

R

2 e−r

tan−1 (1/k).

0

0

y 6

x

5. The cone z = k x 2 + y 2 has semi-vertical angle

V =

dx

0

1

dθ

π d) I = 3

Z

2

Z

dy

4−x 2

2

π/3

dy

Z √6−y

f (x, y) d A,

3x

0

Z

6

R

r =2

Z

Z

where R is as shown in the figure. Thus

1 2 Fig. R-14.4 1

f (x, y) d x +

0

R

b) I =

z = kr

dx

where R is as shown in the figure. y

Z

φ0

0 ≤ z ≤ 1,

0 ≤ y ≤ z,

0 ≤ x ≤ y,

which can also be expressed in the form 0 ≤ x ≤ 1, x ≤ y ≤ 1, y ≤ z ≤ 1. Z 1 Z 1 Z 1 Thus J = dx dy f (x, y, z) dz. 0

x

y

8. A horizontal slice of the object at height z above the base, and having thickness dz, is a disk of radius r = 12 (10 − z) m. Its volume is

π a3 , 3

dV = π

(10 − z)2 dz m3 . 4

577 Copyright © 2014 Pearson Canada Inc.

REVIEW EXERCISES 14 (PAGE 857)

ADAMS and ESSEX: CALCULUS 8

The density of the slice is ρ = kz 2 kg/m3 . Since ρ = 3, 000 when z = 10, we have k = 30. a) The mass of the object is 10

π 30z 2 (10 − z)2 dz 4 0 11. Z 10 15π 2 3 4 = (100z − 20z + z ) dz 2 0   15π 100, 000 = − 50, 000 + 20, 000 ≈ 78, 540 kg. 2 3

m=

Z

b) The moment of inertia (about its central axis) of the disk-shaped slice at height z is d I = 30z 2 dz

Z

2π

dθ

0

Z

(10−z)/2

I =

9.

f (t) =

Z

a

10

30z 2 dz

0

Z

0

2π

dθ 0

Z

(10−z)/2

r 3 dr.

0

2

e−x d x

t

Z Z Z a 1 a 1 a 2 f¯ = f (t) dt = dt e−x d x a 0 a 0 t Z Z x Z 1 a −x 2 1 a −x 2 e dt = xe = dx dx a 0 a 0 0 ! a 2 2 1 − e−a 1 e−x = = − a 2 2a

10. If f (x, y) = ⌊x + y⌋, then f = 0, 1, or 2, in parts of the quarter disk Q, as shown in the figure. y

3

2 f =2 f =1 f =0 1

=

2a

dθ

0

= 2π

Z

√ 2a

0

r 3 dr

6a 2 −r 2

dz

r 2 /a

0

"

p r5 r 3 6a 2 − r 2 − a

#

dr

Let u = 6a 2 − r 2 du = −2r dr Z 6a 2 √ π √ 6 =π (6a 2 − u) u du − ( 2a) 3a 4a 2  6a 2  8 2 = π 4a 2 u 3/2 − u 5/2 − π a 5 5 3 2 4a √ 8π = (18 6 − 41)a 5 15

12. The solid S lies above the region in the x y-plane

0

1

The sphere x 2 + y 2 + z 2 = 6a 2 and the paraboloid z = (x 2 + y 2)/a intersect where z 2 + az − 6a 2 = 0, that is, where (z + 3a)(z − 2a) = 0. Only z = 2a is possible; the plane z = −3a does not intersect the sphere. If z = 2a, 2 then x 2 + y 2 = r 2 = 6a 2 − 4a 2 = 2a √ , so the intersection is on the vertical cylinder of radius 2a with axis on the z-axis. We have, ZZZ (x 2 + y 2) d V D Z Z √ Z √ 2π

r 3 dr.

Thus the moment of inertia about the whole solid cone is Z

Thus     ZZ 3 5 1 +1 + 2 (π − 2) = 2π − , f (x, y) d A = 0 2 2 2 Q   1 5 5 and f¯ = 2π − =2− . π 2 2π

Q

bounded by the circle x 2 + y 2 = 2ay, which has polar equation r = 2a sin θ , (0 ≤ θ ≤ π ). It lies below the conep z = x 2 + y 2 = r . The moment of inertia of S about the z-axis is ZZZ Z π Z 2a sin θ Z r I = (x 2 + y 2 ) d V = dθ r 3 dr dz S 0 0 0 Z π Z 2a sin θ Z 32a 5 π 5 = dθ r 4 dr = sin θ dθ 5 0 0 0 Z 32a 5 π (1 − cos2 θ )2 sin θ dθ Let u = cos θ = 5 0 du = − sin θ dθ Z 32a 5 1 = (1 − 2u 2 + u 4 ) du 5 −1   64a 5 2 1 512a 5 = 1− + = . 5 3 5 75

13. A horizontal slice of D at height z is a right triangle 2

Fig. R-14.10

3

x

with legs (2 − z)/2 and 2 − z. Thus the volume of D is Z 1 1 7 V = (2 − z)2 dz = . 4 0 12

578 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 14 (PAGE 857)

z Its moment about z = 0 is Mz=0

(0, 0, 1)

Z 1 1 = z(2 − z)2 dz 4 0 Z 1 1 11 = (4z − 4z 2 + z 3 ) dz = . 4 0 48

y+z =1 S

(0, 1, 0) 2

The z-coordinate of the centroid of D is z¯ =

11 48



x

7 11 = . 12 28

(2, 0, 0)

15. (0, 0, 1)

y+z =2 (0, 1, 1)

2x + z = 2 (0, 0, 0) (1, 0, 0) x

x = 2 − y − 2z Fig. R-14.14

z

( 12 , 0, 1)

y

(1, 1, 0)

(0, 2, 0)

2x + y + z = 2

y

Z 1 Z 1+z Z 1+z−y z dV = z dz dy dx S 0 0 0 Z 1 Z 1+z = z dz (1 + z − y) d y 0 0  Z 1  (1 + z)2 = z (1 + z)2 − dz 2 0 Z 1 1 17 = (z + 2z 2 + z 3 ) dz = 2 0 24 z

ZZZ

(0, 0, 1)

Fig. R-14.13

(0, 2, 1)

14.

Z 1 Z 1−y Z 2−y−2z dV = dy dz dx S 0 0 0 Z 1 Z 1−y (2 − y − 2z) dz dy = 0 0 Z 1 = [(2 − y)(1 − y) − (1 − y)2 ] d y 0 Z 1 1 = (1 − y) d y = 2 0 ZZZ Z 1 Z 1−y Z 2−y−2z M x=0 = x dV = dy dz x dx S 0 0 0 Z Z 1−y 1 1 = dy [(2 − y)2 − 4(2 − y)z + 4z 2 ] dz 2 0 0 Z  1 1 = (2 − y)2 (1 − y) − 2(2 − y)(1 − y)2 2 0  4 + (1 − y)3 d y Let u = 1 − y 3 du = −d y  Z 1 1 4 = (u + 1)2 u − 2(u + 1)u 2 + u 3 du 2 0 3  Z  1 1 1 3 7 = u + u du = 2 0 3 24  7 1 7 x¯ = = 24 2 12 V =

ZZZ

(2, 0, 1)

S

y =1+z (0, 1, 0)

(1, 0, 0)

y x +y−z =1

x

Fig. R-14.15

16. The plane z = 2x intersects the paraboloid z = x 2 + y 2

on the circular cylinder x 2 + y 2 = 2x, (that is, 2 + y 2 = 1), which has radius 1. Since (x − 1)√ √ d S = 1 + 22 d A = 5 d A on the plane, the area of the part of the plane inside √ the paraboloid (and therefore inside the cylinder) √ is 5 times the area of a circle of radius 1, that is, 5π square units.

17. As noted in the previous exercise, the part of the paraboloid z = x 2 + y 2 that lies below the plane z = 2x is inside the vertical cylinder x 2 + y 2 = 2x, which has polar equation r = 2 cos θ (−π/2 ≤ θ ≤ π/2). On the paraboloid: dS =

q p 1 + (2x)2 + (2y)2 d A = 1 + 4r 2 r dr dθ. 579

Copyright © 2014 Pearson Canada Inc.

REVIEW EXERCISES 14 (PAGE 857)

ADAMS and ESSEX: CALCULUS 8

to the region S inside the sphere u 2 + v 2 + w2 = 1 and above the plane u + v + w = 1. The distance from the 1 origin to this plane is √ , so, by symmetry, the volume 3 of S is equal to the √ volume inside the sphere and above the plane w = 1/ 3, that is,

The area of that part of the paraboloid is Z

π/2

S= =

1 8

Z

dθ

−π/2

Z

2 cos θ

0

π/2

dθ

−π/2 Z π/2 2

Z

p 1 + 4r 2 r dr

1+16 cos2 θ

Let u = 1 + 4r 2 du = 8r dr

u 1/2 du

 1 w3 √ π(1 − w ) dw = π w − 3 1√3 1/ 3 √ 2π(9 − 4 3) cu. units. = 27

1

Z

1 [(1 + 16 cos2 θ )3/2 − 1] dθ 4 0 3 Z 1 π/2 = [(1 + 16 cos2 θ )3/2 − 1] dθ 6 0 ≈ 7.904 sq. units. =

(using a TI-85 numerical integration function).

y2 z2 x2 + + = 1 36 9 4 and above the plane x + y + z = 1 is transformed by the change of variables

18. The region R inside the ellipsoid

x = 6u,

y = 3v,

z = 2w

to the region S inside the sphere u 2 + v 2 + w2 = 1 and above the plane 6u + 3v + 2w = 1. The distance from the origin to this plane is D= √

1 62

+ 32

+ 22

=

1 , 7

so, by symmetry, the volume of S is equal to the volume inside the sphere and above the plane w = 1/7, that is,  1  180π w3 = units3 . π(1 − w2 ) dw = π w − 3 1/7 343 1/7

Z

1

Since |∂(x, y, z)/∂(u, v, w)| = 6 · 3 · 2 = 18, the volume of R is 18 × (180π/343) = 3240π/343 ≈ 29.68 cu. units.

Challenging Problems 14

(page 858)

1. This problem is similar to Review Exercise 18 above. x2 y2 z2 The region R inside the ellipsoid 2 + 2 + 2 = 1 and a b c x y z above the plane + + = 1 is transformed by the a b c change of variables x = au,

y = bv,

z = cw

2.

1

2



Since |∂(x, √ y, z)/∂(u, v, w)| = abc, the volume of R is 2π(9 − 4 3) abc cu. units. 27 The plane (x/a) + (y/b) + (z/c) = 1 intersects the ellipsoid (x/a)2 + (y/b)2 + (z/c)2 = 1 above the region R in the x y-plane bounded by the ellipse x2 y2  x y 2 + + 1 − − = 1, a2 b2 a b

or, equivalently,

x y x2 y2 xy − − = 0. + + a2 b2 ab a b Thus the area of the part of the plane lying inside the ellipsoid is s ZZ c2 c2 S= 1 + 2 + 2 dx dy a b √R 2 2 2 2 a b + a c + b2 c2 (area of R). = ab Under the transformation x = a(u + v), y = b(u − v), R corresponds to the ellipse in the uv-plane bounded by (u + v)2 + (u − v)2 + (u 2 − v 2 ) − (u + v) − (u − v) = 0 3u 2 + v 2 − 2u = 0   2 1 1 3 u2 − u + + v2 = 3 9 3 (u − 1/3)2 v2 + = 1, 1/9 1/3

√ √ an ellipse with area π(1/3)(1/ 3) = π/(3 3) sq. units. Since a a | du dv = 2ab du dv, d x d y = | b −b we have

2π p S = √ a 2 b2 + a 2 c2 + b2 c2 sq. units. 3 3

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INSTRUCTOR’S SOLUTIONS MANUAL

3.

a) Z

0

1Z 1 0

CHALLENGING PROBLEMS 14 (PAGE 858)

∞ X 1 = 1 + x y + (x y)2 + · · · = (x y)n−1 1 − xy n=1 Z 1 ∞ Z 1 X dx dy = x n−1 d x y n−1 d y 1 − xy 0 n=1 0

=

∞ X 1 . 2 n n=1

Remark: The series for 1/(1 − x y) converges for |x y| < 1.R Therefore the outer integral is improper (i.e., c limc→1− 0 d x). We cannot do a detailed analysis of the P convergence here, but the convergence of 1/n 2 shows that the iterated double integral must converge. b) Similarly, ∞ X 1 = 1 − x y + (x y)2 − · · · = (−x y)n−1 1 + xy n=1 Z 1Z 1 dx dy 0 0 1 + xy Z 1 Z 1 ∞ X = (−1)n−1 x n−1 d x y n−1 d y

= Z

n=1 ∞ X

0

1 3 n n=1 Z 1Z 1Z =

1

y n−1 d y

0

=

n=1

1

dx π/4

dθ

0

Z

0

x

0

0

Z

Z

π/4

dθ

Z

Z

q 4 − x 2 − y2 d y sec θ

0

4

p 4 − r 2 r dr

Let u = 4 − r 2 du = −2r dr

u 1/2 du

4−sec2 θ

Z i 16 π/4 h 8 − (4 − sec2 θ )3/2 dθ 3 0 Z 16 π/4 (4 cos2 θ − 1)3/2 32π − dθ. = 3 3 0 cos3 θ

Z

1

z n−1 dz

Now the volume of the whole ball is (4π/3)23 = 32π/3, so the volume remaining after the hole is cut is

0

0

32π − V0 3 Z π/4 16 (3 − 4 sin2 θ )3/2 = cos θ dθ 3 0 (1 − sin2 θ )2 Z √ 16 1/ 2 (3 − 4v 2 )3/2 = dv. 3 0 (1 − v 2 )2

V =

1

0

Z

=

dx dy 0 0 0 1 + x yz Z Z 1 ∞ X n−1 n−1 = (−1) x dx n=1 ∞ X

=8

n2

n=1 0 ∞ X

part in the first octant, which is itself split into two equal parts by the plane x = y:

= 16

(−1)n−1

dx dy 1 − x yz 0 0 0 Z ∞ Z 1 X = x n−1 d x

5. The volume V0 removed from the ball is eight times the

V0 = 16

0

n=1 1Z 1Z 1

and we have ZZZ (a • r)(b • r)(c • r) d x d y dz P ZZZ uvw du dv dw = R |a • (b × c)| Z d1 Z d2 Z d3 1 = u du v dv w dw |a • (b × c)| 0 0 0 d12 d22 d32 = . 8|a • (b × c)|

1

y

n−1

dy

Z

1

z

n−1

dz

0

(−1)n−1 . n3

Let v = sin θ dv = cos θ dθ

We submitted this last integral to Mathematica to obtain

4. Under the transformation u = a • r, v = b • r, w = c • r,

where r = xi + yj + zk, the parallelepiped P corresponds to the rectangle R specified by 0 ≤ u ≤ d1 , 0 ≤ v ≤ d2 , 0 ≤ w ≤ d3 . If a = a1 i+a2 j+a3 k and similar expressions hold for b and c, then ∂(u, v, w) a1 a2 a3 = b1 b2 b3 = a • (b × c). ∂(x, y, z) c1 c2 c3

Therefore

∂(x, y, z) du dv dw = du dv dw , d x d y dz = ∂(u, v, w) |a • (b × c)|

r  4 2 −1 V = 32sin − 23/2 + 11tan−1 (3 − 23/2 ) 3 3  3/2 −1 − 11tan (3 + 2 ) ≈ 18.9349.

6. Under the transformation x = u 3 , y = v 3 ,

z = w3 , the region R bounded by the surface x 2/3 + y 2/3 + z 2/3 = a 2/3 gets mapped to the ball B bounded by u 2 + v 2 + w2 = a 2/3 . Assume that a > 0. Since ∂(x, y, z) = 27u 2 v 2 w2 , ∂(u, v, w)

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CHALLENGING PROBLEMS 14 (PAGE 858)

the volume of R is V = 27

ADAMS and ESSEX: CALCULUS 8

the required volume is ZZZ

u 2 v 2 w2 du dv dw.

V = 8(63 )

B

Now switch to spherical coordinates [R, φ, θ ] in uvwspace. Since

ZZZ

u 5 v 5 w5 du dv dw. B

Now switch to spherical coordinates [R, φ, θ ] in uvwspace. Since

uvw = (R sin φ cos θ )(R sin φ sin θ )(R cos φ), uvw = (R sin φ cos θ )(R sin φ sin θ )(R cos φ),

we have Z π Z a 1/3 cos2 θ sin2 θ dθ sin5 φ cos2 φ dφ R8 d R 0 0 0 Z 2π Z π sin2 (2θ ) = 3a 3 dθ (1 − cos2 φ)2 cos2 φ sin φ dφ 4 0 0 Let t = cos φ, dt = − sin φ dφ Z 2π Z 1 1 − cos(4θ ) = 3a 3 dθ (1 − t 2 )2 t 2 dt 8 0 −1 Z 1 3a 3 4π a 3 = (2π )2 (t 2 − 2t 4 + t 6 ) dt = cu. units. 8 35 0

V = 27

Z

2π

7. One-eighth of the required volume lies in the first octant. Under the transformation x = u 6 , y = v 6 , z = w6 , the region first-octant R bounded by the surface x 1/3 + y 1/3 + z 1/3 = a 1/3 and the coordinate planes gets mapped to the first octant part B of the ball bounded by u 2 + v 2 + w2 ≤ a 1/3 . Assume that a > 0. Since ∂(x, y, z) = 63 u 5 v 5 w5 , ∂(u, v, w)

we have

V = 1, 728

Z

π/2

(cos θ sin θ )5 dθ

0

×

Z

a 1/6

Z

π/2

(sin2 φ cos φ)5 sin φ dφ

0

R 17 d R

0

π/2

Z π/2 sin5 (2θ ) dθ sin11 φ(1 − sin2 φ)2 cos φ dφ 32 0 0 Let s = sin φ, ds = cos φ dφ Z π/2 Z 1 = 3a 3 (1 − cos2 (2θ ))2 sin(2θ ) dθ s 11 (1 − s 2 )2 ds = 96a 3

Z

0

0

Let t = cos(2θ ), dt = −2 sin(2θ ) dθ Z Z 1 3a 3 1 (1 − 2t 2 + t 4 ) dt (s 11 − 2s 13 + s 15 ) ds = 2 −1 0    2 1 1 1 a3 1 3 = 3a 1 − + − + = cu. units. 3 5 12 7 16 210

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.1 (PAGE 865)

CHAPTER 15. VECTOR FIELDS

4. F = i + sin xj.

dy . sin x dy Thus = sin x. The field lines are the curves dx y = − cos x + C. The field lines satisfy d x =

Section 15.1 Vector and Scalar Fields (page 865)

y

1. F = xi + xj.

dx dy = , i.e., d y = d x. The field x x lines are y = x + C, straight lines parallel to y = x.

The field lines satisfy

y

x

x

Fig. 15.1.4

5. F = e x i + e−x j.

Fig. 15.1.1

The field lines satisfy

2. F = xi + yj.

dx dy The field lines satisfy = . x y Thus ln y = ln x + ln C, or y = C x. The field lines are straight half-lines emanating from the origin.

dx dy = −x . ex e

dy = e−2x . The field lines are the curves dx 1 y = − e−2x + C. 2

Thus

y

y

x x

Fig. 15.1.5 Fig. 15.1.2

3. F = yi + xj.

dx dy = . The field lines satisfy y x Thus x d x = y d y. The field lines are the rectangular hyperbolas (and their asymptotes) given by x 2 − y 2 = C.

6. F = ∇(x 2 − y) = 2xi − j. The field lines satisfy 1 y = − ln x + C. 2

y

dy dx = . They are the curves 2x −1 y

x

x

Fig. 15.1.3

Fig. 15.1.6

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SECTION 15.1 (PAGE 865)

ADAMS and ESSEX: CALCULUS 8

2xi + 2yj . x 2 + y2 dx dy The field lines satisfy = . Thus they are radial x y lines y = C x (and x = 0)

7. F = ∇ ln(x 2 + y 2 ) =

y

x + C2 . The streamlines C1 are the spirals in which the surfaces x = C1 sin(z − C2 ) intersect the cylinders x 2 + y 2 = C12 . This implies that z = sin−1

12. v =

xi + yj . (1 + z 2 )(x 2 + y 2 )

dy dx = . Thus x y z = C1 and y = C2 x. The streamlines are horizontal half-lines emanating from the z-axis. The streamlines satisfy dz = 0 and

x

13. v = x zi + yzj + xk. The field lines satisfy dx dy dz = = , xz yz x Fig. 15.1.7

or, equivalently, d x/x = d y/y and d x = z dz. Thus the field lines have equations y = C1 x, 2x = z 2 + C2 , and are therefore parabolas.

8. F = cos yi − cos xj.

dx dy =− , that is, cos y cos x cos x d x + cos y d y = 0. Thus they are the curves sin x + sin y = C. The field lines satisfy

14. v = e xyz (xi + y 2 j + zk). The field lines satisfy dx dy dz = 2 = , x y z

y

so they are given by z = C1 x, ln |x| = ln |C2 | − (1/y) (or, equivalently, x = C2 e−1/y ).

15. v(x, y) = x 2 i − yj. The field lines sat-

x

isfy d x/x 2 = −d y/y, so they are given by ln |y| = (1/x) + ln |C|, or y = Ce1/x .

16. v(x, y) = xi + (x + y)j. The field lines satisfy dy dx = x x+y dy x+y = dx x

Fig. 15.1.8

9. v(x, y, z) = yi − yj − yk.

Let y = xv(x) dv dy =v +x dx dx dv x(1 + v) v +x = = 1 + v. dx x

The streamlines satisfy d x = −d y = −dz. Thus y + x = C1 , z + x = C2 . The streamlines are straight lines parallel to i − j − k.

10. v(x, y, z) = xi + yj − xk.

dy dz dx = = − . Thus x y x z + x = C1 , y = C2 x. The streamlines are straight halflines emanating from the z-axis and perpendicular to the vector i + k. The streamlines satisfy

11. v(x, y, z) = yi − xj + k.

dz 1 1 = = q . dx y 2 C1 − x 2

ˆ The field lines satisfy dr = dθ , so they are 17. F = rˆ + r θ. the spirals r = θ + C.

ˆ The field lines satisfy dr = r dθ/θ , or 18. F = rˆ + θ θ.

dx dy =− = dz. Thus y x 2 2 2 x d x + y d y = 0, so x + y = C1 . Therefore,

The streamlines satisfy

Thus dv/d x = 1/x, and so v(x) = ln |x| + C. The field lines have equations y = x ln |x| + C x.

dr/r = dθ/θ , so they are the spirals r = Cθ .

ˆ The field lines satisfy dr/2 = r dθ/θ , or 19. F = 2ˆr + θ θ. dr/r = 2dθ/θ , so they are the spirals r = Cθ 2 .

ˆ The field lines satisfy dr/r = −r dθ , or 20. F = r rˆ − θ.

−dr/r 2 = dθ , so they are the spirals 1/r = θ + C, or r = 1/(θ + C).

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.2 (PAGE 874)

= x 2 − x y + y 2 , then
2 V (x, y) = x − 21 y + 34 y 2 > 0 except at (0, 0) and

21. If V (x, y)

  dV = (2x − y)y + (2y − x) −x + y(1 − x 2 ) dt = −y 2 + x 2 + 2y 2 (1 − x 2 ) − x y(1 − x 2 ).

If x 2 < 1/2, then 1 − x 2 > 1/2 and |x y(1 − x 2 )| ≤ |x y|. Therefore,

Therefore, F may be conservative. If F = ∇ φ, then ∂φ = x, ∂x

22. For µ = 0 the corresponding vector field is F = yi − xj.

x2 3z 2 − y2 + is a potential for F. 2 2 3 Thus F is conservative on R .

2. F = yi + xj + z 2 k, F1 = y, F2 = x, F3 = z 2 . We have ∂ F2 ∂ F1 =1= , ∂y ∂x ∂ F1 ∂ F3 =0= , ∂z ∂x ∂ F3 ∂ F2 =0= . ∂z ∂y Therefore, F may be conservative. If F = ∇ φ, then ∂φ = y, ∂x

For V (x, y) = x 2 + y 2 we have

dV = ∇ V • F = 2x y − 2x y = 0. dt

φ(x, y, z) =

Section 15.2 Conservative Fields (page 874) 1. F = xi − 2yj + 3zk, F1 = x, F2 = −2y, F3 = 3z. We have

∂ F2 ∂ F1 =0= , ∂y ∂x ∂ F1 ∂ F3 =0= , ∂z ∂x ∂ F2 ∂ F3 =0= . ∂z ∂y

Z

y d x = x y + C1 (y, z)

∂φ ∂C1 ∂C1 =x+ ⇒ =0 ∂y ∂y ∂y C1 (y, z) = C2 (z), φ(x, y, z) = x y + C2 (z) z2 =

x . x2 − 1

V ′ = 2rr ′ = 2y 2 (x 2 − 1) ≤ 0 when |x| < 1, thus r (t) is a decreasing function except at critical points where y = 0, which is the nulcline for x ′ = 0. Since r(t) decreases to the critical point and continues to decrease after passing the critical point, the concavity has to be opposite in sign on opposite sides of the critical point. Thus the critical points of r (t) are also points of inflection, so they do not halt the asymptotic decline in distance to (0,0). Therefore the fixed point is asymptotically stable.

∂φ = z 2. ∂z

x=

23. The second nulcline corresponds to y ′ = 0, that is 24.

∂φ = x, ∂y

Therefore,

All trajectories are everywhere tangent to the level curves of V . Thus the trajectories are circles about (0, 0) and the fixed point is weakly stable, but not asymptotically stable. For µ = 0, the Van der Pol equation is just the equation of simple harmonic motion. −x + y(x 2 − 1) = 0, or y =

∂φ = 3z. ∂z

Evidently φ(x, y, z) =

  dV |y| 2 3y 2 + > x 2 + y 2 − |x y| = |x| − >0 dt 2 4 provided (x, y) 6= (0, 0). This implies that the trajectories of F cross all the level curves of V that are sufficiently close to the origin (they are ellipses) in an outward direction, so the origin is an unstable fixed point of F and therefore of the Van der Pol equation with µ = 1.

∂φ = −2y, ∂y

∂φ z3 = C2′ (z) ⇒ C2 (z) = . ∂z 3

Thus φ(x, y, z) = x y + conservative on R3 .

3. F =

z3 is a potential for F, and F is 3

xi − yj x y , F1 = 2 , F2 = − 2 . We have x 2 + y2 x + y2 x + y2 ∂ F1 2x y =− 2 , ∂y (x + y 2 )2

2x y ∂ F2 = 2 . ∂x (x + y 2 )2

Thus F cannot be conservative.

4. F =

xi + yj x y , F1 = 2 , F2 = 2 . We have x 2 + y2 x + y2 x + y2 ∂ F1 2x y ∂ F2 =− 2 = . 2 2 ∂y (x + y ) ∂x

Therefore, F may be conservative. If F = ∇ φ, then ∂φ x = 2 , ∂x x + y2

∂φ y = 2 . ∂y x + y2

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SECTION 15.2 (PAGE 874)

ADAMS and ESSEX: CALCULUS 8

Therefore, x ln(x 2 + y 2 ) d x = + C1 (y) x 2 + y2 2 y ∂φ y = = 2 + c1′ (y) ⇒ c1′ (y) = 0. ∂y x 2 + y2 x + y2 φ(x, y) =

Z

7.

Thus we can choose C1 (y) = 0, and φ(x, y) =

1 ln(x 2 + y 2 ) 2

is a scalar potential for F, and F is conservative everywhere on R2 except at the origin.

Since similar formulas hold for the other first partials of φ, we have

5. F = (2x y − z 2 )i + (2yz + x 2 )j − (2zx − y 2 )k,

F = ∇φ

F1 = 2x y − z 2 , F2 = 2yz + x 2 , F3 = y 2 − 2zx. We have

h i 2 (x − x )i + (y − y )j + (z − z )k 0 0 0 |r − r0 |4 r − r0 . = −2 |r − r0 |4

=−

∂ F1 ∂ F2 = 2x = , ∂y ∂x ∂ F1 ∂ F3 = −2z = , ∂z ∂x ∂ F3 ∂ F2 = 2y = . ∂z ∂y

This is the vector field whose scalar potential is φ.

Therefore, F may be conservative. If F = ∇ φ, then ∂φ = 2x y − z 2 , ∂x ∂φ = y 2 − 2zx. ∂z

∂φ = 2yz + x 2 , ∂y

φ(x, y, z) =

8.

∂r ∂ 1 r • ∂x x ln |r| = = 2 ∂x |r| |r| |r| r xi + yj + zk = 2. ∇ ln |r| = 2 |r| |r|

2y x 2 + y2 2x i+ j− k, z z z2 2y x 2 + y2 2x , F2 = , F3 = − F1 = . We have z z z2

9. F =

Therefore, Z

Thus F cannot be conservative. 1 φ(r) = |r − r0 |2 ∂φ 2 ∂ =− |r − r0 | 3 ∂x |r − r0 | ∂ x ∂r (r − r0 ) • 2 ∂ x =− |r − r0 |3 |r − r0 | 2(x − x0 ) =− . |r − r0 |4

∂ F2 ∂ F1 =0= , ∂y ∂x ∂ F1 2x ∂ F3 =− 2 = , ∂z z ∂x 2y ∂ F2 ∂ F3 =− 2 = . ∂z z ∂y

(2x y − z 2 ) d x = x 2 y − x z 2 + C1 (y, z)

∂φ ∂C1 = x2 + ∂y ∂y ∂C1 ⇒ = 2yz ⇒ C1 (y, z) = y 2 z + C2 (z) ∂y φ(x, y, z) = x 2 y − x z 2 + y 2 z + C2 (z) ∂φ = −2x z + y 2 + C2′ (z) y 2 − 2zx = ∂z ⇒ C2′ (z) = 0. 2yz + x 2 =

Thus φ(x, y, z) = x 2 y − x z 2 + y 2 z is a scalar potential for F, and F is conservative on R3 .

6. F = e

x 2 +y 2 +z 2

(x zi + yzj + x yk). 2 2 2 , F2 = yze x +y +z , F1 = x ze 2 2 2 F3 = x ye x +y +z . We have

Therefore, F may be conservative in R3 except on the plane z = 0 where it is not defined. If F = ∇φ, then ∂φ 2x = , ∂x z

∂ F1 ∂ F2 2 2 2 = 2x yze x +y +z = , ∂y ∂x ∂ F1 2 2 2 = (x + 2x z 2 )e x +y +z , ∂z ∂ F3 ∂ F1 2 2 2 = (y + 2x 2 y)e x +y +z 6= . ∂x ∂z

∂φ x 2 + y2 =− . ∂z z2

Therefore, x2 2x dx = + C1 (y, z) z z 2y ∂φ ∂C1 y2 = = ⇒ C1 (y, z) = + C2 (z) z ∂y ∂y z x 2 + y2 φ(x, y, z) = + C2 (z) z x 2 + y2 ∂φ x 2 + y2 − = =− + C2′ (z) 2 z ∂z z2 ⇒ C2 (z) = 0. φ(x, y, z) =

x 2 +y 2 +z 2

∂φ 2y = , ∂y z

586 Copyright © 2014 Pearson Canada Inc.

Z

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.2 (PAGE 874)

x 2 + y2 is a potential for F, and F is z conservative on R3 except on the plane z = 0. Thus φ(x, y, z) =

The equipotential surfaces have equations x2

+ z

y2

= C,

or

C z = x 2 + y 2.

Thus the equipotential surfaces are circular paraboloids. The field lines of F satisfy dy dz dx = = . 2 2x 2y x + y2 − z z z2

dz z2 − x 2 − y2 z 2 − (1 + A2 )x 2 = = . dx 2x z 2zx This first order DE is of homogeneous type (see Section 9.2), and can be solved by a change of dependent variable: z = xv(x). We have dv dz x 2 v 2 − (1 + A2 )x 2 = = dx dx 2x 2 v dv v 2 − (1 + A2 ) v 2 + (1 + A2 ) x = −v =− dx 2v 2v 2v dv dx =− v 2 + (1 + A2 ) x   2 2 ln v + (1 + A ) = − ln x + ln B

v+x

dy dx = , so y = Ax for an x y arbitrary constant A. Therefore From the first equation,

z dz z dz dx = = , 2x −(x 2 + y 2 ) −x 2 (1 + A2 ) so −(1 + A2 )x d x = 2z dz. Hence 1 + A2 2 B x + z2 = , 2 2 or x 2 + y 2 + 2z 2 = B, where B is a second arbitrary constant. The field lines of F are the ellipses in which the vertical planes containing the z-axis intersect the ellipsoids x 2 + y 2 + 2z 2 = B. These ellipses are orthogonal to all the equipotential surfaces of F. 2y x 2 + y2 2x i+ j− k = G + k, z z z2 where G is the vector field F of Exercise 9. Since G is conservative (except on the plane z = 0), so is F, which has scalar potential

10. F =

φ(x, y, z) =

As in Exercise 9, the first equation has solutions y = Ax, representing vertical planes containing the z-axis. The remaining equations can then be written in the form

x 2 + y2 x 2 + y2 + z2 +z = , z z

x 2 + y2 is a potential for G and z is a potential for z the vector k.

B x z2 B 2 +1+ A = x2 x z 2 + x 2 + y 2 = Bx.

v 2 + 1 + A2 =

These are spheres centred on the x-axis and passing through the origin. The field lines are the intersections of the planes y = Ax with these spheres, so they are vertical circles passing through the origin and having centres in the x y-plane. (The technique used to find these circles excludes those circles with centres on the y-axis, but they are also field lines of F.) Note: In two dimensions, circles passing through the origin and having centres on the x-axis intersect perpendicularly circles passing through the origin and having centres on the y-axis. Thus the nature of the field lines of F can be determined geometrically from the nature of the equipotential surfaces.

since

or

The equipotential surfaces of F are φ(x, y, z) = C, x 2 + y2 + z2 = C z

which are spheres tangent to the x y-plane having centres on the z-axis. The field lines of F satisfy dx dy dz = = . 2x 2y x 2 + y2 1− z z z2

11. The scalar potential for the two-source system is φ(x, y, z) = φ(r) = −

m m − . |r − ℓk| |r + ℓk|

Hence the velocity field is given by v(r) = ∇φ(r) m(r − ℓk) m(r + ℓk) = + |r − ℓk|3 |r + ℓk|3 m(xi + yj + (z − ℓ)k) m(xi + yj + (z + ℓ)k) = 2 + 2 . [x + y 2 + (z − ℓ)2 ]3/2 [x + y 2 + (z − ℓ)2 ]3/2

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SECTION 15.2 (PAGE 874)

ADAMS and ESSEX: CALCULUS 8

Observe that v 1 = 0 if and only if x = 0, and v 2 = 0 if and only if y = 0. Also v(0, 0, z) = m



z−ℓ z+ℓ + |z − ℓ|3 |z + ℓ|3



k,

which is 0 if and only if z = 0. Thus v = 0 only at the origin. At points in the x y-plane we have v(x, y, 0) =

2m(xi + yj) . + y 2 + ℓ2 )3/2

(x 2

The velocity is radially away from the origin in the x y-plane, as is appropriate by symmetry. The speed at (x, y, 0) is v(x, y, 0) = where s =

p

where K = 22/3 /(22/3 −1). This latter equation represents a sphere, S, since K 2 − K > 0. The velocity is vertical at all points on S, as well as at all points on the z-axis. Since the source at the origin is twice as strong as the sink at (0, 0, 1), only half the fluid it emits will be sucked into the sink. By symmetry, this half will the half emitted into the half-space z > 0. The rest of the fluid emitted at the origin will flow outward to infinity. There is one point where v = 0.√This point (which is easily calculated to be (0, 0, 2 + 2)) lies inside S. Streamlines emerging from the origin parallel to the x y-plane lead to this point. Streamlines emerging into z > 0 cross S and approach the sink. Streamlines emerging into z < 0 flow to infinity. Some of these cross S twice, some others are tangent to S, some do not intersect S anywhere. z

p 2m x 2 + y 2 2ms = 2 = g(s), (x 2 + y 2 + ℓ2 )3/2 (s + ℓ2 )3/2

x 2 + y 2 . For maximum g(s) we set

3 (s 2 + ℓ2 )3/2 − s(s 2 + ℓ2 )1/2 2s 2 0 = g (s) = 2m (s 2 + ℓ2 )3 2 2 2m(ℓ − 2s ) = . (s 2 + ℓ2 )5/2 ′

Thus, the speed in the x y-plane is greatest at points of the circle x 2 + y 2 = ℓ2 /2.

x

12. The scalar potential for the source-sink system is φ(x, y, z) = φ(r) = −

2 1 + . |r| |r − k|

Thus, the velocity field is 2r r−k v = ∇φ = 3 − |r| |r − k|3 2(xi + yj + zk) xi + yj + (z − 1)k = 2 − 2 . 2 2 3/2 (x + y + z ) (x + y 2 + (z − 1)2 )3/2

Fig. 15.2.12

13. Fluid emitted by interval 1z in time interval [0, t] occupies, at time t, a cylinder of radius r , where πr 2 1Z = vol. of cylinder = 2π mt1z. dr = m. The surface of this dt cylinder is moving away from the z-axis at rate Thus r 2 = 2mt, and r

dr m m = = p , 2 dt r x + y2

For vertical velocity we require 2x x = 2 , (x 2 + y 2 + z 2 )3/2 (x + y 2 + (z − 1)2 )3/2 and a similar equation for y. Both equations will be satisfied at all points of the z-axis, and also wherever  3/2  3/2 2 x 2 + y 2 + (z − 1)2 = x 2 + y2 + z2   22/3 x 2 + y 2 + (z − 1)2 = x 2 + y 2 + z 2

so the velocity at any point (x, y, z) is m × unit vector in direction xi + yj v= p 2 x + y2 m(xi + yj) = . x 2 + y2

14. For v(x, y) =

x 2 + y 2 + (z − K )2 = K 2 − K ,

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m(xi + yj) , we have x 2 + y2 ∂v 1 2mx y ∂v 2 =− 2 = , ∂y (x + y 2 )2 ∂x

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.2 (PAGE 874)

17. All circles tangent to the y-axis at the origin intersect all

so v may be conservative, except at (0, 0). We have

circles tangent to the x-axis at the origin at right angles, so they must be the streamlines of the two-dimensional dipole.

Z

x dx m φ(x, y) = m = ln(x 2 + y 2 ) + C1 (y) x 2 + y2 2 ∂φ my dC1 my = = 2 + . 2 2 2 x +y ∂y x +y dy

As an alternative derivation of this fact, the streamlines must satisfy dy dx = 2 , 2x y y − x2

Thus we may take C1 (y) = 0, and obtain φ(x, y) =

or, equivalently,

m ln(x 2 + y 2 ) = m ln |r|, 2

dy y2 − x 2 = . dx 2x y

as a scalar potential for the velocity field v of a line source of strength of m.

15. The two-dimensional dipole of strength µ has potential

This homogeneous DE can be solved (as was that in Exercise 10) by a change in dependent variable. Let y = xv(x). Then dv dy v2x2 − x2 = = dx dx 2v x 2 2 dv v −1 v2 + 1 x = −v = − dx 2v 2v 2v dv dx =− v2 + 1 x 2 ln(v + 1) = − ln x + ln C

φ(x, y) "

2 !

  m ℓ ℓ ln x 2 + y − − ln x 2 + y + ℓ→0 2 2 2 mℓ=µ   !   ! ℓ 2 ℓ 2 2 2 ln x + y − − ln x + y + 2 2 µ lim = 2 ℓ→0 ℓ (apply l’Hˆopital’s Rule)     ℓ ℓ − y− y+ µ 2 2 = lim  2 −   2 ℓ→0 ℓ ℓ 2 x2 + y − x2 + y + 2 2 µy µy =− 2 =− 2. x + y2 r = lim

v +x

2 !#

C ⇒ x 2 2 x + y = Cx

v2 + 1 =

y2 C +1= x2 x

(x − C)2 + y 2 = C 2 . These streamlines are circles tangent to the y-axis at the origin.

18. The velocity field for a point source of strength m dt at (0, 0, t) is

Now 2µy ∂r 2µx y ∂φ = 3 = ∂x r ∂x r4 y 2 r − 2yr 2 2 ∂φ r = µ(y − x ) . = −µ 4 4 ∂y r r Thus   µ 2 2 F = ∇φ = 2 2x yi + (y − x )j . (x + y 2 )2

16. The equipotential curves for the two-dimensional dipole have equations y = 0 or µy 1 = x 2 + y2 C x 2 + y 2 + µC y = 0   µC 2 µ2 C 2 x2 + y + = . 2 4 −

These equipotentials are circles tangent to the x-axis at the origin.

  m xi + yj + (z − t)k vt (x, y, z) =  3/2 . x 2 + y 2 + (z − t)2

Hence we have Z ∞ vt (x, y, z) dt −∞ Z ∞ xi + yj + (z − t)k =m  3/2 dt −∞ x 2 + y 2 + (z − t)2 Z ∞ dt = m(xi + yj)  3/2 −∞ x 2 + y 2 + (z − t)2 p Let z − t = x 2 + y 2 tan θ p −dt = x 2 + y 2 sec2 θ dθ Z m(xi + yj) π/2 = cos θ dθ x 2 + y 2 −π/2 2m(xi + yj) = , x 2 + y2

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SECTION 15.2 (PAGE 874)

ADAMS and ESSEX: CALCULUS 8

which is the velocity field of a line source of strength 2m along the z-axis. The definition of strength of a point source in 3-space was made to ensure that the velocity field of a source of strength 1 had speed 1 at distance 1 from the source. This corresponds to fluid being emitted from the source at a volume rate of 4π . Similarly, the definition of strength of a line source guaranteed that a source of strength 1 gives rise to fluid speed of 1 at unit distance 1 from the line source. This corresponds to a fluid emission at a volume rate 2π per unit length along the line. Thus, the integral of a 3-dimensional source gives twice the volume rate of a 2-dimensional source, per unit length along the line. The potential of a point source m dt at (0, 0, t) is φ(x, y, z) = − p

m x 2 + y 2 + (x − t)2

.

20. If F = Fr (r, θ )ˆr + Fθ (r, θ )θˆ is conservative, then F = ∇ φ for some scalar field φ(r, θ ), and by Exercise 19, ∂φ = Fr , ∂r

For the equality of the mixed second partial derivatives of φ, we require that ∂ Fr ∂ ∂ Fθ = (r Fθ ) = Fθ + r , ∂θ ∂r ∂r that is,

∂ Fθ ∂ Fr −r = Fθ . ∂θ ∂r

21. If F = r sin(2θ )ˆr + r cos(2θ )θˆ = ∇φ(r, θ ), then we must have

∂φ 1 ∂φ = r sin(2θ ), = r cos(2θ ). ∂r r ∂θ Both of these equations are satisfied by

This potential cannot be integrated to give the potential for a line source along the z-axis because the integral −m

Z

∞ −∞

φ(r θ ) =

1 2 r sin(2θ ) + C, 2

so F is conservative and this φ is a potential for it.

dt p

1 ∂φ = Fθ . r ∂θ

x 2 + y 2 + (z − t)2

does not converge, in the usual sense in which convergence of improper integrals was defined.

19. Since x = r cos θ and y = r sin θ , we have ∂φ ∂φ ∂φ = cos θ + sin θ ∂r ∂x ∂y ∂φ ∂φ ∂φ = −r sin θ + r cos θ . ∂θ ∂x ∂y Also, xi + yj = (cos θ )i + (sin θ )j r −yi + xj = −(sin θ )i + (cos θ )j. θˆ = r

rˆ =

Therefore,

22. If F = r 2 cos θ rˆ + αr β sin θ θˆ = ∇ φ(r, θ ), then we must have

∂φ = r 2 cos θ, ∂r From the first equation φ(r, θ ) =

r3 cos θ + C(θ ). 3

The second equation then gives C ′ (θ ) −

r3 ∂φ sin θ = = αr β+1 sin θ. 3 ∂θ

This equation can be solved for a function C(θ ) independent of r only if α = −1/3 and β = 2. In this case, C(θ ) = C (a constant). F is conservative if α and β have these values, and a potential for it is φ = 31 r 3 cos θ + C.

Section 15.3

1 ∂φ ˆ ∂φ rˆ + θ ∂r  r ∂θ  ∂φ ∂φ = cos2 θ + sin θ cos θ i ∂x ∂y   ∂φ 2 ∂φ + cos θ sin θ j + sin θ ∂x ∂y   ∂φ ∂φ + sin2 θ − sin θ cos θ i ∂x ∂y   ∂φ ∂φ + − cos θ sin θ + cos2 θ j ∂x ∂y ∂φ ∂φ = i+ j = ∇φ. ∂x ∂y

1 ∂φ = αr β sin θ. r ∂θ

Line Integrals

1. C is given by r = ati + btj + ctk,

(page 878) 0 ≤ t ≤ m. Thus

dr = ai + bj + ck dt p dr = a 2 + b2 + c2 dt Z Z m p (x + y) ds = (at + bt) a 2 + b2 + c2 dt C 0 Z m p = (a + b) a 2 + b2 + c2 t dt 0 √ (a + b) a 2 + b2 + c2 2 = m 2

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.3 (PAGE 878)

2. C is given by r = t 2 i + tj + t 2 k, for 0 ≤ t ≤ 1. Thus dr = 2ti + j + 2tk dt p p dr = 4t 2 + 1 + 4t 2 = 1 + 8t 2 dt Z Z 1 p y ds = t 1 + 8t 2 dt Let 1 + 8t 2 = u C 0 Z 9 √ 1 = u du 16 1 27 − 1 1 2 3/2 9 13 = · ·u = = 16 3 24 12 1

3. C: r = a cos t sin ti + a sin2 tj + a cos tk, 0 ≤ t ≤ π/2. Since

|r|2 = a 2 (cos2 t sin2 t + sin4 t + cos2 t) = a 2 for all t, C must lie on the sphere of radius a centred at the origin. We have p ds = a (cos2 t − sin2 t)2 + 4 sin2 t cos2 t + sin2 t dt p = a cos2 2t + sin2 2t + sin2 t dt p = a 1 + sin2 t dt. Thus Z

C

z ds =

Z

π/2

0

= a2

Z

= a2

Z

0

p a cos t a 1 + sin2 t dt

1p

1 + u 2 du

π/4

Let u = sin t du = cos t dt

Let u = tan φ du = sec2 φ dφ

sec3 φ dφ

0

i π/4 a2 h sec φ tan φ + ln | sec φ + tan φ| = 2 0 √  a 2 √ = 2 + ln(1 + 2) . 2

4. C: x = t cos t, y = t sin t, z = t, (0 ≤ t ≤ 2π ). We have p (cos t − t sin t)2 + (sin t + t cos t)2 + 1 dt p = 2 + t 2 dt.

ds = Thus Z

C

Z

2π

z ds = =

1 2

Z

=

0

p t 2 + t 2 dt

2+4π 2

Let u = 2 + t 2 du = 2t dt

u 1/2 du

1 3/2 u 3 2

=

2

(2 + 4π 2 )3/2 − 23/2 . 3

(0 ≤ t ≤ 1)

v = 3i + 6tj + 6t k p v = 3 1 + 4t 2 + 4t 4 = 3(1 + 2t 2 ). If the wire has density δ(t) = 1 + t g/unit length, then its mass is m=3

Z

1 0

(1 + 2t 2 )(1 + t) dt

 1  2t 3 t 4 t2 = 8 g. + + =3 t+ 2 3 2 0

6. The wire of Example 3 lies in the first octant on the surfaces z = x 2 and z = 2 − x 2 − 2y 2 , and, therefore, also on the surface x 2 = 2 − x 2 − 2y 2 , or x 2 + y 2 = 1, a circular cylinder. Since it goes from (1, 0, 1) to (0, 1, 0) it can be parametrized r = cos ti + sin tj + cos2 k, (0 ≤ t ≤ π/2) v = − sin ti + cos tj − 2 cos t sin tk p p v = 1 + sin2 (2t) = 2 − cos2 (2t).

Since the wire has density δ = x y = sin t cos t = its mass is m=

1 2

Z

=

1 4

Z

0

π/2 p

2 − cos2 (2t) sin(2t) dt

1

1 2

sin(2t),

Let v = cos(2t) dv = −2 sin(2t) dt

Z p 1 1p 2 − v 2 dv = 2 − v 2 dv, 2 0 −1

which is the same integral obtained in Example 3, and has value (π + 2)/8.

7. C: r = et cos ti + et sin tj + tk, 0 ≤ t ≤ 2π ). p e2t (cos t − sin t)2 + e2t (sin t + cos t)2 + 1 dt p = 1 + 2e2t dt.

ds =

The moment of inertia of C about the z-axis is I =δ =δ =

2 2+4π 2

5. Wire: r = 3ti + 3t 2 j + 2t 3 k,

=

Z

(x 2 + y 2 ) ds

C Z 2π

δ 4

0

Z

p e2t 1 + 2e2t dt

1+2e4π

√

Let u = 1 + 2e2t du = 4e2t dt

u du

3 1+2e4π

δ 3/2 u 6 3

=

i δh (1 + 2e4π )3/2 − 33/2 . 6

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SECTION 15.3 (PAGE 878)

ADAMS and ESSEX: CALCULUS 8

8. C is the same curve as in Exercise 5. We have Z

C

e z ds =

Z

2π

0

1 = √ 2

p et 1 + 2e2t dt

Z

t=2π

11.

√ Let 2et = tan θ √ t 2e dt = sec2 θ dθ

sec θ dθ

t=0

9. The line of intersection of the planes x − y + z = 0

M x=0 M y=0 Mz=0

(We have omitted the details of the  evaluation of these 1 4π integrals.) The centre of mass is 0, − , . π 3

12. Here the wire of Exercise 9 extends only from t = 0 to t = π:

and x + y + 2z = 0 from (0, 0, 0) to (3, 1, −2) can be parametrized (0 ≤ t ≤ 1).

r = 3ti + tj − 2tk, √ 14 dt and Z

C

x 2 ds =

√

14

Z

0

1

√ 9t 2 dt = 3 14.

10. The curve C of intersection of x 2 + z 2 = 1 and y = x 2

13.

can be parametrized

r = cos ti + cos2 tj + sin tk,

(0 ≤ t ≤ 2π ).

Thus ds =

p

sin2 t + 4 sin2 t cos2 t + cos2 t dt =

p 1 + sin2 2t dt.

√ √ Z π π2 2 t dt = m= 2 2 0 Z π √ √ M x=0 = 2 t cos t dt = −2 2 0 √ √ Z π t sin t dt = π 2 M y=0 = 2 0 √ √ Z π 2 π3 2 Mz=0 = 2 t dt = . 3 0   4 2 2π The centre of mass is − 2 , , . π π 3 √ r = et i + 2tj + e−t k, (0 ≤ t ≤ 1) √ v = et i + 2j − e−t k p v = e2t + 2 + e−2t = et + e−t Z Z 1 (x 2 + z 2 ) ds = (e2t + e−2t )(et + e−t ) dt C 0 Z 1 (e3t + et + e−t + e−3t ) dt = 0

e3 1 1 = +e − − 3. 3 e 3e

We have Z p 1 + 4x 2 z 2 ds C Z 2π p p = 1 + 4 cos2 t sin2 t 1 + sin2 2t dt 0 Z 2π = (1 + sin2 2t) dt 0  Z 2π  1 − cos 4t = 1+ dt 2 0 3 = (2π ) = 3π. 2

√ √ Z 2π 2 t dt = 2π 2 2 0 √ Z 2π = 2 t cos t dt = 0 0 √ Z 2π √ = 2 t sin t dt = −2π 2 0 √ √ Z 2π 2 8π 3 2 = 2 t dt = . 3 0

m=

3

i t=2π 1 h = √ sec θ tan θ + ln | sec θ + tan θ | 2 2 t=0 √ √ t√ √ 2π 2e 1 + 2e2t + ln( 2et + 1 + 2e2t ) = √ 2 2 0 √ √ e2π 1 + 2e4π − 3 = 2 √ √ 1 2e2π + 1 + 2e4π + √ ln . √ √ 2 2 2+ 3

Thus ds =

(0 ≤ t ≤ 2π ) √ v = − sin ti + cos tj + k, v = 2. If the density is δ = z = t, then r = cos ti + sin tj + tk,

14.

e2 − 1 e 0 Z 1 2+1 e et (et + e−t ) dt = M x=0 = 2 0 √ Z 1√ 2(e − 1) 2 t −t 2t (e + e ) dt = M y=0 = e 0 Z 1 2−1 3e Mz=0 = e−t (et + e−t ) dt = 2e2 0 ! √ 3 e + e 2 2 3e2 − 1 The centroid is , , . 2e2 − 2 e + 1 2e3 − 2e m=

Z

592 Copyright © 2014 Pearson Canada Inc.

1

(et + e−t ) dt =

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.4 (PAGE 886)

15. The first octant part C of the curve x 2 + y 2 = a 2 , z = x, can be parametrized

=a

2

t=π/2

sec θ dθ t=0

q p p z = 1 − x 2 − y 2 = 1 − x 2 − (1 − x)2 = 2(x − x 2 ).

Thus C can be parametrized p r = ti + (1 − t)j + 2(t − t 2 )k,

(0 ≤ t ≤ 1).

Hence

C

s

1+1+

z ds =

Z

0

(1 − 2t)2 dt dt = p . 2(t − t 2 ) 2(t − t 2 )

1p

2(t − t 2 ) p

dt 2(t − t 2 )

= 1.

17. The parabola z 2 = x 2 + y 2 , x +z = 1, can be parametrized in terms of y = t since

(1 − x)2 = z 2 = x 2 + y 2 = x 2 + t 2 ⇒

1 − 2x = t 2

⇒

x=

1 − t2 2

1 + t2 . 2 √ √ Thus ds = t 2 + 1 + t 2 dt = 1 + 2t 2 dt, and Z Z ∞ √ 1 + 2t 2 ds = dt 2 3/2 2 3/2 C (2y + 1) −∞ (2t + 1) Z ∞ dt =2 1 + 2t 2 0 ∞ √ √ √ π π = 2 tan−1 ( 2t) = 2 = √ . 2 2 0 ⇒

z =1−x =

(0 ≤ t ≤ 2).

√ 1 + 4t 2 + 16t 6 dt, we have

C

x yz ds =

Z

3

16. On C, we have

We have Z

Z

Let sin t = tan θ cos t dt = sec2 θ dθ

i t=π/2 a2 h = sec θ tan θ + ln | sec θ + tan θ | 2 t=0 i π/2 h 2 p p a = sin t 1 + sin2 t + ln | sin t + 1 + sin2 t| 2 0 √ i a 2 h√ = 2 + ln(1 + 2) . 2

ds =

r = ti + t 2 j + t 4 k, Since ds =

0

Z

Parametrize C by

(0 ≤ t ≤ π/2).

r = a cos ti + a sin tj + a cos tk, √ We have ds = a 1 + sin2 t dt, so Z Z π/2 p x ds = a 2 cos t 1 + sin2 t dt C

18. C: y = x 2 , z = y 2 , from (0, 0, 0) to (2, 4, 16).

0

2

p t 7 1 + 4t 2 + 16t 6 dt.

19. Helix: x = a cos t, y = b sin t, z = ct (0 < a < b). p a 2 sin2 t + b2 cos2 t + c2 dt p = c2 + b2 − (b2 − a 2 ) sin2 t dt p p b2 − a 2 (k 2 = 2 = b2 + c2 1 − k 2 sin2 t dt ). b + c2

ds =

One complete revolution of the helix corresponds to 0 ≤ t ≤ 2π , and has length Z 2π p p b2 + c2 1 − k 2 sin2 t dt 0 Z π/2 p p = 4 b2 + c2 1 − k 2 sin2 t dt 0 s  2 − a2 p p b  units. = 4 b2 + c2 E(k) = 4 b2 + c2 E  b2 + c2

L=

The length of the part of the helix from t = 0 to t = T < π/2 is L=

p

b2

=

p

b2

+ c2

Z

0

T

p 1 − k 2 sin2 t dt

+ c2 E(k, T )

s  2 − a2 p b 2 2 = b +c E , T  units. b2 + c2

20. The straight line L with Ax + By = C, (C 6= 0), √ √ equation lies at distance D = |C|/ A2 + B 2 from the origin. So does the line L 1 with equation y = D. Since x 2 + y 2 depends only on distance from the origin, we have, by symmetry, Z

L

ds = x 2 + y2

Z

ds + y2 dx = 2 + D2 x −∞ ∞  2 x 2 π = tan−1 = −0 D D D 2 √ 0 π π A2 + B 2 = = . D |C| L1 x Z ∞

2

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SECTION 15.4 (PAGE 886)

ADAMS and ESSEX: CALCULUS 8

Section 15.4 Line Integrals of Vector Fields (page 886)

1.

F = x yi − x 2 j.

r = ti + t 2 j, (0 ≤ t ≤ 1). Z 1 Z F • dr = [t 3 − t 2 (2t)] dt = −

C: Z

0

C

2.

3.

0

1

1 t 3 dt = − . 4

  y2 F = cos xi − yj = ∇ sin x − . 2 C : y = sin x from (0,0) to (π, 0).   (π,0) Z y 2 F • dr = sin x − = 0. 2 (0,0) C

7.

F = (x + y)i + (x − z)j + (z − y)k  2  x + z2 =∇ + y(x − z) . 2 The work done by F in moving an object from (1, 0, −1) to (0, −2, 3) is  2  (0,−2,3) Z x + z2 W = F • dr = + y(x − z) 2 C (1,0,−1) 9 19 = − 2(−3) − (1 + 0) = units. 2 2

8. C is made up of four segments as shown in the figure. On C1 , On C2 , On C3 , On C4 , Thus

C1

C

Z

C2

Z

ZC3 C4

C: r = ti + t 2 j + t 3 k, (0 ≤ t ≤ 1).

Z

= 0, = 1, = 1, = 0,

Z

F = yi + zj − xk. C : r = ti + tj + tk, (0 ≤ t ≤ 1). 1 Z Z 1 1 t 2 F • dr = (t + t − t) dt = = . 2 2 C 0 0

4. F = zi − yj + 2xk.

y x y x

F • dr =

Z

=

Z

1

0

0

1

[t 3 − t 2 (2t) + 2t (3t 2 )] dt

dy dx dy dx

= 0, = 0, = 0, = 0,

and and and and

x y x y

goes goes goes goes

from from from from

0 0 1 1

to to to to

1. 1. 0. 0.

x 2 y2 d x + x 3 y d y = 0 x 2 y2 d x + x 3 y d y = x 2 y2 d x + x 3 y d y = 2 2

Z

Z

1

y dy =

0 0 1

1 2

x2 dx = −

1 3

3

x y d x + x y d y = 0.

Finally, therefore, Z 1 1 1 x 2 y2 d x + x 3 y d y = 0 + − + 0 = . 2 3 6 C

1 5 5t 4 = . 5t dt = 4 0 4

y

3

C3

(1,1)

C2 C4

5. F = yzi + x zj + x yk = ∇ (x yz).

C: a curve from (−1, 0, 0) to (1, 0, 0). (Since F is conservative, it doesn’t matter what curve.)

C1 x

Z

6.

C

(1,0,0) F • dr = x yz = 0 − 0 = 0.

F = (x − z)i + (y − z)j − (x + y)k  2  x + y2 =∇ − (x + y)z . 2 C is a given polygonal path from (0,0,0) to (1,1,1) (but any other piecewise smooth path from the first point to the second would do as well). Z

C

F • dr =

Fig. 15.4.8

(−1,0,0)



 (1,1,1) x 2 + y2 − (x + y)z = 1 − 2 = −1. 2 (0,0,0)

9. Observe that if φ = e x+y sin(y + z), then 



∇ φ = e x+y sin(y + z)i + e x+y sin(y + z) + cos(y + z) j + e x+y cos(y + z)k.

Thus, for
any piecewise smooth path from (0, 0, 0) to 1, π4 , π4 , we have Z   e x+y sin(y + z) d x + e x+y sin(y + z) + cos(y + z) d y C

+ e x+y cos(y + z) dz

=

Z

C

(1,π/4,π/4) ∇φ • dr = φ(x, y, z) = e1+(π/4) .

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(0,0,0)

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.4 (PAGE 886)

10. F = (ax y + z)i + x 2 j + (bx + 2z)k is conservative if ∂ F1 ∂ F2 = ∂y ∂x ∂ F1 ∂ F3 = ∂z ∂x ∂ F3 ∂ F2 = ∂z ∂y

⇔

13. For z = ln(1 + x), y = x, from x = 0 to x = 1, we have Z h

a=2

C

⇔

b=1

⇔

0 = 0.

i + (π x 2 cos(π y) − 3e z ) d y − xe z dz Z Z   = ∇ x 2 sin(π y) − xe z • dr − 3 e z d y C C Z 1   (1,1,ln 2) −3 (1 + x) d x = x 2 sin(π y) − xe z

If a = 2 and b = 1, then F = ∇φ where Z φ = (2x y + z) d x = x 2 y + x z + C2 (y, z)

(0,0,0)



= −2 − 3 x +

∂C1 + x 2 = F2 = x 2 ⇒ C1 (y, z) = C2 (z) ∂y dC2 + x = F3 = x + 2z ⇒ C2 (z) = z 2 + C. dz

11.

Thus φ = x 2 y + x z + z 2 + C is a potential for F.  2  x F = Ax ln zi + By 2 zj + + y 3 k is conservative if z ∂ F1 ∂ F2 = ∂y ∂x ∂ F1 ∂ F3 = ∂z ∂x ∂ F3 ∂ F2 = ∂z ∂y

12.

14.

0

 1 x2 2

= −2 − 9 = − 13 . 2 2 0

a) S = {(x, y) : x > 0, y ≥ 0} is a simply connected domain. b) S = {(x, y) : x = 0, y ≥ 0} is not a domain. (It has empty interior.) c) S = {(x, y) : x 6= 0, y > 0} is a domain but is not connected. There is no path in S from (−1, 1) to (1, 1).

⇔

0=0

⇔

A=2

⇔

B = 3.

If A = 2 and B = 3, then F = ∇ φ where φ = x 2 ln z + y 3 z. If C is the straight line x = t + 1, y = 1, z = t + 1, (0 ≤ t ≤ 1), from (1, 1, 1) to (2, 1, 2), then Z 2x ln z d x + 2y 2 z d y + y 3 dz C Z Z x2 dz = ∇ φ • dr − y 2 z d y + z C C (2,1,2) Z 1 − [(t + 1)(0) + (t + 1)] dt = (x 2 ln z + y 3 z) 0 (1,1,1)  2  1 t 1 = 4 ln 2 + 2 − 1 − + t = 4 ln 2 − . 2 2 0 F = (y 2 cos x + z 3 )i + (2y sin x − 4)j + (3x z 2 + 2)k 2

(2x sin(π y) − e z ) d x

3

= ∇ (y sin x + x z − 4y + 2z). The curve C: x = sin−1 t, y = 1 − 2t, z = 3t − 1, (0 ≤ t ≤ 1), goes from (0, 1, −1) to (π/2, −1, 2). The work done by F in moving a particle along C is Z W = F • dr C (π/2,−1,2) = (y 2 sin x + x z 3 − 4y + 2z) (0,1,−1)

= 1 + 4π + 4 + 4 − 0 − 0 + 4 + 2 = 15 + 4π.

d) S = {(x, y, z) : x 2 > 1} is a domain but is not connected. There is no path in S from (−2, 0, 0) to (2, 0, 0). e) S = {(x, y, z) : x 2 + y 2 > 1} is a connected domain but is not simply connected. The circle x 2 + y 2 = 2, z = 0 lies in S, but cannot be shrunk through S to a point since it surrounds the cylinder x 2 + y 2 ≤ 1 which is outside S. f) S = {(x, y, z) : x 2 + y 2 + z 2 > 1} is a simply connected domain even though it has a ball-shaped “hole” in it.

15. C is the curve r = a cos ti + a sin tj, (0 ≤ t ≤ 2π ). I

I

C

x dy =

C

y dx =

Z

2π

0

Z

2π

0

a cos t a cos t dt = π a 2 a sin t (−a sin t) dt = −π a 2 .

16. C is the curve r = a cos ti + b sin tj, (0 ≤ t ≤ 2π ). I

I

C

C

x dy = y dx =

Z

Z

2π

a cos t b cos t dt = π ab

0 2π 0

b sin t (−a sin t) dt = −π ab.

595 Copyright © 2014 Pearson Canada Inc.

SECTION 15.4 (PAGE 886)

ADAMS and ESSEX: CALCULUS 8

17. C consists of two parts:

19. C is made up of three segments as shown in the figure.

On C1 , y = 0, d y = 0, and x goes from −a to a. On C2 , x = a cos t, y = a sin t, t goes from 0 to π . I

C

x dy =

Z

Z

x dy +

C1

Z

π

Z

π

x dy

C2

C

C1

C

π a2 , 2

=0+ a 2 cos2 t dt = 0 I Z Z y dx = y dx + y dx

On C1 , y = 0, d y = 0, and x goes from 0 to a. On C2 , y = bt, x = a(1 − t), and t goes from 0 to 1. On C3 , x = 0, d x = 0, and y goes from b to 0. I Z Z Z x dy = + + Z

0

(−a 2 cos2 t) dt = −

C2 1

C3

C2 1

C3

ab =0+ a(1 − t) b dt + 0 = 2 0Z I Z Z y dx = + +

C2

=0+

C1

C

π a2 . 2

C1

=0+

Z

0

bt (−a dt) + 0 = −

ab . 2

y

y

b

C2 C2

C3 C1

C1 a x

−a

a

x

Fig. 15.4.19

Fig. 15.4.17

20. Conjecture: If D is a domain in R2 whose boundary is a closed, non-self-intersecting curve C, oriented counterclockwise, then

18. C is made up of four segments as shown in the figure. On On On On

C1 , C2 , C3 , C4 ,

y x y x

= 0, = 1, = 1, = 0, I

C

I

C

dy dx dy dx

= 0, = 0, = 0, = 0,

and and and and

x y x y

goes goes goes goes

Z

+

Z

+

x dy =

C1

Z

C2 1

from from from from

Z

C3

+

0 0 1 1

to to to to

Z

C4

dy + 0 + 0 = 1 =0+ 0Z Z Z Z + y dx = + + C1

C2

= 0+0+

Z

C3

C4

0

1

I

1. 1. 0. 0.

d x + 0 = −1.

IC C

Let C consist of the four parts shown in the figure. On C1 and C3 , d y = 0. On C2 , x = g(y), where y goes from c to d. On C2 , x = f (y), where y goes from d to c. Thus

C

C3

(1,1)

x dy =

C2 C4 C1 x

The proof that

Z

C1

+

Z

C2 d

+

Z

C3

I

Z

Copyright © 2014 Pearson Canada Inc.

C4

y d x = −(area of D) is similar, and

uses the fact that D is y-simple.

596

+

Z

Z c =0+ g(y) d y + 0 + f (y) d y c d   = g(y) − f (y) d y = area of D. C

Fig. 15.4.18

y d x = − area of D.

Proof for a domain D that is x-simple and y-simple: Since D is x-simple, it can be specified by the inequalities c ≤ y ≤ d, f (y) ≤ x ≤ g(y).

I

y

x d y = area of D,

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.4 (PAGE 886)

y

b) See the figure. C has four parts. On C1 , x = 1, d x = 0, y goes from 1 to −1. On C2 , y = −1, d y = 0, x goes from 1 to −1. On C3 , x = −1, d x = 0, y goes from −1 to 1. On C4 , x = 1, d x = 0, y goes from 1 to −1. I 1 x dy − y dx 2π C x 2 + y 2 Z −1 Z −1 dx dy 1 + = 2+1 2π 1 1 + y 2 x 1  Z 1 Z 1 −d x −d y + 2 2 −1 x + 1 −1 1 + y Z 1 2 dt =− π −1 1 + t 2 1 2 π π 2 + = − tan−1 t = − = −1. π π 4 4

C3 d x= f (y)

C2 D

C4

x=g(y)

C1

c

x

Fig. 15.4.20

21.

−1

   ∂f ∂g ∂f ∂g + g i+ f + g j ∇( f g) = + f ∂x ∂x ∂y ∂y   ∂g ∂f + f + g k ∂z ∂z = g ∇ f + f ∇ g. Thus, since C goes from P to Q, 

y

C2 C4 C1

C3

Z

f ∇ g • dr +

C

=

Z

C

Z

C

−2

g ∇ f • dr

I

x dy − y dx x 2 + y2 Z 2π 2 1 a cos2 t + a 2 sin2 t = dt = 1. 2π 0 a 2 cos2 t + a 2 sin2 t C

y

23. Although

y

∂ ∂y

1

C4

C

C1 a x

−1

1 x

C3 −1

Fig. 15.4.22(a)

x

c) See the figure. C has four parts. On C1 , y = 0, d y = 0, x goes from 1 to 2. On C2 , x = 2 cos t, y = 2 sin t, t goes from 0 to π . On C3 , y = 0, d y = 0, x goes from −2 to −1. On C4 , x = cos t, y = sin t, t goes from π to 0. I x dy − y dx 1 2π C x 2 + y 2  Z π 1 4 cos2 t + 4 sin2 t = 0+ dt 2 2 2π 0 4 cos t + 4 sin t  Z 0 cos2 t + sin2 t +0+ dt 2 2 π cos t + sin t 1 = (π − π ) = 0. 2π

P

a) C: x = a cos t, x = a sin t, 0 ≤ t ≤ 2π . 1 2π

2

Fig. 15.4.22

Q ∇ ( f g) • dr = ( f g)

= f (Q)g(Q) − f (P)g(P).

22.

1

−1

C2



−y x 2 + y2



=

∂ ∂x



x x 2 + y2



for all (x, y) 6= (0, 0), Theorem 1 does not imply that I x dy − y dx is zero for all closed curves C in R2 . x 2 + y2 C The set consisting of points in R except the origin is not simply connected, and the vector field

Fig. 15.4.22(b)

F=

−yi + xj x 2 + y2

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SECTION 15.4 (PAGE 886)

ADAMS and ESSEX: CALCULUS 8

is not conservative on any domain in R2 that contains the origin in its interior. (See Example 5.) However, the integral will be 0 for any closed curve that does not contain the origin in its interior. (An example is the curve in Exercise 22(c).)

2. The area element d S is bounded by the curves in which the coordinate planes at θ and θ + dθ and the coordinate cones at φ and φ + dφ intersect the sphere R = a. (See the figure.) The element is rectangular with sides a dφ and a sin φ dθ . Thus

24. If C is a closed, piecewise smooth curve in R2 having

d S = a 2 sin φ dφ dθ.

equation r = r(t), a ≤ t ≤ b, and if C does not pass through the  origin, then the polar angle function  θ = θ x(t), y(t) = θ (t) can be defined so as to vary continuously on C. Therefore,

z φ dφ

t=b = 2π × w(C), θ (x, y)

a sin φ

t=a

where w(C) is the number of times C winds around the origin in a counterclockwise direction. For example, w(C) equals 1, −1 and 0 respectively, for the curves C in parts (a), (b) and (c) of Exercise 22. Since

dS a θ y

∂θ ∂θ ∇θ = i + j ∂x ∂y −yi + xj , = 2 x + y2

x

Fig. 15.5.2

we have 1 2π

I

C

x dy − y dx 1 = x 2 + y2 2π =

I

C

dθ

3. The plane Ax + By + C z = D has normal ∇ θ • dr

t=b 1 = w(C). θ (x, y) 2π t=a

Section 15.5 Surfaces and Surface Integrals (page 897)

n = Ai + Bj + Ck, and so an area element on it is given by |n| dS = dx dy = |n • k|

y = g(θ ) sin θ.

x2 y2 + 2 =1 2 a b is given by

Hence its arc length element is s 

  2 dx 2 dy ds = + dθ dθ dθ r  2  2 = g ′ (θ ) cos θ − g(θ ) sin θ + g ′ (θ ) sin θ + g(θ ) cos θ dθ r  2  2 4. g(θ ) + g ′ (θ ) dθ. = The area element on the vertical cylinder r = g(θ ) is d S = ds dz =

r 

A2 + B 2 + C 2 d x d y. |C|

Hence the area S of that part of the plane lying inside the elliptic cylinder

1. The polar curve r = g(θ ) is parametrized by x = g(θ ) cos θ,

√

S=

ZZ

√ x2 a2

+

y2 2

≤1

A2 + B 2 + C 2 dx dy |C|

√b π ab A2 + B 2 + C 2 sq. units. = |C|

One-quarter of the required area is shown in the figure. It lies above the semicircular disk R bounded by x 2 + y 2 = 2ay, or, in terms of polar coordinates, r = 2a sin θ . On the sphere x 2 + y 2 + z 2 = 4a 2 , we have

2  2 g(θ ) + g ′ (θ ) dθ dz.

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2z

∂z = −2x, ∂x

or

∂z x =− . ∂x z

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.5 (PAGE 897)

∂z y = − , so the surface area element on the ∂y z sphere can be written Similarly,

dS =

s

1+

x 2 + y2 2a d x d y dx dy = p . 2 z 4a 2 − x 2 − y 2

The required area is ZZ

2a p dx dy 2 R 4a − x 2 − y 2 Z π/2 Z 2a sin θ r dr = 8a √ Let u = 4a 2 − r 2 dθ 4a 2 − r 2 0 0 du = −2r dr Z π/2 Z 4a 2 = 4a dθ u −1/2 du 0 4a 2 cos2 θ Z π/2 = 8a (2a − 2a cos θ ) dθ 0 π/2 = 8a 2 (π − 2) sq. units. = 16a 2 (θ − sin θ )

S=4

0

z 2a

z 2 =4a 2 −x 2 −y 2

The is 4 that 0≤

area of the part of the cylinder inside the sphere times the part shown in Figure 15.23 in the text, is, 4 times the double p integral of d S over the region y ≤ 2a, 0 ≤ z ≤ 4a 2 − 2ay, or S=4

2a

Z

p dz 2ay − y 2 0 √ 2a √ 3/2 Z 2a d y 2a(2a − y) d y = 4 2a = 4a √ √ y y(2a − y) 0 0 2a √ 3/2 √ = 4 2a (2 y) = 16a 2 sq. units. 0

Z

0

7. On the surface S with equation z = x 2 /2 we have ∂z/∂ x = x and ∂z/∂ y = 0. Thus dS =

y 2a

5.

1 + x 2 d x d y.

ZZ

S

ZZ

p x 1 + x2 dx dy R Z √1−x 2 Z 1 p = dy x 1 + x2 dx 0 0 Z 1 p = x 1 − x 4 d x Let u = x 2 0 du = 2x d x Z 1 1p 1π π = 1 − u 2 du = = . 2 0 24 8

x dS =

2 2 ◦ 8. The normal to the cone z 2 = √ x + y makes a 45 angle

with the vertical, so d S = 2 d x d y is a surface area element for the cone. Both nappes (halves) of the cone pass through the interior of the cylinder x 2 + y 2 = 2ay, so√the area of that part of the cone inside the cylinder is 2 2π a 2 square units, since the cylinder has a circular cross-section of radius a.

r=2a sin θ

Fig. 15.5.4 ∇ F(x, y, z) d x dz d S = F (x, y, z) 2 ∇ F(x, y, z) d y dz dS = F (x, y, z)

p

If R is the first quadrant part of the disk x 2 + y 2 ≤ 1, then the required surface integral is

2a x

Z √4a 2 −2ay

a dy

9. One-quarter of the required area lies in the first octant.

1

6. The cylinder x 2 + y 2 = 2ay intersects the sphere

x 2 + y 2 + z 2 = 4a 2 on the parabolic cylinder 2ay + z 2 = 4a 2 . By Exercise 5, the area element on x 2 + y 2 − 2ay = 0 is 2xi + (2y − 2a)j d y dz d S = 2x s (y − a)2 = 1+ d y dz 2ay − y 2 s 2ay − y 2 + y 2 − 2ay + a 2 a = d y dz = p d y dz. 2ay − y 2 2ay − y 2

(See the figure.) In polar coordinates, the Cartesian equation x 2 + y 2 = 2ay becomes r = 2a sin θ . The arc length element on this curve is s  2 dr dθ = 2a dθ. ds = r 2 + dθ p Thus d S = x 2 + y 2 ds = 2ar dθ = 4a 2 sin θ dθ on the cylinder. The area of that part of the cylinder lying between the nappes of the cone is 4

Z

π/2 0

4a 2 sin θ dθ = 16a 2 sq. units..

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SECTION 15.5 (PAGE 897)

ADAMS and ESSEX: CALCULUS 8

11. Let the sphere be x 2 + y 2 + z 2 = R 2 , and the cylinder be

z

x 2 + y 2 = R 2 . Let S1 and S2 be the parts of the sphere and the cylinder, respectively, lying between the planes z = a and z = b, where −R ≤ a ≤ b ≤ R. Evidently, the area of S2 is S2 = 2π R(b−a) square units. An area element on the sphere is given in terms of spherical coordinates by

z 2 =x 2 +y 2

dS

d S = R 2 sin φ dφ dθ. ds

x

y x 2 +y 2 =2y

On S1 we have z = R cos φ, so S1 lies between φ = cos−1 (b/R) and φ = cos−1 (a/R). Thus the area of S1 is

Fig. 15.5.9

S1 = R 2

10. One-eighth of the required area lies in the first octant, above the triangle T with vertices (0, 0, 0), (a, 0, 0) and (a, a, 0). (See the figure.) The surface x 2 + z 2 = a 2 has normal n = xi + zk, so an area element on it can be written dS =

Z

2π

dθ 0

Z

cos−1 (a/R)

sin φ dφ

cos−1 (b/R)

cos−1 (a/R) = 2π R(b − a) sq. units. = 2π R 2 (− cos φ) cos−1 (b/R)

Observe that S1 and S2 have the same area. z

|n| a a dx dy dx dy = dx dy = √ . |n • k| z a2 − x 2

z=R z=b

The area of the part of that cylinder lying inside the cylinder y 2 + z 2 = a 2 is Z a Z x a dx dy dx √ √ = 8a dy 2 2 a2 − x 2 0 0 ZTa a − x x dx √ = 8a a 2 − x 2 0 p a = −8a a 2 − x 2 = 8a 2 sq. units.

S=8

ZZ

z=a y x

z=−R

0

Fig. 15.5.11

z y 2 +z 2 =a 2

x 2 +z 2 =a 2

12. We want to find A1 , the area of that part of the cylinder x 2 + z 2 = a 2 inside the cylinder y 2 + z 2 = b2 , and A2 , the area of that part of y 2 + z 2 = b2 inside x 2 + z 2 = a 2 . We have A1 = 8 × (area of S1 ), A2 = 8 × (area of S2 ), T

y

x (a,a,0)

Fig. 15.5.10

where S1 and S2 are the parts of these surfaces lying in the first octant, as shown in the figure. A normal to S1 is n1 = xi + zk, and the area element on S1 is |n1 | a d y dz . d S1 = d y dz = √ |n1 • i| a2 − z 2

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.5 (PAGE 897)

13. The intersection of the plane z = 1 + y and the cone p

z

x 2 +z 2 =a 2

z = 2(x 2 + y 2 ) has projection onto the x y-plane the elliptic disk E bounded by

b y 2 +z 2 =b2

S1 a

(1 + y)2 = 2(x 2 + y 2 )

S2

1 + 2y + y 2 = 2x 2 + 2y 2

2x 2 + y 2 − 2y + 1 = 2

R2 R1

x

y

Fig. 15.5.12 A normal to S2 is n2 = xj + zk, and the area element on S2 is b d x dz |n2 | d x dz = √ . d S2 = |n2 • j| b2 − z 2 Let R1 be the region of the first quadrant of the yz-plane bounded by y 2 + z 2 = b2 , y = 0, z = 0, and z = a. Let R2 be the quarter-disk in the first quadrant of the x zplane bounded by x 2 + z 2 = a 2 , x = 0, and z = 0. Then

ZZ

Z

a

Z √b2 −z 2

dz d S1 = 8a dy √ 2 a − z2 0 0 R1 Z a √ 2 b − z2 √ = 8a dz Let z = a sin t a2 − z 2 0 dz = a cos t dt Z π/2 p b2 − a 2 sin2 t dt = 8a 0 s Z π/2 a2 = 8ab 1 − 2 sin2 t dt b 0 a  = 8abE sq. units. b ZZ Z a Z √a 2 −z 2 dz A2 = 8 d S2 = 8b √ dx b2 − z 2 0 0 R2 Z a√ 2 a − z2 = 8b dz Let z = b sin t √ b2 − z 2 0 dz = b cos t dt Z sin−1 (a/b) p = 8b a 2 − b2 sin2 t dt 0 s Z sin−1 (a/b) b2 = 8ab 1 − 2 sin2 t dt a 0   b −1 a = 8abE , sin sq. units. a b A1 = 8

(y − 1)2 = 1. 2 √ Note that E has area A = π(1)( 2) and centroid (0, 1). If S is the part of the plane lying inside the cone, then the area element on S is s  2 √ ∂z dS = 1 + d x d y = 2 d x d y. ∂y x2 +

Thus ZZ

S

y dS =

√ ZZ √ 2 y d x d y = 2 A y¯ = 2π. E

14. Continuing the above solution, the cone z = has area element s dS = =

1+ s 1+



∂z ∂x

2

+



∂z ∂y

2

p 2(x 2 + y 2 )

dx dy

√ 4(x 2 + y 2 ) d x d y = 3 d x d y. 2 z

If S is the part of the cone lying below the plane z = 1 + y, then ZZ √ √ ZZ √ y d x d y = 3 A y¯ = 6π. y dS = 3 S

E

15. If S is the part of z = x 2 in the first octant and inside

(that is, below) z = 1 − 3x 2 − y 2 , then S has projection E onto the x y-plane bounded by x 2 = 1 − 3x 2 − y 2 , or 4x 2 +√ y 2 = 1, an ellipse. Since z = x 2 has area element d S = 1 + 4x 2 d x d y, we have ZZ ZZ p xz dS = x 3 1 + 4x 2 d x d y S E Z 1/2 p Z √1−4x 2 = x 3 1 + 4x 2 d x dy 0 0 Z 1/2 p = x 3 1 − 16x 4 d x Let u = 1 − 16x 4 0 du = −64x 3 d x Z 1 1 1 = u 1/2 du = . 64 0 96

601 Copyright © 2014 Pearson Canada Inc.

SECTION 15.5 (PAGE 897)

16. The surface z =

ADAMS and ESSEX: CALCULUS 8

√ 2x y has area element

and, similarly,

r y x 1+ + dx dy 2x 2y s |x + y| 2x y + y 2 + x 2 = dx dy = √ d x d y. 2x y 2x y

c2 y ∂z = − 2 , the area element on the ∂y a z

spheroid is

dS =

c4 x 2 + y 2 1+ 4 dx dy a z2 s c2 x 2 + y2 = 1+ 2 2 dx dy a a − x 2 − y2 s a 4 + (c2 − a 2 )r 2 r dr dθ = a 2 (a 2 − r 2 )

dS =

If its density is kz, the mass of the specified part of the surface is m=

Z

=k =k

5

dx

0

Z

2

0

Z

5

dx

0

Z

0

Z

0

5

p x+y k 2x y √ dy 2x y 2

(x + y) d y

(2x + 2) d x = 35k units.

in polar coordinates. Thus the area of the spheroid is

z = u, for 0 ≤ u ≤ 1, 0 ≤ v ≤ π . Since

the area element on S is dS =

p

p e2u cos2 v + e2u sin2 v + e4u du dv = eu 1 + e2u du dv.

√ If the charge density on S is 1 + e2u , then the total charge is ZZ p S

1 + e2u d S =

Z

1 0

eu (1 + e2u ) du

Z

π

dv

0

  1 π e3u = (3e + e3 − 4). = π eu + 3 0 3

2π

a

s

a 4 + (c2 − a 2 )r 2 r dr a2 − r 2 0 0 Let u 2 = a 2 − r 2 Zu du = −r dr 4π a p 4 = a + (c2 − a 2 )(a 2 − u 2 ) du a 0 Z 4π a p 2 2 = a c − (c2 − a 2 )u 2 du a 0 s Z a c2 − a 2 1 − 2 2 u 2 du. = 4π c a c 0

2 S= a

17. The surface S is given by x = eu cos v, y = eu sin v, ∂(y, z) eu sin v eu cos v = −eu cos v = 0 ∂(u, v) 1 ∂(z, x) 1 0 = −eu sin v = ∂(u, v) eu cos v −eu sin v ∂(x, y) eu cos v −eu sin v = e2u = u e sin v eu cos v ∂(u, v)

s

Z

dθ

Z

For the case of a prolate spheroid 0 < a < c, let c2 − a 2 k2 = . Then a 2 c2

S = 4π c =

4π c k

Z

a

0

p 1 − k 2 u 2 du

Z sin−1 (ka)

Let ku = sin v k du = cos v dv

cos2 v dv

0

sin−1 (ka) 2π c (v + sin v cos v) = k √ 0 2π ac2 c2 − a 2 = √ sin−1 + 2π a 2 sq. units. 2 2 c c −a

x2 y2 z2 + + = 1 has a a2 a2 c2 circular disk of radius a as projection onto the x y-plane. Since

18. The upper half of the spheroid

19. We continue from the formula for the surface area of a 2x 2z ∂z + 2 =0 a2 c ∂x

⇒

∂z c2 x =− 2 , ∂x a z

spheroid developed part way through the solution above. For the case of an oblate spheroid 0 < c < a, let

602 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

k2 =

SECTION 15.5 (PAGE 897)

a 2 − c2 . Then a 2 c2 Z

S = 4π c 4π c = k

a 0

z

p 1 + k 2 u 2 du

Z tan−1 (ka)

2π b

Let ku = tan v k du = sec2 v dv

sec3 v dv

0

−1

 tan (ka) 2π c  sec v tan v + ln(sec v + tan v) k 0 !# " √ √ a 2π ac2 a a 2 − c2 a 2 − c2 + ln + =√ c2 c c a 2 − c2 ! √ 2π ac2 a + a 2 − c2 = 2π a 2 + √ ln sq. units. 2 2 c a −c =

a y

x

Fig. 15.5.20

21. The distance from the origin to the plane P with equation Ax + By + C z = D, (D 6= 0) is δ= √

20.

x = au cos v, y = au sin v, z = bv, (0 ≤ u ≤ 1, 0 ≤ v ≤ 2π ). This surface is a spiral (helical) ramp of radius a and height 2π b, wound around the z-axis. (It’s like a circular staircase with a ramp instead of stairs.) We have ∂(x, y) a cos v −au sin v = a2 u = ∂(u, v) a sin v au cos v ∂(y, z) a sin v au cos v = = ab sin v 0 b ∂(u, v) ∂(z, x) 0 b = −ab cos v = a cos v −au sin v ∂(u, v) p d S = a 4 u 2 + a 2 b2 sin2 v + a 2 b2 cos2 v du dv p = a a 2 u 2 + b2 du dv. The area of the ramp is

A2

|D|

+ B2 + C 2

.

If P1 is the plane z = δ, then, since the integrand depends only on distance from the origin, we have ZZ dS 2 (x + y 2 + z 2 )3/2 ZZ P dS = 2 2 2 3/2 P1 (x + y + z ) Z 2π Z ∞ r dr = dθ Let u = r 2 + δ 2 2 + δ 2 )3/2 (r 0 0 du = 2r dr Z 1 ∞ du = 2π × 2 δ2 u 3/2   ∞ 2 = π −√ u δ2 √ 2π 2π A2 + B 2 + C 2 = = . δ |D|

22. Use spherical coordinates. The area of the eighth-sphere S is

A=a

Z

0

1p

= 2π a

a 2 u 2 + b2 du

Z

= 2π b2

1p

Z

Z

u=1

Let au = b tan θ a du = b sec2 θ dθ

sec3 θ dθ

u=0

  u=1 = π b sec θ tan θ + ln | sec θ + tan θ | u=0 ! √ au + √a 2 u 2 + b2 1 2 u 2 + b2 a au 2 = πb + ln 0 b2 b ! √ p a + a 2 + b2 = π a a 2 + b2 + π b2 ln sq. units. b 2

π a2 1 (4π a 2 ) = sq. units. 8 2 The moment about z = 0 is ZZ z dS Mz=0 = S Z π/2 Z π/2 dθ a cos φ a 2 sin φ dφ = 0 0 Z π a 3 π/2 sin 2φ π a3 = dφ = . 2 0 2 4 A=

dv

0

a 2 u 2 + b2 du

0

2π

Mz=0 a Thus z¯ = = . By symmetry, x¯ = y¯ = z¯ , A 2 so the centroid of that part of the surface of the  a sphere a a 2 2 2 2 x + y + z = a lying in the first octant is , , . 2 2 2 603

Copyright © 2014 Pearson Canada Inc.

SECTION 15.5 (PAGE 897)

23. The cone z = h 1 −

p

x 2 + y2 a

ADAMS and ESSEX: CALCULUS 8

!

Thus, the total force on m is

has normal

F = kmσ

∂z ∂z n=− i− j+k ∂x ∂y ! xi + yj h p =− + k, a x 2 + y2

Z

2π

Z

dθ

0

π/2

sin φ cos φ dφ = π kmσ units.

0

z

so its surface area element is

dS =

s

h2 + 1 dx dy = a2

√ a2 + h 2 d x d y. a

dS φ

The mass of the conical shell is

a

m

a

a

x

√ ZZ p σ a2 + h 2 (π a 2 ) = π σ a a 2 + h 2 . m=σ dS = a x 2 +y 2 ≤a 2

y

Fig. 15.5.24

The moment about z = 0 is Mz=0

!√ x 2 + y2 a2 + h 2 =σ h 1− dx dy a a x 2 +y 2 ≤a 2 √ Z 2π σ h a 2 + h 2 a  r 1− = r dr a a 0 √ π σ ha a 2 + h 2 . = 3 p

ZZ

h . By symmetry, x¯ = y¯ = 0. The centre 3 of mass is on the axis of the cone, one-third of the way from the base towards the vertex. Thus z¯ =

25. The surface element d S = a dθ dz at the point with

cylindrical coordinates (a, θ, z) attracts mass m at point (0, 0, b) with a force whose vertical component (see the figure) is kmσ a(b − z) dθ dz kmσ d S cos ψ = D2 D3 kmσ a(b − z) dθ dz =  3/2 . a 2 + (b − z)2

dF =

z

h h√ z=h− a x 2 +y 2

The total force exerted by the cylindrical surface on the mass m is

F =−

x

a

a

y

Fig. 15.5.23

2π

shown in the figure on the mass m at the origin is vertical. The vertical component of the force exerted by area element d S = a 2 sin φ dφ dθ at the position with spherical coordinates (a, φ, θ ) is kmσ d S cos φ = kmσ sin φ cos φ dφ dθ. a2

dθ

0

Z

h



0

= 2π kmσ a Z

Z

z=h

z=0 z=h

kmσ a(b − z) dz 3/2 a 2 + (b − z)2

Let b − z = a tan t −dz = a sec2 t dt

a tan t a sec2 t dt a 3 sec3 t

sin t dt z=h = 2π kmσ (− cos t) = 2π kmσ

24. By symmetry, the force of attraction of the hemisphere

dF =

Z

z=0

z=0

h = 2π kmσ p 2 2 a + (b − z) 0 a

= 2π kmσ a

604 Copyright © 2014 Pearson Canada Inc.

1

p

a 2 + (b − h)2

−√

1 a 2 + b2

!

.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.5 (PAGE 897)

The mass of S, which has areal density √ σ , was also determined in that exercise: m = π σ a a 2 + h 2 . The moment of inertia of S about the z-axis is

z

m h

(0,0,b)

ψ

D

I =σ

a

a

a

(x 2 + y 2 ) d S

√S σ a2

+ h2

Z

2π

y

dθ

The radius of gyration is D¯ =

Fig. 15.5.25

Z

a

r 2 r dr a 0 0 √ √ π σ a3 a2 + h 2 2π σ a 2 + h 2 a 4 = . = a 4 2

=

dS

x

ZZ

√

a I /m = √ . 2

26. S is the cylindrical surface x 2 + y 2 = a 2 , 0 ≤ z ≤ h,

with areal density σ . Its mass is m = 2π ahσ . Since all surface elements are at distance a from the z-axis, the radius of gyration of the cylindrical surface about the zaxis is D¯ = a. Therefore the moment of inertia about that axis is I = m D¯ 2 = ma 2 = 2π σ a 3 h.

27. S is the spherical shell, x 2 + y 2 + z 2 = a 2 , with areal den-

29. By Exercise 27, the moment of inertia of a spherical 2 shell of radius a about its diameter is I = ma 2 . Fol3 lowing the argument given in Example 4(b) of Section 14.7, the kinetic energy of the sphere, rolling with speed v down a plane inclined at angle α above the horizontal (and therefore rotating with angular speed  = v/a) is

sity σ . Its mass is 4π σ a 2 . Its moment of inertia about the z-axis is ZZ

(x 2 + y 2 ) d S Z Z π =σ dθ a 2 sin2 φ a 2 sin φ dφ 0 0 Z π = 2π σ a 4 sin φ(1 − cos2 φ) dφ Let u = cos φ 0 du = − sin φ dφ Z 1 4 8π σ a = 2π σ a 4 (1 − u 2 ) du = . 3 −1

I =σ

1 2 mv + 2 1 = mv 2 + 2 5 = mv 2 . 6

K .E. =

S 2π

The radius of gyration is D¯ =

√

I /m =

r

2 a. 3

1 2 I 2 1 2 2 v2 ma 2 2 3 a

The potential energy is P.E. = mgh, so, by conservation of total energy, 5 2 mv + mgh = constant. 6 Differentiating with respect to time t, we get

28. The surface area element for a conical surface S, z =h 1−

p

x2

+ a

y2

!

0=

,

5 dv dh 5 dv m 2v + mg = mv + mgv sin α. 6 dt dt 3 dt

Thus the sphere rolls with acceleration

having base radius a and height h, was determined in the solution to Exercise 23 to be √ a2 + h 2 dS = d x d y. a

dv 3 = g sin α. dt 5

605 Copyright © 2014 Pearson Canada Inc.

SECTION 15.5 (PAGE 897)

ADAMS and ESSEX: CALCULUS 8

Section 15.6 Oriented Surfaces and Flux Integrals (page 903)

ˆ = a on S. Thus the flux If F = xi + yj + zk, then F • N of F out of S is ZZ

1. F = xi + zj.

The surface S of the tetrahedron has four faces: ˆ = −i, F • N ˆ = 0. On S1 , x = 0, N ˆ ˆ = −z, d S = d x dz. On S2 , y = 0, N = −j, F • N ˆ = −k, F • N ˆ = 0. On S3 , z = 0, N 2j + 3k + 2z ˆ = i +√ ˆ = x√ On S4 , x +2y +3z = 6, N , F• N , 14 14 √ 14 dx dy = d x dz. dS = ˆ • j| 2 |N We have ZZ ZZ ˆ dS = ˆ dS = 0 F•N F•N S1

ZZ

S2

ZZ

S4

S3

ˆ dS = − F•N

2

Z

z dz

0

=− √

ˆ dS = F•N

= = =

0

6−3z

dx

0

2

Z

Z

3. F = xi + yj + zk.

ˆ = 0 on the three faces The box has six faces. F • N x = 0, y = 0, and z = 0. On the face x = a, we have ˆ = i, so F • N ˆ = a. Thus the flux of F out of that face N is a × (area of the face) = abc.

By symmetry, the flux of F out of the faces y = b and z = c are also each abc. Thus the total flux of F out of the box is 3abc. z c

Z 2 Z 6−3z 14 1 √ dz (x + 2z) d x 2 14 0 0   Z 1 2 (6 − 3z)2 + 2z(6 − 3z) dz 2 0 2 Z 2 1 (6 − 3z)(6 + z) dz 4 0 2 1 (36z − 6z 2 − z 3 ) = 10. 4

b

a

y

x

Fig. 15.6.3

0

S

4. F = yi + zk. Let S1 be the conical surface and S2 be the base disk. The flux of F outward through the surface of the cone is ZZ ZZ ZZ ˆ = . + F•N

ˆ = √1 On S1 : N 2 Thus

2

S1

S2 S4

ZZ

3 y

S3

Fig. 15.6.1

S1

!

√ xi + yj p + k , d S = 2 d x d y. 2 2 x +y

ˆ dS F•N

ZZ

xy

ˆ = xi + yj + zk . N a

q

+ 1 − x 2 + y2 x 2 + y2 Z 2π Z 1 = 0 + π × 12 − dθ r 2 dr =

x 2 +y 2 ≤1

p

0

2. On the sphere S with equation x 2 + y 2 + z 2 = a 2 we have

S2

S1

S

z

6

ˆ d S = a × 4π a 2 = 4π a 3 . F•N

(6z − 3z 2 ) dz = −4

The flux of F out of the tetrahedron is ZZ ˆ d S = 0 − 4 + 0 + 10 = 6. F•N

x

S

=π−

!

dx dy

0

2π π = . 3 3

ˆ = −k and z = 0, so F • N ˆ = 0. Thus, the total On S2 : N flux of F out of the cone is π/3.

606 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.6 (PAGE 903)

Note that D has area 4π and centroid (−1, 0). For z = 4 − x 2 − y 2 , the downward vector surface element is

z

ˆ d S = −2xi − 2yj − k d x d y. N 1

1

ˆ N √

S1

z=1−

Thus the flux of F = y 3 i + z 2 j + xk downward through S is ZZ ZZ   ˆ dS = − F•N 2x y 3 + 2y(4 − x 2 − y 2 )2 + x d x d y

x 2 +y 2

D

S

S2

ˆ N

1

x

(use the symmetry of D about the x-axis) ZZ =− x d A = −(4π )(−1) = 4π.

y

D

Fig. 15.6.4

5. The part S of z = a − x 2 − y 2 lying above z = b < a lies inside the vertical cylinder x 2 + y 2 = a − b. For z = a − x 2 − y 2 , the upward vector surface element is ˆ d S = 2xi + 2yj + k d x d y. N 1 Thus the flux of F = xi + yj + zk upward through S is ZZ

Z

ˆ d S = 2xi + 2yj + 2zk d x d y = N 2z

dθ



 xi + yj + k d x d y. z

The flux of F = z 2 k upward through the first octant part S of the sphere is ZZ

S

[2(x 2 + y 2 ) + a − x 2 − y 2 ] d x d y

x 2 +y 2 ≤a−b Z √a−b 2π

(r 2 + a)r dr 0 0   a(a − b) π (a − b)2 + = (a − b)(3a − b). = 2π 4 2 2

=

x 2 + y 2 + z 2 = a 2 is

ˆ dS F•N

SZZ

=

8. The upward vector surface element on the top half of

ˆ dS = F•N

Z

π/2

dθ 0

Z

a 0

(a 2 − r 2 )r dr =

π a4 . 8

9. The upward vector surface element on z = 2 − x 2 − 2y 2 is

ˆ d S = 2xi + 4yj + k d x d y. N 1

x2 + y 2 = 1, then the 2 flux of F = xi + yj through the required surface S is If E is the elliptic disk bounded by

6. For z = x 2 − y 2 the upward surface element is ˆ d S = −2xi + 2yj + k d x d y. N 1 The flux of F = xi + xj + k upward through S, the part of z = x 2 − y 2 inside x 2 + y 2 = a 2 is ZZ

S

ˆ dS = F•N

ZZ

2

x 2 +y 2 ≤a 2 Z 2π

= −2

(−2x + 2x y + 1) d x d y

cos2 θ dθ

0

= π a 2 − 2(π )

Z

a 0

r 3 dr + 0 + π a 2

a4 π = a 2 (2 − a 2 ). 4 2

7. The part S of z = 4 − x 2 − y 2 lying above z = 2x + 1 has projection onto the x y-plane the disk D bounded by 2x + 1 = 4 − x 2 − y 2 ,

or (x + 1)2 + y 2 = 4.

ZZ =

ˆ dS F•N

SZZ

E

(2x 2 + 4y 2 ) d x d y

√ ZZ =4 2

√ Z =4 2

0

√ Let x = 2u, y = v √ d x d y = 2 du dv

(u 2 + v 2 ) du dv

u 2 +v 2 ≤1 Z 1 2π

dθ

0

(now use polars)

√ r 3 dr = 2 2π.

10. S: r = u 2 vi + uv 2 j + v 3 k, (0 ≤ u ≤ 1, 0 ≤ v ≤ 1), has upward surface element

∂r ∂r × du dv ∂u ∂v = (2uvi + v 2 j) × (u 2 i + 2uvj + 3v 2 k) du dv

ˆ dS = N

= (3v 4 i − 6uv 3 j + 3u 2 v 2 k) du dv.

607 Copyright © 2014 Pearson Canada Inc.

SECTION 15.6 (PAGE 903)

ADAMS and ESSEX: CALCULUS 8

y

The flux of F = 2xi + yj + zk upward through S is ZZ

S

a

ˆ dS F•N Z 1 Z 1 du (6u 2 v 5 − 6u 2 v 5 + 3u 2 v 5 ) dv = 0 0 Z 1 1 2 1 = u du = . 2 0 6

R −a

a x

−a

Fig. 15.6.13

11. S: r = u cos vi + u sin vj + uk, (0 ≤ u ≤ 2, 0 ≤ v ≤ π ), has upward surface element

∂r ∂r × du dv ∂u ∂v = (−u cos vi − u sin vj + uk) du dv.

ˆ dS = N

6ma

The flux of F = xi + yj + z 2 k upward through S is ZZ

S

ˆ dS F•N Z 2 Z π = du (−u 2 cos2 v − u 2 sin2 v + u 3 ) dv 0 0 Z π Z 2 4π 3 2 dv = (u − u ) du = . 3 0 0

12. S: r = eu cos vi + eu sin vj + uk, (0 ≤ u ≤ 1, 0 ≤ v ≤ π ), has upward surface element

∂r ∂r × du dv ∂u ∂v u = (−e cos vi − eu sin vj + e2u k) du dv.

ˆ dS = N

The flux of F = yzi − x zj + (x 2 + y 2 )k upward through S is ZZ

ˆ dS F•N Z 1 Z π = du (−ue2u sin v cos v + ue2u sin v cos v + e4u ) dv 0 0 Z 1 Z π (e4 − 1) = e4u du dv = π . 4 0 0 S

13. F =

mr m(xi + yj + zk) = 2 . |r|3 (x + y 2 + z 2 )3/2

By symmetry, the flux of F out of the cube −a ≤ x, y, z ≤ a is 6 times the flux out of the top ˆ = k and d S = d x d y. The total flux face, z = a, where N is

Z

dx dy (x 2 + y 2 + a 2 )3/2 ZZ r dr dθ = 48ma 2 2 3/2 R (r + a ) (R as shown in the figure) Z π/4 Z a sec θ r dr = 48ma dθ (r 2 + a 2 )3/2 0 0 Let u = r 2 + a 2 du = 2r dr Z a 2 (1+sec2 θ ) Z π/4 du dθ = 24ma u 3/2 a2 0  Z π/4  1 1 = 48ma − √ dθ a a 1 + sec2 θ 0   Z π/4 π cos θ dθ = 48m − √ 4 cos2 θ + 1 0   Z π/4 cos θ dθ π − = 48m √ 4 2 − sin2 θ √ 0 Let 2 sin v = sin θ √ 2 cos v dv = cos θ dθ ! Z π/6 √ 2 cos v dv π − √ = 48m 4 2 cos v 0 π π − = 48m = 4π m. 4 6 −a≤x ≤a −a≤y≤a

mr out of the cube 1 ≤ x, y, z ≤ 2 |r|3 is equal to three times the total flux out of the pair of opposite faces z = 1 and z = 2, which have outward normals −k and k respectively. This latter flux is

14. The flux of F =

608 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 15.6 (PAGE 903)

2m I2 − m I1 , where Z 2 Z 2 Ik = dx 1

2

=

Z

2

=

Z

2

=

Z

Z

2

1

1

1

where

Using the identities

dy 2 + y 2 + k 2 )3/2 (x 1 √ Let y = x 2 + k 2 tan u √ d y = x 2 + k 2 sec2 u du Z y=2 dx cos u du x 2 + k 2 y=1  y=2 dx  sin u x 2 + k2 y=1 2 ! dx y = Jk2 − Jk1 , p 2 2 x +k x 2 + y 2 + k2 1

dx √ (x 2 + k 2 ) x 2 + n 2 + k 2 √ Let x = n 2 + k 2 tan v √ d x = n 2 + k 2 sec2 v dv Z x=2 sec2 v dv h i =n x=1 (n 2 + k 2 ) tan2 v + k 2 sec v Z x=2 cos v dv =n 2 + k 2 ) sin2 v + k 2 cos2 v (n x=1 Z x=2 cos v dv =n Let w = n sin v 2 2 2 x=1 k + n sin v dw = n cos v dv x=2 Z x=2 dw 1 −1 w = = tan 2 2 k k x=1 x=1 k + w x=2 1 n sin v = tan−1 k k x=1 2 1 nx = tan−1 √ 2 + n2 + k2 k k x 1   1 2n n −1 = tan √ . − tan−1 √ k k 4 + n2 + k2 k 1 + n2 + k2

Jkn = n

1

Thus Ik =

 1 4 2 tan−1 √ − 2 tan−1 √ k k 8 + k2 k 5 + k2  1 −1 + tan √ . k 2 + k2

The contribution to the total flux from the pair of surfaces z = 1 and z = 2 of the cube is 2m I2 − m I1  1 1 1 = m tan−1 √ − 2 tan−1 + tan−1 √ 3 3 2 6  4 1 2 − tan−1 + 2 tan−1 √ − tan−1 √ . 3 6 3

2a , and 1 − a2 π 1 tan−1 a = − tan−1 , 2 a

2 tan−1 a = tan−1

we calculate 1 3 π 4 = − tan−1 = − + tan−1 3 4 2 3 1 2 12 π √ = tan−1 √ = − tan−1 √ . 2 6 6 2 6

− 2 tan−1 2 tan−1

Thus the net flux out of the pair of opposite faces is 0. By symmetry this holds for each pair, and the total flux out of the cube is 0. (You were warned this would be a difficult calculation!)

15. The flux of the plane vector field F across the piecewise ˆ to smooth curve C, in the direction of the unit normal N the curve, is Z F • n ds. C The flux of F = xi + yj outward across a) the circle x 2 + y 2 = a 2 is I

C



F•

xi + yj a



ds =

a2 × 2π a = 2π a 2 . a

b) the boundary of the square −1 ≤ x, y ≤ 1 is 4

Z

1

−1

16. F = −

(i + yj) • i d y = 4

Z

1

−1

d y = 8.

xi + yj . x 2 + y2

a) The flux of F inward across the circle of Exercise 7(a) is  I  xi + yj xi + yj − − • ds a2 a C I a2 1 = ds = × 2π a = 2π. 3 a a C b) The flux of F inward across the boundary of the square of Exercise 7(b) is four times the flux inward across the edge x = 1, −1 ≤ y ≤ 1. Thus it is −4

1

  Z 1 i + yj dy − • i d y = 4 2 2 1+y −1 −1 1 + y 1 = 4 tan−1 y = 2π.

Z

−1

609 Copyright © 2014 Pearson Canada Inc.

SECTION 15.6 (PAGE 903)

ADAMS and ESSEX: CALCULUS 8

17. The flux of Nˆ across S is ZZ

S

ˆ •N ˆ dS = N

ZZ

S

d S = area of S.

18. Let F = F1 i + F2 j + F3 k be a constant vector field. a) If R is a rectangular box, we can choose the origin and coordinate axes in such a way that the box is 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c. On the faces ˆ = −i and N ˆ = i x = 0 and x = a we have N respectively. Since F1 is constant, the total flux out of the box through these two faces is ZZ (F1 − F1 ) d y dz = 0. 0≤y≤b 0≤z≤c

The flux out of the other two pairs of opposite faces is also 0. Thus the total flux of F out of the box is 0. b) If S is a sphere of radius a we can choose the origin so that S has equation x 2 + y 2 + z 2 = a 2 , and so its outward normal is ˆ = xi + yj + zk . N a Thus the flux out of S is ZZ 1 (F1 x + F2 y + F3 z) ds = 0, a S since the sphere S is symmetric about the origin.

Review Exercises 15 1.

(page 904)

x = t, y = 2et , z = e2t , (−1 ≤ t ≤ 1) p v = 1 + 4e2t + 4e4t = 1 + 2e2t Z Z 1 ds 1 + 2e2t = dt 2et C y −1  1  −t 3(e2 − 1) e = − + et = . 2 2e −1 C:

2. C can be parametrized x = t, y = 2t, z = t + 4t 2 , (0 ≤ t ≤ 2). Thus Z 2y d x + x d y + 2 dz C Z 2 = [4t (1) + t (2) + 2(1 + 8t)] dt 0 Z 2 = (22t + 2) dt = 48. 0

p

x 2 + y 2 has area element s √ x 2 + y2 d x d y = 2 d x d y. dS = 1 + 2 z

3. The cone z =

If S is the part of the cone in the region 0 ≤ x ≤ 1 − y 2 (which itself lies between y = −1 and y = 1), then ZZ

S

x dS =

√ Z 2

√ =2 2

1

dy

−1 Z 1 0

Z

1−y 2

x dx 0

√ 1 − 2y 2 + y 4 8 2 dy = . 2 15

4. The plane x + y + z = 1 has area element d S =

√

3 d x d y. If S is the part of the plane in the first octant, then the projection of S on the x y-plane is the triangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x. Thus ZZ Z 1−x √ Z 1 x yz d S = 3 x dx y(1 − x − y) d y S 0 0 √ Z 1 x(1 − x)3 = 3 d x Let u = 1 − x 6 0 du = −d x √ √  √ Z 1  3 3 1 1 3 3 = u (1 − u) du = − . = 6 0 6 4 5 120

5. For z = x y, the upward vector surface element is ˆ d S = −yi − xj + k d x d y. N 1 The flux of F = x 2 yi−10x y 2j upward through S, the part of z = x y satisfying 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 is ZZ Z 1 Z 1 ˆ dS = F•N dx (−x 2 y 2 + 10x 2 y 2 ) d y S 0 0 Z 1 Z 1 3y 2 d y = 1. 3x 2 d x = 0

0

6. The plane x + 2y + 3z = 6 has downward vector surface element

ˆ d S = −i − 2j − 3k d x d y. N 3 If S is the part of the plane in the first octant, then the projection of S on the x y-plane is the triangle 0 ≤ y ≤ 3, 0 ≤ x ≤ 6 − 2y. Thus ZZ ˆ dS (xi + yj + zk) • N S Z Z 6−2y 1 3 =− dy (x + 2y + 6 − x − 2y) d x 3 0 0 Z 3 = −2 (6 − 2y) = −36 + 18 = −18. 0

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INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 15 (PAGE 904)

7. r = a sin ti + a cos tj + btk, (0 ≤ t ≤ 6π )

The field line passes through (1, 1, 0) provided C1 = 0 and C2 = 0. In this case the field line also passes through (e, e, 1), and the segment from (1, 1, 0) to (e, e, 1) can be parametrized r(t) = et i + et j + tk, (0 ≤ t ≤ 1). Then

r(0) = aj, r(6π ) = aj + 6π bk.

a) The force F = −mgk = −∇(mgz) is conservative, so the work done by F as the bead moves from r(6π ) to r(0) is

W=

Z

t=0

t=6π

z=0 F • dr = −mgz

z=6π b

= 6π mgb.

√ b) v = a cos ti − a sin tj + bk, |v| = a 2 + b2 . A force of constant magnitude R opposing the motion of the bead is in the direction of −v, so it is F = −R

v R = −√ v. |v| a 2 + b2

Since dr = v dt, the work done against the resistive force is W=

Z

6π 0

√

R a 2 + b2

Z

p |v|2 dt = 6π R a 2 + b2 .

C

1

Z

(e2t + e2t + 1) dt 1 = (e2t + t) = e2 .

F • dr =

0

0

10.

a) F = (1 + x)e x+y i + (xe x+y + 2y)j − 2zk = ∇ (xe x+y + y 2 − z 2 ). Thus F is conservative.

b) G = (1 + x)e x+y i + (xe x+y + 2z)j − 2yk = F + 2(z − y)(j + k). C : r = (1 − t)et i + tj + 2tk, (0 ≤ t ≤ 1). r(0) = (1, 0, 0), r(1) = (0, 1, 2). Thus Z

C

G • dr =

Z

C

F • dr +

= (xe x+y

8.

R

CF

• dr can be determined using only the endpoints of C, provided

Thus we need a = 2, b = 3, and c = 1. With these values, F = ∇(x 2 y + 3x yz + y 3 z). Thus Z

C

(2,1,1,) = 11 − (−1) = 12. F • dr = (x 2 y + 3x yz + y 3z)

y dx dy = = dz. Thus x2 y 2 2 d x/x = d y/y and the field lines are given by

The field lines satisfy

1 1 = + C1 , x y

ln y = z + C2 .

1

0

11. Since the field lines of F are x y = C, and so satisfy y d x + x d y = 0,

or

dx dy =− , x y

thus F = λ(x, y)(xi − yj). Since p |F(x, y)| = 1 if (x, y) 6= (0, 0), λ(x, y) = ±1/ x 2 + y 2 , and xi − yj F(x, y) = ± p . x 2 + y2

√ Since F(1, 1) = (i − j)/ 2, we need the plus sign. Thus

(0,1,−1)

9. F = (x 2 /y)i + yj + k.

(1,0,0)

Z

0

is conservative, that is, if ∂ F1 ∂ F2 = = 2x + 3z ∂y ∂x ∂ F1 ∂ F3 3y = = = by ∂z ∂x ∂ F3 ∂ F2 = = bx + 3cy 2 . 3x + by 2 = ∂z ∂y

2(z − y)(j + k) • dr (0,1,2) 2 2 + y − z ) C

(2t − t)(1 + 2) dt 1 = −3 − e + 3t 2 = −e. +2

F = (ax y + 3yz)i + (x 2 + 3x z + by 2 z)j + (bx y + cy 3 )k

ax + 3z =

Z

xi − yj F(x, y) = p , x 2 + y2

which is continuous everywhere except at (0, 0).

12. The first octant part of the cylinder y 2 + z 2 = 16 has outward vector surface element ˆ d S = 2yj + 2zk d x d y = N 2z

y

!

p j+k 16 − y 2

d x d y.

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REVIEW EXERCISES 15 (PAGE 904)

ADAMS and ESSEX: CALCULUS 8

The flux of 3z 2 xi − xj − yk outward through the specified surface S is

Therefore the surface S is traced out twice as u goes from 0 to 2π . (It is a M¨obius band. See Figure 15.28 in the text.) If S1 is the part of the surface corresponding to 0 ≤ u ≤ π , and S2 is the part corresponding to π ≤ u ≤ 2π , then S1 and S2 coincide as point sets, ˆ 2 = −N ˆ 1 at but their normals are oppositely oriented: N corresponding points on the two surfaces. Hence ZZ ZZ ˆ 1 dS = − ˆ 2 d S, F•N F•N

! xy 0− p − y dy 0 0 16 − y 2  y=4 Z 5 q y 2 dx = x 16 − y 2 − 2 y=0 0 Z 5 =− (4x + 8) d x = −90.

ˆ dS = F•N

5

Z

dx

Z

4

S1

0

for any smooth vector field, and ZZ ZZ ZZ ˆ dS = ˆ 1 dS + F•N F•N S

Challenging Problems 15

(page 904)

S2

S1

(0, 0, b)

z = sin v for 0 ≤ u ≤ 2π , 0 ≤ v ≤ π . The cylindrical coordinate r satisfies

b − a cos φ

(r − 2)2 + z 2 = 1.

0

dS φ

Mz=0 = 2π

0

π

(2 + cos v) sin v dv



= 2π −2 cos v − Thus z¯ =

a

Fig. C-15.3 2π(2 + cos v) dv = 4π 2 .

The strip has moment z d S = 2π(2 + cos v) sin v dv about z = 0, so the moment of the whole surface about z = 0 is Z

D

a cos φ

This equation represents the surface of a torus, obtained by rotating about the z-axis the circle of radius 1 in the x z-plane centred at (2, 0, 0). Since 0 ≤ v ≤ π implies that z ≥ 0, the given surface is only the top half of the toroidal surface. By symmetry, x¯ = 0 and y¯ = 0. A ring-shaped strip on the surface at angular position v with width dv has radius 2+cos v, and so its surface area is d S = 2π(2 + cos v) dv. The area of the whole given surface is π

m ψ

r 2 = x 2 + y 2 = (2 + cos v)2 r = 2 + cos v

Z

ˆ 2 d S = 0. F•N

3.

1. Given: x = (2 + cos v) cos u, y = (2 + cos v) sin u,

S=

S2

 π 1 cos(2v) = 8π. 4 0

8π 2 = . The centroid is (0, 0, 2/π ). 2 4π π

2. This is a trick question. Observe that the given parametrization r(u, v) satisfies r(u + π, v) = r(u, −v).

The mass element σ d S at p position [a, φ, θ ] on the sphere is at distance D = a 2 + b2 − 2ab cos φ from the mass m located at (0, 0, b), and thus it attracts m with a force of magnitude d F = kmσ d S/D 2 . By symmetry, the horizontal components of d F coresponding to mass elements on opposite sides of the sphere (i.e., at [a, φ, θ ] and [a, φ, θ + π ]) cancel, but the vertical components kmσ d S b − a cos φ D2 D reinforce. The total force on the mass m is the sum of all such vertical components. Since d S = a 2 sin φ dφ dθ , it is Z 2π Z π (b − a cos φ) sin φ dφ 2 F = kmσ a dθ 2 + b 2 − 2ab cos φ)3/2 (a 0 0 Z 1 (b − at)dt . = 2π kmσ a 2 2 2 3/2 −1 (a − 2abt + b )

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d F cos ψ =

INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 15 (PAGE 904)

We have made the change of variable t = cos φ to get the last integral. This integral √ can be evaluated by using another substitution. Let u = a 2 − 2abt + b2 . Thus t=

a 2 + b2 − u 2 , 2ab

dt = −

u du , ab

b−at =

u 2 + b2 − a 2 . 2b

When t = −1 and t = 1 we have u = a + b and u = |a − b| respectively. Therefore   Z |a−b| 2 u + b2 − a 2 u du F = 2π kmσ a 2 − 2bu 3 ab a+b  Z a+b  π kmσ a b2 − a 2 = 1 + du b2 u2 |a−b|   a+b π kmσ a b2 − a 2 = . u − b2 u |a−b|

There are now two cases to consider. If the mass m is

outside the sphere, so that b > a and |a − b| = b − a,

then

F=

π kmσ a b2



 a2 (a+b)−(b−a)−(b−a)+(b+a) = 4π kmσ 2 . b

However, if m is inside the sphere, so that b < a and |a − b| = a − b, then π kmσ a F= b2



 (a + b) + (a − b) − (a − b) − (a + b) = 0.

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SECTION 16.1 (PAGE 914)

ADAMS and ESSEX: CALCULUS 8

CHAPTER 16. VECTOR CALCULUS

7.

Section 16.1 Gradient, Divergence, and Curl (page 914) 1.

2.

3.

F = xi + yj ∂ ∂ ∂ div F = (x) + (y) + (0) = 1 + 1 = 2 ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = =0 ∂ x ∂ y ∂z x y 0 F = yi + xj ∂ ∂ ∂ div F = (y) + (x) + (0) = 0 + 0 = 0 ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = (1 − 1)k = 0 ∂ x ∂ y ∂z y x 0

F = f (z)i − f (z)j  ∂  ∂ div F = f (z) + − f (z) = 0 ∂x ∂y i j k ∂ ∂ ∂ curl F = = f ′ (z)(i + j) ∂x ∂y ∂z f (z) − f (z) 0

9. Since x = r cos θ , and y = r sin θ , we have r 2 = x 2 + y 2 , and so

F = yi + zj + xk ∂ ∂ ∂ (y) + (z) + (x) = 0 div F = ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = −i − j − k ∂ x ∂ y ∂z y z x

4.

F = yzi + x zj + x yk ∂ ∂ ∂ div F = (yz) + (x z) + (x y) = 0 ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = ∂ x ∂ y ∂z yz x z x y = (x − x)i + (y − y)j + (z − z)k = 0

5.

F = xi + xk ∂ ∂ ∂ div F = (x) + (0) + (x) = 1 ∂x ∂y ∂z i j k ∂ ∂ ∂ curl F = = −j ∂ x ∂ y ∂z x 0 x

6.

8.

F = f (x)i + g(y)j + h(z)k ∂ ∂ ∂ div F = f (x) + g(y) + h(z) ∂x ∂y ∂z = f ′ (x) + g ′ (y) + h ′ (z) i j k ∂ ∂ ∂ curl F = =0 ∂x ∂y ∂z f (x) g(y) h(z)

∂r x = = cos θ ∂x r y ∂r = = sin θ ∂y r ∂ ∂ y −x y cos θ sin θ sin θ = = 3 =− ∂x ∂x r r r ∂ y 1 y2 ∂ sin θ = = − 3 ∂y ∂y r r r x2 cos2 θ = 3 = r r ∂ ∂ x 1 x2 cos θ = = − 3 ∂x ∂x r r r sin2 θ y2 = 3 = r r ∂ ∂ x −x y cos θ sin θ cos θ = = 3 =− . ∂y ∂y r r r (The last two derivatives are not needed for this exercise, but will be useful for the next two exercises.) For F = r i + sin θ j,

F = x y 2 i − yz 2j + zx 2 k  ∂  2 ∂  ∂  2 div F = xy + −yz 2 + zx ∂x ∂y ∂z = y2 − z2 + x 2 j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 x y −yz 2 zx 2 = 2yzi − 2x zj − 2x yk

we have ∂r ∂ cos2 θ + sin θ = cos θ + ∂x ∂y r i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z r sin θ 0   sin θ cos θ − sin θ k. = − r

div F =

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INSTRUCTOR’S SOLUTIONS MANUAL

10.

SECTION 16.1 (PAGE 914)

F = rˆ = cos θ i + sin θ j sin2 θ

It follows that

cos2 θ

1 1 + = = p r r r x 2 + y2 i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z cos θ sin θ 0   cos θ sin θ cos θ sin θ =− − k=0 r r

div F =

11.

F = θˆ = − sin θ i + cos θ j cos θ sin θ cos θ sin θ div F = − =0 r r i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z − sin θ cos θ 0  2  sin θ 1 cos2 θ 1 = k= k= p k + 2 r r r x + y2

ZZ 1 ˆ d S = div F(0, 0, 0).

F•N a,b,c→0+ 8abc Ba,b,c lim

13. This proof just mimics that of Theorem 1. F can be expanded in Maclaurin series F = F0 + F1 x + F2 y + · · · , where F0 = F(0, 0)

 ∂ ∂ F1 = F(x, y) i+ ∂x ∂x (0,0)  ∂ ∂ F1 F2 = = F(x, y) i+ ∂y ∂y

F1 =

12. We use the Maclaurin expansion of F, as presented in the proof of Theorem 1:

(0,0)

F = F0 + F1 x + F2 y + F3 z + · · · , where F0 = F(0, 0, 0)

  ∂ F1 ∂ F2 ∂ F3 ∂ = F(x, y, z) i+ j+ k F1 = ∂x ∂x ∂x ∂x (0,0,0) (0,0,0)   ∂ ∂ F ∂ F ∂ F 1 2 3 F2 = = F(x, y, z) i+ j+ k ∂y ∂y ∂y ∂y (0,0,0) (0,0,0)   ∂ ∂ F ∂ F ∂ F 1 2 3 = F3 = F(x, y, z) i+ j+ k ∂z ∂z ∂z ∂z (0,0,0) (0,0,0)

and where · · · represents terms of degree 2 and higher in x, y, and z. ˆ = k. On the top of the box Ba,b,c , we have z = c and N ˆ = −k. On the bottom of the box, we have z = −c and N On both surfaces d S = d x d y. Thus ZZ  ZZ ˆ dS F•N + Z

top a

Z

bottom  b

 d y cF3 • k − cF3 • (−k) + · · · −a −b ∂ = 8abcF3 • k + · · · = 8abc F3 (x, y, z) + ···, ∂z (0,0,0) =

dx

where · · · represents terms of degree 4 and higher in a, b, and c. Similar formulas obtain for the two other pairs of faces, and the three formulas combine into ZZ ˆ d S = 8abcdiv F(0, 0, 0) + · · · .

F•N

 ∂ F2 j ∂x (0,0)  ∂ F2 j ∂y (0,0)

and where · · · represents terms of degree 2 and higher in x and y. On the curve Cǫ of radius ǫ centred at (0, 0), we have ˆ = 1 (xi + yj). Therefore, N ǫ  ˆ = 1 F0 • ix + F0 • jy + F1 • ix 2 F•N ǫ  + F1 • jx y + F2 • ix y + F2 • jy 2 + · · · where · · · represents terms of degree 3 or higher in x and y. Since I

Cǫ I Cǫ

x ds =

I

x 2 ds =

Cǫ I

y ds =

Cǫ

I

Cǫ Z y 2 ds =

x y ds = 0

0

2π

ǫ 2 cos2 θ ǫ dθ = π ǫ 3 ,

we have I 3 1 ˆ ds = 1 π ǫ (F1 • i + F2 • j) + · · · F•N 2 2 π ǫ Cǫ πǫ ǫ = div F(0, 0) + · · · where · · · represents terms of degree 1 or higher in ǫ. Therefore, taking the limit as ǫ → 0 we obtain

Ba,b,c

lim

ǫ→0

I 1 ˆ ds = div F(0, 0). F•N π ǫ 2 Cǫ

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SECTION 16.1 (PAGE 914)

ADAMS and ESSEX: CALCULUS 8

14. We use the same Maclaurin expansion for F as in Exer-

3. Theorem 3(d):

cises 12 and 13. On Cǫ we have r = ǫ cos θ i + ǫ sin θ j, (0 ≤ θ ≤ 2π ) dr = −ǫ sin θ i + ǫ cos θ j  F • dr = −ǫ sin θ F0 • i + ǫ cos θ F0 • j

− ǫ 2 sin θ cos θ F1 • i + ǫ 2 cos2 θ F1 • j

 − ǫ 2 sin2 θ F2 • i + ǫ 2 sin θ cos θ F2 • j + · · · ds,

where · · · represents terms of degree 3 or higher in ǫ. Since Z

2π

sin θ dθ =

0

Z

0

2π

Z

cos2 θ dθ =

2π

cos θ dθ =

0

Z

0

2π

Z

2π 0

sin θ cos θ dθ = 0

sin2 θ dθ = π,

we have I 1 F • dr = F1 • j − F2 • i + · · · , π ǫ 2 Cǫ where · · · represents terms of degree at least 1 in ǫ. Hence I 1 lim F • dr = F1 • j − F2 • i ǫ→0+ π ǫ 2 Cǫ ∂ F2 ∂ F1 = − ∂x ∂y ˆ = curl F • k = curl F • N.

Section 16.2 Some Identities Involving Grad, Div, and Curl (page 920) 1. Theorem 3(a): ∂ ∂ ∂ (φψ) + (φψ) + (φψ) ∂x ∂y ∂z     ∂ψ ∂φ ∂ψ ∂φ = φ + ψ i +··· + φ + ψ k ∂x ∂x ∂z ∂z = φ ∇ ψ + ψ ∇φ.

∇ (φψ) =

2. Theorem 3(b): ∂ ∂ ∂ (φ F1 ) + (φ F2 ) + (φ F3 ) ∂x ∂y ∂z ∂φ ∂ F1 ∂φ ∂ F3 = F1 + φ + ··· + F3 + φ +··· ∂x ∂x ∂z ∂z = ∇ φ • F + φ ∇ • F.

∇ • (φF) =

∂ (F2 G 3 − F3 G 2 ) + · · · ∂x ∂ F2 ∂ F3 ∂G 3 ∂G 2 = G 3 + F2 − G 2 − F3 + ··· ∂x ∂x ∂x ∂x = (∇ × F) • G − F • (∇ × G).

∇ • (F × G) =

4. Theorem 3(f). The first component of ∇ (F • G) is ∂ F1 ∂G 1 ∂ F2 ∂G 2 ∂ F3 ∂G 3 G 1 + F1 + G 2 + F2 + G 3 + F3 . ∂x ∂x ∂x ∂x ∂x ∂x We calculate the first components of the four terms on the right side of the identity to be proved. The first component of F × (∇ × G) is     ∂G 2 ∂G 1 ∂G 3 ∂G 1 F2 − − F3 − . ∂x ∂y ∂z ∂x The first component of G × (∇ × F) is     ∂ F1 ∂ F2 ∂ F1 ∂ F3 G2 − G3 . − − ∂x ∂y ∂z ∂x The first component of (F • ∇)G is F1

∂G 1 ∂G 1 ∂G 1 + F2 + F3 . ∂x ∂y ∂z

The first component of (G • ∇ )F is G1

∂ F1 ∂ F1 ∂ F1 + G2 + G3 . ∂x ∂y ∂z

When we add these four first components, eight of the fourteen terms cancel out and the six remaining terms are the six terms of the first component of ∇ (F • G), as calculated above. Similar calculations show that the second and third components of both sides of the identity agree. Thus

∇ (F•G) = F×(∇ ×G)+G×(∇ ×F)+(F•∇)G+(G•∇ )F. 5. Theorem 3(h). By equality of mixed partials, i j k ∂ ∂ ∂ ∇ × ∇ φ = ∂ x ∂ y ∂z ∂φ ∂φ ∂φ ∂ x ∂ y ∂z   ∂ ∂φ ∂ ∂φ = − i + · · · = 0. ∂ y ∂z ∂z ∂ y

6. Theorem 3(i). We examine the first components of the terms on both sides of the identity

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∇ × (∇ × F) = ∇(∇ • F) − ∇ 2 F.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.2 (PAGE 920)

The first component of ∇ × (∇ × F) is    ∂ F2 ∂ F1 ∂ ∂ F1 ∂ F3 − − − ∂x ∂y ∂z ∂z ∂x ∂ 2 F2 ∂ 2 F1 ∂ 2 F1 ∂ 2 F3 = − − + . ∂ y∂ x ∂ y2 ∂z 2 ∂z∂ x

∂ ∂y



If f (r )r is solenoidal then ∇ • u = f (r ) satisfies

f (r )r



= 0, so that

du + 3u = 0 dr 3 dr du =− u r ln |u| = −3 ln |r | + ln |C|

r

The first component of ∇ (∇ • F) is

u = Cr −3 .

∂ 2 F1 ∂ 2 F2 ∂ 2 F3 ∂ ∇•F= + + . ∂x ∂ x∂ y ∂ x∂z ∂x2

Thus f (r ) = Cr −3 , for some constant C.

10. Given that div F = 0 and curl F = 0, Theorem 3(i) implies that ∇ 2 F = 0 too. Hence the components of F

The first component of −∇ 2 F is −∇2 F1 = −



are harmonic functions. If F = ∇ φ, then

∂ 2 F1 ∂ 2 F1 ∂ 2 F1 − − . 2 2 ∂x ∂y ∂z 2

Evidently the first components of both sides of the given identity agree. By symmetry, so do the other components.

7. If the field lines of F(x, y, z) are parallel straight lines, in

∇ 2 φ = ∇ • ∇ φ = ∇ • F = 0, so φ is also harmonic.

11. By Theorem 3(e) and 3(f),

the direction of the constant nonzero vector a say, then F(x, y, z) = φ(x, y, z)a for some scalar field φ, which we assume to be smooth. By Theorem 3(b) and (c) we have

div F = div (φa) = ∇ φ • a curl F = curl (φa) = ∇ φ × a.

8. If r = xi + yj + zk and r = |r|, then ∇ × r = 0,

If r = xi + yj + zk, then ∇ • r = 3 and ∇ × r = 0. Also, (F • ∇ )r = F1

Since ∇ φ is an arbitrary gradient, div F can have any value, but curl F is perpendicular to a, and thereofore to F.

∇ • r = 3,

∇ × (F × r) = (∇ • r)F + (r • ∇ )F − (∇ • F)r − (F • ∇ )r ∇ (F • r) = F × (∇ × r) + r × (∇ × F) + (F • ∇ )r + (r • ∇ )F.

r r

∇r = .

Combining all these results, we obtain

∇ × (F × r) − ∇ (F • r) = 3F − 2(F • ∇)r − (∇ • F)r − r × (∇ × F) = F − (∇ • F)r − r × (∇ × F). In particular, if ∇ • F = 0 and ∇ × F = 0, then

If c is a constant vector, then its divergence and curl are both zero. By Theorem 3(d), (e), and (f) we have

∇ • (c × r) = (∇ × c) • r − c • (∇ × r) = 0 ∇ × (c × r) = (∇ • r)c + (r • ∇ )c − (∇ • c)r − (c • ∇)r = 3c + 0 − 0 − c = 2c

∇(c • r) = c × (∇ × r) + r × (∇ × c) + (c • ∇ )r + (r • ∇)c = 0 + 0 + c + 0 = c.

9.









∇ • f (r )r = ∇ f (r ) • r + f (r )(∇ • r) r•r = f (r ) + 3 f (r ) r = r f ′ (r ) + 3 f (r ). ′

∂r ∂r ∂r + F2 + F3 = F. ∂x ∂y ∂z

∇ × (F × r) − ∇ (F • r) = F. 12. If ∇ 2 φ = 0 and ∇ 2 ψ = 0, then ∇ • (φ ∇ ψ − ψ ∇φ) = ∇ φ • ∇ψ + φ ∇ 2 ψ − ∇ ψ • ∇ φ − ψ ∇ 2 φ = 0, so φ ∇ ψ − ψ ∇ φ is solenoidal.

13. By Theorem 3(c) and (h), ∇ × (φ ∇ ψ) = ∇ φ × ∇ψ + φ ∇ × ∇ ψ = ∇ φ × ∇ ψ −∇ × (ψ ∇ φ) = −∇ψ × ∇ φ − ψ ∇ × ∇ φ = ∇ φ × ∇ ψ.

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SECTION 16.2 (PAGE 920)

ADAMS and ESSEX: CALCULUS 8

14. By Theorem 3(b), (d), and (h), we have

Look for a solution with G 2 = 0. We have



 ∇ • f (∇ g × ∇ h)

= ∇ f • (∇ g × ∇ h) + f ∇ • (∇ g × ∇ h)   = ∇ f • (∇ g × ∇ h) + f (∇ × ∇ g) • ∇ h − ∇ g • (∇ × ∇ h)

= ∇ f • (∇ g × ∇ h) + 0 − 0 = ∇ f • (∇ g × ∇ h).

G3 =

∂G 1 ∂G 3 = ye2z + = 2ye2z . ∂z ∂x

by Theorem 3(h). Therefore, by Theorem 3(d) we have

Thus

Thus F × G is solenoidal. By Exercise 13,

∇ × (φ ∇ ψ) = ∇ φ × ∇ ψ = F × G, so φ ∇ψ is a vector potential for F × G. (So is −ψ ∇φ.)

xe2z d y = x ye2z + M(x, z).

Try M(x, z) = 0. Then G 3 = x ye2z , and

15. If F = ∇ φ and G = ∇ ψ, then ∇ × F = 0 and ∇ × G = 0 ∇ • (F × G) = (∇ × F) • G + F • (∇ × G) = 0.

Z

G1 =

Z

2ye2z dz = ye2z + N (x, y).

Since −e2z = −

16. If ∇ × G = F = −yi + xj, then

∂N ∂G 1 = −e2z − , ∂y ∂y

∂G 3 ∂G 2 − = −y ∂y ∂z ∂G 1 ∂G 3 − =x ∂z ∂x ∂G 1 ∂G 2 − = 0. ∂x ∂y

we can take N (x, y) = 0. Thus G = ye2z i + x ye2z k is a vector potential for F.

As in Example 1, we try to find a solution with G 2 = 0. Then Z y2 G 3 = − y d y = − + M(x, z). 2

18. For (x, y, z) in D let v = xi + yj + zk. The line segment r(t) = tv, (0 ≤ t ≤ 1), lies in D, so div F = 0 on the

y2 Again we try M(x, z) = 0, so G 3 = − . Thus 2 ∂G 3 = 0 and ∂x Z G 1 = x dz = x z + N (x, y).

∂G 1 Since = 0 we may take N (x, y) = 0. ∂y 1 G = x zi − y 2 k is a vector potential for F. (Of course, 2 this answer is not unique.)

17. If F = xe2z i + ye2z j − e2z k, then div F = e2z + e2z − 2e2z = 0, so F is solenoidal. If F = ∇ × G, then ∂G 3 ∂G 2 − = xe2z ∂y ∂z ∂G 1 ∂G 3 − = ye2z ∂z ∂x ∂G 2 ∂G 1 − = −e2z . ∂x ∂y

path. We have

G(x, y, z) = =

Z

Z

1 0 1 0

  tF r(t) × v dt

  tF ξ(t), η(t), ζ (t) × v dt

where ξ = t x, η = t y, ζ = t z. The first component of

curl G is

(curl G)1 Z 1   = t curl (F × v) dt 1 0  Z 1  ∂ ∂ = t (F × v)3 − (F × v)2 dt ∂y ∂z 0  Z 1  ∂ ∂ = t (F1 y − F2 x) − (F3 x − F1 z) dt ∂y ∂z 0 Z 1 ∂ F ∂ F2 ∂ F3 1 = t F1 + t 2 y − t2x − t2x ∂η ∂η ∂ζ 0  ∂ F 1 + t F1 + t 2 z dt ∂ζ  Z 1 ∂ F1 ∂ F1 ∂ F1 = 2t F1 + t 2 x + t2 y + t 2z dt. ∂ξ ∂η ∂ζ 0

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.3 (PAGE 924)

To get the last line we used the fact that di vF = 0 to ∂ F2 ∂ F3 ∂ F1 replace −t 2 x − t2x with t 2 x . Continuing the ∂η ∂ζ ∂ξ calculation, we have Z

(d) LHS := Del . (F &x G):

RHS := (Del &x F) . G - F . (Del &x G): simplify(LHS - RHS); 0

1

 d2 t F1 (ξ, η, ζ ) dt 0 dt 1 = t 2 F1 (t x, t y, t z) = F1 (x, y, z).

(curl G)1 =

(e) LHS := Del &x (F &x G):

RHS1 := (Del . G)*F: RHS2 := G[1]*diff(F,x) +G[2]*diff(F,y)+G[3]*diff(F,z): RHS3 := (Del . F)*G: RHS4 := F[1]*diff(G,x) +F[2]*diff(G,y)+F[3]*diff(G,z): RHS := RHS1 + RHS2 - RHS3 - RHS4: simplify(LHS - RHS);

0

Similarly, (curl G)2 = F2 and (curl G)3 = F3 . Thus curl G = F, as required.

19. In the following we suppress output (which for some calculations can be quite lengthy) except for the final check on each inequality. You may wish to use semicolons instead of colons to see what the output actually looks like. >

0 e¯ x (f) LHS := Del(F . G):

RHS1 := F &x (Del &x G): RHS2 := G &x (Del &x F): RHS3 := F[1]*diff(G,x) +F[2]*diff(G,y)+F[3]*diff(G,z): RHS4 := G[1]*diff(F,x) +G[2]*diff(F,y)+G[3]*diff(F,z): RHS := RHS1 + RHS2 + RHS3 + RHS4: simplify(LHS - RHS);

with(VectorCalculus):

>

SetCoordinates(’cartesian’[x,y,z]): >

F := VectorField ():

0 e¯ x

>

G := VectorField ():

All these zero outputs indicate that the inequalities (a)–(f) of the theorem are valid.

Section 16.3 (page 924)

(a) LHS := Del(phi(x,y,z)*psi(x,y,z)):

RHS := phi(x,y,z)*Del(psi(x,y,z)) + psi(x,y,z)*Del(phi(x,y,z)): simplify(LHS - RHS); 0 e¯ x (b) LHS := Del . (F*phi(x,y,z)):

RHS := (Del(phi(x,y,z))).F + phi(x,y,z)*(Del.F): simplify(LHS - RHS);

1.

Green’s Theorem in the Plane

I

2

(sin x + 3y 2 ) d x + (2x − e−y ) d y CZZ   ∂ ∂ 2 (2x − e−y ) − (sin x + 3y 2 ) d A = ∂x ∂y ZZ R = (2 − 6y) d A Z πR Z a = dθ (2 − 6r sin θ )r dr 0 0Z Z a π 2 = πa − 6 sin θ dθ r 2 dr 0

0

0

= π a 2 − 4a 3 .

y

(c) LHS := Del &x (phi(x,y,z)*F):

RHS := RHS := (Del(phi(x,y,z))) &x F + phi(x,y,z)*(Del &x F): simplify(LHS - RHS); 0 e¯ x

C

R a

−a

x

Fig. 16.3.1

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SECTION 16.3 (PAGE 924)

2.

ADAMS and ESSEX: CALCULUS 8

I

(x 2 − x y) d x + (x y − y 2 ) d y C ZZ   ∂ ∂ 2 =− (x y − y 2 ) − (x − x y) d A ∂x ∂y ZZT =− (y + x) d A T   4 1 +1 ×1= − . = −( y¯ + x) ¯ × (area of T ) = − 3 3

5. By Example 1, I 1 x dy − y dx 2 C Z 2π h 1 = a cos3 t 3b sin2 t cos t 2 0

Area =

i − b sin3 t (−3a cos2 t sin t) dt

Z 3ab 2π 2 sin t cos2 t dt 2 0 Z 3ab 2π sin2 (2t) 3π ab = dt = . 2 0 4 8 =

y

(1,1)

C

T

6. Let R, C, and F be as in the statement of Green’s The2

x

orem. As noted in the proof of Theorem 7, the unit ˆ satisfy tangent Tˆ to C and the unit exterior normal N ˆ = Tˆ × k. Let N

Fig. 16.3.2

3.

I

(x sin y 2 − y 2 ) d x + (x 2 y cos y 2 + 3x) d y CZZ h i = 2x y cos y 2 + 3 − (2x y cos y 2 − 2y) d A ZZT ZZ = (3 + 2y) d A = 3 d A + 0 = 3 × 3 = 9. T

T

y

2

7. r = sin ti + sin 2tj,

(1,1)

T

G = F2 (x, y)i − F1 (x, y)j. ˆ ˆ Applying the 2-dimensional DiverThen F • T = G • N. gence Theorem to G, we obtain Z Z Z ˆ ds F1 d x + F2 d y = F • Tˆ ds = G•N C C C ZZ = div G d A  ZZ R  ∂ F2 ∂ F1 = − dA ∂x ∂y R as required (0 ≤ t ≤ 2π ) y

C

C x

R1

R2 x

(1,−1)

−2

Fig. 16.3.3 Fig. 16.3.7 2

4. Let D be the region x 2 + y 2 ≤ 9, y ≥ 0. Since C is the clockwise boundary of D, I

C

x 2 y d x − x y2 d y  ZZ  ∂ 2 ∂ (−x y 2 ) − (x y) d x d y =− ∂y D ∂x ZZ Z π Z 3 81π = (y 2 + x 2 ) d A = dθ r 3 dr = . 4 D 0 0

F = ye x i + x 3 e y j i j k ∂ ∂ ∂ 2 = (3x 2 e y − e x )k. curl F = ∂ x ∂y ∂z x2 ye x 3e y 0

Observe that C bounds two congruent regions, R1 and R2 , one counterclockwise and the other clockwise. ˆ = k; for R2 , N ˆ = −k. Since R1 and R2 For R1 , N are mirror images of each other in the y-axis, and since curl F is an even function of x, we have ZZ ZZ curl F • Nˆ d S = − curl F • Nˆ d S. R1

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R2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.4 (PAGE 930)

Thus I

8.

C

ZZ

F • dr =

R1

+

ZZ  R2

curl F • Nˆ d S = 0.

a) F = x 2 j I I ZZ F • dr = x2 dy = 2x d A = 2 A x¯ . C C R b) F = x yi I I ZZ F • dr = xy dx = − x d A = − A x. ¯ C C R c) F = y 2 i + 3x yj I I F • dr = y 2 d x + 3x y d y CZZ C = (3y − 2y) d A = A y¯ . R

parametrization (0 ≤ t ≤ 2π ).

ˆ the unit normal to Note that dr/dt = cos ti + sin tj = N, Cr exterior to the disk Dr of which Cr is the boundary. The average value of u(x, y) on Cr is u¯ r =

1 2π

Z

0

ZZ ZZZ ˆ dS =

F•N 0 d V = 0. S B

3. If F = (x 2 + y 2 )i + (y 2 − z 2 )j + zk, then div F = 2x + 2y + 1, and ZZ ZZZ ZZZ 4 ˆ dS =

F•N (2x+2y+1) d V = 1 d V = π a3 . 3 S B B

4. If F = x 3 i + 3yz 2 j + (3y 2 z + x 2 )k, then div F = 3x 2 + 3z 2 + 3y 2 , and

9. The circle Cr of radius r and centre at r0 has r = r0 + r cos ti + r sin tj,

2. If F = ye z i + x 2 e z j + x yk, then div F = 0, and

2π

u(x0 + r cos t, y0 + r sin t) dt,

ZZ ZZZ ˆ dS = 3

F•N (x 2 + y 2 + z 2 ) d V S B Z 2π Z π Z a =3 dθ sin φ dφ R4 d R 0

since ds = r dt. By the (2-dimensional) divergence theorem, and since u is harmonic, ZZ 1 d u¯ r = ∇ • ∇u d x d y dr 2πr Dr  ZZ  2 1 ∂ u ∂2u = + 2 d x d y = 0. 2πr ∂x2 ∂y Dr Thus u¯ r = limr→0 u¯ r = u(x0 , y0 ).

Section 16.4 The Divergence Theorem in 3Space (page 930) 1. In this exercise, the sphere S bounds the ball B of radius a centred at the origin. If F = xi − 2yj + 4zk, then div F = 1 − 2 + 4 = 3. Thus ZZ ZZZ ˆ dS =

F•N 3 d V = 4π a 3 . S B

0

5. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z). Therefore the flux of F out of any solid region R is

and so  Z 2π  1 ∂u ∂u d u¯ r = cos t + sin t dt dr 2π 0 ∂x ∂y I 1 = ∇ u • Nˆ ds 2πr Cr

0

12 5 = πa . 5

Flux =

ZZZ

=2

R ZZZ

div F d V

R

(x + y + z) d V = 2(x¯ + y¯ + z¯ )V

where (x, ¯ y¯ , z¯ ) is the centroid of R and V is the volume of R. If R is the ball (x − 2)2 + y 2 + (z − 3)2 ≤ 9, then x¯ = 2, y¯ = 0, z¯ = 3, and V = (4π/3)33 = 36π . The flux of F out of R is 2(2 + 0 + 3)(36π ) = 360π .

6. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z). Therefore the flux of F out of any solid region R is Flux =

ZZZ

=2

R ZZZ

div F d V

R

(x + y + z) d V = 2(x¯ + y¯ + z¯ )V

where (x, ¯ y¯ , z¯ ) is the centroid of R and V is the volume of R. If R is the ellipsoid x 2 + y 2 + 4(z − 1)2 ≤ 4, then x¯ = 0, y¯ = 0, z¯ = 1, and V = (4π/3)(2)(2)(1) = 16π/3. The flux of F out of R is 2(0 + 0 + 1)(16π/3) = 32π/3.

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SECTION 16.4 (PAGE 930)

ADAMS and ESSEX: CALCULUS 8

7. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z).

can be calculated directly by the methods of Section 6.6. We will do it here by using the Divergence Theorem instead. S is one face of a tetrahedral domain D whose other faces are in the coordinate planes, as shown in the figure. Since φ = x y + z 2 , we have

Therefore the flux of F out of any solid region R is ZZZ Flux = div F d V R ZZZ =2 (x + y + z) d V = 2(x¯ + y¯ + z¯ )V

∇ • ∇ φ = ∇2 φ = 2.

∇ φ = yi + xj + 2zk,

R

where (x¯ , y¯ , z¯ ) is the centroid of R and V is the volume of R.

Thus

If R is the tetrahedron with vertices (3, 0, 0), (0, 3, 0), (0, 0, 3), and (0, 0, 0), then x¯ = y¯ = z¯ = 3/4, and V = (1/6)(3)(3)(3) = 9/2. The flux of F out of R is 2((3/4) + (3/4) + (3/4))(9/2) = 81/4.

ZZZ

D

∇ • ∇φ d V = 2 ×

the volume of the tetrahedron D being abc/6 cubic units. z c

8. If F = x 2 i + y 2 j + z 2 k, then div F = 2(x + y + z). Therefore the flux of F out of any solid region R is ZZZ Flux = div F d V R ZZZ =2 (x + y + z) d V = 2(x¯ + y¯ + z¯ )V

back side

S D

R

x

ˆ = −k and z = 0, so ∇φ • N ˆ = 0, and On the bottom, N the flux out of the bottom face is 0.

region having volume V , then ZZZ div F d V = 3V .

ˆ = −j, so ∇φ • N ˆ = −x. The On the side, y = 0 and N flux out of the side face is

C

ZZ

The region C described in the statement of the problem is the part of a solid cone with vertex at the origin that lies inside a ball of radius R with centre at the origin. The surface S of C consists of two parts, the conical wall S1 , and the region D on the spherical boundary of the ball. At any point P on S1 , the outward normal ˆ is perpendicular to the line O P, that is, to F, so field N ˆ = 0. At any point P on D, N ˆ is parallel to F, in F•N ˆ = F/|F| = F/R. Thus fact N ZZ ZZ ZZ ˆ dS = ˆ dS + ˆ dS

F•N F•N F•N S S1 D ZZ ZZ F•F R2 =0+ dS = d S = AR R R D D

∇ φ • Nˆ d S = −

side

ZZ

side

x d x dz = −

ac a a2 c × =− . 2 3 6

(We used the fact that M x=0 = area × x¯ and x¯ = a/3 for that face.) ˆ = −i, so the flux out of On the back face, x = 0 and N that face is ZZ

back

∇ φ • Nˆ d S = −

ZZ

back

y d y dz = −

b2 c bc b × =− . 2 3 6

Therefore, by the Divergence Theorem I−

where A is the area of D. By the Divergence Theorem, 3V = AR, so V = AR/3.

so

10. The required surface integral, S

bottom

The flux of ∇ φ out of D is the sum of its fluxes out of the four faces of the tetrahedron.

9. If F = xi + yj + zk, then div F = 3. If C is any solid

∇ φ • Nˆ d S,

a

Fig. 16.4.10

If R is the cylinder x 2 + y 2 ≤ 2y (or, equivalently, x 2 + (y − 1)2 ≤ 1), 0 ≤ z ≤ 4, then x¯ = 0, y¯ = 1, z¯ = 2, and V = (π 12 )(4) = 4π . The flux of F out of R is 2(0 + 1 + 2)(4π ) = 24π .

I =

y

b

where (x¯ , y¯ , z¯ ) is the centroid of R and V is the volume of R.

ZZ

abc abc = , 6 3

ZZ

S

11.

a 2 c b2 c abc − +0 = , 6 6 3

∇φ • Nˆ d S = I =

abc c(a 2 + b2 ) + . 3 6

F = (x + y 2 )i + (3x 2 y + y 3 − x 3 )j + (z + 1)k

div F = 1 + 3(x 2 + y 2 ) + 1 = 2 + 3(x 2 + y 2 ).

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.4 (PAGE 930)

z

z

a b

back

ˆ N

side

S S

D

D B x

a

x

bottom

−k Fig. 16.4.11

Fig. 16.4.12 ˆ = −j, F • N ˆ = 0, so On the side, y = 0, N ZZ ˆ d S = 0. F•N

Let D be the conical domain, S its conical surface, and B its base disk, as shown in the figure. We have ZZZ

D

div F d V =

Z

2π

0

dθ Z a

Z

a

r dr 0

Z

b(1−(r/a)) 0

side

(2 + 3r 2 ) dz

ˆ = −i, F • N ˆ = −y, so On the back, x = 0, N Z a ZZ Z π/2 ˆ dS = − dθ r cos θ r dr F•N

r



r (2 + 3r 2 ) 1 − dr a   Z a 2r 2 3r 4 3 = 2π b 2r + 3r − − dr a a 0 2π a 2 b 3π a 4 b = + . 3 10 = 2π b

0

B

ˆ = −k, F • N ˆ = 2x, so On the bottom, z = 0, N Z a Z π/2 ZZ 2a 3 ˆ dS = 2 r cos θ r dr = dθ F•N . 3 0 0 bottom By the Divergence Theorem ZZ 3 3 3 ˆ d S + 0 − a + 2a = π a . F•N 3 3 6 S

ˆ d S = −area of B = −π a 2 . F•N

S

ˆ dS + F•N

ZZ

B

ˆ dS = F•N

ZZZ

Hence the flux of F upward through S is ZZ 3 3 ˆ dS = πa − a . F•N 6 3 S

div F d V , D

so the flux of F upward through the conical surface S is 2π a 2 b 3π a 4 b = + + π a2 . 3 10 S

ZZ

0

π/2 a3 a3 = − sin θ × =− . 3 3 0

By the Divergence Theorem, ZZ

0

back

ˆ = −k, F • N ˆ = −1, so On B we have z = 0, N ZZ

y

a

y

a

a

13.

F = (x + yz)i + (y − x z)j + (z − e x sin y)k div F = 1 + 1 + 1 = 3. z

ˆ N

12. F = (y + x z)i + (y + yz)j − (2x + z 2 )k div F = z + (1 + z) − 2z = 1. Thus

2a a

S2

D y

ZZZ

π a3 div F d V = volume of D = , 6 D

where D is the region in the first octant bounded by the sphere and the coordinate planes. The boundary of D consists of the spherical part S and the four planar parts, called the bottom, side, and back in the figure.

x

ˆ N

S1

Fig. 16.4.13

623 Copyright © 2014 Pearson Canada Inc.

SECTION 16.4 (PAGE 930)

ADAMS and ESSEX: CALCULUS 8

z

a) The flux of F out of D through S = S1 ∪ S2 is

1

side

ZZ ZZZ ˆ dS =

F•N div F d V S D Z 2π Z 2a Z √4a 2 −r 2 =3 dθ r dr 2 dz 0 a 0 Z 2a p = 12π r 4a 2 − r 2 dr

back

S front

D x

1

a

4a 2

= 6π

Z

3a 2

1

− r2

Let u = du = −2r dr

0

bottom

Fig. 16.4.14

√ u 1/2 du = 12 3π a 3 .

ˆ = −j, F • N ˆ = x, so On the side, y = 0, N ZZ Z 1 Z 1 1 ˆ dS = F•N x dx dz = . 2 side 0 0

ˆ = − xi + yj , d S = a dθ dz. The flux of F b) On S1 , N a out of D through S1 is ZZ

S1

ˆ = −k, F • N ˆ = y, so On the bottom, z = 0, N ZZ Z 1 Z 1 1 ˆ dS = F•N y dy dx = . 2 bottom 0 0

−x 2 − x yz − y 2 + x yz a dθ dz a S1 √ Z 2π Z 3a √ = −a 2 dθ √ dz = −4 3π a 3 .

ˆ dS = F•N

ZZ

0

ˆ = −i, F • N ˆ = 0, so On the back, x = 0, N ZZ ˆ d S = 0. F•N

− 3a

back

ˆ = i, F • N ˆ = 3z 2 , so On the front, x = 1, N Z 1 Z π/2 ZZ 3π ˆ dS = 3 r 2 cos2 θ r dr = dθ . F•N 16 0 0 front

c) The flux of F out of D through the spherical part S2 is ZZ

y

ZZ ZZ ˆ dS ˆ dS − ˆ dS = F • N F•N F•N S1 S S2 √ √ √ = 12 3π a 3 + 4 3π a 3 = 16 3π a 3 .

Hence, ZZ ˆ d S = 3π − 1 − 1 −0− 3π = −1. (3x z 2 i− xj− yk)• N 16 2 2 16 S

15. 14. Let D be the domain bounded by S, the coordinate planes, and the plane x = 1. If F = 3x z 2 i − xj − yk, then div F = 3z 2 , so the total flux of F out of D is

F = (x 2 − x − 2y)i + (2y 2 + 3y − z)j − (z 2 − 4z + x y)k div F = 2x − 1 + 4y + 3 − 2z + 4 = 2x + 4y − 2z + 6. The flux of F out of R through its surface S is ZZ ZZZ ˆ dS = (2x + 4y − 2z + 6) d V .

F•N R S ZZZ x d V = M x=0 = V x, ¯ where R has volume Now R

ZZ

bdry of D

ˆ dS = F•N

ZZZ

=3

Z

0

=3×

3z 2 d V D 1

dx

Z

π/2

dθ 0

Z

1

r 2 cos2 θ r dr

0

3π 1 π × = . 4 4 16

V and centroid (x, ¯ y¯ z¯ ). Similar formulas obtain for the other variables, so the required flux is ZZ ˆ d S = 2V x¯ + 4V y¯ − 2V z¯ + 6V .

F•N S

16. F = xi + yj + zk implies that div F = 3. The total flux of

The boundary of D consists of the cylindrical surface S and four planar surfaces, the side, bottom, back, and front.

F out of D is ZZ

624 Copyright © 2014 Pearson Canada Inc.

bdry of D

ˆ dS = 3 F•N

ZZZ

D

d V = 12,

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.4 (PAGE 930)

since the volume of D is half that of a cube of side 2, that is, 4 square units. D has three triangular faces, three pentagonal faces, and a hexagonal face. By symmetry, the flux of F out of each triangular face is equal to that out of the triangular face ˆ = k • k = 1 on that T in the plane z = 1. Since F • N face, these fluxes are ZZ

d x d y = area of T =

T

18. φ = x 2 − y 2 + z 2 , G = 13 (−y 3 i + x 3 j + z 3 k). F = ∇φ + µcurl G. Let R be the region of 3-space occupied by the sandpile. Then R is bounded by the upper surface S of the sandpile and by the disk D: x 2 + y 2 ≤ 1 in the plane z = 0. The outward (from R) normal on D is −k. The flux of F out of R is given by

1 . 2

ZZ

Similarly, the flux of F out of each pentagonal face is equal to the flux out of the pentagonal face P in the ˆ = −k • (−k) = 1; that plane z = −1, where F • N flux is ZZ 1 7 d x d y = area of P = 4 − = . 2 2 P

S

ZZZ

(0,−1,1)

z

H y

19.

P x

R

div F d V =

above z = 0 and the disk D: x 2 + y 2 = 3a 2 in the x yplane form the boundary of a region R in 3-space. The outward normal from R on D is −k. If + y 2 )j + (3 + x)k,

then di vF = 2x + 2y. By the Divergence Theorem, ZZ

S

ˆ dS + F•N

ZZ

D

F • (−k) d x d y =

R

div F d V . R

(2 + µ × 0) d V = 2(5π ) = 10π.

i 1 ∂ curl G = 3 ∂x −y 3

j ∂ ∂y x3

ZZ

Z

D

k ∂ ∂z z3

F • k d A = 3µ

2π

dθ 0

= 3(x 2 + y 2 )k,

Z

0

1

r 3 dr =

3π µ . 2

The flux of F out of S is 10π + (3π µ)/2. ZZ ZZZ ˆ dS =

curl F • N div curl F = 0, by Theorem S D 3(g).

ZZ ZZZ 1 ˆ dS = 1

r•N 3 dV = V. 3 S 3 D

17. The part of the sphere S: x 2 + y 2 + (z − a)2 = 4a 2

F = (x 2 + y + 2 + z 2 )i + (e

ZZZ

20. If r = xi + yj + zk, then div r = 3 and

Fig. 16.4.16

x2

ZZZ

and ∇ φ • k = 2z = 0 on D, so

(−1,0,1)

D

D

F • (−k) d A =

In addition,

(This can also be seen directly, since F radiates from the origin, so is everywhere tangent to the plane of the hexagonal face, the plane x + y + z = 0.) T

ZZ

Now div curl G = 0 by Theorem 3(g). Also div ∇φ = div (2xi−2yj+2zk) = 2−2+2 = 2. Therefore

Thus the flux of F out of the remaining hexagonal face H is   1 7 12 − 3 × + = 0. 2 2

(−1,−1,1)

ˆ dS + F•N

ZZZ

R

div F d V = 0

because R is symmetric about x = 0 and y = 0. Thus the flux of F outward across S is ZZ ZZ ˆ dS = F•N (3 + x) d x d y = 3π(3a 2 ) = 9π a 2 . S D

21. We use Theorem 7(b), the proof of which is given in Exercise 29. Taking φ(x, y, z) = x 2 + y 2 + z 2 , we have ZZ ZZ 1 ˆ d S = 1 φN ˆ dS

(x 2 + y 2 + z 2 )N 2V S 2V S ZZZ 1 = grad φ d V 2V ZZZ D 1 = (xi + yj + zk) d V V = r¯ , since

ZZ

x d V = M x=0 = V x. ¯

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SECTION 16.4 (PAGE 930)

ADAMS and ESSEX: CALCULUS 8

22. Taking F = ∇φ in the first identity in Theorem 7(a), we have

ZZ ZZZ ˆ dS = −

∇φ × N curl ∇ φ d V = 0, S D since ∇ × ∇ φ = 0 by Theorem 3(h).

so φ is constant on any connected component of S, and u and v can only differ by a constant on S.

27. Apply the Divergence Theorem to F = ∇ φ: ZZZ

23. div (φF) = φ div F + ∇ φ • F by Theorem 3(b). Thus ZZZ

D

φ div F d V +

ZZZ

D

ZZZ

28. By Theorem 3(b),

24. If F = ∇φ in the previous exercise, then div F = ∇2 φ

D

φ ∇2 φ d V +

ZZZ

ZZ 2 ˆ d S. |∇ φ| d V = φ ∇ φ • N D S

If ∇2 φ = 0 in D and φ = 0 on S, then ZZZ

D

ZZZ

∇ • ∇φ d V ZZ D ZZ ˆ d S = ∂φ d S. = ∇φ • N S S ∂n

div (φF) d V ZZ D ˆ dS = φF • N S

∇φ • F d V =

by the Divergence Theorem. and ZZZ

D

∇2 φ d V =

div (φ ∇ ψ − ψ ∇ φ) = ∇φ • ∇ ψ + φ ∇ 2 ψ − ∇ ψ • ∇ φ − ψ ∇ 2 φ = φ ∇2 ψ − ψ ∇2 φ. Hence, by the Divergence Theorem,

|∇φ|2 d V = 0.

ZZZ

D

(φ ∇ 2 ψ − ψ ∇2 φ) d V =

ZZZ

div (φ ∇ ψ − ψ ∇ φ) d V ZZ D ˆ dS = (φ ∇ ψ − ψ ∇ φ) • N S  ZZ  ∂φ ∂ψ −ψ d S. = φ ∂n ∂n S

Since φ is assumed to be smooth, ∇ φ = 0 throughout D, and therefore φ is constant on each connected component of D. Since φ = 0 on S, these constants must all be 0, and φ = 0 on D.

25. If u and v are two solutions of the given Dirichlet problem, and φ = u − v, then

∇ 2 φ = ∇ 2 u − ∇ 2 v = f − f = 0 on D φ = u − v = g − g = 0 on S.

By the previous exercise, φ = 0 on D, so u = v on D. That is, solutions of the Dirichlet problem are unique.

29. If F = φc, where c is an arbitrary, constant vector, then div F = ∇ φ • c, and by the Divergence Theorem, c•

26. Re-examine the solution to Exercise 24 above. If ∇ 2 φ = 0 in D and ∂φ/∂n = ∇ φ • Nˆ = 0 on S, then we

ZZZ

D

can again conclude that ZZZ

D

|∇ φ| d V = 0

Thus

and ∇ φ = 0 throughout D. Thus φ is constant on the connected components of D. (We can’t conclude the constant is 0 because we don’t know the value of φ on S.) If u and v are solutions of the given Neumann problem, then φ = u − v satisfies 2

2

ZZZ

div F d V ZZ D ˆ dS = F•N ZZS ZZ ˆ d S. ˆ = φc • N d S = c • φ N S S

∇φ d V =

c•

ZZZ

D

ZZ



∇ φ d V − φ Nˆ d S = 0. S

Since c is arbitrary, the vector in the large parentheses must be the zero vector. Hence

2

∇ φ = ∇ u − ∇ v = f − f = 0 on D ∂φ ∂u ∂v = − = g − g = 0 on S, ∂n ∂n ∂n

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ZZZ

D

ZZ

∇ φ d V = φ Nˆ d S. S

INSTRUCTOR’S SOLUTIONS MANUAL

30.

SECTION 16.5 (PAGE 934)

z

ZZ ZZZ 1 1 ˆ dS =

F•N div F d V vol(Dǫ ) Sǫ vol(Dǫ ) Dǫ ZZZ 1 = div F(P0 ) d V vol(Dǫ ) Dǫ  ZZZ   + div F − div F(P0 ) d V Dǫ ZZZ   1 div F − div F(P0 ) d V . = div F(P0 ) + vol(Dǫ ) Dǫ Thus

1

C

T 1 y x

1

Fig. 16.5.1

ZZ 1 ˆ d S − div F(P0 )

F • N vol(Dǫ ) Sǫ ZZZ 1 ≤ |div F − div F(P0 )| d V vol(Dǫ ) Dǫ

2. Let S be the part of the surface z = y 2 lying inside the

ˆ cylinder x 2 + y 2 = 4, and having upward normal N. Then C is the oriented boundary of S. Let D be the disk x 2 + y 2 ≤ 4, z = 0, that is, the projection of S onto the x y-plane.

≤ max |div F − div F(P0 )| P in Dǫ

z

→ 0 as ǫ → 0 + assuming div F is continuous. ZZ 1 ˆ d S = div F(P0 ). lim

F•N ǫ→0+ vol(Dǫ ) Sǫ

C

ˆ N

S

D

Section 16.5 Stokes’s Theorem

y

(page 934) x

1. The triangle T lies in the plane x + y + z = 1. We use the downward normal

If F = yi −

ˆ = − i +√j + k N 3 on T , because of the given orientation of its boundary. If F = x yi + yzj + zxk, then curl F =

Therefore I

C

i ∂ ∂x xy

j ∂ ∂y yz

k ∂ ∂z zx

= −yi − zj − xk. I

x y d x + yz dz + zx dz = F • dr C ZZ ZZ y+z+x = curl F • Nˆ d S = dS √ 3 T T ZZ 1 1 = √ d S = √ × (area of T ) 3 T 3 √ ! √ 1 1 3 1 = √ × × 2× √ = . 2 2 3 2

xj + z 2 k,

Fig. 16.5.2 then

i ∂ curl F = ∂x y

j ∂ ∂y −x

k ∂ ∂z z2

= −2k.

dx dy on S, we have ˆ k•N I I ZZ curl F • Nˆ d S y d x − x d y + z 2 dz = F • dr = S C C ZZ ˆ d x d y = −8π. = −2k • N ˆ k•N D

Since d S =

3. Let C be the circle x 2 + y 2 = a 2 , z = 0, oriented

counterclockwise as seen from the positive z-axis. Let D be the disk bounded by C, with normal k. We have F = 3yi − 2x zj + (x 2 − y 2 )k j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 3y −2x z x 2 − y 2 = 2(x − y)i − 2xj − (2z + 3)k.

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SECTION 16.5 (PAGE 934)

ADAMS and ESSEX: CALCULUS 8

Applying Stokes’s Theorem (twice) we calculate ZZ

ZZ F • dr = curl F • k d A S CZZ D =− 3 d A = −3π a 2 . =

I

D

z

since D has area π a 2 .

6. The curve C:

S

k

r = cos ti + sin tj + sin 2tk, D

y

C

x

Fig. 16.5.3

4. The surface S with equation x 2 + y 2 + 2(z − 1)2 = 6,



 ∂ 3 z ∂ 3 x e − (x z − y cos z) ∂x ∂y z=0   2 2 z 2 = 3(x + y 2 ). = 3x e + 3y cos z z=0

S

curl F • Nˆ d S =

I

ZZ

F • dr = curl F • k d A C D Z 2π Z 2 = dθ 3r 2 r dr = 24π. 0

0

5. The circle C of intersection of x 2 + y 2 + z 2 = a 2 and

x + y + z = 0 is the boundary of a circular disk of radius a in the plane x + y + z = 0. If F = yi + zj + xk, then i ∂ curl F = ∂x y

j ∂ ∂y z

k ∂ ∂z x

= −(i + j + k).

0 ≤ t ≤ 2π,

lies on the surface z = 2x y, since sin 2t = 2 cos t sin t. It also lies on the cylinder x 2 +y 2 = 1, so it is the boundary of that part of z = 2x y lying inside that cylinder. Since C is oriented counterclockwise as seen from high on the z-axis, S should be oriented with upward normal, −2yi − 2xj + k ˆ = p N , 1 + 4(x 2 + y 2 )

z ≥ 0,

ˆ is that part of an ellipsoid of with outward normal N, revolution about the z-axis, centred at (0, 0, 1), and lying above the x y-plane. The boundary of S is the circle C: x 2 + y 2 = 4, z = 0, oriented counterclockwise as seen from the positive z-axis. C is also the oriented boundary ˆ = k. of the disk x 2 + y 2 ≤ 4, z = 0, with normal N 2 +y 2 +z 2 3 3 z x If F = (x z − y cos z)i + x e j + x yze k, then, on z = 0, we have

Thus ZZ

ˆ = − i +√j + k , N 3 √ ˆ = 3 on D, so then curl F • N I I ZZ y d x + z d y + x dz = F • dr = curl F • Nˆ d S C C ZZ D √ √ = 3 d S = 3π a 2 , D

ˆ N

curl F • k =

If C is oriented so that D has normal

and has area element q d S = 1 + 4(x 2 + y 2 ) d x d y.

If F = (e x − y 3 )i + (e y + x 3 )j + e z k, then i j k ∂ ∂ ∂ curl F = = 3(x 2 + y 2 )k. ∂y ∂z ∂x x e − y 3 e y + x 3 ez

If D is the disk x 2 + y 2 ≤ 1 in the x y-plane, then I ZZ ZZ F • dr = curl F • Nˆ d S = 3(x 2 + y 2 ) d x d y C S D Z 2π Z 1 3π =3 dθ r 2 r dr = . 2 0 0

7. The part of the paraboloid z = 9 − x 2 − y 2 lying above

ˆ has boundary the the x y-plane having upward normal N circle C: x 2 + y 2 = 9, oriented counterclockwise as seen from above. C is also the oriented boundary of the plane disk x 2 + y 2 ≤ 9, z = 0, oriented with normal field ˆ = k. N If F = −yi + x 2 j + zk, then j i ∂ ∂ curl F = ∂x ∂y −y x 2

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k ∂ ∂z z

= (2x + 1)k.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.5 (PAGE 934)

By Stokes’s Theorem, the circulation of F around C is I

C

F • dr = =

ZZ

ZZ D

D

(curl F • k) d A (2x + 1) d A = 0 + π(32 ) = 9π.

8. The closed curve r = (1 + cos t)i + (1 + sin t)j + (1 − cos t − sin t)k, (0 ≤ t ≤ 2π ), lies in the plane x + y + z = 3 and is oriented counterclockwise as seen from above. Therefore it is the boundary of a √ region S in that plane with normal ˆ = (i + j + k)/ 3. The projection of S onto the field N x y-plane is the circular disk D of radius 1 with centre at (1, 1). If F = ye x i + (x 2 + e x )j + z 2 e z k, then i ∂ curl F = ∂x x ye

j ∂ ∂y x 2 + ex

By Stokes’s Theorem, I

C

F • dr =

ZZ

ZZS

ZZ

D

2x √ √ ( 3) d x d y 3

where x¯ = 1 is the x-coordinate of the centre of D, and A = π 12 = π is the area of D.

9. If S1 and S2 are two surfaces joining C1 to C2 , each having upward normal, then the closed surface S3 consisting of S1 and −S2 (that is, S2 with downward normal) bound a region R in 3-space. Then ZZ

S1

ˆ dS − F•N

ZZ

ZZS2

ˆ dS F•N ˆ dS + F•N

ZZ

ˆ dS F•N S −S2 1 ZZZ ZZ ˆ dS = ± = F•N div F d V = 0, S3 R =

10. The curve C: (x − 1)2 + 4y 2 = 16, 2x + y + z = 3,

oriented counterclockwise as seen from above, bounds an elliptic disk S on the √ plane 2x + y + z = 3. S has normal ˆ = (2i + j + k)/ 6. Since its projection onto the x yN plane is an elliptic disk with centre at√(1, 0, 0) and area π(4)(2) = 8π , therefore S has area 8 6π and centroid (1, 0, 1). If

then

k ∂ = 2xk. ∂z 2 z z +e

curl F • Nˆ d S

2x √ dS = S 3 = 2x¯ A = 2π, =

we have div F = 2αx + x + 3y 2 + βy 2 = 0RR if α = −1/2 ˆ d S for and β = −3. In this case we can evaluate S F• N any such surface S by evaluating the special case where S is the half-disk H : x 2 + y 2 ≤ 1, z = 0, y ≥ 0, with ˆ = k. We have upward normal N ZZ ZZ ˆ d S = −3 F•N y2 d x d y S H Z π Z 1 3π = −3 sin2 θ dθ r 3 dr = − . 8 0 0

provided that div F = 0 identically. Since F = (αx 2 − z)i + (x y + y 3 + z)j + βy 2 (z + 1)k,

F = (z 2 + y 2 + sin x 2 )i + (2x y + z)j + (x z + 2yz)k,

i ∂ ∂x z 2 + y 2 + sin x 2 = (2z − 1)i + zj.

curl F =

j ∂ ∂y 2x y + z

k ∂ ∂z x z + 2yz

By Stokes’s Theorem, I ZZ F • dr = curl F • Nˆ d S C S ZZ 1 =√ (2(2z − 1) + z) d S 6 S 5¯z − 2 √ = √ (8 6π ) = 24π. 6

11. As was shown in Exercise 13 of Section 7.2, ∇ × (φ ∇ ψ) = −∇ × (ψ × φ) = ∇ φ × ∇ ψ. Thus, by Stokes’s Theorem, I ZZ φ ∇ψ = ∇ × (φ ∇ ψ) • Nˆ d S C ZZS ˆ dS = (∇ φ × ∇ ψ) • N I ZZS ˆ dS − ψ ∇φ = −∇ × (ψ ∇ φ) • N C ZZS ˆ d S. = (∇ φ × ∇ ψ) • N S

∇ φ× ∇ ψ is solenoidal, with potential φ ∇ ψ, or −ψ ∇φ.

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SECTION 16.5 (PAGE 934)

ADAMS and ESSEX: CALCULUS 8

12. We are given that C bounds a region R in a plane P ˆ = ai + bj + ck. Therefore, with unit normal N 2 2 2 a + b + c = 1. If F = (bz − cy)i + (cx − az)j + (ay − bx)k, then i ∂ curl F = ∂x bz − cy

j ∂ ∂y cx − az = 2ai + 2bj + 2ck.

k ∂ ∂z ay − bx

ˆ = 2(a 2 + b2 + c2 ) = 2. We have Hence curl F • N I

1 (bz − cy) d x + (cx − az) d y + (ay − bx) dz 2 C I ZZ 1 1 = F • dr = curl F • Nˆ d S 2 C 2 R ZZ 1 = 2 d S = area of R. 2 R

Section 16.6 Some Physical Applications of Vector Calculus (page 942)

1. a) If we measure depth in the liquid by −z, so that the zaxis is vertical and z = 0 at the surface, then the pressure at depth −z is p = −ρgz, where ρ is the density of the liquid. Thus

∇ p = −ρgk = ρg, where g = −gk is the constant downward vector acceleration of gravity. The force of the liquid on surface element d S of the ˆ is solid with outward (from the solid) normal N ˆ d S = −(−ρgz)N ˆ d S = ρgz N ˆ d S. dB = − pN Thus, the total force of the liquid on the solid (the buoyant force) is

13. The circle Cǫ of radius ǫ centred at P is the oriented boundary of the disk Sǫ of area π ǫ 2 having constant norˆ By Stokes’s Theorem, mal field N. I

Cǫ

F • dr = =

ZZ

ZZSǫ

curl F • Nˆ d S

+

R

curl F(P) • Nˆ d S

SǫZZ

ZZ ˆ dS B = ρgz N S ZZZ ∇ (ρgz) d V (see Theorem 7) = R ZZZ =− ρg d V = −Mg,



 curl F − curl F(P) • Nˆ d S

Sǫ 2 ˆ = π ǫ curl F(P) • N ZZ   + curl F − curl F(P) • Nˆ d S. Sǫ

where M =

I 1 π ǫ2

ˆ F • dr − curl F(P) • N CZZ ǫ   1 ˆ d S ≤ curl F − curl F(P) • N π ǫ 2 Sǫ ≤ max |curl F(Q) − curl F(P)| Q on Sǫ → 0 as ǫ → 0+.

ǫ→0+

I

Cǫ

ρ d V is the mass of the liquid R

which would occupy the same space as the solid. Thus B = −F, where F = Mg is the weight of the liquid displaced by the solid.

Since F is assumed smooth, its curl is continuous at P. Therefore

Thus lim

ZZZ

ˆ F • dr = curl F(P) • N.

z

y x

z

dS R

Fig. 16.6.1

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ˆ N S

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.6 (PAGE 942)

b) The above argument extends to the case where the solid is only partly submerged. Let R ∗ be the part of the region occupied by the solid that is below the surface of the liquid. Let S∗ = S1 ∪ S2 be the boundary of R ∗ , with S1 ⊂ S and S2 in the plane of the surface of the liquid. Since p = −ρgz = 0 on S2 , we have ZZ ˆ d S = 0. ρgz N S2 Therefore the buoyant force on the solid is ZZ ˆ dS B= ρgz N S 1 ZZ ZZ ˆ dS + ˆ dS = ρgz N ρgz N S2 ZZ S1 ˆ dS = ∗ ρgz N SZZZ =− ρg d V = −M ∗ g, R∗

where M ∗ =

ZZZ

ρ d V is the mass of the liquid R∗

which would occupy R ∗ . Again we conclude that the buoyant force is the negative of the weight of the liquid displaced.

3. Suppose the closed surface S bounds a region R in which charge is distributed with density ρ. Since the electric field E due to the charge satisfies div E = kρ, the total flux of E out of R through S is, by the Divergence Theorem, ZZ ZZZ ZZZ ˆ dS =

E•N div E d V = k ρ d V = k Q, S R R RRR where Q = R ρ d V is the total charge in R.

4. If f is continuous and vanishes outside a bounded

region (say the ball of radius a centred at r), then | f (ξ, η, ζ )| ≤ K , and, if (R, φ, θ ) denote spherical coordinates centred at r, then Z 2π Z π Z a 2 | f (s)| R d V ≤ K dθ sin φ dφ dR s R3 |r − s| 0 0 0 R = 2π K a 2 a constant.

ZZZ

5. This derivation is similar to that of the continuity equation for fluid motion given in the text. If S is an (imaginary) surface bounding an arbitrary region D, then the rate of change of total charge in D is ∂ ∂t

ZZZ

D

ρ dV =

ZZZ

D

∂ρ dV, ∂t

where ρ is the charge density. By conservation of charge, this rate must be equal to the rate at which charge is crossing S into D, that is, to

S2

R∗

I

S

S1

ˆ dS = − (−J) • N

ZZZ

div J d V . D

ˆ is the outward (from (The negative sign occurs because N D) normal on S.) Thus we have ZZZ  D

Fig. 16.6.1 ˆ is (F1 G) • N. ˆ Applying 2. The first component of F(G • N) the Divergence Theorem and Theorem 3(b), we obtain ZZ ZZZ ˆ dS =

(F1 G) • N div (F1 G) d V S ZZZ D   = ∇ F1 • G + F1 ∇ • G d S. D

But ∇ F1 • G is the first component of (G • ∇ )F, and F1 ∇ • G is the first component of Fdiv G. Similar results obtain for the other components, so ZZ ZZZ   ˆ dS =

F(G • N) Fdiv G + (G • ∇ )F d V . S D

 ∂ρ + div J d V = 0. ∂t

Since D is arbitrary and we are assuming the integrand is continuous, it must be 0 at every point: ∂ρ + div J = 0. ∂t

6. Since r = xi + yj + zk and b = b1 i + b2 j + b3 k, we have |r − b|2 = (x − b1 )2 + (y − b2 )2 + (z − b3 )2 ∂ 2|r − b| |r − b| = 2(x − b1 ) ∂x ∂ x − b1 |r − b| = . ∂x |r − b|

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SECTION 16.6 (PAGE 942)

ADAMS and ESSEX: CALCULUS 8

Similar formulas hold for the other first partials of |r − b|, so   1 ∇ |r − b|   −1 ∂ ∂ = |r − b|i + · · · + |r − b|k |r − b|2 ∂ x ∂z −1 (x − b1 )i + (y − b2 )j + (z − b3 )k = |r − b|2 |r − b| r−b =− . |r − b|3

10. The first component of (ds • ∇)F(s) is ∇ F1 (s)• ds. Since F is closed and ∇ F1 is conservative, i•

I

F

(ds • ∇ )F(s) =

I

F

∇ F1 (s) • ds = 0.

Similarly, the other components have zero line integrals, so I (ds • ∇ )F(s) = 0. F

7. Using the result of Exercise 4 and Theorem 3(d) and (h), we calculate, for constant a,   r−b div a × |r − b|3   1 = −div a × ∇ |r − b| 1 1 +a•∇×∇ = 0 + 0 = 0. = −(∇ × a) • ∇ |r − b| |r − b|

11. Using the results of Exercises 7 and 8, we have curl

  I ds × (r − s) r−s = curl ds × =0 |r − s|3 F |r − s|3 F

I

for r not on F. (Again, this is because the curl is taken with respect to r, so s and ds can be regarded as constant for the calculation of the curl.)

8. For any element ds on the filament F, we have 

div ds ×

r−s |r − s|3



12. By analogy with the filament case, the current in volume

=0

by Exercise 5, since the divergence is taken with respect to r, and so s and ds can be regarded as constant. Hence   I I r−s ds × (r − s) = 0. = div ds × div |r − s|3 F F |r − s|3

9. By the result of Exercise 4 and Theorem 3(e), we calculate 

 r−b |r − b|3   1 = −curl a × ∇ |r − b|     1 1 a− ∇ •∇ a =− ∇•∇ |r − b| |r − b| 1 1 + (∇ • a)∇ + (a • ∇ )∇ . |r − b| |r − b|

element d V at position s is J(s) d V , which gives rise at position r to a magnetic field dB(r) =

If R is a region of 3-space outside which J is identically zero, then at any point r in 3-space, the total magnetic field is

curl a ×

1 Observe that ∇ • ∇ = 0 for r 6= b, either by direct |r − b| 1 calculation or by noting that ∇ is the field of a |r − b| point source at r = b and applying the result of Example 3 of Section 7.1.   1 Also − ∇ • ∇ a = 0 and ∇ • a = 0, since a is |r − b| constant. Therefore we have   r−b 1 = (a • ∇ )∇ curl a × 3 |r − b| |r − b| r−b = −(a • ∇ ) . |r − b|3

1 J(s) × (r − s) dV. 4π |r − s|3

B(r) =

1 4π

ZZZ

R

J(s) × (r − s) dV. |r − s|3

Now A(r) was defined to be A(r) =

1 4π

ZZZ

R

J(s) dV. |r − s|

We have   ZZZ 1 1 J(s) d V ∇r × 4π |r − s| ZZZ R 1 1 = ∇r × J(s) d V 4π |r − s| R (by Theorem 3(c)) ZZZ 1 (r − s) × J(s) =− dV 4π |r − s|3 R (by Exercise 4) = B(r).

curl A(r) =

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INSTRUCTOR’S SOLUTIONS MANUAL

13.

14.

I 4π I div A(r) = 4π I = 4π A(r) =

SECTION 16.6 (PAGE 942)

I

ds |r − s|  IF

Thus U = E and U = B both satisfy the wave equation

 1 ds |r − s|  IF  1 ∇ • ds |r − s| F (by Theorem 3(b)) = 0 for r not on F, since ∇(1/|r − s|) is conservative. ZZZ 1 J(s) d V A(r) = , where R is a region of 34π R |r − s| space such that J(s) = 0 outside R. We assume that J(s) is continuous, so J(s) = 0 on the surface S of R. In the following calculations we use subscripts s and r to denote the variables with respect to which derivatives are taken. By Theorem 3(b),

div s

∂2U = c2 ∇ 2 U, ∂t 2

div r

  J(s) 1 1 = ∇s • J(s) + ∇s • J(s) |r − s| |r − s| |r − s|   1 • J(s) + 0 = −∇ r |r − s|

c2 =

where

1 . µ0 ǫ0

∂B 6= 0. Thus in the ∂t non-static case we cannot have E = −∇φ.

16. curl (−∇φ) = 0, but curl E = −

On the other hand, we can still that curl A = B since div curl A = div B = 0 because there are no magnetic monopoles.

17. If E = −∇ φ −

∂A , then ∂t

curl E = −curl ∇ φ −

∂ curl A ∂B ∂B =0− =− . ∂t ∂t ∂t

18. The internal energy of an arbitrary region R (with surface S) at time t is

H (t) = ρc

ZZZ

T (x, y, z, t) d V .

because ∇ r |r − s| = −∇s |r − s|, and because ∇ • J = ∇ • (∇ × B) = 0 by Theorem 3(g). Hence

This internal energy increases at (time) rate

ZZZ 

dH = ρc dt

1 div A(r) = 4π

1 =− 4π 1 =− 4π

 1 ∇r • J(s) d V |r − s| R ZZZ J(s) dV ∇s • |r − s| ZZ R J(s) ˆ dS = 0 •N

S |r − s|

since J(s) = 0 on S. By Theorem 3(i), J = ∇ × B = ∇ × (∇ × A) = ∇ (∇ • A) − ∇ 2 A = −∇ 2 A.

∂B curl E = −µ0 ∂t

div B = 0 curl B = ǫ0

∂E ∂t

Therefore,

curl curl E = grad div E − ∇ 2 E = −∇ 2 E ∇ 2 E = −curl curl E = µ0

∂ ∂ 2E curl B = µ0 ǫ0 2 . ∂t ∂t

ZZZ

R

∂T dV. ∂t

If heat is not “created” or “destroyed” (by chemical or other means) within R, then the increase in internal energy must be due to heat flowing into R across S. The rate of flow of heat into R across surface element ˆ is d S with outward normal N ˆ d S. −k ∇ T • N Therefore, the rate at which heat enters R through S is ZZ ˆ d S. k ∇T • N S

15. By Maxwell’s equations, since ρ = 0 and J = 0, div E = 0

R

By conservation of energy and the Divergence Theorem we have ZZ ZZZ ∂T ˆ dS d V = k ∇T • N ρc S R ∂t ZZZ =k ∇ • ∇T d V ZZZ R =k ∇2 T d V . R

Similarly, ∂ 2B ∇ B = µ0 ǫ0 2 . ∂t 2

Thus,

ZZZ  R

∂T k 2 − ∇ T ∂t ρc



d V = 0.

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SECTION 16.6 (PAGE 942)

ADAMS and ESSEX: CALCULUS 8

Since R is arbitrary, and the temperature T is assumed to be smooth, the integrand must vanish everywhere. Thus

6.

  ∂T k 2 k ∂2T ∂2T ∂2T = ∇ T= + + . ∂t ρc ρc ∂ x 2 ∂ y2 ∂z 2

7. Section 16.7 Orthogonal Curvilinear Coordinates (page 952) 1.

f (r, θ, z) = r θ z 9,

 ∂  2 R =0 ∂θ ˆ R R φˆ R sin φ θˆ ∂ ∂ ∂ 1 curl F = 2 ∂ R ∂φ ∂θ R sin φ 2 0 0 R sin φ ˆ ˆ − 2 φ. = cot φ R

∇f =

2.

f (R, φ, θ ) = Rφθ (spherical coordinates). By Example 10, 1 ∂f ˆ 1 ∂f ˆ ∂f ˆ R+ φ+ θ ∇f = ∂R R ∂φ R sin φ ∂θ ˆ ˆ + θ φˆ + φ θ. = φθ R sin φ

F(R, φ, θ ) = R θˆ

div F =

(cylindrical coordinates). By Example ∂f 1 ∂f ˆ ∂f rˆ + θ+ k ∂r r ∂θ ∂z = θ z rˆ + z θˆ + r θ k.

F(R, φ, θ ) = R φˆ   ∂  2 1 div F = 2 R sin φ = cot φ R sin φ ∂φ ˆ R R φˆ R sin φ θˆ ∂ ∂ ∂ 1 = 2 θ. ˆ curl F = 2 ∂θ R sin φ ∂ R ∂φ 0 R2 0

8.

1 R 2 sin φ



ˆ F(R, φ, θ ) = R 2 R   1 ∂  4 div F = 2 R sin φ = 4R R sin φ ∂ R ˆ R R φˆ R sin φ θˆ ∂ ∂ ∂ 1 = 0. curl F = 2 ∂θ R sin φ ∂ R ∂φ 2 R 0 0

9. Let r = x(u, v) i + y(u, v) j. The scale factors are 3.

4.

5.

F(r, θ, z) = r rˆ   1 ∂ 2 (r ) = 2 div F = r ∂r rˆ r θˆ k ∂ ∂ 1 ∂ curl F = r ∂r ∂θ ∂z r 0 0 F(r, θ, z) = r θˆ   1 ∂ div F = (r ) = 0 r ∂θ rˆ r θˆ k ∂ ∂ 1 ∂ curl F = ∂r ∂θ ∂z r 0 r2 0

∂r h u = ∂u

∂r and h v = . ∂v

The local basis consists of the vectors = 0.

= 2k.

uˆ =

1 ∂r h u ∂u

and vˆ =

1 ∂r . h v ∂v

The area element is d A = h u h v du dv.

10. Since (u, v, z) constitute orthogonal curvilinear coordinates in R3 , with scale factors h u , h v and h z = 1, we have, for a function f (u, v) independent of z, 1 hu 1 = hu

∇ f (u, v) =

ˆ F(R, φ, θ ) = sin φ R   1 2 sin φ ∂  2 2  div F = 2 R sin φ = R sin φ ∂ R R ˆ ˆ R sin φ θˆ R φ R ∂ 1 ∂ ∂ curl F = 2 R sin φ ∂ R ∂φ ∂θ sin φ 0 0 cos φ ˆ =− θ. R

∂f 1 uˆ + ∂u hv 1 ∂f uˆ + ∂u hv

∂f 1 ∂f vˆ + k ∂v 1 ∂z ∂f vˆ . ∂v

For F(u, v) = Fu (u, v) uˆ + Fv (u, v) vˆ (independent of z and having no k component), we have   1 ∂ ∂ div F(u, v) = (h u Fu ) + (h v Fv ) h u h v ∂u ∂v h u uˆ ˆ h v k v 1 ∂ ∂ ∂ curl F(u, v) = h u h v ∂u ∂v ∂z h F h F 0 v v  u u  ∂ 1 ∂ (h v Fv ) − (h u Fu ) k. = h u h v ∂u ∂v

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 16.7 (PAGE 952)

11. We can use the expressions calculated in the text for

The scale factors are p ∂r h u = = a sinh2 u cos2 v + cosh2 u sin2 v ∂u p ∂r h v = = a sinh2 u cos2 v + cosh2 u sin2 v = h u . ∂v

cylindrical coordinates, applied to functions independent of z and having no k components: ∂f 1 ∂f ˆ rˆ + θ ∂r r ∂θ ∂ Fr Fr 1 ∂ Fθ div F(r, θ ) = + + ∂r r r ∂θ   Fθ 1 ∂ Fr ∂ Fθ + − k. curl F(r, θ ) = ∂r r r ∂θ

∇ f (r, θ ) =

12.

x = a cosh u cos v,

A2

d A = h u h v du dv   = a 2 sinh2 u cos2 v + cosh2 u sin2 v du dv.

y = a sinh u sin v.

a) u-curves: If A = a cosh u and B = a sinh u, then x2

The area element is

+

y2 B2

13.

= cos2 v + sin2 v = 1.

Since A2 − B 2 = a 2 (cosh2 u − sinh2 u) = a 2 , the u-curves are ellipses with foci at (±a, 0). b) v-curves: If A = a cos v and B = a sin v, then x2 y2 − 2 = cosh2 u − sinh2 u = 1. 2 A B

The coordinate curves are u-curves: the horizontal hyperbolas in which the v = v 0 cylinders intersect the z = z 0 planes. v-curves: the horizontal ellipses in which the u = u 0 cylinders intersect the z = z 0 planes. z-curves: sets of four vertical straight lines where the elliptic cylinders u = u 0 and hyperbolic cylinders v = v 0 intersect.

Since A2 + B 2 = a 2 (cos2 v + sin2 v) = a 2 , the v-curves are hyperbolas with foci at (±a, 0). c) The u-curve u = u 0 has parametric equations x = a cosh u 0 cos v,

y = a sinh u 0 sin v,

and therefore has slope at (u 0 , v 0 ) given by

14. mu =

dy dy = dx dv



a sinh u 0 cos v 0 d x = . dv (u 0 ,v0 ) −a cosh u 0 sin v 0

The v-curve v = v 0 has parametric equations x = a cosh u cos v 0 ,

y = a sinh u sin v 0 ,

and therefore has slope at (u 0 , v 0 ) given by mv =

dy dy = dx du



a cosh u 0 sin v 0 d x = . du (u 0 ,v0 ) a sinh u 0 cos v 0

Since the product of these slopes is m u m v = −1, the curves u = u 0 and v = v 0 intersect at right angles. d)

r = a cosh u cos v i + a sinh u sin v j ∂r = a sinh u cos v i + a cosh u sin v j ∂u ∂r = −a cosh u sin v i + a sinh u cos v j. ∂v

x = a cosh u cos v y = a sinh u sin v z = z. Using the result of Exercise 12, we see that the coordinate surfaces are u = u 0 : vertical elliptic cylinders with focal axes x = ±a, y = 0. v = v 0 : vertical hyperbolic cylinders with focal axes x = ±a, y = 0. z = z 0 : horizontal planes.

15.

1 ∂f ˆ ∂f ∂f rˆ + k θ+ ∂r  r ∂θ ∂z ∇ 2 f (r, θ, z) = div ∇ f (r, θ, z)        1 ∂ ∂f ∂ 1 ∂f ∂ ∂f = r + + r r ∂r ∂r ∂θ r ∂θ ∂z ∂z ∂2 f 1 ∂f 1 ∂2 f ∂2 f = + + 2 + 2. ∂r 2 r ∂r r ∂θ 2 ∂z

∇ f (r, θ, z) =

∂f ˆ 1 ∂f ˆ 1 ∂f ˆ R+ φ+ θ ∂R R ∂φ R sin φ ∂θ   ∇ 2 f (R, φ, θ ) = div f (R, φ, θ )      1 ∂ ∂f ∂ 1 ∂f 2 R sin φ + R sin φ = 2 R sin φ ∂ R ∂R ∂φ R ∂φ   ∂ R ∂f + ∂θ R sin φ ∂θ ∂2 f 2 ∂f 1 ∂2 f = + + R ∂R ∂ R2 R 2 ∂φ 2 cot φ ∂ f 1 ∂2 f + + 2 2 . 2 R ∂φ R sin φ ∂θ 2

∇ f (R, φ, θ ) =

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SECTION 16.7 (PAGE 952)

16.

ADAMS and ESSEX: CALCULUS 8

1 ∂f 1 ∂f 1 ∂f ˆ uˆ + vˆ + w h u ∂u h v ∂v h w ∂w   ∇ 2 f (u, v, w) = div ∇ f (u, v, w)      ∂ hu hw ∂ f 1 ∂ hv hw ∂ f = + h u h v h w ∂u h u ∂u ∂v h v ∂v   ∂ hu hv ∂ f + ∂w h w ∂w  2    1 ∂ f 1 ∂h v 1 ∂h w 1 ∂h u ∂ f = 2 + + − h u ∂u 2 h v ∂u h w ∂u h u ∂u ∂u     1 ∂2 f 1 ∂h u 1 ∂h w 1 ∂h v ∂ f + 2 + + − h v ∂v 2 h u ∂v h w ∂v h v ∂v ∂v  2    1 ∂ f 1 ∂h u 1 ∂h v 1 ∂h w ∂ f + 2 + + − . h u ∂w h v ∂w h w ∂w ∂w h w ∂w2

∇ f (u, v, w) =

Review Exercises 16

ZZ

Also

ZZ F • (−k) d A = − e0 d A = −π a 2 D2 D2 ZZ ZZ F • kdA = eb d A = π a 2 eb . D1

By the Divergence Theorem ZZ

S

ˆ dS + F•N =

Therefore,

ZZ

S

ZZ

D1

ZZZ

D1

F • kdA +

ZZ

D2

F • (−k) d A

div F d V = π a 2 b + π a 2 (eb − 1).

R

ˆ d S = π a 2 b. F•N

z

ˆ1 = k N

b

D1

(page 952)

1. The semi-ellipsoid S with upward normal Nˆ specified in

the problem and the disk D given by x 2 + y 2 ≤ 16, z = 0, with downward normal p −k together bound the solid region R: 0 ≤ z ≤ 21 16 − x 2 − y 2 . By the Divergence Theorem: ZZ ZZ ZZZ ˆ dS + F•N F • (−k) d A = div F d V . S D R For F = ZZZ

x 2 zi + (y 2 z

S

D2

R

3 4 2 π 4 2 = 64π. = 2 3

The flux of F across S is ZZ ZZ ˆ d S = 64π + F•N F • kdA S ZZ D = 64π +

Z

D 2π

cos2 θ dθ

0

y

Fig. R-16.2

3.

I

2

2

(3y 2 + 2xe y ) d x + (2x 2 ye y ) d y CZZ 2 2 [4x ye y − (6y + 4x ye y )] d A = PZZ = −6 y d A = −6 y¯ A = −6, P

since P has area A = 2 and its centroid has y-coordinate y¯ = 1/2. y

x2 d A

= 64π +

ˆ 2 = −k N

2a

x

x 2k

+ 3y)j + we have ZZZ div F d V = (2x z + 2yz + 3) d V R R ZZZ d V = 3 × (volume of R) = 0+0+3

R

ˆ N

(1, 1) Z

0

4

r 3 dr = 128π.

(3, 1)

P

C

2. Let R be the region inside the cylinder S and between the planes z = 0 and z = b. The oriented boundary of R ˆ 1 = k and consists of S and the disks D1 with normal N ˆ 2 = −k as shown in the figure. For D2 with normal N F = xi + cos(z 2 )j + e z k we have div F = 1 + e z and ZZZ ZZ Z b div F d V = dx dy (1 + e z ) dz R D2 0 ZZ = [b + (eb − 1)] d x d y D2 2

= π a b + π a 2 (eb − 1).

(2, 0) Fig. R-16.3

4. If F = −zi + xj + yk, then i ∂ curl F = ∂x −z

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j ∂ ∂y x

k ∂ ∂z y

= i − j + k.

x

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 16 (PAGE 952)

ˆ to a region in the plane The unit normal N 2x + y + 2z = 7 is ˆ = ± 2i + j + 2k . N 3

8. If curl F = µF on R3 , where µ 6= 0 is a constant, then div F =

by Theorem 3(g) of Section 7.2. By part (i) of the same theorem,

If C is the boundary of a disk D of radius a in that plane, then I

C

F • dr =

ZZ

=±

curl F • Nˆ d S

D ZZ

D

2−1+2 d S = ±π a 2 . 3

5. If Sa is the sphere of radius a centred at the origin, then ZZ ˆ dS

F•N a→0+ 4 π a 3 Sa 3 3 3 (π a 3 + 2a 4 ) = . = lim a→0+ 4π a 3 4

div F(0, 0, 0) = lim

1

6. If S is any surface with upward normal Nˆ and boundary

the curve C: x 2 + y 2 = 1, z = 2, then C is oriented counterclockwise as seen from above, and it has parametrization r = cos ti + sin tj + 2k (0 ≤ 2 ≤ 2π ).

Thus dr = (− sin ti + cos tj) dt, and if F = −yi + x cos(1 − x 2 − y 2 )j + yzk, then the flux of curl F upward through S is ZZ

S

curl F • Nˆ d S =

I

F • dr C Z 2π = (sin2 t + cos2 t + 0) dt = 2π. 0

7. F(r) = r λ r where r = xi + yj + zk and r = |r|. Since r 2 = x 2 + y 2 + z 2 , therefore ∂r/∂ x = x/r and

x2 ∂ λ (r x) = λr λ−1 + r λ = r λ−2 (λx 2 + r 2 ). ∂x r Similar expressions hold for (∂/∂ y)(r λ y) and (∂/∂z)(r λ z), so

div F(r) = r λ−2 (λr 2 + 3r 2 ) = (λ + 3)r λ . F is solenoidal on any set in R3 that excludes the origin if an only if λ = −3. In this case F is not defined at r = 0. There is no value of λ for which F is solenoidal on all of R3 .

1 div curl F = 0 µ

∇ 2 F = ∇ (div F) − curl curl F = 0 − µcurl F = −µ2 F. Thus ∇ 2 F + µ2 F = 0.

9. Apply the variant of the Divergence Theorem given in Theorem 7(b) of Section 7.3, namely ZZZ ZZ grad φ d V = φ Nˆ d S, P S to the scalar field φ = 1 over the polyhedron P. Here n [ S = Fi is the surface of P, oriented with outward i=1

ˆ i on the face Fi . If Ni = Ai N ˆ i , where Ai normal field N is the area of Fi , then, since grad φ = 0, we have ZZ n n n ZZ X X X Ni Ni ˆ dS = Ai = Ni . 0 = N dS = A S Fi Ai i=1 i i=1 i=1

10. Let C be a simple, closed curve in the x y-plane bounding a region R. If F = (2y 3 − 3y + x y 2)i + (x − x 3 + x 2 y)j, then by Green’s Theorem, the circulation of F around C is I F • dr CZZ   ∂ ∂ (x − x 3 + x 2 y) − (2y 3 − 3y + x y 2 ) d A = ∂x ∂y ZZ R = (1 − 3x 2 + 2x y − 6y 2 + 3 − 2x y) d A ZZ R = (4 − 3x 2 − 6y 2 ) d x d y. R

The last integral has a maximum value when the region R is bounded by the ellipse 3x 2 + 6y 2 = 4, oriented counterclockwise; this is the largest region in the x yplane where the integrand is nonnegative.

11. Let S be a closed, oriented surface in R3 bounding a ˆ If region R, and having outward normal field N.

F = (4x + 2x 3 z)i − y(x 2 + z 2 )j − (3x 2 z 2 + 4y 2 z)k, then by the Divergence Theorem, the flux of F through S is ZZ ZZZ ZZZ ˆ dS =

F•N div F d V = (4−x 2 −4y 2 −z 2 ) d V . S R R

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REVIEW EXERCISES 16 (PAGE 952)

ADAMS and ESSEX: CALCULUS 8

ZZ ˆ d S = 0 by Theorem 7(b) of Section 7.3 But N(r) Sǫ with φ = 1. Also, since v satisfies

The last integral has a maximum value when the region R is bounded by the ellipsoid x 2 + 4y 2 + z 2 = 4 with outward normal; this is the largest region in R3 where the integrand is nonnegative.

v(r2 ) − v(r1 ) = C|r2 − r1 |2 ,

12. Let C be a simple, closed curve on the plane

we have

x + y + z = 1, oriented counterclockwise as seen from above, and bounding a plane region √ S on x + y + z = 1. ˆ = (i + j + k)/ 3. If Then S has normal N F = x y 2i + (3z − x y 2 )j + (4y − x 2 y)k, then

ZZ   r−r 1

v(r) − v(r1 ) • dS ǫ Sǫ ZZ Cǫ 2 = d S = 4π Cǫ 3 . Sǫ ǫ

j k i ∂ ∂ ∂ curl F = ∂y ∂z ∂x 2 x y 3z − x y 2 4y − x 2 y = (1 − x 2 )i + 2x yj − (y 2 + 2x y)k.

Thus

div v(r1 ) = lim

ǫ→0+

The divergence of the large-scale velocity field of matter in the universe is three times Hubble’s constant C.

By Stokes’s Theorem we have

2. I

C

F • dr =

ZZ

S

curl F • Nˆ d S =

ZZ

1 − x 2 − y2 √ d S. 3 S

The last integral will be maximum if the projection of S onto the x y-plane is the disk x 2 + y 2 ≤ 1. This maximum value is

(page 953)

1. By Theorem 1 of Section 7.1, we have

ǫ→0+

ZZ 3 ˆ

v(r) • N(r) d S. 4π ǫ 3 Sǫ

= area of S1 .

Here Sǫ is the sphere of radius ǫ centred at the point (with position vector) r1 and having outward normal field ˆ N(r). If r is (the position vector of) any point on Sǫ , ˆ then r = r1 + ǫ N(r), and ZZ ˆ

v(r) • N(r) dS SZǫZ h  i ˆ = v(r1 ) + v(r) − v(r1 ) • N(r) dS Sǫ ZZ ˆ = v(r1 ) • N(r) dS S ǫ ZZ   r−r 1 + v(r) − v(r1 ) • d S. ǫ Sǫ

0

b) If S is the intersection of a smooth surface with the general half-cone K , and is oriented with normal ˆ pointing away from the vertex P of K , and field N if Sa is the intersection with K of a sphere of radius a centred at P, with a chosen so that S and Sa do not intersect in K , then S, Sa , and the walls of K bound a solid region R that does not contain the origin. If F = r/|r|3 , then div F = 0 in R (see ˆ = 0 on the Example 3 in Section 7.1), and F • N walls of K . It follows from the Divergence Theorem applied to F over R that ZZ ZZ r ˆ dS = dS F• F•N |r| S Sa ZZ a2 1 = 4 d S = 2 (area of Sa ) a a Sa

1 − x 2 − y2 √ √ 3 dx dy 3 x 2 +y 2 ≤1   Z 1 Z 2π 1 1 π (1 − r 2 )r dr = 2π dθ = − = . 2 4 2 0 0

div v(r1 ) = lim

a) The steradian measure of a half-cone of semi-vertical angle α is Z 2π Z α dθ sin φ dφ = 2π(1 − cos α). 0

ZZ

Challenging Problems 16

3 (0 + 4π Cǫ 3 ) = 3C. 4π ǫ 3

The area of S1 (the part of the sphere of radius 1 in K ) is the measure (in steradians) of the solid angle subtended by K at its vertex P. Hence this measure is given by ZZ r ˆ d S. •N |r| S 3

3.

a) Verification of the identity     ∂ ∂r ∂ ∂r G• − G• ∂t ∂s ∂s ∂t   ∂F ∂r ∂r ∂r = • + (∇ × F) × • . ∂t ∂s ∂t ∂s

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INSTRUCTOR’S SOLUTIONS MANUAL

CHALLENGING PROBLEMS 16 (PAGE 953)

can be carried out using the following MapleV commands: > > > > >

4.

a) Verification of the identity       ∂r ∂r ∂ ∂r ∂r G• × − G• × ∂u ∂v ∂u ∂t ∂v    ∂ ∂r ∂r − × G• ∂v ∂u ∂t     ∂F ∂r ∂r ∂r ∂r ∂r = • × × + (∇ • F) • . ∂t ∂u ∂v ∂t ∂u ∂v

∂ ∂t

with(linalg): F:=(x,y,z,t)-> [F1(x,y,z,t), F2(x,y,z,t),F3(x,y,z,t)]; r:=(s,t)->[x(s,t),y(s,t),z(s,t)];

G:=(s,t)->F(x(s,t),y(s,t),z(s,t),t); > g:=(s,t)-> dotprod(G(s,t), > map(diff,r(s,t),s)); > h:=(s,t)-> dotprod(G(s,t), > map(diff,r(s,t),t)); > LH1:=diff(g(s,t),t); > LH2:=diff(h(s,t),s); > LHS:=simplify(LH1-LH2); >

RH1:=dotprod(subs(x=x(s,t),y=y(s,t), > z=z(s,t),diff(F(x,y,z,t),t)), > diff(r(s,t),s)); >

RH2:=dotprod(crossprod(subs(x=x(s,t), > y=y(s,t),z=z(s,t), > curl(F(x,y,z,t),[x,y,z])), > diff(r(s,t),t)),diff(r(s,t),s)); > RHS:=RH1+RH2; LHS-RHS; simplify(%);

can be carried out using the following MapleV commands: > > > > > > > > > > >

with(linalg): F:=(x,y,z,t)->[F1(x,y,z,t), F2(x,y,z,t),F3(x,y,z,t)]; r:=(u,v,t)->[x(u,v,t),y(u,v,t), z(u,v,t)]; ru:=(u,v,t)->diff(r(u,v,t),u); rv:=(u,v,t)->diff(r(u,v,t),v); rt:=(u,v,t)->diff(r(u,v,t),t); G:=(u,v,t)->F(x(u,v,t), y(u,v,t),z(u,v,t),t);

ruxv:=(u,v,t)->crossprod(ru(u,v,t), > rv(u,v,t)); >

We omit the output here; some of the commands produce screenfulls of output. The output of the final command is 0, indicating that the identity is valid. b) As suggested by the hint,

d dt

Z

Ct b

F • dr =

a

b

∂ ∂t 

  ∂r G• ds ∂s

 ∂ ∂r = G• ∂s ∂t a      ∂ ∂r ∂ ∂r + G• − G• ds ∂t ∂s ∂s ∂t s=b ∂r = G • ∂t s=a   Z b ∂F ∂r ∂r + + (∇ × F) × • ds ∂t ∂t ∂s a    = F r(b, t), t • vC (b, t) − F r(a, t), t • vC (a, t) Z Z   ∂F + • dr + (∇ × F) × vC • dr. Ct ∂t Ct Z



Z

rtxv:=(u,v,t)->crossprod(rt(u,v,t), > rv(u,v,t)); >

ruxt:=(u,v,t)->crossprod(ru(u,v,t), > rt(u,v,t)); > LH1:=diff(dotprod(G(u,v,t), > ruxv(u,v,t)),t); > LH2:=diff(dotprod(G(u,v,t), > rtxv(u,v,t)),u); > LH3:=diff(dotprod(G(u,v,t), > ruxt(u,v,t)),v); > LHS:=simplify(LH1-LH2-LH3); > RH1:=dotprod(subs(x=x(u,v,t), > y=y(u,v,t),z=z(u,v,t), >

diff(F(x,y,z,t),t)),ruxv(u,v,t)); > RH2:=(divf(u,v,t))* >

(dotprod(rt(u,v,t),ruxv(u,v,t))); > RHS:=simplify(RH1+RH2); > simplify(LHS-RHS);

Again the final output is 0, indicating that the identity is valid. b) If Ct is the oriented boundary of St and L t is the corresponding counterclockwise boundary of the

639 Copyright © 2014 Pearson Canada Inc.

CHALLENGING PROBLEMS 16 (PAGE 953)

ADAMS and ESSEX: CALCULUS 8

parameter region R in the uv-plane, then  I  ∂r • dr F× ∂t Ct    I  ∂r ∂r ∂r = G× • du + dv ∂t ∂u ∂v L     I t ∂r ∂r ∂r ∂r × +G• × dt = −G • ∂u ∂t ∂t ∂v Lt    ZZ  ∂ ∂r ∂r = G• × ∂u ∂t ∂v R    ∂ ∂r ∂r + G• × du dv, ∂v ∂u ∂t by Green’s Theorem. c) Using the results of (a) and (b), we calculate    ZZ ZZ d ∂ ∂r ∂r ˆ dS = F•N G• × du dv dt St ∂u ∂v R ∂t   ZZ ∂F ∂r ∂r = • × du dv ∂u ∂v R ∂t   ZZ ∂r ∂r ∂r × du dv + (div F) • ∂t ∂u ∂v R    ZZ  ∂ ∂r ∂r + G• × ∂u ∂t ∂v R    ∂ ∂r ∂r + G• × du dv ∂v ∂u ∂t ZZ ZZ ∂F ˆ ˆ dS (div F)v S • N • N dS + = St St ZZ∂t + Ct (F × vC ) • dr.

5. We have 1 1t

ZZZ

Dt+1t

f (r, t + 1t) d V −

ZZZ

f (r, t) d V Dt



f (r, t + 1t) − f (r, t) dV 1t Dt ZZZ 1 + f (r, t + 1t) d V 1t Dt+1t −Dt ZZZ 1 f (r, t + 1t) d V − 1t Dt −Dt+1t = I1 + I2 − I3 . =

ZZZ

Evidently I1 →

ZZZ

Dt

∂f d V as 1t → 0. ∂t

I2 and I3 are integrals over the parts of 1Dt where the surface §t is moving outwards and inwards, respectively, ˆ is, respectively, positive and negative. that is, where v S • N ˆ Since d V = |v S • N| d S 1T , we have

I2 − I3 = =

ZZ

ZZ

+

St

ˆ dS f (r, t + 1t)v S • N

St

ˆ dS f (r, t)v S • N

ZZ  St

 ˆ d S. f (r, t + 1t) − f (r, t) v S • N

The latter integral approaches 0 as 1t → 0 because ZZ   ˆ d S f (r, t + 1t) − f (r, t) v • N S St ∂f ≤ max |v S | (area of St )1t. ∂t

640 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 17.2 (PAGE 967)

CHAPTER 17. DIFFERENTIAL FORMS AND EXTERIOR CALCULUS Section 17.1 k-Forms

6. φ is a sum of wedge products of k differentials, and ψ is

(page 962)

1. There are six terms in the expanded wedge product of φ and ψ; each is a 4-form. Five of them have at least one of the basic forms d xi repeated twice, and so are zero. The only remaining nonzero form is

7.

φ ∧ ψ = a2 b1 d x3 ∧ d x4 ∧ d x1 ∧ d x2 = a2 b1 d x1 ∧ d x2 ∧ d x3 ∧ d x4 because {3, 4, 1, 2} is an even permutation of {1, 2, 3, 4}.

2. Only the first term of ψ contributes to the product; the other two give zero when wedged with φ. Thus φ ∧ ψ = d x2 ∧ d x3 ∧ d x4 ∧ d x1 = −d x1 ∧ d x2 ∧ d x3 ∧ d x4

8.

because {2, 3, 4, 1} is an odd permutation of {1, 2, 3, 4}.

3. When we calculate φ ∧ ψ the five terms of φ combine

with the two terms of ψ to give 10 terms. Eight of them have repeated differential factors and so are zero. The remaining two are

Section 17.2 Differential Forms and the Exterior Derivative (page 967)

5 d x5 ∧ d x1 ∧ d x2 ∧ d x3 ∧ d x4 and 2 d x1 ∧ d x2 ∧ d x3 ∧ d x4 ∧ d x5 . Since {5, 1, 2, 3, 4} is an even permutation of {1, 2, 3, 4, 5}, these two terms combine to give φ ∧ ψ = 7 d x1 ∧ d x2 ∧ d x3 ∧ d x4 ∧ d x5 .

4. Half of the 16 terms resulting from the wedge product

1. 2.

5. π = (12)(13)(14) · · · (1k), which involves k − 1 reversals. Thus π is odd if k is even and even if k is odd.

8 = x 2 d x + y 2 dz d8 = 2x d x ∧ d x + 2y d y ∧ dz = 2y d y ∧ dz f = x e2y si n(3z)

d f = e2y sin(3z) d x + 2x e2y si n(3z) d y + 3x e2y cos(3z) dz

3.

9 = x1 d x2 ∧ d x3 + x2 d x1 ∧ d x4 + (x3 + x4 ) d x1 ∧ d x2 d9 = d x1 ∧ d x2 ∧ d x3 + d x2 ∧ d x1 ∧ d x4 + d x3 ∧ d x1 ∧ d x2 + d x4 ∧ d x1 ∧ d x2 = 2 d x1 ∧ d x2 ∧ d x3 .

4.

2 = x1 x2 x3 d x1 ∧ d x3 ∧ d x5 + x3 x4 x5 d x2 ∧ d x4 ∧ d x5 d2 = d x2 ∧ d x1 ∧ d x3 ∧ d x5 + x4 x5 d x3 ∧ d x2 ∧ d x4 ∧ d x5 = −x1 x3 d x1 ∧ d x2 ∧ d x3 ∧ d x5 − x4 x5 d x2 ∧ d x3 ∧ d x4 ∧ d x5 .

5.

8 = e2y sin(3z) d x + 2x e2y si n(3z) d y + 3x e2y cos(3z) dz

of these two 4-term forms are zero because of repeated differentials. The remaining ones are φ ∧ ψ = a1 b1 d x1 ∧ d x2 ∧ d x3 + a1 b2 d x1 ∧ d x3 ∧ d x4 + a2 b2 d x2 ∧ d x3 ∧ d x4 + a2 b3 d x2 ∧ d x4 ∧ d x1 + a3 b3 d x3 ∧ d x4 ∧ d x1 + a3 b4 d x3 ∧ d x1 ∧ d x2 + a4 b1 d x4 ∧ d x2 ∧ d x3 + a4 b4 d x4 ∧ d x1 ∧ d x2 = (a1 b1 + a3 b4 )d x1 ∧ d x2 ∧ d x3 + (a1 b2 + a3 b3 )d x1 ∧ d x3 ∧ d x4 + (a4 b1 + a2 b2 )d x2 ∧ d x3 ∧ d x4 + (a2 b3 + a4 b4 )d x1 ∧ d x2 ∧ d x4 .

a sum of wedge products of ℓ differentials. To transform φ ∧ ψ to ψ ∧ φ, in each term of the product each of the ℓ differentials from ψ must pass leftward through all k differentials from φ. This requires k reversals for each of the ℓ differentials from ψ, and so a total of kℓ reversals in total (for each term of the product). Thus φ ∧ ψ = (−1)kl ψ ∧ φ 1 1 = −1 d x1 ∧ d x2 (u, v) = 1 0 1 1 1 1 1 =1 d x1 ∧ d x2 ∧ d x3 (u, v, w) = 1 0 0 = − 1 0 0 1 0 0 1 0 0 1 =1 d x3 ∧ d x4 ∧ d x1 (u, v, w) = 0 0 1 = − 1 1 1 1 1 0 1 0 d x3 ∧ d x2 ∧ d x4 (u, v, w) = 1 0 0 = −1 0 0 1 3 −4 1 2 2 3 −4 0 φ(v1 . . . , v4 ) = 0 3 −4 0 4 0 0 0 3 4 2 3 −4 = −(−4) 3 −4 0 = 16 −4 0 −4 0 0 = (16)(−16) = −256.

d8 = 2e2y sin(3z) d y ∧ d x + 3e2y cos(3z) dz ∧ d x

+ 2e2y sin(3z) d x ∧ d y + 6xe2y cos(3z) dz ∧ d y

+ 3e2y cos(3z) d x ∧ dz + 6xe2y cos(3z) d y ∧ dz = 0 the zero differential 2-form.

Not surprising, since 8 = d f , where f is the differential 0-form in Exercise 2, and so d8 = d 2 f = 0.

641 Copyright © 2014 Pearson Canada Inc.

SECTION 17.2 (PAGE 967)

6.

ADAMS and ESSEX: CALCULUS 8

8 = x1 x3 d x1 ∧ d x2 ∧ d x3 ∧ d x5 + x4 x5 d x2 ∧ d x3 ∧ d x4 ∧ d x14. 5 d8 = 0 each term has a repeated differential. Not surprising, since 8 = −d2, where 2 is the differential 3-form in Exercise 4, and so d8 = −d 2 2 = 0, the zero differential 4-form.

To make the calculations a bit more compact we can assume (without loss of generality) that x0 is the origin in Rk . If x ∈ D, then the line segment {t x : 0 ≤ t ≤ 1} lies in D and we want to show that f (x) =

7. Since a and b are constants, da = db = 0 and so

∂ f (x) = ∂ xj

8. Each term in 8 ∧ 2 is of the form

=

ai1 ···ik b j1 ··· jk d xi1 ∧ · · · ∧ d xik ∧ d x j1 ∧ · · · ∧ d jℓ . Now 

d ai1 ···ik b j1 ··· jℓ =

n X

m=1



∂b j1··· jℓ ∂ai1 ···ik b j1··· jℓ + ai1 ···ik ∂ xm ∂ xm



d xm

If we adjoin d xi1 ∧ · · · ∧ d xik ∧ d x j1 ∧ · · · ∧ d jℓ , the result will the corresponding term of (d8)∧2+(−1)k 8∧(d2). The (−1)k accounts for the fact that in the second term of (∗) the d xm must be shifted to the right of all the d xi s (k of them) to a position just before the first d x j to give d2.

9.

(∗)

d(81 ∧ · · · ∧ 8m ) m X = (−1)k1 +···+kj −1 81 ∧ · · · ∧ 8 j −1 ∧ (d8 j ) ∧ 8 j +1 ∧ · · · 8m . j =1

11. curl grad f = 0 16.

12. div curl (F1 i + F2 j + F3 k) = 0 k X i=1

(a)

(e) k X i=1

0

Z

0

k X

∂ai (tx) a j (tx) + t xi ∂ xj i=1

1

a j (tx) +

k X

t xi

i=1

∂a j (tx) ∂ xi

!

dt

!

dt,

Pk

j =1 a j (x) d x j

= 8.

∂S ∂V ∂T ∂µ =− (b) =− ∂T ∂N ∂S ∂µ ∂S ∂V ∂S ∂P = (d) = ∂P ∂T ∂V ∂T ∂V ∂T = ∂P ∂S

17.

dai (x) ∧ d xi

(a) We have

k X k X ∂ai (x) d x j ∧ d xi = ∂ xj i=1 j =1 X  ∂ai (x) ∂a j (x)  = − d x j ∧ d xi . ∂ xj ∂ xi 1≤ j 0 and X > 0, then d dx

xn+1 = xn + h

pn = cos(xn2 )

qn = cos((xn + (h/2))2 )



φ(x) ekx



=

ekx φ ′ (x) − kekx φ(x) ≥ 0. e2kx

Thus φ(x)/ekx is increasing on [0, X]. Since its value at x = 0 is φ(0) = A ≥ 0, therefore φ(x)/ekx ≥ A on [0, X], and φ(x) ≥ Aekx there.

2

rn = cos((xn + (h/2)) )

c) For h = 0.05 we get x20 = 1, y20 = 0.904524.

x1 = x0 + h = b    h a+b a+ = f = r0 2 2

which is the Simpson’s Rule approximation to Z b f (x) d x based on 2 subintervals of length h/2.

b) For h = 0.1 we get x10 = 1, y10 = 0.903122.

b) For h = 0.1 we get x10 = 1, y10 = 0.904524.

q0 = f



s0 = f (a + h) = f (b) h y1 = y0 + ( p0 + 2q0 + 2r0 + s0 ) 6    b−a a+b = f (a) + 4 f + f (b) , 6 2

yn + h cos(xn2 )

a) For h = 0.2 we get x5 = 1, y5 = 0.904524.

y0 = 0,

p0 = f (a),

11. We start with x0 = 0, y0 = 0, and calculate

qn = cos((xn + h)2 ) h yn+1 = yn + ( pn + 2qn + 2rn + sn ). 6

t 2 u(t) dt

2

Kutta method with h = b − a gives:

c) For h = 0.05 we get x20 = 1, y20 = 0.915666.

12. We start with x0 = 0, y0 = 0, and calculate

x

15. For the problem y ′ = f (x), y(a) = 0, the 1-step Runge-

b) For h = 0.1 we get x10 = 1, y10 = 0.926107.

c) For h = 0.05 we get x20 = 1, y20 = 0.904174.

Z

du = 3x 2 u(x), u(2) = 1 + 0 = 1 dx du = 3x 2 d x ⇒ ln u = x 3 + C u 0 = ln 1 = ln u(2) = 23 + C ⇒ C = −8

c) For h = 0.05 we get x20 = 1, y20 = 0.865769.

xn+1 = xn + h,

Z x 2 y(x) = 2 + y(t) dt 1  2 dy y(1) = 2 + 0 = 2 = y(x) , dx dy 1 = dx ⇒ − =x +C y2 y(x) 1 3 − =1+C ⇒ C =− 2 2 1 2 y=− = . x − (3/2) 3 − 2x

17.

a) Suppose u ′ = u 2 , y ′ = x + y 2 , and v ′ = 1 + v 2 on [0, X], where u(0) = y(0) = v(0) = 1, and X > 0 is such that v(x) is defined on [0, X]. (In part (b) below, we will show that X < 1, and we assume this fact now.) Since all three functions are increasing on [0, X], we have u(x) ≥ 1, y(x) ≥ 1, and v(x) ≥ 1 on [0, X].

653 Copyright © 2014 Pearson Canada Inc.

SECTION 18.3 (PAGE 1005)

ADAMS and ESSEX: CALCULUS 8

If φ(x) = y(x) − u(x), then φ(0) = 0 and 2

′

2

2

φ (x) = x + y − u ≥ y − u ≥ (y + u)(y − u) ≥ 2φ

2

on [0, X]. By Exercise 16, φ(x) ≥ 0 on [0, X], and so u(x) ≤ y(x) there. Similarly, since X < 1, if φ(x) = v(x) − y(x), then φ(0) = 0 and φ ′ (x) = 1 + v 2 − x − y 2 ≥ v 2 − y 2 ≥ (v + y)(v − y) ≥ 2φ

on [0, X], so y(x) ≤ v(x) there.

b) The IVP u ′ = u 2 , u(0) = 1 has solution 1 u(x) = , obtained by separation of variables. 1−x This solution is valid for x < 1. The IVP v ′ = 1 + v
2 , v(0) = 1 has solution v(x) = tan x + π4 , also obtained by separation of variables. It is valid only for −3π/4 < x < π/4. Observe that π/4 < 1, proving the assertion made about v in part (a). By the result of part (a), the solution of the IVP y ′ = x + y 2 , y(0) = 1, increases on an interval [0, X] and → ∞ as x → X from the left, where X is some number in the interval [π/4, 1]. c) Here are some approximations to y(x) for values of x near 0.9 obtained by the Runge-Kutta method with x0 = 0 and y0 = 1: For h = 0.05 n = 17 n = 18 n = 19

For h = 0.02 n n n n n

= 43 = 44 = 45 = 46 = 47

For h = 0.01 n n n n n n n n

= 86 = 87 = 88 = 89 = 90 = 91 = 92 = 93

n = 94

xn = 0.85 xn = 0.90 xn = 0.95

yn = 12.37139 yn = 31.777317 yn = 4071.117315.

xn xn xn xn xn

= 0.86 = 0.88 = 0.90 = 0.92 = 0.94

yn yn yn yn yn

= 14.149657 = 19.756061 = 32.651029 = 90.770048 = 34266.466629.

xn xn xn xn xn xn xn xn

= 0.86 = 0.87 = 0.88 = 0.89 = 0.90 = 0.91 = 0.92 = 0.93

yn yn yn yn yn yn yn yn

= 14.150706 = 16.493286 = 19.761277 = 24.638758 = 32.703853 = 48.591332 = 94.087476 = 636.786465

xn = 0.94

yn = 2.8399 × 1011 .

The values are still in reasonable agreement at x = 0.9, but they start to diverge quickly thereafter. This suggests that X is slightly greater than 0.9.

Section 18.4 Differential Equations of Second Order (page 1009) 1. If y1 = e x , then y1′′ − 3y1′ + 2y1 = e x (1 − 3 + 2) = 0, so y1 is a solution of the DE y ′′ − 3y ′ + 2y = 0. Let y = e x v. Then y ′ = e x (v ′ + v), y ′′ = e x (v ′′ + 2v ′ + v) ′′ ′ x ′′ y − 3y + 2y = e (v + 2v ′ + v − 3v ′ − 3v + 2v) = e x (v ′′ − v ′ ). y satisfies y ′′ − 3y ′ + 2y = 0 provided w = v ′ satisfies w′ − w = 0. This equation has solution v ′ = w = C1 e x , so v = C1 e x + C2 . Thus the given DE has solution y = e x v = C1 e2x + C2 e x .

2. If y1 = e−2x , then y1′′ − y1′ − 6y1 = e−2x (4 + 2 − 6) = 0, so y1 is a solution of the DE y ′′ − y ′ − 6y = 0. Let y = e−2x v. Then y ′ = e−2x (v ′ − 2v),

y ′′ = e−2x (v ′′ − 4v ′ + 4v)

y ′′ − y ′ − 6y = e−2x (v ′′ − 4v ′ + 4v − v ′ + 2v − 6v) = e x (v ′′ − 5v ′ ).

y satisfies y ′′ − y ′ − 6y = 0 provided w = v ′ satisfies w′ − 5w = 0. This equation has solution v ′ = w = (C1 /5)e5x , so v = C1 e5x + C2 . Thus the given DE has solution y = e−2x v = C1 e3x + C2 e−2x .

3. If y1 = x on (0, ∞), then x 2 y1′′ + 2x y1′ − 2y1 = 0 + 2x − 2x = 0, so y1 is a solution of the DE x 2 y ′′ + 2x y ′ − 2y = 0. Let y = xv(x). Then y ′ = xv ′ + v,

y ′′ = xv ′′ + 2v ′

x 2 y ′′ + 2x y ′ − 2y = x 3 v ′′ + 2x 2 v ′ + 2x 2 v ′ + 2xv − 2xv = x 2 (xv ′′ + 4v ′ ).

y satisfies x 2 y ′′ + 2x y ′ − 2y = 0 provided w = v ′ satisfies xw′ + 4w = 0. This equation has solution v ′ = w = −3C1 x −4 (obtained by separation of variables), so v = C1 x −3 + C2 . Thus the given DE has solution y = xv = C1 x −2 + C2 x.

654 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.4 (PAGE 1009)

4. If y1 = x 2 on (0, ∞), then x 2 y1′′ − 3x y1′ + 4y1 = 2x 2 − 6x 2 + 4x 2 = 0, so y1 is a solution of the DE x 2 y ′′ − 3x y ′ + 4y = 0. Let y = x 2 v(x). Then y ′ = x 2 v ′ + 2xv,

y ′′ = x 2 v ′′ + 4xv ′ + 2v

x 2 y ′′ − 3x y ′ + 4y = x 4 v ′′ + 4x 3 v ′ + 2x 2 v

− 3x 3 v ′ − 6x 2 v + 4x 2 v

= x 3 (xv ′′ + v ′ ).

y satisfies x 2 y ′′ − 3x y ′ + 4y = 0 provided w = v ′ satisfies xw′ + w = 0. This equation has solution v ′ = w = C1 /x (obtained by separation of variables), so v = C1 ln x + C2 . Thus the given DE has solution y = x 2 v = C1 x 2 ln x + C2 x 2 .

5. If y = x, then y ′ = 1 and y ′′ = 0. Thus

x 2 y ′′ − x(x + 2)y ′ + (x + 2)y = 0. Now let y = xv(x). Then y ′ = v + xv ′ ,

y ′′ = 2v ′ + xv ′′ .

Therefore y = x −1/2 cos x is a solution of the Bessel equation   1 x 2 y ′′ + x y ′ + x 2 − y = 0. (∗) 4 Now let y = x −1/2 (cos x)v(x). Then 1 y ′ = − x −3/2 (cos x)v − x −1/2 (sin x)v + x −1/2 (cos x)v ′ 2 3 −5/2 ′′ y = x (cos x)v + x −3/2 (sin x)v − x −3/2 (cos x)v ′ 4 − x −1/2 (cos x)v − 2x −1/2 (sin x)v ′ + x −1/2 (cos x)v ′′ . If we substitute these expressions into the equation (∗), many terms cancel out and we are left with the equation (cos x)v ′′ − 2(sin x)v ′ = 0. Substituting u = v ′ , we rewrite this equation in the form du (cos x) = 2(sin x)u d x Z Z du = 2 tan x d x ⇒ ln |u| = 2 ln | sec x| + C0 . u Thus v ′ = u = C1 sec2 x, from which we obtain v = C1 tan x + C2 .

Substituting these expressions into the differential equation we get

Thus the general solution of the Bessel equation (∗) is 2x 2 v ′ + x 3 v ′′ − x 2 v − 2xv − x 3 v ′ − 2x 2 v ′ + x 2 v + 2xv = 0

x 3 v ′′ − x 3 v ′ = 0,

y = x −1/2 (cos x)v = C1 x −1/2 sin x + C2 x −1/2 cos x.

or v ′′ − v ′ = 0,

which has solution v = C1 + C2 e x . Hence the general solution of the given differential equation is y = C1 x + C2 xe x .

7. If y1 = y and y2 = y ′ where y satisfies y ′′ + a1 (x)y ′ + a0 (x)y = f (x), then y1′ = y2 and y2′ = −a0 y1 − a1 y2 + f . Thus

6. If y = x −1/2 cos x, then 1 y = − x −3/2 cos x − x −1/2 sin x 2 3 y ′′ = x −5/2 cos x + x −3/2 sin x − x −1/2 cos x. 4

d dx

′

Thus



y1 y2



=



0 −a0

1 −a1



y1 y2



+



0 f



.

8. If y satisfies y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y ′ + a0 (x)y = f (x), then let



 1 x y + xy + x − y 4 3 = x −1/2 cos x + x 1/2 sin x − x 3/2 cos x 4 1 1 − x −1/2 cos x − x 1/2 sin x + x 3/2 cos x − x −1/2 cos x 2 4 = 0. 2 ′′

′

2

y1 = y,

y2 = y ′ ,

y3 = y ′′ ,

...

yn = y (n−1) .

Therefore y1′ = y2 ,

yn′

y2′ = y3 ,

...

′ yn−2 = yn−1 ,

and

= −a0 y1 − a1 y2 − a2 y3 − · · · − an−1 yn + f,

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SECTION 18.4 (PAGE 1009)

ADAMS and ESSEX: CALCULUS 8

and we have 0 0  y2   d   .  =  .. . d x  ..    0 yn −a0   0 0 . . +  . . 0 f

1 0 .. .

0 1 .. .

... ...

0 −a1

0 −a2

... ...



y  1

 0 y  1 0  y   ..   2   .   ..  .  1 yn −an

Section 18.5 Linear Differential Equations with Constant Coefficients (page 1013) 1.

y ′′′ − 4y ′′ + 3y ′ = 0 Auxiliary: r 3 − 4r 2 + 3r = 0 r (r − 1)(r − 3) = 0 ⇒ r = 0, 1, 3 General solution: y = C1 + C2 et + C3 e3t .

2.

y (4) − 2y ′′ + y = 0 Auxiliary: r 4 − 2r 2 + 1 = 0

(r 2 − 1)2 = 0 ⇒ r = −1, −1, 1, 1 General solution: y = C1 e−t + C2 te−t + C3 et + C4 tet .

9. If y = C1 eλx v, then y′ = C1 λeλx v = C1 eλx Av = Ay

10.

3.

provided λ and v satisfy Av = λv. 2 − λ 1 = 6 − 5λ + λ2 − 2 2 3 − λ

(r 2 + 1)2 = 0 ⇒ r = −i, −i, i, i General solution: y = C1 cos t + C2 sin t + C3 t cos t + C4 t sin t.

= λ2 − 5λ + 4 = (λ − 1)(λ − 4) = 0

4.

5. If y = e2t , then y ′′′ − 2y ′ − 4y = e2t (8 − 4 − 4) = 0.

If λ = 1 and Av = v, then 

=



v1 v2

Thus we may take v = v1 =



 1 . −1



v1 v2

2 1 2 3



v1 v2





The auxiliary equation for the DE is r 3 − 2r − 4 = 0, for which we already know that r = 2 is a root. Dividing the left side by r − 2, we obtain the quotient r 2 + 2r + 2. Hence the other two auxiliary roots are −1 ± i . General solution: y = C1 e2t + C2 e−t cos t + C3 e−t sin t.

⇔ v 1 + v 2 = 0.

6. Aux. eqn: (r 2 − r − 2)2 (r 2 − 4)2 = 0

(r + 1)2 (r − 2)2 (r − 2)2 (r + 2)2 = 0 r = 2, 2, 2, 2, −1, −1, −2, −2. The general solution is

If λ = 4 and Av = 4v, then A=



2 2

1 3



v1 v2



=4



⇔ 2v 1 − v 2 = 0.

y = e2t (C1 + C2 t + C3 t 2 + C4 t 3 ) + e−t (C5 + C6 t)

  1 Thus we may take v = v2 = . 2

+ e−2t (C7 + C8 t).

By the result of Exercise 9, y = e x v1 and y = e4x v2 are solutions of the homogeneous linear system y′ = Ay. Therefore the general solution of the system is

7.

(r − 1)2 = 0, r = 1, 1. Thus y = Ax + Bx ln x.

that is y1 y2



= C1 e x

y1 =

 x

1 −1



+ C2 e4x

C1 e + C2 e

4x

y2 = −C1 e x + 2C2 e4x .

x 2 y ′′ − x y ′ + y = 0 aux: r (r − 1) − r + 1 = 0 r 2 − 2r + 1 = 0

y = C1 e x v1 + C2 e4x v2 ,



y (4) + 4y (3) + 6y ′′ + 4y ′ + y = 0 Auxiliary: r 4 + 4r 3 + 6r 2 + 4r + 1 = 0

(r + 1)4 = 0 ⇒ r = −1, −1, −1, −1 General solution: y = e−t (C1 + C2 t + C3 t 2 + C4 t 3 ).

if λ = 1 or λ = 4.   2 1 Let A = . 2 3

A=

y (4) + 2y ′′ + y = 0 Auxiliary: r 4 + 2r 2 + 1 = 0

  1 , 2

or

8.

x 2 y ′′ − x y ′ − 3y = 0

r (r − 1) − r − 3 = 0 ⇒ r 2 − 2r − 3 = 0 ⇒(r − 3)(r + 1) = 0 ⇒ r1 = −1 and r2 = 3 Thus, y = Ax −1 + Bx 3 .

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INSTRUCTOR’S SOLUTIONS MANUAL

9.

x 2 y ′′ + x y ′ − y = 0 aux: r (r − 1) + r − 1 = 0 B y = Ax + . x

SECTION 18.6 (PAGE 1019)

Accordingly, z = z(t) satisfies

⇒ r = ±1

dz d2z + (b − a) + cz dt 2 dt 2 d y dy dy =ax 2 2 + ax + (b − a)x + cy = 0. dx dx dx a

10. Consider x 2 y ′′ − x y ′ + 5y = 0. Since a = 1, b = −1, and c = 5, therefore (b −a)2 < 4ac. Then k = (a −b)/2a = 1 and ω2 = 4. Thus, the general solution is y = Ax cos(2 ln x) + Bx sin(2 ln x).

11.

x 2 y ′′ + x y ′ = 0 aux: r (r − 1) + r = 0 Thus y = A + B ln x.

15. By the previous exercise, z(t) = y(et ) = y(x) must satisfy the constant coefficient equation

⇒ r = 0, 0.

d2z dz −2 + 2z = 0. 2 dt dt

12. Given that x 2 y ′′ + x y ′ + y = 0. Since a = 1, b = 1, c = 1 therefore (b − a)2 < 4ac. Then k = (a − b)/2a = 0 and ω2 = 1. Thus, the general solution is y = A cos(ln x) + B sin(ln x).

13.

The auxiliary equation for this equation is r 2 −2r +2 = 0, which has roots r = 1 ± i . Thus z = C1 et cos t + C2 et sin t.

x 3 y ′′′ + x y ′ − y = 0. Trying y = x r leads to the auxiliary equation

Since t = ln x, the given Euler equation has solution

r (r − 1)(r − 2) + r − 1 = 0

y = C1 x cos(ln x) + C2 x sin(ln x).

r 3 − 3r 2 + 3r − 1 = 0

(r − 1)3 = 0 ⇒ r = 1, 1, 1.

Section 18.6 Nonhomogeneous Linear Equations (page 1019)

Thus y = x is a solution. To find the general solution, try y = xv(x). Then y ′ = xv ′ + v,

y ′′ = xv ′′ + 2v ′ ,

y ′′′ = xv ′′′ + 3v ′′ .

1.

Now x 3 y ′′′ + x y ′ − y = x 4 v ′′′ + 3x 3 v ′′ + x 2 v ′ + xv − xv = x 2 (x 2 v ′′′ + 3xv ′′ + v ′ ), and y is a solution of the given equation if v ′ = w is a solution of x 2 w′′ + 3xw′ + w = 0. This equation has auxiliary equation r (r − 1) + 3r + 1 = 0, that is (r + 1)2 = 0, so its solutions are

y ′′ + y ′ − 2y = 1. The auxiliary equation for y ′′ + y ′ − 2y = 0 is r 2 + r − 2 = 0, which has roots r = −2 and r = 1. Thus the complementary function is yh = C1 e−2x + C2 e x . For a particular solution y p of the given equation try y = A. This satisfies the given equation if A = −1/2. Thus the general solution of the given equation is

C2 2C3 ln x + x x v = C1 + C2 ln x + C3 (ln x)2 .

v′ = w =

1 y = − + C1 e−2x + C2 e x . 2

The general solution of the given equation is, therefore, y = C1 x + C2 x ln x + C3 x(ln x)2 .

14. Since

dx = et = x, we have dt dz dy dx dy = =x , dt d x dt dx 2 2 d z d y dx dx dy = +x 2 dt d x dt 2 d x dt dy d2 y =x + x2 2 . dx dx

2.

y ′′ + y ′ − 2y = x. The complementary function is yh = C1 e−2x + C2 e x , as shown in Exercise 1. For a particular solution try y = Ax + B. Then y ′ = A and y ′′ = 0, so y satisfies the given equation if x = A − 2(Ax + B) = A − 2B − 2 Ax. We require A − 2B = 0 and −2 A = 1, so A = −1/2 and B = −1/4. The general solution of the given equation is y=−

2x + 1 + C1 e−2x + C2 e x . 4

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SECTION 18.6 (PAGE 1019)

3.

ADAMS and ESSEX: CALCULUS 8

y ′′ + y ′ − 2y = e−x . The complementary function is yh = C1 e−2x + C2 e x , as shown in Exercise 1. For a particular solution try y = Ae−x . Then y ′ = − Ae−x and y ′′ = Ae−x , so y satisfies the given equation if

Thus 2 A + 4C = 0, 4 A = 1, 4B = 0, and we have 1 1 A = , B = 0, and C = − . The given equation has 4 8 general solution y=

e−x = e−x (A − A − 2 A) = −2 Ae−x . We require A = −1/2. The general solution of the given equation is

7.

1 y = − e−x + C1 e−2x + C2 e x . 2

4.

y ′′ + y ′ − 2y = e x . The complementary function is yh = C1 e−2x + C2 e x , as shown in Exercise 1. For a particular solution try y = Axe x . Then y ′ = Ae x (1 + x),

For a particular solution, try y = Axe−2x . Then y ′ = e−2x (A − 2 Ax) and y ′′ = e−2x (−4 A + 4 Ax). We have

y ′′ = Ae x (2 + x),

e−2x = y ′′ − y ′ − 6y

= e−2x (−4 A + 4 Ax − A + 2 Ax − 6 Ax) = −5 Ae−2x .

e x = Ae x (2 + x + 1 + x − 2x) = 3 Ae x . We require A = 1/3. The general solution of the given equation is

5.

y ′′ − y ′ − 6y = e−2x . The homogeneous equation has auxiliary equation r 2 − r − 6 = 0 with roots r = −2 and r = 3. Thus the complementary function is yh = C1 e−2x + C2 e3x .

so y satisfies the given equation if

y=

Thus we require A = −1/5. The given equation has general solution

1 x xe + C1 e−2x + C3 e x . 3

y ′′ + 2y ′ + 5y = x 2 . The homogeneous equation has auxiliary equation r 2 + 2r + 5 = 0 with roots r = −1 ± 2i . Thus the complementary function is yh = C1 e−x cos(2x) + C2 e−x sin(2x).

1 y = − xe−2x + C1 e−2x + C2 e3x . 5

8.

y ′′ + 4y ′ + 4y = e−2x . The homogeneous equation has auxiliary equation r 2 + 4r + 4 = 0 with roots r = −2, −2. Thus the complementary function is

For a particular solution, try y = Ax 2 + Bx + C. Then y ′ = 2 Ax + B and y ′′ = 2 A. We have x 2 = y ′′ + 2y ′ + 5y

= 2 A + 4 Ax + 2B + 5 Ax 2 + 5Bx + 5C.

Thus we require 5 A = 1, 4 A + 5B = 0, and 2 A + 2B + 5C = 0. This gives A = 1/5, B = −4/25, and C = −2/125. The given equation has general solution y=

4x 2 x2 − − + e−x (C1 cos(2x) + C2 sin(2x)). 5 25 125

1 2 1 x − + C1 cos(2x) + C2 sin(2x). 4 8

yh = C1 e−2x + C2 xe−2x . For a particular solution, try y = Ax 2 e−2x . Then y ′ = e−2x (2 Ax −2 Ax 2 ) and y ′′ = e−2x (2 A−8 Ax +4 Ax 2 ). We have e−2x = y ′′ + 4y ′ + 4y

= e−2x (2 A − 8 Ax + 4 Ax 2 + 8 Ax − 8 Ax 2 + 4 Ax 2 )

= 2 Ae−2x .

Thus we require A = 1/2. The given equation has general solution

6.

y ′′ + 4y = x 2 . The complementary function is y = C1 cos(2x) + C2 sin(2x). For the given equation, try y = Ax 2 + Bx + C. Then x 2 = y ′′ + 4y = 2 A + 4 Ax 2 + 4Bx + 4C

658 Copyright © 2014 Pearson Canada Inc.

y=e

−2x



 x2 + C1 + C2 x . 2

INSTRUCTOR’S SOLUTIONS MANUAL

9.

SECTION 18.6 (PAGE 1019)

y ′′ + 2y ′ + 2y = e x sin x. The homogeneous equation has auxiliary equation r 2 + 2r + 2 = 0 with roots r = −1 ± i . Thus the complementary function is

This satisfies the nonhomogeneous DE if 4 + 2x + e−x = y ′′ + y ′ = A + 2B + 2Bx − Ce−x .

yh = C1 e−x cos x + C2 e−x sin x.

Thus we require A + 2B = 4, 2B = 2, and −C = 1, that is, A = 2, B = 1, C = −1. The given equation has general solution

For a particular solution, try y = Ae x cos x + Be x sin x. Then

y = 2x + x 2 − xe−x + C1 + C2 e−x .

y ′ = (A + B)e x cos x + (B − A)e x sin x y ′′ = 2Be x cos x − 2 Ae x sin x.

12.

This satisfies the nonhomogeneous DE if e x sin x = y ′′ + 2y ′ + 2y = e x cos x(2B + 2(A + B) + 2 A) + e x sin x(−2 A + 2(B − A) + 2B) x = e cos x(4 A + 4B) + e x sin x(4B − 4 A).

y ′′ + 2y ′ + y = xe−x . The homogeneous equation has auxiliary equation r 2 + 2r + 1 = 0 with roots r = −1 and r = −1. Thus the complementary function is yh = C1 e−x + C2 xe−x . For a particular solution, try y = e−x (Ax 2 + Bx 3 ). Then y ′ = e−x (2 Ax + (3B − A)x 2 − Bx 3 )

y ′′ = e−x (2 A + (6B − 4 A)x − (6B − A)x 2 + Bx 3 ).

Thus we require A + B = 0 and 4(B − A) = 1, that is, B = − A = 1/8. The given equation has general solution

This satisfies the nonhomogeneous DE if

ex y = (sin x − cos x) + e−x (C1 cos x + C2 sin x). 8

10.

xe−x = y ′′ + 2y ′ + y = e−x (2 A + 6Bx).

y ′′ + 2y ′ + 2y = e−x sin x. The complementary function is the same as in Exercise 9, but for a particular solution we try y = Axe−x cos x + Bxe−x sin x y ′ = e−x cos x(A − Ax + Bx) + e−x sin x(B − Bx − Ax) y ′′ = e−x cos x(2B − 2Bx − 2 A) + e−x sin x(2 Ax − 2 A − 2B). This satisfies the nonhomogeneous DE if e−x sin x = y ′′ + 2y ′ + 2y = 2Be−x cos x − 2 Ae−x sin x.

Thus we require A = 0 and B = 1/6. The given equation has general solution y=

13.

y ′′ + y ′ − 2y = e−x . The complementary function is yh = C1 e−2x + C2 e x . For a particular solution use y p = e−2x u 1 (x) + e x u 2 (x), where the coefficients u 1 and u 2 satisfy

Thus we require B = 0 and A = −1/2. The given equation has general solution 1 y = − xe−x cos x + e−x (C1 cos x + C2 sin x). 2

11.

y ′′ + y ′ = 4 + 2x + e−x . The homogeneous equation has auxiliary equation r 2 + r = 0 with roots r = 0 and r = −1. Thus the complementary function is yh = C1 + C2 e−x . For a particular solution, try y = Ax + Bx 2 + C xe−x . Then

1 3 −x x e + C1 e−x + C2 xe−x . 6

−2e−2x u ′1 + e x u ′2 = e−x e−2x u ′1 + e x u ′2 = 0.

Thus

1 u ′1 = − e x 3 1 x u1 = − e 3

1 −2x e 3 1 u 2 = − e−2x . 6

u ′2 =

1 1 1 Thus y p = − e−x − e−x = − e−x . The general 3 6 2 solution of the given equation is

y ′ = A + 2Bx + e−x (C − C x) y ′′ = 2B + e−x (−2C + C x).

1 y = − e−x + C1 e−2x + C2 e x . 2

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SECTION 18.6 (PAGE 1019)

14.

ADAMS and ESSEX: CALCULUS 8

y ′′ + y ′ − 2y = e x . The complementary function is yh = C1 e−2x + C2 e x . For a particular solution use

17.

x = x2

y p = e−2x u 1 (x) + e x u 2 (x),

−2e−2x u ′1 + e x u ′2 = e x e−2x u ′1 + e x u ′2 = 0.

1 = − e3x 3 1 3x u1 = − e 9

18.

1 = 3 1 u 2 = x. 3

u ′1

u ′2

1 1 Thus y p = − e x + xe x . The general solution of the 9 3 given equation is 1 1 y = − e x + xe x + C1 e−2x + C2 e x 9 3 1 x = xe + C1 e−2x + C3 e x . 3

15.

x 2 y ′′ + x y ′ − y = x 2 . If y = Ax 2 , then y ′ = 2 Ax and y ′′ = 2 A. Thus x 2 = x 2 y ′′ + x y ′ − y 2

2

2

A + x A(ln x + 1) − Ax ln x = 2 Ax. x

Thus A = 1/2. The complementary function was obtained in Exercise 15. The given equation has general solution 1 C2 . y = x ln x + C1 x + 2 x

where the coefficients u 1 and u 2 satisfy

Thus

x 2 y ′′ + x y ′ − y = x. Try y = Ax ln x. Then y ′ = A(ln x + 1) and y ′′ = A/x. We have

x 2 y ′′ + x y ′ − y = x. 1 Try y = xu 1 (x) + u 2 (x), where u 1 and u 2 satisfy x

x u ′2 = − , 2 Thus u 1 =

u ′2 1 = . 2 x x

u ′1 =

1 . 2x

x2 1 ln x and u 2 = − . A particular solution is 2 4 y=

1 x x ln x − . 2 4

The term −x/4 can be absorbed into the term C1 x in the complementary function, so the general solution is y=

= 2 Ax + 2 Ax − Ax = 3 Ax ,

y=

u ′1 −

Solving these equations for u ′1 and u ′2 , we get

2

so A = 1/3. A particular solution of the given equation is y = x 2 /3. The auxiliary equation for the homogeneous equation x 2 y ′′ + x y ′ − y = 0 is 4r (r − 1) + r − 1 = 0, or r 2 − 1 = 0, which has solutions r = ±1. Thus the general solution of the given equation is

u ′2 = 0, x

xu ′1 +

1 C2 x ln x + C1 x + . 2 x

19. The homogeneous DE y ′′ − 2y ′ + y = 0 has auxiliary

equation r 2 − 2r + 1 = 0 with roots r = 1, 1. Therefore, its general solution is y = C1 e x + C2 xe x . Accordingly, we look for a particular solution of the given equation having the form

1 2 C2 x + C1 x + . 3 x

y p = u 1 (x)e x + u 2 (x) xe x .

16.

x 2 y ′′ + x y ′ − y = x r has a solution of the form y = Ax r provided r 6= ±1. If this is the case, then

According to the procedure developed in the text, u ′1 and u ′2 can be determined by solving the pair of equations u ′1 (x)e x + u ′2 (x)xe x = 0

  x = Ax r (r − 1) + r − 1 = Ax r (r 2 − 1). r

r

Thus A = 1/(r 2 − 1) and a particular solution of the DE is 1 y= 2 xr . r −1

u ′1 (x)e x + u ′2 (x)(1 + x)e x =

ex x

or, equivalently,

660 Copyright © 2014 Pearson Canada Inc.

u ′1 (x) + xu ′2 (x) = 0 u ′1 (x) + (1 + x)u ′2 (x) =

1 . x

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.6 (PAGE 1019)

21.

The solution is u ′1 (x) = −1,

u ′2 (x) =

1 . x

Thus u 1 (x) = −x and u 2 (x) = ln x. A particular solution of the DE is y p = −xe x + xe x ln x, or, since the first term is a solution of the homogeneous equation, the simpler form y p = xe x ln x will do. The given equation has general solution

x 2 y ′′ − (2x + x 2 )y ′ + (2 + x)y = x 3 . Since x and xe x are independent solutions of the corresponding homogeneous equation, we can write a solution of the given equation in the form y = xu 1 (x) + xe x u 2 (x), where u 1 and u 2 are chosen to satisfy xu ′1 + xe x u ′2 = 0,

y = C1 e x + C2 xe x + xe x ln x.

Solving these equations for u ′1 and u ′2 , we get u ′1 = −1 and u ′2 = e−x . Thus u 1 = −x and u 2 = −e−x . The particular solution is y = −x 2 − x. Since −x is a solution of the homogeneous equation, we can absorb that term into the complementary function and write the general solution of the given DE as

20. The homogeneous DE y ′′ + 4y ′ + 4y = 0 has auxiliary equation r 2 + 4r + 4 = 0 with roots r = −2, −2. Therefore, its general solution is

y = −x 2 + C1 x + C2 xe x .

y = C1 e−2x + C2 xe−2x . Accordingly, we look for a particular solution of the given equation having the form y p = u 1 (x)e−2x + u 2 (x) xe−2x . u ′1

According to the procedure developed in the text, and u ′2 can be determined by solving the pair of equations u ′1 (x)e−2x + u ′2 (x)xe−2x = 0 − 2u ′1 (x)e−2x + (1 − 2x)u ′2 (x)e−2x =

e−2x x2

or, equivalently, u ′1 (x) + xu ′2 (x) = 0 − 2u ′1 (x) + (1

− 2x)u ′2 (x)

1 = 2. x

22.

  1 x 2 y ′′ + x y ′ + x 2 − y = x 3/2 . 4 A particular solution can be obtained in the form y = x −1/2 (cos x)u 1 (x) + x −1/2 (sin x)u 2 (x), where u 1 and u 2 satisfy x −1/2 (cos x)u ′1 + x −1/2 (sin x)u ′2 = 0   1 − x −3/2 cos x − x −1/2 sin x u ′1 2   1 −3/2 −1/2 − x sin x − x cos x u ′2 = x −1/2 . 2 We can simplify these equations by dividing the first by x −1/2 , and adding the first to 2x times the second, then dividing the result by 2x 1/2 . The resulting equations are (cos x)u ′1 + (sin x)u ′2 = 0 −(sin x)u ′1 + (cos x)u ′2 = 1,

The solution is 1 u ′1 (x) = − , x

u ′2 (x) =

u ′1 + (1 + x)e x u ′2 = x.

1 . x2

1 Thus u 1 (x) = − ln x and u 2 (x) = − . A particular x solution of the DE is y p = −e−2x ln x − e−2x , or, since the last term is a solution of the homogeneous equation, the simpler form y p = −e−2x ln x will do. The given equation has general solution

which have solutions u ′1 = − sin x, u ′2 = cos x, so that u 1 = cos x and u 2 = sin x. Thus a particular solution of the given equation is y = x −1/2 cos2 x + x −1/2 sin2 x = x −1/2 . The general solution is

y = C1 e−2x + C2 xe−2x − xe−2x ln x.

  y = x −1/2 1 + C2 cos x + C2 sin x .

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SECTION 18.6 (PAGE 1019)

ADAMS and ESSEX: CALCULUS 8

Section 18.7 Series Solutions of Differential Equations (page 1024) 1.

2.

y ′′ = (x − 1)2 y. Try y= y ′′ = =

∞ X

n=0 ∞ X n=2

∞ X n=0 ′′

an (x − 1)n .

y′ =

n(n − 1)an (x − 1)n−2

y ′′ = n

(n + 2)(n + 1)an+2 (x − 1)

n=0

n=0

∞ X

∞ X

n=0

n

(n + 2)(n + 1)an+2 (x − 1) −

n=2

n

an−2 (x − 1)

n=2

an−2 = for n ≥ 2. (n + 1)(n + 2)

nan x n−1 =

n=1

(x − 1)4n 4n n! × 3 × 7 × · · · × (4n − 1)

+ a1 x − 1 +

∞ X n=1

nan x n−1

n=1

n(n − 1)an x n−2 =

∞ X (n + 2)(n + 1)an+2 x n . n=0

0 = y ′′ − x y ∞ ∞ X X = (n + 2)(n + 1)an+2 x n − an x n+1 n=0

(x − 1)4n+1

+ 1)

n=1

∞ h X n=1

i (n + 2)(n + 1)an+2 − an−1 x n .

an−1 for n ≥ 1. (n + 2)(n + 1) Given a0 and a1 , we have

Thus a2 = 0 and an+2 =

a0 2×3 a3 a0 1 × 4 × a0 a6 = = = 5×6 2×3×5×6 6! 1 × 4 × 7 × a0 a6 = a9 = 8×9 9! .. . 1 × 4 × · · · × (3n − 2)a0 a3n = (3n)! a1 2 × a1 a4 = = 3×4 4! a4 2 × 5 × a1 = a7 = 6×7 7! .. . 2 × 5 × · · · × (3n − 1)a1 a3n+1 = (3n + 1)! 0 = a2 = a5 = a8 = · · · = a3n+2 . Thus the general solution of the given equation is

y = a0

!

4n n! × 5 × 9 × · · · × (4n

n=0

∞ ∞ X X = (n + 2)(n + 1)an+2 x n − an−1 x n

The solution is y = a0 1 +

∞ X

a3 =

a0 a4 = 3×4 a4 a0 a8 = = 7×8 3×4×7×8 .. . a0 a4n = 3 × 4 × 7 × 8 × · · · × (4n − 1)(4n) a0 = n 4 n! × 3 × 7 × · · · × (4n − 1) a1 a5 = 4×5 a5 a1 a9 = = 8×9 4×5×8×9 .. . a1 a4n+1 = 4 × 5 × 8 × 9 × · · · × (4n)(4n + 1) a1 = n 4 n! × 5 × 9 × · · · × (4n + 1) a4n+3 = a4n+2 = · · · = a3 = a2 = 0.

∞ X

n=0

= 2a2 +

n=2

Given a0 and a1 we have

n=0 ∞ X

an x n . Then

n=0

= 2a2 + 6a3 (x − 1) ∞ h i X + (n + 2)(n + 1)an+2 − an−2 (x − 1)n . Thus a2 = a3 = 0, and an+2

∞ X

∞ X

Thus we have

0 = y − (x − 1)2 y ∞ ∞ X X = (n + 2)(n + 1)an+2 (x − 1)n − an (x − 1)n+2 =

y ′′ = x y. Try

!

.

662 Copyright © 2014 Pearson Canada Inc.

∞ X 1 × 4 × · · · × (3n − 2) 3n 1+ x (3n)! n=1

+ a1

!

∞ X 2 × 5 × · · · × (3n − 1) 3n+1 x . (3n + 1)! n=1

INSTRUCTOR’S SOLUTIONS MANUAL

3.

  

Let

SECTION 18.7 (PAGE 1024)

y ′′ + x y ′ + 2y = 0 y(0) = 1 y ′ (0) = 2

y= y ′′ =

∞ X

n=0 ∞ X n=2

an x n

Thus,

y′ =

∞ X

0 = y ′′ + x y ′ + y ∞ ∞ ∞ X X X = (n + 2)(n + 1)an+2 x n + x nan x n−1 + an x n n=0

nan x n−1

n=1

= 2a2 + a0 +

∞ X

n(n − 1)an x n−2 =

n=0

(n + 2)(n + 1)an+2 x n .

n=0

(n + 2)(n + 1)an+2 x n + +2

∞ X

n

an x = 0,

∞ X

that is,

nan x n

n=1

n=1

[(n + 2)(n + 1)an+2 + (n + 2)an ]x n = 0.

y =1−

It follows that a2 = −1,

an+2 = −

an , n+1

n = 1, 2, 3, . . . .

Since a0 = y(0) = 1, and a1 = y ′ (0) = 2, we have a1 = 2

a0 = 1 a2 = −1 1 a4 = 3 1 3×5 1 a8 = 3×5×7 a6 = −

2 2×4×6 2 a9 = . 2×4×6×8

a7 = −

The patterns here are obvious: (−1)n 3 × 5 × · · · × (2n − 1) (−1)n 2n n! = (2n)!

Thus y =

4. If y = y ′′ =

P∞

∞ X n=0

n n=0 (−1)



∞ X n=2

5.

P∞

a2n+1 =

n(n − 1)an x n−2 =

n=1 nan x

n−1

∞ X 1 2 1 4 (−1)n 2n 1 6 x + x − x + ··· = x . 2 8 48 2n · n! n=0

y ′′ + (sin x)y = 0, y(0) = 1, y ′ (0) = 0. Try y = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + · · · .

(−1)n 2 2n n!

y ′′ = 2a2 + 6a3 x + 12a4 x 2 + 20a5 x 3 + · · · ! x5 x3 + −··· (sin x)y = x − 6 120 × (1 + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + · · ·)   1 = x + a2 − x 3 + a3 x 4 6   1 1 + a4 − a2 + x5 + · · · . 6 120 Hence we must 1 20a5 + a2 − 6 1 a3 = − , a5 = 6

 2n n!x 2n x 2n+1 . + n−1 (2n)! 2 n!

an x n , then y ′ =

(n + 2)(n + 1)an+2 + (n + 1)an = 0, −an an+2 = . n+2

Then a0 = 1 and a1 = 0. We have

2 a3 = − 2 2 a5 = 2×4

a2n =

i (n + 2)(n + 1)an+2 + (n + 1)an x n .

−1 1 If y(0) = 1, then a0 = 1, a2 = , a4 = 2 , 2 2 · 2! −1 1 a6 = 3 , a8 = 4 ,. . .. If y ′ (0) = 0, then 2 · 3! 2 · 4! a1 = a3 = a5 = . . . = 0. Hence,

so

n=0 ∞ X

2a2 + 2 +

n=1

n=0

Since coefficients of all powers of x must vanish, therefore 2a2 + a0 = 0 and, for n ≥ 1,

Substituting these expressions into the differential equation, we get ∞ X

n=1

∞ h X

have 2a2 = 0, 6a3 + 1 = 0, 12a4 = 0, = 0, . . . . That is, a2 = 0, a4 = 0,

1 . The solution is 120

y =1−

1 3 1 5 x + x +···. 6 120

and

6. (1 − x 2 )y ′′ − x y ′ + 9y = 0, y(0) = 0, y ′ (0) = 1. Try

∞ X (n + 2)(n + 1)an+2 x n . n=0

y=

∞ X

an x n .

n=0

663 Copyright © 2014 Pearson Canada Inc.

SECTION 18.7 (PAGE 1024)

ADAMS and ESSEX: CALCULUS 8

an−1 for 3(n + µ)2 − (n + µ) n ≥ 1. There are two cases: µ = 0 and µ = 1/3. an−1 CASE I. µ = 0. Then an = − . Since a0 = 1 n(3n − 1) we have

Thus 3µ2 − µ = 0 and an = −

Then a0 = 0 and a1 = 1. We have y′ = y ′′ =

∞ X

nan x n−1

n=1

∞ X n=2

n(n − 1)an x n−2

a1 = −

0 = (1 − x 2 )y ′′ − x y ′ + 9y ∞ ∞ X X = (n + 2)(n + 1)an+2 x n − n(n − 1)an x n n=0

−

a3 = −

n=2

∞ X n=1

n

nan x + 9

∞ X

an x

.. .

n

an =

n=0

= 2a2 + 9a0 + (6a3 + 8a1 )x ∞ h i X + (n + 2)(n + 1)an+2 − (n 2 − 9)an x n .

a2 =

1 1×2×2×5

1 1×2×2×5×3×8

(−1)n . n! × 2 × 5 × · · · × (3n − 1)

One series solution is

n=2

y =1+

Thus 2a2 + 9a0 = 0, 6a3 + 8a1 = 0, and an+2 =

1 , 1×2

(n 2

− 9)an . (n + 1)(n + 2)

CASE II. µ =

∞ X n=1

(−1)n x n . n! × 2 × 5 × · · · × (3n − 1)

1 . Then 3

Therefore we have an =

a2 = a4 = a6 = · · · = 0 4 a3 = − , a5 = 0 = a7 = a9 = · · · . 3

3 n+

Since a0 = 1 we have

The initial-value problem has solution y=x−

a1 = −

4 3 x . 3

a3 = − .. .

7. 3x y ′′ + 2y ′ + y = 0.

Since x = 0 is a regular singular point of this equation, try ∞ X y= an x n+µ (a0 = 1) y′ = y ′′ =

n=0

y=x

(n + µ)(n + µ − 1)an x n+µ−2 .

0 = 3x y ′′ + 2y ′ + y ∞ h ∞ i X X = 3(n + µ)2 − (n + µ) an x n+µ−1 + an−1 x n+µ−1 n=1

= (3µ2 − µ)x µ−1 ∞ h  i X 3(n + µ)2 − (n + µ) an + an−1 x n+µ−1 . + n=1

1 , 1×4

1 3

a2 =

1 1×4×2×7

1 1 × 4 × 2 × 7 × 3 × 10

(−1)n . n! × 1 × 4 × 7 × · · · × (3n + 1)

A second series solution is

(n + µ)an x n+µ−1

Then we have

n=0

an =

n=0 ∞ X n=0 ∞ X

−an−1
= n(3n + 1) . − n+

−an−1


1 2 3

8.

1/3

1+

∞ X n=1

(−1)n x n n! × 1 × 4 × 7 × · · · × (3n + 1)

!

.

x y ′′ + y ′ + x y = 0. Since x = 0 is a regular singular point of this equation, try ∞ X y= an x n+µ (a0 = 1) n=0

∞ X (n + µ)an x n+µ−1 y′ = n=0

∞ X y = (n + µ)(n + µ − 1)an x n+µ−2 .

664 Copyright © 2014 Pearson Canada Inc.

′′

n=0

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 18 (PAGE 1024)

Then we have

4.

0 = x y ′′ + y ′ + x y ∞ h i X = (n + µ)(n + µ − 1) + (n + µ) an x n+µ−1 n=0

+ =

∞ X

an x n+µ+1

n=0

∞ ∞ X X (n + µ)2 an x n+µ−1 + an−2 x n+µ−1 n=0

n=2

= µ2 x µ−1 + (1 + µ)2 a1 x µ ∞ h i X + (n + µ)2 an + an−2 x n+µ−1 .

5.

n=2

an−2 for n ≥ 2. n2 It follows that 0 = a1 = a3 = a5 = · · ·, and, since a0 = 1, Thus µ = 0, a1 = 0, and an = −

1 1 , a4 = 2 2 , . . . 22 2 4 (−1)n (−1)n = 2n . = 2 2 2 2 4 · · · (2n) 2 (n!)2

a2 = − a2n

6.

One series solution is y =1+

Review Exercises 18

1.

3.

7.

(page 1024)

dy = 2x y dx dy = 2x d x ⇒ ln |y| = x 2 + C1 y y = Ce x

2.

∞ X (−1)n x 2n . 22n (n!)2 n=1

8.

2

x 2 + y2 dy = (let y = xv(x)) dx 2x y dv 1 + v2 v+x = dx 2v dv 1 + v2 1 − v2 x = −v = dx 2v 2v 2v dv dx =− x v2 − 1 C 1 2 ln(v − 1) = ln + ln C = ln x x y2 C 2 2 −1= ⇒ y − x = Cx x2 x x+y dy = dx y−x (x + y) d x + (x − y) d y = 0  2  x y2 d + xy − =0 2 2 x 2 + 2x y − y 2 = C

(exact)

dy y + ex =− dx x + ey (y + e x ) d x + (x + e y ) d y = 0 (exact)
d x y + ex + e y = 0 x y + ex + e y = C  2 d2 y dy = (let p = d y/dt) dt 2 dt dp dp = p2 ⇒ 2 = dt dt p 1 = C1 − t p 1 dy =p= dt C −t 1 Z dt y= = − ln |t − C1 | + C2 C1 − t

d2 y dy + 2y = 0 +5 dt dt 2 2 Aux: 2r + 5r + 2 = 0 ⇒ r = −1/2, −2 2

y = C1 e−t/2 + C2 e−2t

dy = e−y sin x dx e y d y = sin x d x ⇒ e y = − cos x + C y = ln(C − cos x)

9.

4y ′′ − 4y ′ + 5y = 0

Aux: 4r 2 − 4r + 5 = 0

dy dy = x + 2y ⇒ − 2y = x dx dx  d −2x −2x d y (e y) = e − 2y = xe−2x dx dx Z 1 x e−2x y = xe−2x d x = − e−2x − e−2x + C 2 4 x 1 y = − − + Ce2x 2 4

1 ±i 2 x/2 x/2 y = C1 e cos x + C2 e sin x (2r − 1)2 + 4 = 0 ⇒ r =

10.

2x 2 y ′′ + y = 0 Aux: 2r (r − 1) + 1 = 0

1 2r 2 − 2r + 1 = 0 ⇒ r = (1 ± i )
2 y = C1 |x|1/2 cos 21 ln |x| + C2 |x|1/2 sin

1 2

ln |x|


665

Copyright © 2014 Pearson Canada Inc.

REVIEW EXERCISES 18 (PAGE 1024)

11.

ADAMS and ESSEX: CALCULUS 8

d2 y dy −t + 5y = 0 dt 2 dt Aux: r (r − 1) − r + 5 = 0

Thus A = 1, B = −4, C = 6. The general solution is

t2

y = x 2 − 4x + 6 + C1 e−x + C2 xe−x .

2

(r − 1) + 4 = 0 ⇒ r = 1 ± 2i y = C1 t cos(2 ln |t|) + C2 t sin(2 ln |t|)

12.

16.

d3 y d2 y dy + 8 2 + 16 =0 3 dt dt dt Aux: r 3 + 8r 2 + 16r = 0

r (r + 4)2 = 0 ⇒ r = 0, −4, −4

y = C1 + C2 e−4t + C3 te−4t

13.

d2 y dy −5 + 6y = e x + e3x dx2 dx Aux: r 2 − 5r + 6 = 0 ⇒ r = 2, 3. Complementary function: y = C1 e2x + C2 e3x . Particular solution: y = Ae x + Bxe3x ′

x

y = Ae + B(1 + 3x)e

d2 y − 2y = x 3 . dx2 The corresponding homogeneous equation has auxiliary equation r (r − 1) − 2 = 0, with roots r = 2 and r = −1, so the complementary function is y = C1 x 2 + C2 /x. A particular solution of the nonhomogeneous equation can have the form y = Ax 3 . Substituting this into the DE gives x2

6 Ax 3 − 2 Ax 3 = x 3 , so that A = 1/4. The general solution is

3x

y=

y ′′ = Ae x + B(6 + 9x)e3x e x + e3x = Ae x (1 − 5 + 6)

+ Be3x (6 + 9x − 5 − 15x + 6x)

17.

= 2 Ae x + Be3x .

y 3 = x 3 − 7 ⇒ y = (x 3 − 7)1/3

1 x e + xe3x + C1 e2x + C2 e3x . 2

18. 14.

d2 y dy −5 + 6y = xe2x dx dx2 Same complementary function as in Exercise 13: C1 e2x + C2 e3x . For a particular solution we try y = (Ax 2 + Bx)e2x . Substituting this into the given DE leads to xe2x = (2 A − B)e2x − 2 Axe2x , so that we need A = −1/2 and B = 2 A = −1. The general solution is 

 1 2 x + x e2x + C1 e2x + C2 e3x . y=− 2

15.

d2 y dy +2 + y = x2 dx2 dx Aux: r 2 + 2r + 1 = 0 has solutions r = −1, −1. Complementary function: y = C1 e−x + C2 xe−x . Particular solution: try y = Ax 2 + Bx + C. Then x 2 = 2 A + 2(2 Ax + B) + Ax 2 + Bx + C.

dy x2 = 2 , y(2) = 1 dx y y2 d y = x 2 d x

y3 = x 3 + C 1 = 8 + C ⇒ C = −7

Thus A = 1/2 and B = 1. The general solution is y=

1 3 C2 x + C1 x 2 + . 4 x

19.

dy y2 = 2 , y(2) = 1 dx x dy 1 dx 1 = 2 ⇒ − = − −C y2 x y x 1 1 1= +C ⇒ C = 2 2   1 −1 2x 1 y= + = x 2 x +2

xy dy = 2 , y(0) = 1. Let y = xv(x). Then dx x + y2 v dv = dx 1 + v2 dv v v3 x = −v =− 2 dx 1+v 1 + v2 2 1+v dx − dv = v3 x 1 − ln |v| = ln |x| + ln C 2v 2 x2 1 = 2 = ln(Cv x)2 = ln(C 2 y 2 ) y2 v

v+x

C 2 y2 = ex y2 = e

666 Copyright © 2014 Pearson Canada Inc.

2 /y 2

x 2 /y 2

,

,

y(0) = 1 ⇒ C 2 = 1

or y = e x

2 /(2y 2 )

.

INSTRUCTOR’S SOLUTIONS MANUAL

20.

REVIEW EXERCISES 18 (PAGE 1024)

dy + (cos x)y = 2 cos x, y(π ) = 1 dx     d sin x sin x d y + (cos x)y = 2 cos xesin x e y =e dx dx esin x y = 2esin x + C y = 2 + Ce− sin x 1 = 2 + Ce0 ⇒ C = −1 y = 2 − e− sin x

21.

y ′′ + 3y ′ + 2y = 0, y(0) = 1, y ′ (0) = 2 Aux: r 2 + 3r + 2 = 0 ⇒ r = −1, −2. y = Ae−x + Be−2x

y ′ = − Ae−x − 2Be−2x

⇒

1= A+ B

⇒

2 = − A − 2B.

Thus B = −3, A = 4. The solution is y = 4e−x − 3e−2x .

22.

y ′′ + 2y ′ + (1 + π 2 )y = 0, y(1) = 0, y ′ (1) = π Aux: r 2 + 2r + 1 + π 2 = 0 ⇒ r = −1 ± πi . y = Ae−x cos(π x) + Be−x sin(π x) y ′ = e−x cos(π x)(− A + Bπ ) + e−x sin(π x)(−B − Aπ ). − Ae−1

We require 1 = y(0) = 1 + C1 and −2 = y ′ (0) = 2 + 2C2 . Thus C1 = 0 and C2 = −2. The solution is y = e2t − 2 sin(2t). d2 y dy +5 − 3y = 6 + 7e x/2 , y(0) = 0, y ′ (0) = 1 dx2 dx Aux: 2r 2 + 5r − 3 = 0 ⇒ r = 1/2, −3. Complementary function: y = C1 e x/2 + C2 e−3x . Particular solution: y = A + Bxe x/2  x y ′ = Be x/2 1 + 2  x ′′ x/2 y = Be 1+ . 4 We need   x 5x Be x/2 2 + + 5 + − 3x − 3 A = 6 + 7e x/2 . 2 2

26. 2

This is satisfied if A = −2 and B = 1. The general solution of the DE is y = −2 + xe x/2 + C1 e x/2 + C2 e−3x . Now the initial conditions imply that 0 = y(0) = −2 + C1 + C2 C1 1 = y ′ (0) = 1 + − 3C2 , 2

Bπ )e−1

Thus = 0 and (A − = π , so that A = 0 and B = −e. The solution is y = −e1−x sin(π x).

23.

y ′′ + 10y ′ + 25y = 0, y(1) = e−5 , y ′ (1) = 0 Aux: r 2 + 10r + 25 = 0 ⇒ r = −5, −5. y = Ae−5x + Bxe−5x

y ′ = −5 Ae−5x + B(1 − 5x)e−5x . We require e−5 = (A + B)e−5 and 0 = e−5 (−5 A − 4B). Thus A + B = 1 and −5 A = 4B, so that B = 5 and A = −4. The solution is y = −4e−5x + 5xe−5x .

24.

1 y = −2 + xe x/2 + (12e x/2 + 2e−3x ). 7

27. [(x + A)e x sin y + cos y] d x + x[e x cos y + B sin y] d y = 0 is M d x + N d y. We have

∂M = (x + A)e x cos y − sin y ∂y ∂N = e x cos y + B sin y + xe x cos y. ∂x

x 2 y ′′ − 3x y ′ + 4y = 0, y(e) = e2 , y ′ (e) = 0 Aux: r (r − 1) − 3r + 4 = 0, or (r − 2)2 = 0, so that r = 2, 2. y = Ax 2 + Bx 2 ln x y ′ = 2 Ax + 2Bx ln x + Bx. We require e2 = Ae2 + Be2 and 0 = 2 Ae + 3Be. Thus A + B = 1 and 2 A = −3B, so that A = 3 and B = −2. The solution is y = 3x 2 − 2x 2 ln x, valid for x > 0.

25.

which give C1 = 12/7, C2 = 2/7. Thus the IVP has solution

d2 y + 4y = 8e2t , y(0) = 1, y ′ (0) = −2 dt 2 Complementary function: y = C1 cos(2t) + C2 sin(2t). Particular solution: y = Ae2t , provided 4 A + 4 A = 8, that is, A = 1. Thus y = e2t + C1 cos(2t) + C2 sin(2t)

These expressions are equal (and the DE is exact) if A = 1 and B = −1. If so, the left side of the DE is dφ(x, y), where φ(x, y) = xe x sin y + x cos y. The general solution is xe x sin y + x cos y = C.

28. (x 2 + 3y 2 ) d x + x y d y = 0. Multiply by x n :

x n (x 2 + 3y 2 ) d x + x n+1 y d y = 0 is exact provided 6x n y = (n + 1)x n y, that is, provided n = 5. In this case the left side is dφ, where

y ′ = 2e2t − 2C1 sin(2t) + 2C2 cos(2t).

φ(x, y) =

1 6 2 1 8 x y + x . 2 8

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REVIEW EXERCISES 18 (PAGE 1024)

ADAMS and ESSEX: CALCULUS 8

The general solution of the given DE is

Divide the first equation by x and subtract from the second equation to get

4x 6 y 2 + x 8 = C.

29.

−x sin xu ′2 = x sin x.

x 2 y ′′ − x(2 + x cot x)y ′ + (2 + x cot x)y = 0 If y = x, then y ′ = 1 and y ′′ = 0, so the DE is clearly satisfied by y. To find a second, independent solution, try y = xv(x). Then y ′ = v + xv ′ , and y ′′ = 2v ′ + xv ′′ . Substituting these expressions into the given DE, we obtain 2x 2 v ′ + x 3 v ′′ − (xv + x 2 v ′ )(2 + x cot x) + xv(2 + x cot x) = 0 x 3 v ′′ − x 3 v ′ cot x = 0,

dw cos x d x = w sin x ln w = ln sin x + ln C2 v ′ = w = C2 sin x ⇒ v = C1 − C2 cos x. A second solution of the DE is x cos x, and the general solution is y = C1 x + C2 x cos x.

30.

cot x)y ′

− x(2 + x + (2 + x cot x)y = Look for a particular solution of the form y = xu 1 (x) + x cos xu 2 (x), where u ′1

y = x sin x − x 2 cos x + C1 x + C2 x cos x.

31. Suppose y ′ = f (x, y) and y(x0 ) = y0 , where f (x, y) is continuous on the whole x y-plane and satisfies | f (x, y)| ≤ K there. By the Fundamental Theorem of Calculus, we have

or, putting w = v ′ , w′ = (cot x)w, that is,

x 2 y ′′

Thus u ′2 = −1 and u 2 = −x. The first equation now gives u ′1 = cos x, so that u 1 = sin x. The general solution of the DE is

x 3 sin x

xu ′1 + x cos xu ′2 = 0 + (cos x − x sin x)u ′2 = x sin x.

y(x) − y0 = y(x) − y(x0 ) Z x Z = y ′ (t) dt = x0

x x0

  f t, y(t) dt.

Therefore, |y(x) − y0 | ≤ K |x − x0 |. Thus y(x) is bounded above and below by the lines y = y0 ± K (x − x0 ), and cannot have a vertical asymptote anywhere. Remark: we don’t seem to have needed the continuity of ∂ f /∂ y, only the continuity of f (to enable the use of the Fundamental Theorem).

668 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.1 (PAGE 992)

CHAPTER 18. ORDINARY DIFFERENTIAL EQUATIONS

Since the DE is linear and homogeneous, y = Ay1 + By2 = A cos(kx) + B sin(kx)

Section 18.1 Classifying Differential Equations (page 992) 1. 2. 3. 4. 5. 6. 7.

dy = 5y: 1st order, linear, homogeneous. dx d2 y + x = y: 2nd order, linear, nonhomogeneous. dx2 dy = x: 1st order, nonlinear. y dx y ′′′ + x y ′ = x sin x: 3rd order, linear, nonhomogeneous.

10.

3 = y(π/k) = A cos(π ) + B sin(π ) = − A 3 = y ′ (π/k) = − Ak sin(π ) + Bk cos(π ) = −Bk, provided A = −3 and B = −3/k. The required solution is 3 y = −3 cos(kx) − sin(kx). k

y ′′ + x sin x y ′ = y: 2nd order, linear, homogeneous.

y ′′ + 4y ′ − 3y = 2y 2 : 2nd order, nonlinear.

14. Given that y1 = ekx is a solution of y ′′ − k 2 y = 0, we

d3 y dy +t + t 2 y = t 3: dt 3 dt 3rd order, linear, nonhomogeneous.

8. cos x 9.

is a solution for any constants A and B. It will satisfy

suspect that y2 = e−kx is also a solution. This is easily verified since

dx + x sin t = 0: 1st order, nonlinear, homogeneous. dt

y (4) + e x y ′′ = x 3 y ′ : 4th order, linear, homogeneous.

y2′′ − k 2 y2 = k 2 e−kx − k 2 e−kx = 0. Since the DE is linear and homogeneous,

1 x 2 y ′′ + e x y ′ = : 2nd order, nonlinear. y

y = Ay1 + By2 = Aekx + Be−kx

11. If y = cos x, then y ′′ + y = − cos x + cos x = 0.

If y = sin x, then y ′′ + y = − sin x + sin x = 0. Thus y = cos x and y = sin x are both solutions of y ′′ + y = 0. This DE is linear and homogeneous, so any function of the form y = A cos x + B sin x,

is a solution for any constants A and B. It will satisfy

where A and B are constants, is a solution also. Therefore sin x − cos x is a solution (A = −1, B = 1), and

provided A = e−k /k and B = −ek /k. The required solution is 1 1 y = ek(x−1) − e−k(x−1) . k k

sin(x + 3) = sin 3 cos x + cos 3 sin x is a solution, but sin 2x is not since it cannot be represented in the form A cos x + B sin x.

12. If y = e x , then y ′′ − y = e x − e x = 0; if y = e−x , then

y ′′ − y = e−x − e−x = 0. Thus e x and e−x are both solutions of y ′′ − y = 0. Since y ′′ − y = 0 is linear and homogeneous, any function of the form

0 = y(1) = Aek + Be−k

2 = y ′ (1) = Akek − Bke−k ,

15. By Exercise 11, y = A cos x + B sin x is a solution of y ′′ + y = 0 for any choice of the constants A and B. This solution will satisfy 0 = y(π/2) − 2y(0) = B − 2 A, B A 3 = y(π/4) = √ + √ , 2 2

y = Ae x + Be−x is also a solution. Thus cosh x = 21 (e x + e−x ) is a solution, but neither cos x nor x e is a solution.

13. Given that y1 = cos(kx) is a solution of

y ′′

k2 y

+ = 0, we suspect that y2 = sin(kx) is also a solution. This is easily verified since

provided A =

√ √ 2 and B = 2 2. The required solution is y=

√

√ 2 cos x + 2 2 sin x.

y2′′ + k 2 y2 = −k 2 sin(kx) + k 2 sin(kx) = 0.

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SECTION 18.1 (PAGE 992)

16.

ADAMS and ESSEX: CALCULUS 8

y = er x is a solution of the equation y ′′ − y ′ − 2y = 0 if r 2 er x − r er x − 2er x = 0, that is, if r 2 − r − 2 = 0. This quadratic has two roots, r = 2, and r = −1. Since the DE is linear and homogeneous, the function y = Ae2x + Be−x is a solution for any constants A and B. This solution satisfies 1 = y(0) = A + B,

3.

4.

′

2 = y (0) = 2 A − B,

provided A = 1 and B = 0. Thus, the required solution is y = e2x .

17. If y = y1 (x) = x, then y1′ = 1 and y1′′ = 0. Thus

y1′′ + y1 = 0 + x = x. By Exercise 11 we know that y2 = A cos x + B sin x satisfies the homogeneous DE y ′′ + y = 0. Therefore, by Theorem 2, y = y1 (x) + y2 (x) = x + A cos x + B sin x

5.

is a solution of y ′′ + y = x. This solution satisfies 1 = y(π ) = π − A,

′

0 = y (π ) = 1 − B,

provided A = π − 1 and B = 1. Thus the required solution is y = x + (π − 1) cos x + sin x.

6.

18. If y = y1 (x) = −e, then y1′ = 0 and y1′′ = 0. Thus

y1′′ − y1 = 0 + e = e. By Exercise 12 we know that y2 = Ae x + Be−x satisfies the homogeneous DE y ′′ − y = 0. Therefore, by Theorem 2,

0 = y(1) = Ae +

B − e, e

1 = y ′ (1) = Ae −

7. B , e

provided A = (e + 1)/(2e) and B = e(e − 1)/2. Thus the required solution is y = −e + 21 (e +1)e x−1 + 12 (e −1)e1−x .

Section 18.2 Solving First-Order Equations (page 1004) 1.

dy y = dx 2x dy dx 2 = y x 2 ln y = ln x + C1

2.

⇒

8.

⇒

y2 = C x

3y − 1 dy = dZ x x Z dy dx = 3y − 1 x 1 1 ln |3y − 1| = ln |x| + ln C 3 3 3y − 1 =C x3 1 y = (1 + C x 3 ). 3

y2 d y = x 2 d x x 3 − y3 = C

or

dy = x 2 y2 Z dx Z dy = x2 dx y2 1 1 1 − = x3 + C y 3 3 3 ⇒ y=− 3 . x +C dY = tY ⇒ dt 2 t ln Y = + C1 , 2

dY = t dt Y or

Y = Cet

2 /2

dx = e x sin t dt Z Z e−x d x = sin t dt

−e−x = − cos t − C ⇒ x = − ln(cos t + C).

y = y1 (x) + y2 (x) = −e + Ae x + Be−x is a solution of y ′′ − y = e. This solution satisfies

x2 dy = 2 ⇒ dx y y3 x3 = + C1 , 3 3

9.

dy dy = 1 − y2 ⇒ = dx dx 1 − y2   1 1 1 + dy = dx 2 1+ y 1− y 1 1 + y = x + C1 ln 2 1 − y 1+y Ce2x − 1 = Ce2x or y = 1−y Ce2x + 1 dy d x Z dy 1 + y2 tan−1 y ⇒ y

= 1 + y2 Z = dx

=x +C = tan(x + C).

dy dy = 2 + ey ⇒ = dt dt 2 + ey Z Z e−y d y = dt 2e−y + 1 1 − ln(2e−y + 1) = t + C1 2 2e−y + 1 = C2 e−2t ,

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or

  1 y = − ln Ce−2t − 2

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.2 (PAGE 1004)

10. We have

13. dy = y 2 (1 − y) d x Z Z dy = dx = x + K. y 2 (1 − y)

14.

Expand the left side in partial fractions: A B C 1 = + 2+ y 2 (1 − y) y y 1− y A(y − y 2 ) + B(1 − y) + C y 2 = y 2 (1 − y) ( − A + C = 0; ⇒ A − B = 0; ⇒ A = B = C = 1. B = 1.

Hence, y =

Hence,

15.  Z  dy 1 1 1 = + + dy y 2 (1 − y) y y2 1−y 1 = ln |y| − − ln |1 − y|. y

Z

xe x d x = xe x − e x + C

y = x − 1 + Ce−x y 1 − = x + K. ln 1− y y

16. We have

(linear) Z  1 2 µ = exp − dx = 2 x x 2 1 dy − 3 y=1 x2 dx x d y =1 dx x2 y = x + C, so y = x 3 + C x 2 x2 dy 2y + dx x

Therefore, e

1 . Let 2 x Z 2 µ= d x = 2 ln x = ln x 2 , then eµ = x 2 , and x =

d 2 dy (x y) = x 2 + 2x y dx  d x   dy 2y 1 = x2 + = x2 =1 dx x x2 Z ⇒ x2y = dx = x + C ⇒

y=

1 C + 2. x x

R dy + 2e x y = e x . Let µ = 2e x d x = 2e x , then dx

d  2ex  x dy x e y = e2e + 2e x e2e y dx d x   x dy x = e2e + 2e x y = e2e e x . dx

dy 2 − y = x2 dx x

12. We have

1 x e + Ce−x . 2

Z  dy +y=x µ = exp 1 d x = ex dx d x (e y) = e x (y ′ + y) = xe x dx Z

ex y =

Therefore,

11.

Z  dy + 2y = 3 µ = exp 2 d x = e2x dx d 2x (e y) = e2x (y ′ + 2y) = 3e2x dx 3 3 e2x y = e2x + C ⇒ y = + Ce−2x 2 2 R dy We have + y = e x . Let µ = d x = x, then eµ = e x , dx and   d x dy dy (e y) = e x + ex y = ex + y = e2x dx dx dx Z 1 2x x 2x ⇒ e y = e d x = e + C. 2

Hence, y =

17.

2e x

Z

x

e2e e x d x

Let u = 2e x du = 2e x d x Z 1 1 x = eu du = e2e + C. 2 2

y=

1 x + Ce−2e . 2

dy + 10y = 1, y dt Z µ = 10 dt = 10t

1 10




=

2 10

d 10t dy (e y) = e10t + 10e10t y = e10t dt dt 1 10t e10t y(t) = e +C 10
2e e e 1 2 y 10 = 10 ⇒ = +C ⇒ C = 10 10 10 1 1 1−10t y= + e . 10 10

671 Copyright © 2014 Pearson Canada Inc.

SECTION 18.2 (PAGE 1004)

18.

dy + 3x 2 y = x 2 , dx Z µ = 3x 2 d x = x 3

ADAMS and ESSEX: CALCULUS 8

23.

y(0) = 1

x 2 y ′ + y = x 2 e1/x , y(1) = 3e 1 y ′ + 2 y = e1/x Zx 1 1 µ= dx = − x2 x   1 d  −1/x  −1/x e y =e y′ + 2 y = 1 dx x Z −1/x e y = 1 dx = x + C

24.

Z

x

1

H⇒

y = ln(x + e ).

y(π ) = 0

25. We require

21.

y(x) = 2 +

0

H⇒

22.

y(x) = 1 +

Z

0

x

y(0) = 2

26.

y(0) = 1

dy y2 = , i.e. d y/y 2 = d x/(1 + x 2 ) dx 1 + x2 1 − = tan−1 x + C y − 1 = 0 + C H⇒ C = −1

y = 1/(1 − tan−1 x).

C = e3

(1 + y 2 ) d y =

dy 2y =1+ dx x v+x

H⇒

H⇒

Z

2x d x

1 3 y = x 2 + C. 3

Since (2, 3) lies on the curve, 12 = 4 + C. Thus C = 8 1 and y + y 3 − x 2 = 8, or 3y + y 3 − 3x 2 = 24. 3

C =4

(y(t))2 dt 1 + t2

y(0) = 3

y = ln(x + C)

y+

dy x = , i.e. y d y = x d x dx y y2 = x 2 + C

22 = 02 + C H⇒ p y = 4 + x 2.

H⇒

dy 2x = . Thus dx 1 + y2 Z

y = (x 2 − π 2 )e− sin x .

t dt y(t)

y(1) = 1

H⇒

i.e. e y d y = d x

y(π ) = 0 ⇒ 0 = π 2 + C ⇒ C = −π 2 x

e−y dt

0

2x d x = x 2 + C

Z

y(t) dt t (t + 1)

3

d sin x (e y) = esin x (y ′ + (cos x)y) = 2x dx Z

esin x y =

x

3 = y(0) = ln C

y = (x + 2)e1/x .

y ′ + (cos x)y = 2xe− sin x , Z µ = cos x d x = sin x

y(x) = 3 + dy = e−y , dx ey = x + C

y(1) = 3e ⇒ 3 = 1 + C ⇒ C = 2

20.

Z

dy y = , for x > 0 dx x(x + 1) dy dx dx dx = = − y x(x + 1) x x +1 x + ln C ln y = ln x +1 Cx y= , H⇒ 1 = C/2 x +1 2x y= . x +1

d x3 3 dy 3 3 (e y) = e x + 3x 2 e x y = x 2 e x dx d x Z 1 3 3 3 ex y = x 2ex d x = ex + C 3 1 2 y(0) = 1 ⇒ 1 = + C ⇒ C = 3 3 1 2 −x 3 y= + e . 3 3

19.

y(x) = 1 +

Let y = v x

dv = 1 + 2v dx

dv x =1+v d Z x Z dv dx = 1+v x ln |1 + v| = ln |x| + C1 y 1 + = C x ⇒ x + y = C x 2. x Since (1, 3) lies on the curve, 4 = C. Thus the curve has equation x + y = 4x 2 .

672 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

27.

dy x+y = dx x−y

Let y = v x

SECTION 18.2 (PAGE 1004)

30.

dv x(1 + v) = dx x(1 − v) 1+v 1 + v2 dv = −v = x 1−v Z 1−v Zd x 1−v dx dv = 1 + v2 x 1 −1 tan v − ln(1 + v 2 ) = ln |x| + C1 2 1 x 2 + y2 −1 = ln |x| + C1 tan (y/x) − ln 2 x2 2 tan−1 (y/x) − ln(x 2 + y 2 ) = C.

dy xy = 2 dx x + 2y 2

v+x

Let y = v x

dv vx2 = dx (1 + 2v 2 )x 2 dv v 2v 3 x = − v = − dx 1 + 2v 2 1 + 2v 2 Z Z 1 + 2v 2 dx dv = −2 v3 x 1 − 2 + 2 ln |v| = −2 ln |x| + C1 2v x2 − 2 + 2 ln |y| = C1 2y x 2 − 4y 2 ln |y| = C y 2.

v+x

31.

32.

29.

x 2 + x y + y2 dy = dx x2 x 2 (1 + v + v 2 ) dv v+x = dx x2 Z Z dv dx = 1 + v2 x −1 tan v = ln |x| + C   y = tan ln |x| + C x   y = x tan ln |x| + C .

Let y = v x

Let y = v x

dv x 3 (1 + 3v 2 ) = 3 dx x (3v + v 3 ) 1 + 3v 2 1 − v4 dv = −v = x 3 dx 3v + v v(3 + v 2 ) Z Z 2 (3 + v )v dv dx = Let u = v 2 1 − v4 x du = 2v dv Z 1 3+u du = ln |x| + C1 2 1 − u 2 3 u + 1 1 ln − ln |1 − u 2 | = ln |x| + C1 4 u − 1 4 2 4 4 y + x2 − ln x − y = 4 ln |x| + C2 3 ln 2 x4 y − x2   x 2 + y2 3 1 ln 2 = C2 x − y2 x 4 − y4 2 (x + y 2 )2 = C2 ln 2 (x − y 2 )4 x 2 + y 2 = C(x 2 − y 2 )2 .

v+x

28.

dy x 3 + 3x y 2 = 2 dx 3x y + y 3

y dy = y + x cos2 (let y = v x) dx x dv = v x + x cos2 v xv + x 2 dx dv x = cos2 v dx dx sec2 v dv = x tan v = ln |x| + ln |C| y = ln |C x| tan x y = xtan−1 (ln |C x|). x

dy y = − e−y/x (let y=vx) dx x dv v+x = v − e−v dx dx ev dv = − x ev = − ln |x| + ln |C| C e y/x = ln x C y = x ln ln . x

33. If ξ = x − x0 , η = y − y0 , and dy ax + by + c = , dx ex + f y + g

673 Copyright © 2014 Pearson Canada Inc.

SECTION 18.2 (PAGE 1004)

ADAMS and ESSEX: CALCULUS 8

then

38. dη dy a(ξ + x0 ) + b(η + y0 ) + c = = dξ dx e(ξ + x0 ) + f (η + y0 ) + g aξ + bη + (ax0 + by0 + c) = eξ + f η + (ex0 + f y0 + g) aξ + bη = eξ + f η

39.

provided x0 and y0 are chosen such that ax0 + by0 + c = 0,

and

ex0 + f y0 + g = 0.

34. The system x0 +2y0 −4 = 0, 2x0 − y0 −3 = 0 has solution x0 = 2, y0 = 1. Thus, if ξ = x − 2 and η = y − 1, where dy x + 2y − 4 = , dx 2x − y − 3 then

ξ + 2η dη = dξ 2ξ − η

Let η = vξ

40.

dv 1 + 2v = dξ 2−v dv 1 + 2v 1 + v2 ξ = −v = dξ 2−v 2−v  Z Z  dξ 2−v dv = 1 + v2 ξ 1 −1 2 2 tan v − ln(1 + v ) = ln |ξ | + C1 2 η 4 tan−1 − ln(ξ 2 + η2 ) = C. ξ

v +ξ

Hence the solution of the original equation is 4 tan−1

35.

36.

  y−1 − ln (x − 2)2 + (y − 1)2 = C. x −2

 2y y2 dy = 0 2x + 1 − 2 d x + x x   y2 d x2 + x + =0 x y2 x2 + x + = C. x 

(x 2 + 2y) d x − x d y = 0

M = x 2 + 2y, N = −x   ∂N 1 ∂M 3 − = − (indep. of y) N ∂y ∂x x dµ 3 1 = − dx ⇒ µ = 3 µ x x   2y 1 1 + 3 dx − 2 dy = 0 x x x  y d ln |x| − 2 = 0 x y ln |x| − 2 = C1 x y = x 2 ln |x| + C x 2 .  2  x x (xe + x ln y + y) d x + + x ln x + x sin y d y = 0 y x2 + x ln x + x sin y M = xe x + x ln y + y, N= y ∂M x ∂N 2x = + 1, = + ln x + 1 + sin y ∂y y ∂x y     ∂N 1 1 ∂M 1 x − = − − ln x − sin y = − N ∂y ∂x N y x 1 1 dµ = − dx ⇒ µ = µ x x    y x x e + ln y + dx + + ln x + sin y d y x y
d e x + x ln y + y ln x − cos y = 0 e x + x ln y + y ln x − cos y = C.

41. Since a > b > 0 and k > 0,

lim x(t) = lim

t→∞

(x y 2 + y) d x + (x 2 y + x) d y = 0   1 2 2 d x y + xy = 0 2 x 2 y 2 + 2x y = C. d(e x sin y + x 2 + y 2 ) = 0 2

lim x(t) = lim

t→∞

2

e sin y + x + y = C.

37.

be(b−a)kt − a ab(0 − 1) = = b. 0−a t→∞

42. Since b > a > 0 and k > 0,

(e x sin y + 2x) d x + (e x cos y + 2y) d y = 0 x

  ab e(b−a)kt − 1

t→∞

= lim

  ab e(b−a)kt − 1

(b−a)kt − a be   ab 1 − e(a−b)kt

b − ae(a−b)kt ab(1 − 0) = = a. b−0

e xy (1 + x y) d x + x 2 e xy d y = 0   d xe xy = 0 ⇒ xe xy = C.

t→∞

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.2 (PAGE 1004)

43. The solution given, namely

45. We proceed by separation of variables: dv = mg − kv 2 dt dv k = g − v2 dt m dv = dt k g − v2 Z Z m kt k dv = dt = + C. mg 2 m m −v k

m


ab e(b−a)kt − 1 x= , be(b−a)kt − a is indeterminate (0/0) if a = b. If a = b the original differential equation becomes dx = k(a − x)2 , dt which is separable and yields the solution Z

1 = a−x

dx =k (a − x)2

Since x(0) = 0, we have C =

Solving for x, we obtain

x=

Z

Let a 2 = mg/k, where a > 0. Thus, we have Z kt dv = +C a 2 − v 2 m a + v kt 1 = ln +C 2a a−v m r a + v 2akt kg ln = + C1 = 2 t + C1 a−v m m √ a+v = C2 e2t kg/m . a−v

dt = kt + C.

1 1 1 , so = kt + . a a−x a

a 2 kt . 1 + akt

This solution also results from evaluating the limit of solution obtained for the case a 6= b as b approaches a (using l’Hˆopital’s Rule, say).

44. Given that m

dv = mg − kv, then dt Z

Z dv = dt k g− v m k m − ln g − v = t + C. k m

Since v(0) = 0, therefore C = −

remains positive for all t > 0, so m ln k

g g−

k v m

m k ln g. Also, g − v k m

=t

k v m = e−kt/m g  mg  ⇒ v = v(t) = 1 − e−kt/m . k

g−

mg . This limiting velocity can be k obtained directly from the differential equation by setting dv = 0. dt Note that lim v(t) = t→∞

46.

Assuming v(0) = 0, we get C2 = 1. Thus √ a + v = e2t kg/m (a − v)   √   √ v 1 + e2t kg/m = a e2t kg/m − 1 r  mg  2t √kg/m e −1 = k √ r 2t kg/m mg e −1 v= √ k e2t kg/m + 1 r mg as t → ∞. This also follows from Clearly v → k dv setting = 0 in the given differential equation. dt The balance in the account after t years is y(t) and y(0) = 1000. The balance must satisfy dy y2 = 0.1y − dt 1, 000, 000 dy 105 y − y 2 = 106 Zdt Z dy dt = 6 5 2 10 10 y − y  Z  1 1 1 t C + 5 dy = 6 − 5 5 y 10 10 10 − y 10 t 5 ln |y| − ln |10 − y| = −C 10 5 10 − y = eC−(t/10) y 105 y = C−(t/10) . e +1

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SECTION 18.2 (PAGE 1004)

ADAMS and ESSEX: CALCULUS 8

1000 = y(0) =

105 +1

and y=

Let µ = C = ln 99,

⇒

eC

105 99e−t/10 + 1

i dh dx + 5(500 + t)4 x (500 + t)5 x = (500 + t)5 dt dy   dx 5x = (500 + t)5 + dy 500 + t

.

= 0.12(500 + t)5 .

The balance after 1 year is y=

105 99e−1/10

+1

Hence,

≈ $1, 104.01.

5

(500 + t) x = 0.12

As t → ∞, the balance can grow to 105 lim y(t) = lim (4.60−0.1t) = = $100, 000. t→∞ t→∞ e +1 0+1

x = 0.02(500 + t) + (1.25 × 1015 )(500 + t)−5 .

100, 000 99e−t/10 + 1 99e−t/10 + 1 = 2 t = 10 ln 99 ≈ 46 years.

After 40 min, there will be

50, 000 = y(t) =

dy = 0, y+x dx

or

x = 0.02(540) + (1.25 × 1015 )(540)−5 = 38.023 kg of salt in the tank.

49. If µ(y)M(x, y) d x + µ(y)N (x, y) d y is exact, then   ∂  ∂  µ(y)M(x, y) = µ(y)N (x, y) ∂y ∂x ∂M ∂N ′ µ (y)M + µ =µ ∂y ∂x   µ′ 1 ∂N ∂M = − . µ M ∂x ∂y

dy y =− . dx x

Curves that intersect these hyperbolas at right angles dy x must therefore satisfy = , or x d x = y d y, a sepdx y arated equation with solutions x 2 − y 2 = C, which is also a family of rectangular hyperbolas. (Both families are degenerate at the origin for C = 0.)

Thus M and N must be such that   1 ∂N ∂M − M ∂x ∂y

48. Let x(t) be the number of kg of salt in the solution in the tank after t minutes. Thus, x(0) = 50. Salt is coming into the tank at a rate of 10 g/L × 12 L/min = 0.12 kg/min. Since the contents flow out at a rate of 10 L/min, the volume of the solution is increasing at 2 L/min and thus, at any time t, the volume of the solution is 1000 + 2t L. Therefore the conx(t) centration of salt is L. Hence, salt is being 1000 + 2t removed at a rate x(t) 5x(t) kg/L × 10 L/min = kg/min. 1000 + 2t 500 + t Therefore, dx 5x = 0.12 − dt 500 + t dx 5 + x = 0.12. dt 500 + t

(500 + t)5 dt = 0.02(500 + t)6 + C

Since x(0) = 50, we have C = 1.25 × 1015 and

For the account to grow to $50,000, t must satisfy

47. The hyperbolas x y = C satisfy the differential equation

Z

⇒ x = 0.02(500 + t) + C(500 + t)−5 .

105

⇒ ⇒

Z

5 dt = 5 ln |500 + t| = ln(500 + t)5 for 500 + t t > 0. Then eµ = (500 + t)5 , and

Since y(0) = 1000, we have

depends only on y.

50.

2y 2 (x + y 2 ) d x + x y(x + 6y 2 ) d y = 0

(2x y 2 + 2y 4 )µ(y) d x + (x 2 y + 6x y 3)µ(y) d y = 0 ∂M = (4x y + 8y 3 )µ(y) + (2x y 2 + 2y 4 )µ′ (y) ∂y ∂N = (2x y + 6y 3 )µ(y). ∂x For exactness we require (2x y 2 + 2y 4 )µ′ (y) = [(2x y + 6y 3 ) − (4x y + 8y 3 )]µ(y) y(2x y + 2y 3 )µ′ (y) = −(2x y + 2y 3 )µ(y) 1 yµ′ (y) = −µ(y) ⇒ µ(y) = y (2x y + 2y 3 ) d x + (x 2 + 6x y 2 ) d y = 0 d(x 2 y + 2x y 3 ) = 0

676 Copyright © 2014 Pearson Canada Inc.

⇒

x 2 y + 2x y 3 = C.

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.3 (PAGE 1012)

51. Consider y d x − (2x + y 3 e y ) d y = 0.

∂M ∂N Here M = y, N = −2x − y 3 e y , = 1, and = −2. ∂y ∂x Thus µ′

3 1 =− ⇒ µ= 3 µ y y   1 2x y d x − + e dy = 0 y2 y3   x d − ey = 0 y2 x − e y = C, or x − y 2 e y = C y 2. y2

52. If µ(x y) is an integrating factor for M d x + N d y = 0, then

∂ ∂ (µM) = (µN ), or ∂y ∂x ∂N ∂M = yµ′ (x y)N + µ(x y) . xµ′ (x y)M + µ(x y) ∂y ∂x Thus M and N will have to be such that the right-hand side of the equation   µ′ (x y) 1 ∂N ∂M = − µ(x y) x M − yN ∂x ∂y

53.

depends only on the product x y.     y2 x sin x dx − For x cos x + + y d y we have x y x sin x y2 N =− −y M = x cos x + , x y ∂M 2y ∂N sin x x cos x = , =− − ∂y x ∂x y y   ∂N ∂M x cos x 2y sin x − =− + + ∂x ∂y y y x 2 2 x M − y N = x cos x + y + x sin x + y 2   1 ∂N ∂M 1 − =− . x M − yN ∂x ∂y xy

The solution is x sin x − y 2 = C x y.

Section 18.3 Existence, Uniqueness, and Numerical Methods (page 1012) A computer spreadsheet was used in Exercises 1–12. The intermediate results appearing in the spreadsheet are not shown in these solutions. 1. We start with x0 = 1, y0 = 0, and calculate xn+1 = xn + h,

yn+1 = yn + h(xn + yn ).

a) For h = 0.2 we get x5 = 2, y5 = 1.97664. b) For h = 0.1 we get x10 = 2, y10 = 2.187485. c) For h = 0.05 we get x20 = 2, y20 = 2.306595.

2. We start with x0 = 1, y0 = 0, and calculate xn+1 = xn + h, u n+1 = yn + h(xn + yn ) h yn+1 = yn + (xn + yn + xn+1 + u n+1 ). 2 a) For h = 0.2 we get x5 = 2, y5 = 2.405416. b) For h = 0.1 we get x10 = 2, y10 = 2.428162. c) For h = 0.05 we get x20 = 2, y20 = 2.434382.

3. We start with x0 = 1, y0 = 0, and calculate xn+1 = xn + h pn = xn + yn h h qn = xn + + yn + pn 2 2 h h rn = xn + + yn + qn 2 2 qn = xn + h + yn + hrn h yn+1 = yn + ( pn + 2qn + 2rn + sn ). 6

Thus, an integrating factor is given by µ′ (t) 1 =− µ(t) t

⇒

µ(t) =

1 . t

We multiply the original equation by 1/(x y) to make it exact:     cos x y sin x 1 + 2 dx − + dy = 0 y x y2 x   sin x y d − =0 y x y sin x − = C. y x

a) For h = 0.2 we get x5 = 2, y5 = 2.436502. b) For h = 0.1 we get x10 = 2, y10 = 2.436559. c) For h = 0.05 we get x20 = 2, y20 = 2.436563.

4. We start with x0 = 0, y0 = 0, and calculate xn+1 = xn + h,

yn+1 = hxn e−yn .

a) For h = 0.2 we get x10 = 2, y10 = 1.074160. b) For h = 0.1 we get x20 = 2, y20 = 1.086635.

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SECTION 18.3 (PAGE 1012)

ADAMS and ESSEX: CALCULUS 8

5. We start with x0 = 0, y0 = 0, and calculate

a) For h = 0.2 we get x5 = 1, y5 = 0.865766. b) For h = 0.1 we get x10 = 1, y10 = 0.865769.

xn+1 = xn + h, u n+1 = yn + hxn e−yn h yn+1 = yn + (xn e−yn + xn+1 e−u n+1 . 2 a) For h = 0.2 we get x10 = 2, y10 = 1.097897.

c) For h = 0.05 we get x20 = 1, y20 = 0.865769.

10. We start with x0 = 0, y0 = 0, and calculate xn+1 = xn + h,

b) For h = 0.1 we get x20 = 2, y20 = 1.098401.

6. We start with x0 = 0, y0 = 0, and calculate xn+1 = xn + h pn = xn e−yn   h qn = xn + e−(yn +(h/2) pn 2   h e−(yn +(h/2)qn rn = xn + 2

a) For h = 0.2 we get x5 = 1, y5 = 0.944884. b) For h = 0.1 we get x10 = 1, y10 = 0.926107. c) For h = 0.05 we get x20 = 1, y20 = 0.915666.

11. We start with x0 = 0, y0 = 0, and calculate xn+1 = xn + h, u n+1 = yn + h cos(xn2 ) h 2 yn+1 = yn + (cos(xn2 ) + cos(xn+1 )). 2

sn = (xn + h)e−(yn +hrn ) h yn+1 = yn + ( pn + 2qn + 2rn + sn ). 6

a) For h = 0.2 we get x5 = 1, y5 = 0.898914.

a) For h = 0.2 we get x10 = 2, y10 = 1.098614.

b) For h = 0.1 we get x10 = 1, y10 = 0.903122.

b) For h = 0.1 we get x20 = 2, y20 = 1.098612.

c) For h = 0.05 we get x20 = 1, y20 = 0.904174.

7. We start with x0 = 0, y0 = 0, and calculate xn+1 = xn + h,

yn+1 = yn + h cos(xn2 ).

yn+1 = yn + h cos yn .

12. We start with x0 = 0, y0 = 0, and calculate xn+1 = xn + h

a) For h = 0.2 we get x5 = 1, y5 = 0.89441.

pn = cos(xn2 )

b) For h = 0.1 we get x10 = 1, y10 = 0.87996.

qn = cos((xn + (h/2))2 )

c) For h = 0.05 we get x20 = 1, y20 = 0.872831.

rn = cos((xn + (h/2))2 )

8. We start with x0 = 0, y0 = 0, and calculate

qn = cos((xn + h)2 ) h yn+1 = yn + ( pn + 2qn + 2rn + sn ). 6

xn+1 = xn + h, u n+1 = yn + h cos yn h yn+1 = yn + (cos yn + cos u n+1 ). 2

a) For h = 0.2 we get x5 = 1, y5 = 0.904524.

a) For h = 0.2 we get x5 = 1, y5 = 0.862812.

b) For h = 0.1 we get x10 = 1, y10 = 0.904524.

b) For h = 0.1 we get x10 = 1, y10 = 0.865065.

c) For h = 0.05 we get x20 = 1, y20 = 0.904524.

c) For h = 0.05 we get x20 = 1, y20 = 0.865598.

9. We start with x0 = 0, y0 = 0, and calculate xn+1 pn qn rn qn yn+1

= xn + h = cos yn = cos(yn + (h/2) pn ) = cos(yn + (h/2)qn ) = cos(yn + hrn ) h = yn + ( pn + 2qn + 2rn + sn ). 6

13.

Z x 2 y(x) = 2 + y(t) dt 1  2 dy = y(x) , y(1) = 2 + 0 = 2 dx dy 1 = dx ⇒ − =x +C y(x) y2 1 3 − =1+C ⇒ C =− 2 2 1 2 y=− = . x − (3/2) 3 − 2x

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INSTRUCTOR’S SOLUTIONS MANUAL

14.

u(x) = 1 + 3

Z

x

SECTION 18.3 (PAGE 1012)

on [0, X], so y(x) ≤ v(x) there.

t 2 u(t) dt

2

b) The IVP u ′ = u 2 , u(0) = 1 has solution 1 u(x) = , obtained by separation of variables. 1−x This solution is valid for x < 1.

du = 3x 2 u(x), u(2) = 1 + 0 = 1 dx du = 3x 2 d x ⇒ ln u = x 3 + C u 0 = ln 1 = ln u(2) = 23 + C ⇒ C = −8 u=e

x 3 −8

.

15. For the problem y ′ = f (x), y(a) = 0, the 1-step RungeKutta method with h = b − a gives: x0 = a,

y0 = 0,

p0 = f (a),

q0 = f



x1 = x0 + h = b    h a+b a+ = f = r0 2 2

s0 = f (a + h) = f (b) h y1 = y0 + ( p0 + 2q0 + 2r0 + s0 ) 6    b−a a+b = f (a) + 4 f + f (b) , 6 2

which is the Simpson’s Rule approximation to Z b f (x) d x based on 2 subintervals of length h/2. a

16. If φ(0) = A ≥ 0 and φ ′ (x) ≥ kφ(x) on an interval [0, X],

The IVP v ′ = 1 + v
2 , v(0) = 1 has solution v(x) = tan x + π4 , also obtained by separation of variables. It is valid only for −3π/4 < x < π/4. Observe that π/4 < 1, proving the assertion made about v in part (a). By the result of part (a), the solution of the IVP y ′ = x + y 2 , y(0) = 1, increases on an interval [0, X] and → ∞ as x → X from the left, where X is some number in the interval [π/4, 1]. c) Here are some approximations to y(x) for values of x near 0.9 obtained by the Runge-Kutta method with x0 = 0 and y0 = 1: For h = 0.05 n = 17 n = 18 n = 19

xn = 0.85 xn = 0.90 xn = 0.95

yn = 12.37139 yn = 31.777317 yn = 4071.117315.

xn xn xn xn xn

= 0.86 = 0.88 = 0.90 = 0.92 = 0.94

yn yn yn yn yn

= 14.149657 = 19.756061 = 32.651029 = 90.770048 = 34266.466629.

xn xn xn xn xn xn xn xn

= 0.86 = 0.87 = 0.88 = 0.89 = 0.90 = 0.91 = 0.92 = 0.93

yn yn yn yn yn yn yn yn

= 14.150706 = 16.493286 = 19.761277 = 24.638758 = 32.703853 = 48.591332 = 94.087476 = 636.786465

where k > 0 and X > 0, then d dx



φ(x) ekx



=

ekx φ ′ (x) − kekx φ(x) ≥ 0. e2kx

Thus φ(x)/ekx is increasing on [0, X]. Since its value at x = 0 is φ(0) = A ≥ 0, therefore φ(x)/ekx ≥ A on [0, X], and φ(x) ≥ Aekx there.

17.

a) Suppose u ′ = u 2 , y ′ = x + y 2 , and v ′ = 1 + v 2 on [0, X], where u(0) = y(0) = v(0) = 1, and X > 0 is such that v(x) is defined on [0, X]. (In part (b) below, we will show that X < 1, and we assume this fact now.) Since all three functions are increasing on [0, X], we have u(x) ≥ 1, y(x) ≥ 1, and v(x) ≥ 1 on [0, X]. If φ(x) = y(x) − u(x), then φ(0) = 0 and φ ′ (x) = x + y 2 − u 2 ≥ y 2 − u 2 ≥ (y + u)(y − u) ≥ 2φ on [0, X]. By Exercise 16, φ(x) ≥ 0 on [0, X], and so u(x) ≤ y(x) there. Similarly, since X < 1, if φ(x) = v(x) − y(x), then φ(0) = 0 and φ ′ (x) = 1 + v 2 − x − y 2 ≥ v 2 − y 2 ≥ (v + y)(v − y) ≥ 2φ

For h = 0.02 n n n n n

= 43 = 44 = 45 = 46 = 47

For h = 0.01 n n n n n n n n

= 86 = 87 = 88 = 89 = 90 = 91 = 92 = 93

n = 94

xn = 0.94

yn = 2.8399 × 1011 .

The values are still in reasonable agreement at x = 0.9, but they start to diverge quickly thereafter. This suggests that X is slightly greater than 0.9.

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SECTION 18.3 (PAGE 1012)

ADAMS and ESSEX: CALCULUS 8

Section 18.4 Differential Equations of Second Order (page 1016) 1. If y1 = e x , then y1′′ − 3y1′ + 2y1 = e x (1 − 3 + 2) = 0, so y1 is a solution of the DE y ′′ − 3y ′ + 2y = 0. Let y = e x v. Then y ′ = e x (v ′ + v), y ′′ = e x (v ′′ + 2v ′ + v) ′′ ′ x ′′ y − 3y + 2y = e (v + 2v ′ + v − 3v ′ − 3v + 2v) = e x (v ′′ − v ′ ). y ′′

3y ′

v′

y satisfies − + 2y = 0 provided w = satisfies w′ − w = 0. This equation has solution v ′ = w = C1 e x , so v = C1 e x + C2 . Thus the given DE has solution y = e x v = C1 e2x + C2 e x .

2. If y1 = e−2x , then y1′′ − y1′ − 6y1 = e−2x (4 + 2 − 6) = 0,

y satisfies x 2 y ′′ − 3x y ′ + 4y = 0 provided w = v ′ satisfies xw′ + w = 0. This equation has solution v ′ = w = C1 /x (obtained by separation of variables), so v = C1 ln x + C2 . Thus the given DE has solution y = x 2 v = C1 x 2 ln x + C2 x 2 .

5. If y = x, then y ′ = 1 and y ′′ = 0. Thus

x 2 y ′′ − x(x + 2)y ′ + (x + 2)y = 0. Now let y = xv(x). Then y ′ = v + xv ′ ,

Substituting these expressions into the differential equation we get

so y1 is a solution of the DE y ′′ − y ′ − 6y = 0. Let y = e−2x v. Then ′

y =e

−2x

′

(v − 2v),

′′

y =e

−2x

′′

2x 2 v ′ + x 3 v ′′ − x 2 v − 2xv − x 3 v ′

′

3 ′′

y satisfies y ′′ − y ′ − 6y = 0 provided w = v ′ satisfies w′ − 5w = 0. This equation has solution v ′ = w = (C1 /5)e5x , so v = C1 e5x + C2 . Thus the given DE has solution y = e−2x v = C1 e3x + C2 e−2x .

3. If y1 = x on (0, ∞), then

y ′ = xv ′ + v,

y ′′ = xv ′′ + 2v ′

x 2 y ′′ + 2x y ′ − 2y = x 3 v ′′ + 2x 2 v ′ + 2x 2 v ′ + 2xv − 2xv = x 2 (xv ′′ + 4v ′ ).

y satisfies x 2 y ′′ + 2x y ′ − 2y = 0 provided w = v ′ satisfies xw′ + 4w = 0. This equation has solution v ′ = w = −3C1 x −4 (obtained by separation of variables), so v = C1 x −3 + C2 . Thus the given DE has solution y = xv = C1 x −2 + C2 x.

4. If y1 = x 2 on (0, ∞), then

x 2 y1′′ − 3x y1′ + 4y1 = 2x 2 − 6x 2 + 4x 2 = 0, so y1 is a solution of the DE x 2 y ′′ − 3x y ′ + 4y = 0. Let y = x 2 v(x). Then y ′ = x 2 v ′ + 2xv, 2 ′′

′

y ′′ = x 2 v ′′ + 4xv ′ + 2v

4 ′′

3 ′

2

x y − 3x y + 4y = x v + 4x v + 2x v

− 3x 3 v ′ − 6x 2 v + 4x 2 v

= x 3 (xv ′′ + v ′ ).

or v ′′ − v ′ = 0,

which has solution v = C1 + C2 e x . Hence the general solution of the given differential equation is y = C1 x + C2 xe x .

6. If y = x −1/2 cos x, then 1 y ′ = − x −3/2 cos x − x −1/2 sin x 2 3 y ′′ = x −5/2 cos x + x −3/2 sin x − x −1/2 cos x. 4

x 2 y1′′ + 2x y1′ − 2y1 = 0 + 2x − 2x = 0, so y1 is a solution of the DE x 2 y ′′ + 2x y ′ − 2y = 0. Let y = xv(x). Then

− 2x 2 v ′ + x 2 v + 2xv = 0

x v − x 3 v ′ = 0,

(v − 4v + 4v)

y ′′ − y ′ − 6y = e−2x (v ′′ − 4v ′ + 4v − v ′ + 2v − 6v) = e x (v ′′ − 5v ′ ).

y ′′ = 2v ′ + xv ′′ .

Thus   1 y x 2 y ′′ + x y ′ + x 2 − 4 3 = x −1/2 cos x + x 1/2 sin x − x 3/2 cos x 4 1 1 − x −1/2 cos x − x 1/2 sin x + x 3/2 cos x − x −1/2 cos x 2 4 = 0. Therefore y = x −1/2 cos x is a solution of the Bessel equation   1 2 ′′ ′ 2 x y + xy + x − y = 0. (∗) 4 Now let y = x −1/2 (cos x)v(x). Then 1 y ′ = − x −3/2 (cos x)v − x −1/2 (sin x)v + x −1/2 (cos x)v ′ 2 3 −5/2 ′′ y = x (cos x)v + x −3/2 (sin x)v − x −3/2 (cos x)v ′ 4 − x −1/2 (cos x)v − 2x −1/2 (sin x)v ′ + x −1/2 (cos x)v ′′ .

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.5 (PAGE 1026)

If we substitute these expressions into the equation (∗), many terms cancel out and we are left with the equation

9. If y = C1 eλx v, then y′ = C1 λeλx v = C1 eλx Av = Ay

(cos x)v ′′ − 2(sin x)v ′ = 0. Substituting u = v ′ , we rewrite this equation in the form

10.

du (cos x) = 2(sin x)u d x Z Z du = 2 tan x d x ⇒ ln |u| = 2 ln | sec x| + C0 . u

provided λ and v satisfy Av = λv. 2 − λ 1 = 6 − 5λ + λ2 − 2 2 3 − λ

= λ2 − 5λ + 4 = (λ − 1)(λ − 4) = 0

if λ = 1 or λ = 4.   2 1 Let A = . 2 3

Thus v ′ = u = C1 sec2 x, from which we obtain v = C1 tan x + C2 .

If λ = 1 and Av = v, then      2 1 v1 v1 A= = ⇔ v 1 + v 2 = 0. 2 3 v2 v2

Thus the general solution of the Bessel equation (∗) is y = x −1/2 (cos x)v = C1 x −1/2 sin x + C2 x −1/2 cos x.

Thus we may take v = v1 =

7. If y1 = y and y2 = y ′ where y satisfies ′′

then y1′ = y2 and y2′ = −a0 y1 − a1 y2 + f . Thus d dx

y1 y2



=



0 −a0

1 −a1



y1 y2



+



0 f

 1 . −1

If λ = 4 and Av = 4v, then      2 1 v1 v A= =4 1 ⇔ 2v 1 − v 2 = 0. 2 3 v2 v2

′

y + a1 (x)y + a0 (x)y = f (x),







Thus we may take v = v2 =

.

  1 . 2

By the result of Exercise 9, y = e x v1 and y = e4x v2 are solutions of the homogeneous linear system y′ = Ay. Therefore the general solution of the system is

8. If y satisfies y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y ′ + a0 (x)y = f (x),

y = C1 e x v1 + C2 e4x v2 ,

then let

that is

y1 = y,

y2 = y ′ ,

y3 = y ′′ ,

...

yn = y (n−1) .



Therefore y1′ = y2 ,

y2′ = y3 ,

...

′ yn−2 = yn−1 ,

y1 y2

and we have 0  y2   0 d   .  =  .. . d x  ..    0 yn −a0   0 0 . . +  . . 0 f y  1

1 0 .. .

0 1 .. .

... ...

0 −a1

0 −a2

... ...

0 0 .. .



y  1  y  2   .   .  1  . yn −an

= C1 e

x



1 −1



+ C2 e

C1 e x + C2 e4x

y1 =

4x

  1 , 2

or

y2 = −C1 e x + 2C2 e4x .

and

yn′ = −a0 y1 − a1 y2 − a2 y3 − · · · − an−1 yn + f,





Section 18.5 Linear Differential Equations with Constant Coefficients (page 1026) y ′′ + 7y ′ + 10y = 0

1.

auxiliary eqn r 2 + 7r + 10 = 0 (r + 5)(r + 2) = 0 y = Ae ′′

2.

−5t ′

+ Be

−2t

y − 2y − 3y = 0

auxiliary eqn r 2 − 2r − 3 = 0 y = Ae

−t

+ Be

3t

⇒ r = −5, −2

⇒ r = −1, r = 3

681 Copyright © 2014 Pearson Canada Inc.

SECTION 18.5 (PAGE 1026)

y ′′ + 2y ′ = 0

3.

auxiliary eqn r 2 + 2r = 0

ADAMS and ESSEX: CALCULUS 8

14. Given that y ′′ + 10y ′ + 25y = 0, hence

r 2 + 10r + 25 = 0 ⇒ (r + 5)2 = 0 ⇒ r = −5. Thus,

⇒ r = 0, −2

y = A + Be−2t

4.

y ′ = −5e−5t (A + Bt) + Be−5t .

4y ′′ − 4y ′ − 3y = 0

4r 2 − 4r − 3 = 0 ⇒ (2r + 1)(2r − 3) = 0 Thus, r1 =

− 21 ,

r2 =

3 2,

and y = Ae

y ′′ + 8y ′ + 16y = 0

5.

auxiliary eqn r 2 + 8r + 16 = 0

y = Ae−4t + Bte−4t

6.

y = Ae−5t + Bte−5t

Since

−(1/2)t

+ Be

(3/2)t

15.

r 2 − 2r + 1 = 0 ⇒ (r − 1)2 = 0 Thus, r = 1, 1, and y = Aet + Btet . y ′′ − 6y ′ + 10y = 0

auxiliary eqn r 2 − 6r + 10 = 0

y ′ = (−2 Ae−2t + Be−2t ) cos t − (Ae−2t + 2Be−2t ) sin t.

9y ′′ + 6y ′ + y = 0

9r 2 + 6r + 1 = 0 ⇒ (3r + 1)2 = 0

Thus, r = − 31 , − 31 , and y = Ae−(1/3)t + Bte−(1/3)t . y ′′ + 2y ′ + 5y = 0

9.

auxiliary eqn r 2 + 2r + 5 = 0 ⇒ r = −1 ± 2i y = Ae−t cos 2t + Be−t sin 2t

10. For y ′′ − 4y ′ + 5y = 0 the auxiliary equation is

r 2 − 4r + 5 = 0, which has roots r = 2 ± i . Thus, the general solution of the DE is y = Ae2t cos t + Be2t sin t.

11. For y ′′ + 2y ′ + 3y = 0 the auxiliary equation is√ r 2 + 2r + 3 = 0, which has solutions r = −1 ± the general solution of the given√equation is √ y = Ae−t cos( 2t) + Be−t sin( 2t).

Now 2 = y(0) = A ⇒ A = 2, and 2 = y ′ (0) = −2 A + B ⇒ B = 6. Therefore y = e−2t (2 cos t + 6 sin t).

16. The auxiliary equation r 2 − (2 + ǫ)r + (1 + ǫ) factors

to (r − 1 − ǫ)(r − 1) = 0 and so has roots r = 1 + ǫ and r = 1. Thus the DE y ′′ − (2 + ǫ)y ′ + (1 + ǫ)y = 0 has general solution y = Ae(1+ǫ)t + Bet . The function e(1+ǫ)t − et yǫ (t) = is of this form with A = −B = 1/ǫ. ǫ We have, substituting ǫ = h/t,

a = 1, b = 1 and c = 1, the discriminant is D = b2 − 4ac = −3 < 0 and −(b/2a) = − 21 and √ ω = 3/2. Thus, the general solution is √ √    3 3 −(1/2)t −(1/2)t y = Ae cos t + Be sin t . 2 2   2y ′′ + 5y ′ − 3y = 0 y(0) = 1  ′ y (0) = 0 The DE has auxiliary equation 2r 2 + 5y − 3 = 0, with roots r = 12 and r = −3. Thus y = Aet/2 + Be−3t . A Now 1 = y(0) = A + B, and 0 = y ′ (0) = − 3B. 2 Thus B = 1/7 and A = 6/7. The solution is 6 1 y = et/2 + e−3t . 7 7

e(1+ǫ)t − et ǫ→0 ǫ et+h − et = t lim h→0  h d t =t e = t et dt

lim yǫ (t) = lim

ǫ→0

2i . Thus

12. Given that y ′′ + y ′ + y = 0, hence r 2 + r + 1 = 0. Since

13.

we have A = −2e5 and B = 2e5 . Thus, y = −2e5 e−5t + 2te5 e−5t = 2(t − 1)e−5(t−1) .   y ′′ + 4y ′ + 5y = 0 y(0) = 2  ′ y (0) = 0 The auxiliary equation for the DE is r 2 + 4r + 5 = 0, which has roots r = −2 ± i . Thus y = Ae−2t cos t + Be−2t sin t

⇒ r = 3±i

y = Ae3t cos t + Be3t sin t

8.

2 = y ′ (1) = −5e−5 (A + B) + Be−5 ,

⇒ r = −4, −4

y ′′ − 2y ′ + y = 0

7.

0 = y(1) = Ae−5 + Be−5

.

which is, along with et , a solution of the CASE II DE y ′′ − 2y ′ + y = 0.

17. Given that a > 0, b > 0 and c > 0: Case 1: If D = b2 − 4ac > 0 then the two roots are √ −b ± b2 − 4ac r1,2 = . 2a Since b2 − 4ac < b2 p ± b2 − 4ac < b p −b ± b2 − 4ac < 0

therefore r1 and r2 are negative. The general solution is

682 Copyright © 2014 Pearson Canada Inc.

y(t) = Aer1 t + Ber2 t .

INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.5 (PAGE 1026)

If t → ∞, then er1 t → 0 and er2 t → 0. Thus, lim y(t) = 0.

21.

y(t) = Aer1 t + Bter2 t .

22.

t→∞

(r 2 + 1)2 = 0 ⇒ r = −i, −i, i, i General solution: y = C1 cos t + C2 sin t + C3 t cos t + C4 t sin t.

Case 2: If D = b2 − 4ac = 0 then the two equal roots r1 = r2 = −b/(2a) are negative. The general solution is

If t → ∞, then er1 t → 0 and er2 t → 0 at a faster rate than Bt → ∞. Thus, lim y(t) = 0.

23. If y = e2t , then y ′′′ − 2y ′ − 4y = e2t (8 − 4 − 4) = 0.

The auxiliary equation for the DE is r 3 − 2r − 4 = 0, for which we already know that r = 2 is a root. Dividing the left side by r − 2, we obtain the quotient r 2 + 2r + 2. Hence the other two auxiliary roots are −1 ± i . General solution: y = C1 e2t + C2 e−t cos t + C3 e−t sin t.

y = Ae−(b/2a)t cos(ωt) + Be−(b/2a)t sin(ωt) √

4ac − b2 . If t → ∞, then the amplitude of 2a −(b/2a)t both terms Ae → 0 and Be−(b/2a)t → 0. Thus, lim y(t) = 0. where ω = t→∞

24. Aux. eqn: (r 2 − r − 2)2 (r 2 − 4)2 = 0

(r + 1)2 (r − 2)2 (r − 2)2 (r + 2)2 = 0 r = 2, 2, 2, 2, −1, −1, −2, −2. The general solution is

18. The auxiliary equation ar 2 + br + c = 0 has roots √ −b − D r1 = , 2a

√ −b + D r2 = , 2a

where D = √ b2 − 4ac. Note that a(r2 − r1 ) = D = −(2ar1 + b). If y = er1 t u, then y ′ = er1 t (u ′ + r1 u), and y ′′ = er1 t (u ′′ + 2r1 u ′ + r12 u). Substituting these expressions into the DE ay ′′ + by ′ + cy = 0, and simplifying, we obtain e

r1 t

′′

′

y = e2t (C1 + C2 t + C3 t 2 + C4 t 3 ) + e−t (C5 + C6 t) + e−2t (C7 + C8 t).

25.

(r − 1)2 = 0, r = 1, 1. Thus y = Ax + Bx ln x.

(au + 2ar1 u + bu ) = 0,

26.

Thus, y = Ax −1 + Bx 3 .

which has general solution v = Ce(r2 −r1 )t . Hence u=

Ce

(r2 −r1 )t

dt = Be

(r2 −r1 )t

20.

27.

+ A,

x 2 y ′′ + x y ′ − y = 0 aux: r (r − 1) + r − 1 = 0 B y = Ax + . x

⇒ r = ±1

28. Consider x 2 y ′′ − x y ′ + 5y = 0. Since a = 1, b = −1, and

and y = er1 t u = Aer1 t + Ber2 t .

19.

x 2 y ′′ − x y ′ − 3y = 0

r (r − 1) − r − 3 = 0 ⇒ r 2 − 2r − 3 = 0 ⇒(r − 3)(r + 1) = 0 ⇒ r1 = −1 and r2 = 3

v ′ = (r2 − r1 )v,

Z

x 2 y ′′ − x y ′ + y = 0 aux: r (r − 1) − r + 1 = 0 r 2 − 2r + 1 = 0

′

or, more simply, u ′′ − (r2 − r1 )u ′ = 0. Putting v = u ′ reduces this equation to first order:

y (4) + 4y (3) + 6y ′′ + 4y ′ + y = 0 Auxiliary: r 4 + 4r 3 + 6r 2 + 4r + 1 = 0

(r + 1)4 = 0 ⇒ r = −1, −1, −1, −1 General solution: y = e−t (C1 + C2 t + C3 t 2 + C4 t 3 ).

t→∞

Case 3: If D = b2 − 4ac < 0 then the general solution is

y (4) + 2y ′′ + y = 0 Auxiliary: r 4 + 2r 2 + 1 = 0

c = 5, therefore (b −a)2 < 4ac. Then k = (a −b)/2a = 1 and ω2 = 4. Thus, the general solution is y = Ax cos(2 ln x) + Bx sin(2 ln x).

y ′′′ − 4y ′′ + 3y ′ = 0 Auxiliary: r 3 − 4r 2 + 3r = 0 r (r − 1)(r − 3) = 0 ⇒ r = 0, 1, 3 General solution: y = C1 + C2 et + C3 e3t .

29.

y (4) − 2y ′′ + y = 0 Auxiliary: r 4 − 2r 2 + 1 = 0

30. Given that x 2 y ′′ + x y ′ + y = 0. Since a = 1, b = 1, c = 1

2

2

(r − 1) = 0 ⇒ r = −1, −1, 1, 1 General solution: y = C1 e−t + C2 te−t + C3 et + C4 tet .

x 2 y ′′ + x y ′ = 0 aux: r (r − 1) + r = 0 Thus y = A + B ln x.

⇒ r = 0, 0.

therefore (b − a)2 < 4ac. Then k = (a − b)/2a = 0 and ω2 = 1. Thus, the general solution is y = A cos(ln x) + B sin(ln x).

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SECTION 18.5 (PAGE 1026)

31.

ADAMS and ESSEX: CALCULUS 8

x 3 y ′′′ + x y ′ − y = 0. Trying y = x r leads to the auxiliary equation r (r − 1)(r − 2) + r − 1 = 0

Since t = ln x, the given Euler equation has solution y = C1 x cos(ln x) + C2 x sin(ln x).

34. Because y ′′ + 4y = 0, therefore y = A cos 2t + B sin 2t.

r 3 − 3r 2 + 3r − 1 = 0

Now

(r − 1)3 = 0 ⇒ r = 1, 1, 1.

y(0) = 2 ⇒ A = 2,

y ′ (0) = −5 ⇒ B = − 52 .

Thus y = x is a solution. To find the general solution, try y = xv(x). Then y ′ = xv ′ + v,

y ′′ = xv ′′ + 2v ′ ,

y ′′′ = xv ′′′ + 3v ′′ .

Now x 3 y ′′′ + x y ′ − y = x 4 v ′′′ + 3x 3 v ′′ + x 2 v ′ + xv − xv

= x 2 (x 2 v ′′′ + 3xv ′′ + v ′ ), and y is a solution of the given equation if v ′ = w is a solution of x 2 w′′ + 3xw′ + w = 0. This equation has auxiliary equation r (r − 1) + 3r + 1 = 0, that is (r + 1)2 = 0, so its solutions are

35.

C2 2C3 ln x + x x v = C1 + C2 ln x + C3 (ln x)2 . v′ = w =

The general solution of the given equation is, therefore,

36. y = C1 x + C2 x ln x + C3 x(ln x)2 .

32. Since

y(2) = 3 = A sin 2 + B cos 2 y ′ (2) = −4 = A cos 2 − B sin 2,

dx = et = x, we have dt

therefore A = 3 sin 2 − 4 cos 2 B = 4 sin 2 + 3 cos 2.

dz dy dx dy = =x , dt d x dt dx 2 2 d z d y dx dx dy +x 2 = dt 2 dt d x d x dt dy d2 y =x + x2 2 . dx dx

Thus, y = (3 sin 2 − 4 cos 2) sin t + (4 sin 2 + 3 cos 2) cos t = 3 cos(t − 2) − 4 sin(t − 2).

Accordingly, z = z(t) satisfies

37.

d2z

dz + (b − a) + cz dt 2 dt 2 dy d y dy =ax 2 2 + ax + (b − a)x + cy = 0. dx dx dx a

Thus, y = 2 cos 2t − 52 sin 2t. circular frequency = ω = 2, frequency = ω 1 = ≈ 0.318 2π π 2π period = = π ≈ 3.14 ω q amplitude = (2)2 + (− 25 )2 ≃ 3.20   y ′′ + 100y = 0 y(0) = 0  ′ y (0) = 3 y = A cos(10t) + B sin(10t) A = y(0) = 0, 10B = y ′ (0) = 3 3 y= sin(10t) 10 For y ′′ + y = 0, we have y = A sin t + B cos t. Since,

33. By the previous exercise, z(t) = y(et ) = y(x) must satisfy the constant coefficient equation

d2z dz −2 + 2z = 0. dt dt 2 The auxiliary equation for this equation is r 2 −2r +2 = 0, which has roots r = 1 ± i . Thus z = C1 et cos t + C2 et sin t.

38.

  y ′′ + ω2 y = 0 y(a) = A  ′ y (a) = B   B   y = A cos ω(t − a) + sin ω(t − a) ω     y = A cos ω(t − c) + B sin ω(t − c)

(easy to calculate y ′′ + ω2 y = 0)  y = A cos(ωt) cos(ωc) + sin(ωt) sin(ωc)   + B sin(ωt) cos(ωc) − cos(ωt) sin(ωc)   = A cos(ωc) − B sin(ωc) cos ωt   + A sin(ωc) + B cos(ωc) sin ωt = A cos ωt + B sin ωt where A = A cos(ωc) − B sin(ωc) and B = A sin(ωc) + B cos(ωc)

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SECTION 18.5 (PAGE 1026)

by the previous problem. Therefore ay ′′ + by ′ + cy = 0 has general solution

39. If y = A cos ωt + B sin ωt then y ′′ + ω2 y = − Aω2 cos ωt − Bω2 sin ωt

y = Aekt cos(ωt) + Bekt sin(ωt).

+ ω2 (A cos ωt + B sin ωt) = 0

for all t. So y is a solution of (†).

40. If f (t) is any solution of (†) then f ′′ (t) = −ω2 f (t) for all t. Thus,

44. From Example 9, the spring constant is k = 9 × 104 gm/sec2 . For a frequency of 10 Hz (i.e., a circular √ frequency ω = 20π rad/sec.), a mass m satisfying k/m = 20π should be used. So,

2 i 2  d h 2 ω f (t) + f ′ (t) dt = 2ω2 f (t) f ′ (t) + 2 f ′ (t) f ′′ (t)

= 2ω2 f (t) f ′ (t) − 2ω2 f (t) f ′ (t) = 0

2  2  for all t. Thus, ω2 f (t) + f ′ (t) is constant. (This can be interpreted as a conservation of energy statement.)

41. If g(t) satisfies (†) and also g(0) =

g ′ (0)

Exercise 20,

Since a sum of squares cannot vanish unless each term vanishes, g(t) = 0 for all t.

42. If f (t) is any solution of (†), let g(t) = f (t) − A cos ωt − B sin ωt where A = f (0) and Bω = f ′ (0). Then g is also solution of (†). Also g(0) = f (0) − A = 0 and g ′ (0) = f ′ (0) − Bω = 0. Thus, g(t) = 0 for all t by Exercise 24, and therefore f (x) = A cos ωt + B sin ωt. Thus, it is proved that every solution of (†) is of this form. b 4ac − b2 and ω2 = which is 2a 4a 2 kt positive for Case III. If y = e u, then

43. We are given that k = −

  y ′ = ekt u ′ + ku   y ′′ = ekt u ′′ + 2ku ′ + k 2 u . Substituting into ay ′′ + by ′ + cy = 0 leads to   0 = ekt au ′′ + (2ka + b)u ′ + (ak 2 + bk + c)u   = ekt au ′′ + 0 + ((b2 /(4a) − (b2 /(2a) + c)u   = a ekt u ′′ + ω2 u . Thus u satisfies u ′′ + ω2 u = 0, which has general solution

9 × 104 k = = 22.8 gm. 400π 2 400π 2

The motion is determined by   y ′′ + 400π 2 y = 0 y(0) = −1  ′ y (0) = 2

= 0, then by

2  2  ω2 g(t) + g ′ (t) 2  2  = ω2 g(0) + g ′ (0) = 0.

u = A cos(ωt) + B sin(ωt)

m=

therefore, y = A cos 20π t + B sin 20π t and y(0) = −1 ⇒ A = −1 1 2 = . y ′ (0) = 2 ⇒ B = 20π 10π 1 sin 20π t, with y in cm 10π and t in second, gives the displacement at time t. The r 1 2 ) ≈ 1.0005 cm. amplitude is (−1)2 + ( 10π Thus, y = − cos 20π t +

ω k , ω2 = (k = spring const, m = mass) 2π m Since the spring does not change, ω2 m = k (constant) For m = 400 gm, ω = 2π(24) (frequency = 24 Hz) 4π 2 (24)2 (400) If m = 900 gm, then ω2 = 900 2π × 24 × 2 so ω = = 32π . 3 32π Thus frequency = = 16 Hz 2π 4π 2 (24)2 400 For m = 100 gm, ω = 100 ω so ω = 96π and frequency = = 48 Hz. 2π

45. Frequency =

46. Using the addition identities for cosine and sine, y = ekt [A cos ω(t − t0 )B sin ω(t − t0 )]

= ekt [A cos ωt cos ωt0 + A sin ωt sin ωt0 + B sin ωt cos ωt0 − B cos ωt sin ωt0 ] = ekt [A1 cos ωt + B1 sin ωt],

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ADAMS and ESSEX: CALCULUS 8

where A1 = A cos ωt0 − B sin ωt0 and B1 = A sin ωt0 + B cos ωt0 . Under the conditions of this problem we know that ekt cos ωt and ekt sin ωt are independent solutions of ay ′′ + by ′ + cy = 0, so our function y must also be a solution, and, since it involves two arbitrary constants, it is a general solution.

47. Expanding the hyperbolic functions in terms of exponentials,

The DE has auxiliary equation r 2 + 4r + 3 = 0 with roots r = −2 + 1 = −1 and r = −2 − 1 = −3 (i.e. k ± ω, where k = −2 and ω = 1). By the second previous problem, a general solution can be expressed in the form y = e−2t [A cosh(t − 3) + B sinh(t − 3)] for which y ′ = −2e−2t [A cosh(t − 3) + B sinh(t − 3)]

+ e−2t [A sinh(t − 3) + B cosh(t − 3)].

The initial conditions give

1 = y(3) = e−6 A

kt

y = e [A cosh ω(t − t0 )B sinh ω(t − t0 )]  A ω(t−t0 ) A = ekt e + e−ω(t−t0 ) 2 2  B ω(t−t0 ) B −ω(t−t0 ) + e − e 2 2 = A1 e(k+ω)t + B1 e(k−ω)t (A/2)e−ωt0

(B/2)e−ωt0

where A1 = + and B1 = (A/2)eωt0 − (B/2)eωt0 . Under the conditions of this problem we know that Rr = k ± ω are the two real roots of the auxiliary equation ar 2 +br +c = 0, so e(k±ω)t are independent solutions of ay ′′ + by ′ + cy = 0, and our function y must also be a solution. Since it involves two arbitrary constants, it is a general solution.

48.

The DE has auxiliary equation r 2 + 2r + 5 = 0 with roots r = −1 ± 2i . By the second previous problem, a general solution can be expressed in the form y = e−t [A cos 2(t − 3) + B sin 2(t − 3)] for which y ′ = −e−t [A cos 2(t − 3) + B sin 2(t − 3)] + e−t [−2 A sin 2(t − 3) + 2B cos 2(t − 3)]. The initial conditions give 2 = y(3) = e−3 A

0 = y ′ (3) = −e−3 (A + 2B) Thus A = 2e3 and B = − A/2 = −e3 . The IVP has solution y = e3−t [2 cos 2(t − 3) − sin 2(t − 3)].

49.

Thus A = e6 and B = 2 A = 2e6 . The IVP has solution y = e6−2t [cosh(t − 3) + 2 sinh(t − 3)].

50. Let u(x) = c − k 2 y(x). Then u(0) = c − k 2 a.

Also u ′ (x) = −k 2 y ′ (x), so u ′ (0) = −k 2 b. We have   u ′′ (x) = −k 2 y ′′ (x) = −k 2 c − k 2 y(x) = −k 2 u(x)

This IVP for the equation of simple harmonic motion has solution u(x) = (c − k 2 a) cos(kx) − kb sin(kx) so that  1  c − u(x) k2   c = 2 c − (c − k 2 a) cos(kx) + kb sin(kx) k c b = 2 (1 − cos(kx) + a cos(kx) + sin(kx). k k

y(x) =

  y ′′ + 2y ′ + 5y = 0 y(3) = 2  ′ y (3) = 0

  y ′′ + 4y ′ + 3y = 0 y(3) = 1  ′ y (3) = 0

0 = y ′ (3) = −e−6 (−2 A + B)

51. Since x ′ (0) = 0 and x(0) = 1 > 1/5, the motion will be

governed by x ′′ = −x + (1/5) until such time t > 0 when x ′ (t) = 0 again.

Let u = x − (1/5). Then u ′′ = x ′′ = −(x − 1/5) = −u, u(0) = 4/5, and u ′ (0) = x ′ (0) = 0. This simple harmonic motion initial-value problem has solution u(t) = (4/5) cos t. Thus x(t) = (4/5) cos t + (1/4) and x ′ (t) = u ′ (t) = −(4/5) sin t. These formulas remain valid until t = π when x ′ (t) becomes 0 again. Note that x(π ) = −(4/5) + (1/5) = −(3/5). Since x(π ) < −(1/5), the motion for t > π will be governed by x ′′ = −x − (1/5) until such time t > π when x ′ (t) = 0 again.

Let v = x + (1/5). Then v ′′ = x ′′ = −(x + 1/5) = −v, v(π ) = −(3/5) + (1/5) = −(2/5), and v ′ (π ) = x ′ (π ) = 0. Thius initial-value problem has solution v(t) = −(2/5) cos(t − π ) = (2/5) cos t, so that x(t) = (2/5) cos t − (1/5) and x ′ (t) = −(2/5) sin t. These formulas remain valid for t ≥ π until t = 2π when x ′ becomes 0 again. We have x(2π ) = (2/5) − (1/5) = 1/5 and x ′ (2π ) = 0.

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SECTION 18.6 (PAGE 1033)

The conditions for stopping the motion are met at t = 2π ; the mass remains at rest thereafter. Thus  4   5 cos t + 2 x(t) = 5 cos t −  1 5

1 5 1 5

4.

if 0 ≤ t ≤ π if π < t ≤ 2π if t > 2π

y ′ = Ae x (1 + x),

e x = Ae x (2 + x + 1 + x − 2x) = 3 Ae x .

y ′′ + y ′ − 2y = 1. The auxiliary equation for y ′′ + y ′ − 2y = 0 is r 2 + r − 2 = 0, which has roots r = −2 and r = 1. Thus the complementary function is

We require A = 1/3. The general solution of the given equation is y=

5.

yh = C1 e−2x + C2 e x . For a particular solution y p of the given equation try y = A. This satisfies the given equation if A = −1/2. Thus the general solution of the given equation is

x 2 = y ′′ + 2y ′ + 5y

= 2 A + 4 Ax + 2B + 5 Ax 2 + 5Bx + 5C.

Thus we require 5 A = 1, 4 A + 5B = 0, and 2 A + 2B + 5C = 0. This gives A = 1/5, B = −4/25, and C = −2/125. The given equation has general solution y=

x = A − 2(Ax + B) = A − 2B − 2 Ax. We require A − 2B = 0 and −2 A = 1, so A = −1/2 and B = −1/4. The general solution of the given equation is

3.

y ′′ + 2y ′ + 5y = x 2 . The homogeneous equation has auxiliary equation r 2 + 2r + 5 = 0 with roots r = −1 ± 2i . Thus the complementary function is

For a particular solution, try y = Ax 2 + Bx + C. Then y ′ = 2 Ax + B and y ′′ = 2 A. We have

y ′′ + y ′ − 2y = x. The complementary function is yh = C1 e−2x + C2 e x , as shown in Exercise 1. For a particular solution try y = Ax + B. Then y ′ = A and y ′′ = 0, so y satisfies the given equation if

y=−

1 x xe + C1 e−2x + C3 e x . 3

yh = C1 e−x cos(2x) + C2 e−x sin(2x).

1 y = − + C1 e−2x + C2 e x . 2

2.

y ′′ = Ae x (2 + x),

so y satisfies the given equation if

Section 18.6 Nonhomogeneous Linear Equations (page 1033) 1.

y ′′ + y ′ − 2y = e x . The complementary function is yh = C1 e−2x + C2 e x , as shown in Exercise 1. For a particular solution try y = Axe x . Then

6.

2x + 1 + C1 e−2x + C2 e x . 4

We require A = −1/2. The general solution of the given equation is

y ′′ + 4y = x 2 . The complementary function is y = C1 cos(2x) + C2 sin(2x). For the given equation, try y = Ax 2 + Bx + C. Then x 2 = y ′′ + 4y = 2 A + 4 Ax 2 + 4Bx + 4C Thus 2 A + 4C = 0, 4 A = 1, 4B = 0, and we have 1 1 A = , B = 0, and C = − . The given equation has 4 8 general solution

y ′′ + y ′ − 2y = e−x . The complementary function is yh = C1 e−2x + C2 e x , as shown in Exercise 1. For a particular solution try y = Ae−x . Then y ′ = − Ae−x and y ′′ = Ae−x , so y satisfies the given equation if e−x = e−x (A − A − 2 A) = −2 Ae−x .

x2 4x 2 − − + e−x (C1 cos(2x) + C2 sin(2x)). 5 25 125

y=

7.

1 2 1 x − + C1 cos(2x) + C2 sin(2x). 4 8

y ′′ − y ′ − 6y = e−2x . The homogeneous equation has auxiliary equation r 2 − r − 6 = 0 with roots r = −2 and r = 3. Thus the complementary function is

1 y = − e−x + C1 e−2x + C2 e x . 2

yh = C1 e−2x + C2 e3x .

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SECTION 18.6 (PAGE 1033)

ADAMS and ESSEX: CALCULUS 8

For a particular solution, try y = Axe−2x . Then y ′ = e−2x (A − 2 Ax) and y ′′ = e−2x (−4 A + 4 Ax). We have e

−2x

′′

Thus we require A + B = 0 and 4(B − A) = 1, that is, B = − A = 1/8. The given equation has general solution y=

′

= y − y − 6y

= e−2x (−4 A + 4 Ax − A + 2 Ax − 6 Ax) = −5 Ae−2x . Thus we require A = −1/5. The given equation has general solution

10.

1 y = − xe−2x + C1 e−2x + C2 e3x . 5

8.

This satisfies the nonhomogeneous DE if e−x sin x = y ′′ + 2y ′ + 2y = 2Be−x cos x − 2 Ae−x sin x.

yh = C1 e−2x + C2 xe−2x .

Thus we require B = 0 and A = −1/2. The given equation has general solution

For a particular solution, try y = Ax 2 e−2x . Then y ′ = e−2x (2 Ax −2 Ax 2 ) and y ′′ = e−2x (2 A−8 Ax +4 Ax 2 ). We have

1 y = − xe−x cos x + e−x (C1 cos x + C2 sin x). 2

e−2x = y ′′ + 4y ′ + 4y = 2 Ae−2x .

11.

Thus we require A = 1/2. The given equation has general solution  2  x y = e−2x + C1 + C2 x . 2

9.

This satisfies the nonhomogeneous DE if 4 + 2x + e−x = y ′′ + y ′ = A + 2B + 2Bx − Ce−x . Thus we require A + 2B = 4, 2B = 2, and −C = 1, that is, A = 2, B = 1, C = −1. The given equation has general solution

yh = C1 e−x cos x + C2 e−x sin x. For a particular solution, try y = Ae x cos x + Be x sin x. Then

This satisfies the nonhomogeneous DE if e x sin x = y ′′ + 2y ′ + 2y = e x cos x(2B + 2(A + B) + 2 A) + e x sin x(−2 A + 2(B − A) + 2B) = e x cos x(4 A + 4B) + e x sin x(4B − 4 A).

y ′′ + y ′ = 4 + 2x + e−x . The homogeneous equation has auxiliary equation r 2 + r = 0 with roots r = 0 and r = −1. Thus the complementary function is yh = C1 + C2 e−x . For a particular solution, try y = Ax + Bx 2 + C xe−x . Then y ′ = A + 2Bx + e−x (C − C x) y ′′ = 2B + e−x (−2C + C x).

y ′′ + 2y ′ + 2y = e x sin x. The homogeneous equation has auxiliary equation r 2 + 2r + 2 = 0 with roots r = −1 ± i . Thus the complementary function is

y ′ = (A + B)e x cos x + (B − A)e x sin x y ′′ = 2Be x cos x − 2 Ae x sin x.

y ′′ + 2y ′ + 2y = e−x sin x. The complementary function is the same as in Exercise 9, but for a particular solution we try y = Axe−x cos x + Bxe−x sin x y ′ = e−x cos x(A − Ax + Bx) + e−x sin x(B − Bx − Ax) y ′′ = e−x cos x(2B − 2Bx − 2 A) + e−x sin x(2 Ax − 2 A − 2B).

y ′′ + 4y ′ + 4y = e−2x . The homogeneous equation has auxiliary equation r 2 + 4r + 4 = 0 with roots r = −2, −2. Thus the complementary function is

= e−2x (2 A − 8 Ax + 4 Ax 2 + 8 Ax − 8 Ax 2 + 4 Ax 2 )

ex (sin x − cos x) + e−x (C1 cos x + C2 sin x). 8

y = 2x + x 2 − xe−x + C1 + C2 e−x .

12.

y ′′ + 2y ′ + y = xe−x . The homogeneous equation has auxiliary equation r 2 + 2r + 1 = 0 with roots r = −1 and r = −1. Thus the complementary function is yh = C1 e−x + C2 xe−x . For a particular solution, try y = e−x (Ax 2 + Bx 3 ). Then y ′ = e−x (2 Ax + (3B − A)x 2 − Bx 3 )

y ′′ = e−x (2 A + (6B − 4 A)x − (6B − A)x 2 + Bx 3 ).

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INSTRUCTOR’S SOLUTIONS MANUAL

SECTION 18.6 (PAGE 1033)

1 1 Thus y p = − e x + xe x . The general solution of the 9 3 given equation is

This satisfies the nonhomogeneous DE if xe−x = y ′′ + 2y ′ + y = e−x (2 A + 6Bx).

1 1 y = − e x + xe x + C1 e−2x + C2 e x 9 3 1 x = xe + C1 e−2x + C3 e x . 3

Thus we require A = 0 and B = 1/6. The given equation has general solution y=

1 3 −x x e + C1 e−x + C2 xe−x . 6

15.

x 2 y ′′ + x y ′ − y = x 2 . If y = Ax 2 , then y ′ = 2 Ax and y ′′ = 2 A. Thus x 2 = x 2 y ′′ + x y ′ − y

13.

= 2 Ax 2 + 2 Ax 2 − Ax 2 = 3 Ax 2 ,

y ′′ + y ′ − 2y = e−x . The complementary function is yh = C1 e−2x + C2 e x . For a particular solution use

so A = 1/3. A particular solution of the given equation is y = x 2 /3. The auxiliary equation for the homogeneous equation x 2 y ′′ + x y ′ − y = 0 is 4r (r − 1) + r − 1 = 0, or r 2 − 1 = 0, which has solutions r = ±1. Thus the general solution of the given equation is

y p = e−2x u 1 (x) + e x u 2 (x), where the coefficients u 1 and u 2 satisfy

y=

−2e−2x u ′1 + e x u ′2 = e−x e−2x u ′1 + e x u ′2 = 0.

16.

Thus 1 u ′1 = − e x 3 1 u1 = − ex 3

1 −2x e 3 1 u 2 = − e−2x . 6 u ′2 =

17. 1 y = − e−x + C1 e−2x + C2 e x . 2

18.

x 2 y ′′ + x y ′ − y = x. 1 Try y = xu 1 (x) + u 2 (x), where u 1 and u 2 satisfy x

e−2x u ′1 + e x u ′2 = 0.

xu ′1 +

Thus 1 u ′1 = − e3x 3 1 3x u1 = − e 9

1 3 1 u 2 = x. 3

u ′2 =

A + x A(ln x + 1) − Ax ln x = 2 Ax. x

Thus A = 1/2. The complementary function was obtained in Exercise 15. The given equation has general solution C2 1 . y = x ln x + C1 x + 2 x

y p = e−2x u 1 (x) + e x u 2 (x),

−2e−2x u ′1 + e x u ′2 = e x

x 2 y ′′ + x y ′ − y = x. Try y = Ax ln x. Then y ′ = A(ln x + 1) and y ′′ = A/x. We have x = x2

y ′′ + y ′ − 2y = e x . The complementary function is yh = C1 e−2x + C2 e x . For a particular solution use

where the coefficients u 1 and u 2 satisfy

x 2 y ′′ + x y ′ − y = x r has a solution of the form y = Ax r provided r 6= ±1. If this is the case, then   x r = Ax r r (r − 1) + r − 1 = Ax r (r 2 − 1). Thus A = 1/(r 2 − 1) and a particular solution of the DE is 1 y= 2 xr . r −1

1 1 1 Thus y p = − e−x − e−x = − e−x . The general 3 6 2 solution of the given equation is

14.

1 2 C2 x + C1 x + . 3 x

u ′2 = 0, x

u ′1 −

u ′2 1 = . x2 x

Solving these equations for u ′1 and u ′2 , we get x u ′2 = − , 2

u ′1 =

1 . 2x

689 Copyright © 2014 Pearson Canada Inc.

SECTION 18.6 (PAGE 1033)

Thus u 1 =

ADAMS and ESSEX: CALCULUS 8

1 x2 ln x and u 2 = − . A particular solution is 2 4 y=

Accordingly, we look for a particular solution of the given equation having the form

1 x x ln x − . 2 4

y p = u 1 (x)e−2x + u 2 (x) xe−2x . According to the procedure developed in the text, u ′1 and u ′2 can be determined by solving the pair of equations

The term −x/4 can be absorbed into the term C1 x in the complementary function, so the general solution is y=

1 C2 x ln x + C1 x + . 2 x

u ′1 (x)e−2x + u ′2 (x)xe−2x = 0 − 2u ′1 (x)e−2x + (1 − 2x)u ′2 (x)e−2x =

19. The homogeneous DE y ′′ − 2y ′ + y = 0 has auxiliary

or, equivalently,

equation r 2 − 2r + 1 = 0 with roots r = 1, 1. Therefore, its general solution is

u ′1 (x) + xu ′2 (x) = 0

y = C1 e x + C2 xe x .

− 2u ′1 (x) + (1 − 2x)u ′2 (x) =

Accordingly, we look for a particular solution of the given equation having the form

1 u ′1 (x) = − , x

According to the procedure developed in the text, u ′1 and u ′2 can be determined by solving the pair of equations

u ′1 (x)e x + u ′2 (x)(1 + x)e x =

u ′1 (x) + (1 + x)u ′2 (x) =

ex x

1 . x2

y = C1 e−2x + C2 xe−2x − xe−2x ln x. 1 . x

21.

The solution is u ′1 (x) = −1,

u ′2 (x) =

1 Thus u 1 (x) = − ln x and u 2 (x) = − . A particular x solution of the DE is y p = −e−2x ln x − e−2x , or, since the last term is a solution of the homogeneous equation, the simpler form y p = −e−2x ln x will do. The given equation has general solution

or, equivalently, u ′1 (x) + xu ′2 (x) = 0

1 . x2

The solution is

y p = u 1 (x)e x + u 2 (x) xe x .

u ′1 (x)e x + u ′2 (x)xe x = 0

e−2x x2

u ′2 (x) =

1 . x

Thus u 1 (x) = −x and u 2 (x) = ln x. A particular solution of the DE is y p = −xe x + xe x ln x, or, since the first term is a solution of the homogeneous equation, the simpler form y p = xe x ln x will do. The given equation has general solution y = C1 e x + C2 xe x + xe x ln x.

20. The homogeneous DE y ′′ + 4y ′ + 4y = 0 has auxiliary equation r 2 + 4r + 4 = 0 with roots r = −2, −2. Therefore, its general solution is

x 2 y ′′ − (2x + x 2 )y ′ + (2 + x)y = x 3 . Since x and xe x are independent solutions of the corresponding homogeneous equation, we can write a solution of the given equation in the form y = xu 1 (x) + xe x u 2 (x), where u 1 and u 2 are chosen to satisfy xu ′1 + xe x u ′2 = 0,

u ′1 + (1 + x)e x u ′2 = x.

Solving these equations for u ′1 and u ′2 , we get u ′1 = −1 and u ′2 = e−x . Thus u 1 = −x and u 2 = −e−x . The particular solution is y = −x 2 − x. Since −x is a solution of the homogeneous equation, we can absorb that term into the complementary function and write the general solution of the given DE as y = −x 2 + C1 x + C2 xe x .

y = C1 e−2x + C2 xe−2x .

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INSTRUCTOR’S SOLUTIONS MANUAL

22.

SECTION 18.7 (PAGE 1038)

  1 x 2 y ′′ + x y ′ + x 2 − y = x 3/2 . 4 A particular solution can be obtained in the form

Thus a2 = a3 = 0, and an+2 = Given a0 and a1 we have

an−2 for n ≥ 2. (n + 1)(n + 2)

a0 3×4 a4 a0 a8 = = 7×8 3×4×7×8 .. . a0 a4n = 3 × 4 × 7 × 8 × · · · × (4n − 1)(4n) a0 = n 4 n! × 3 × 7 × · · · × (4n − 1) a1 a5 = 4×5 a1 a5 = a9 = 8×9 4×5×8×9 .. . a1 a4n+1 = 4 × 5 × 8 × 9 × · · · × (4n)(4n + 1) a1 = n 4 n! × 5 × 9 × · · · × (4n + 1) a4n+3 = a4n+2 = · · · = a3 = a2 = 0.

y = x −1/2 (cos x)u 1 (x) + x −1/2 (sin x)u 2 (x),

a4 =

where u 1 and u 2 satisfy x −1/2 (cos x)u ′1 + x −1/2 (sin x)u ′2 = 0   1 − x −3/2 cos x − x −1/2 sin x u ′1 2   1 −3/2 − x sin x − x −1/2 cos x u ′2 = x −1/2 . 2 We can simplify these equations by dividing the first by x −1/2 , and adding the first to 2x times the second, then dividing the result by 2x 1/2 . The resulting equations are (cos x)u ′1 + (sin x)u ′2 = 0 −(sin x)u ′1 + (cos x)u ′2 = 1, which have solutions u ′1 = − sin x, u ′2 = cos x, so that u 1 = cos x and u 2 = sin x. Thus a particular solution of the given equation is y = x −1/2 cos2 x + x −1/2 sin2 x = x −1/2 .

The solution is

The general solution is   y = x −1/2 1 + C2 cos x + C2 sin x .

y = a0 1 +

∞ X n=1

(x − 1)4n n 4 n! × 3 × 7 × · · · × (4n − 1)

+ a1 x − 1 +

∞ X n=1

!

! (x − 1)4n+1 . 4n n! × 5 × 9 × · · · × (4n + 1)

Section 18.7 Series Solutions of Differential Equations (page 1038) 1.

y ′′

= (x

y= y ′′ = =

− 1)2 y.

∞ X

n=0 ∞ X

n=2 ∞ X n=0

2.

Try

an (x − 1)n .

y′ =

n(n − 1)an (x − 1)n−2

′′

y =

(n + 2)(n + 1)an+2 (x − 1)n

0 = y ′′ − (x − 1)2 y ∞ ∞ X X = (n + 2)(n + 1)an+2 (x − 1)n − an (x − 1)n+2 =

n=0

n=0

∞ X

∞ X

n=0

(n + 2)(n + 1)an+2 (x − 1)n −

n=2

an−2 (x − 1)n

= 2a2 + 6a3 (x − 1) ∞ h i X + (n + 2)(n + 1)an+2 − an−2 (x − 1)n . n=2

y ′′ = x y. Try ∞ X n=0

∞ X n=2

∞ X

an x n . Then

n=0

nan x n−1 =

∞ X

nan x n−1

n=1

n(n − 1)an x n−2 =

∞ X (n + 2)(n + 1)an+2 x n . n=0

Thus we have 0 = y ′′ − x y ∞ ∞ X X an x n+1 (n + 2)(n + 1)an+2 x n − = n=0

n=0

∞ ∞ X X an−1 x n = (n + 2)(n + 1)an+2 x n − n=0

= 2a2 +

n=1

∞ h X n=1

i (n + 2)(n + 1)an+2 − an−1 x n .

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SECTION 18.7 (PAGE 1038)

ADAMS and ESSEX: CALCULUS 8

an−1 for n ≥ 1. (n + 2)(n + 1) Given a0 and a1 , we have

Thus a2 = 0 and an+2 =

a0 2×3 a3 a0 1 × 4 × a0 a6 = = = 5×6 2×3×5×6 6! a6 1 × 4 × 7 × a0 a9 = = 8×9 9! .. . 1 × 4 × · · · × (3n − 2)a0 a3n = (3n)! a1 2 × a1 a4 = = 3×4 4! a4 2 × 5 × a1 a7 = = 6×7 7! .. . 2 × 5 × · · · × (3n − 1)a1 a3n+1 = (3n + 1)! 0 = a2 = a5 = a8 = · · · = a3n+2 . a3 =

Thus the general solution of the given equation is ∞ X 1 × 4 × · · · × (3n − 2) 3n 1+ x (3n)! n=1

y = a0

+ a1

!

∞ X 2 × 5 × · · · × (3n − 1) 3n+1 x . (3n + 1)! n=1

It follows that a2 = −1,

an+2 = −

an , n+1

n = 1, 2, 3, . . . .

Since a0 = y(0) = 1, and a1 = y ′ (0) = 2, we have a0 = 1 a2 = −1 1 a4 = 3

a1 = 2

2 2 2 a5 = 2×4 a3 = −

1 3×5 1 a8 = 3×5×7 a6 = −

2 2×4×6 2 a9 = . 2×4×6×8 a7 = −

The patterns here are obvious: (−1)n (−1)n 2 a2n+1 = n 3 × 5 × · · · × (2n − 1) 2 n! (−1)n 2n n! = (2n)!  n  2n P x 2n+1 n 2 n!x Thus y = ∞ + n−1 . n=0 (−1) (2n)! 2 n! a2n =

4. If y = y ′′ =

∞ X n=0

an x n , then y ′ =

∞ X n=2

P∞

n=1 nan x

n(n − 1)an x n−2 =

n−1

and

∞ X (n + 2)(n + 1)an+2 x n . n=0

Thus,

3.

 

y ′′ + x y ′ + 2y = 0 y(0) = 1 y ′ (0) = 2



Let

0 = y ′′ + x y ′ + y ∞ ∞ ∞ X X X = (n + 2)(n + 1)an+2 x n + x nan x n−1 + an x n n=0

y= y ′′ =

∞ X

n=0 ∞ X n=2

an x

n

′

y =

∞ X

nan x

= 2a2 + a0 +

n−1

n=1

n(n − 1)an x n−2 =

∞ X (n + 2)(n + 1)an+2 x n . n=0

∞ ∞ X X (n + 2)(n + 1)an+2 x n + nan x n

+2

n

an x = 0,

n=0 ∞ X

2a2 + 2 +

n=1

i (n + 2)(n + 1)an+2 + (n + 1)an x n .

that is,

(n + 2)(n + 1)an+2 + (n + 1)an = 0, −an . an+2 = n+2 −1 1 , a4 = 2 , 2 2 · 2! ,. . .. If y ′ (0) = 0, then

If y(0) = 1, then a0 = 1, a2 =

−1 1 , a8 = 4 23 · 3! 2 · 4! a1 = a3 = a5 = . . . = 0. Hence,

n=1

∞ X

n=1

n=0

Since coefficients of all powers of x must vanish, therefore 2a2 + a0 = 0 and, for n ≥ 1,

Substituting these expressions into the differential equation, we get

n=0

n=1

∞ h X

a6 =

so

[(n + 2)(n + 1)an+2 + (n + 2)an ]x n = 0.

y =1−

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∞ X 1 2 1 4 1 6 (−1)n 2n x + x − x + ··· = x . 2 8 48 2n · n! n=0

INSTRUCTOR’S SOLUTIONS MANUAL

5.

SECTION 18.7 (PAGE 1038)

y ′′ + (sin x)y = 0, y(0) = 1, y ′ (0) = 0. Try

Therefore we have a2 = a4 = a6 = · · · = 0 4 a3 = − , a5 = 0 = a7 = a9 = · · · . 3

y = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + · · · . Then a0 = 1 and a1 = 0. We have

The initial-value problem has solution

2

′′

3

y = 2a2 + 6a3 x + 12a4 x + 20a5 x + · · · ! x3 x5 (sin x)y = x − + −··· 6 120 × (1 + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + · · ·)   1 = x + a2 − x 3 + a3 x 4 6   1 1 x5 + · · · . + a4 − a2 + 6 120 Hence we must 1 20a5 + a2 − 6 1 a3 = − , a5 = 6

y=x−

7. 3x y ′′ + 2y ′ + y = 0.

Since x = 0 is a regular singular point of this equation, try ∞ X y= an x n+µ (a0 = 1) n=0

∞ X y′ = (n + µ)an x n+µ−1

have 2a2 = 0, 6a3 + 1 = 0, 12a4 = 0,

n=0

= 0, . . . . That is, a2 = 0, a4 = 0,

1 . The solution is 120

y =1−

4 3 x . 3

∞ X y ′′ = (n + µ)(n + µ − 1)an x n+µ−2 . n=0

Then we have

1 3 1 5 x + x +···. 6 120

0 = 3x y ′′ + 2y ′ + y ∞ ∞ h i X X an−1 x n+µ−1 3(n + µ)2 − (n + µ) an x n+µ−1 + = n=1

n=0

6. (1 −

x 2 )y ′′

−

x y′

+ 9y = 0, y(0) = 0, y=

∞ X

y ′ (0)

= 1. Try

an x n .

n=1

n=0

an−1 for 3(n + µ)2 − (n + µ) n ≥ 1. There are two cases: µ = 0 and µ = 1/3. an−1 . Since a0 = 1 CASE I. µ = 0. Then an = − n(3n − 1) we have Thus 3µ2 − µ = 0 and an = −

Then a0 = 0 and a1 = 1. We have y′ = y ′′ =

∞ X n=1 ∞ X n=2

nan x n−1 n(n − 1)an x n−2

a1 = −

0 = (1 − x 2 )y ′′ − x y ′ + 9y ∞ ∞ X X = (n + 2)(n + 1)an+2 x n − n(n − 1)an x n n=0

−

= (3µ2 − µ)x µ−1 ∞ h  i X + 3(n + µ)2 − (n + µ) an + an−1 x n+µ−1 .

a3 = − .. .

n=2

∞ X n=1

n

nan x + 9

∞ X

an x

n

n=0

an =

(n 2 − 9)an . (n + 1)(n + 2)

1 1×2×2×5

1 1×2×2×5×3×8

(−1)n . n! × 2 × 5 × · · · × (3n − 1)

y =1+

n=2

an+2 =

a2 =

One series solution is

= 2a2 + 9a0 + (6a3 + 8a1 )x ∞ h i X + (n + 2)(n + 1)an+2 − (n 2 − 9)an x n . Thus 2a2 + 9a0 = 0, 6a3 + 8a1 = 0, and

1 , 1×2

∞ X n=1

(−1)n x n . n! × 2 × 5 × · · · × (3n − 1)

1 . Then 3 −an−1 −an−1 an =

= n(3n + 1) . 1 2 1 3 n+ 3 − n+ 3

CASE II. µ =

693 Copyright © 2014 Pearson Canada Inc.

SECTION 18.7 (PAGE 1038)

ADAMS and ESSEX: CALCULUS 8

Since a0 = 1 we have a1 = − a3 = − .. .

1 , 1×4

a2 =

1 1×4×2×7

Review Exercises 18 (page 1038)

1 1 × 4 × 2 × 7 × 3 × 10

1.

an =

(−1)n . n! × 1 × 4 × 7 × · · · × (3n + 1)

y = Ce x

A second series solution is ∞ X

(−1)n x n n! × 1 × 4 × 7 × · · · × (3n + 1)

!

.

2.

x y ′′ + y ′ + x y = 0. Since x = 0 is a regular singular point of this equation, try ∞ X y= an x n+µ (a0 = 1)

3.

y=x

8.

1/3

1+

n=1

dy = 2x y dx dy = 2x d x ⇒ ln |y| = x 2 + C1 y

n=0

∞ X y′ = (n + µ)an x n+µ−1 n=0

∞ X y ′′ = (n + µ)(n + µ − 1)an x n+µ−2 .

2

dy = e−y sin x dx e y d y = sin x d x ⇒ e y = − cos x + C y = ln(C − cos x) dy dy = x + 2y ⇒ − 2y = x dx dx  d −2x dy (e y) = e−2x − 2y = xe−2x dx dx Z 1 x −2x −2x e y = xe d x = − e−2x − e−2x + C 2 4 x 1 y = − − + Ce2x 2 4

n=0

Then we have

4.

0 = x y ′′ + y ′ + x y ∞ h i X = (n + µ)(n + µ − 1) + (n + µ) an x n+µ−1 n=0

+

∞ X

an x n+µ+1

n=0

∞ ∞ X X = (n + µ)2 an x n+µ−1 + an−2 x n+µ−1 n=0

n=2

= µ2 x µ−1 + (1 + µ)2 a1 x µ ∞ h i X + (n + µ)2 an + an−2 x n+µ−1 . n=2

5.

an−2 for n ≥ 2. n2 It follows that 0 = a1 = a3 = a5 = · · ·, and, since a0 = 1, Thus µ = 0, a1 = 0, and an = −

1 1 , a4 = 2 2 , . . . 22 2 4 (−1)n (−1)n = 2 2 = . 2 4 · · · (2n)2 22n (n!)2

a2 = − a2n

One series solution is y =1+

∞ X (−1)n x 2n . 22n (n!)2 n=1

6.

dy x 2 + y2 = (let y = xv(x)) dx 2x y dv 1 + v2 v+x = dx 2v dv 1 + v2 1 − v2 x = −v = dx 2v 2v dx 2v dv =− v2 − 1 x 1 C 2 ln(v − 1) = ln + ln C = ln x x y2 C 2 2 −1= ⇒ y − x = Cx x2 x dy x+y = dx y−x (x + y) d x + (x − y) d y = 0  2  x y2 d + xy − =0 2 2 x 2 + 2x y − y 2 = C

(exact)

dy y + ex =− dx x + ey x (y + e ) d x + (x + e y ) d y = 0 (exact)
d x y + ex + e y = 0 x y + ex + e y = C

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INSTRUCTOR’S SOLUTIONS MANUAL

7.

d2 y dt 2 dp dt 1 p dy dt

=



dy dt

2

= p2 ⇒

REVIEW EXERCISES 18 (PAGE 1038)

Thus A = 1/2 and B = 1. The general solution is

(let p = d y/dt) dp = dt p2

y=

= C1 − t

14.

1 = p= C 1−t Z dt y= = − ln |t − C1 | + C2 C1 − t

8.

dy d2 y +5 + 2y = 0 dt 2 dt Aux: 2r 2 + 5r + 2 = 0 ⇒ r = −1/2, −2 y = C1 e

9.

xe2x = (2 A − B)e2x − 2 Axe2x ,

+ C2 e

so that we need A = −1/2 and B = 2 A = −1. The general solution is   1 2 y=− x + x e2x + C1 e2x + C2 e3x . 2

−2t

4y ′′ − 4y ′ + 5y = 0

Aux: 4r 2 − 4r + 5 = 0

1 ±i 2 y = C1 e x/2 cos x + C2 e x/2 sin x (2r − 1)2 + 4 = 0 ⇒ r =

10.

11.

15.

2x 2 y ′′ + y = 0 Aux: 2r (r − 1) + 1 = 0

2d

1 2r 2 − 2r + 1 = 0 ⇒ r = (1 ± i ) 2
y = C1 |x|1/2 cos 21 ln |x| + C2 |x|1/2 sin

d2 y dy +2 + y = x2 2 dx dx Aux: r 2 + 2r + 1 = 0 has solutions r = −1, −1. Complementary function: y = C1 e−x + C2 xe−x . Particular solution: try y = Ax 2 + Bx + C. Then x 2 = 2 A + 2(2 Ax + B) + Ax 2 + Bx + C.

1 2

ln |x|

2y

dy t + 5y = 0 −t 2 dt dt Aux: r (r − 1) − r + 5 = 0

Thus A = 1, B = −4, C = 6. The general solution is




y = x 2 − 4x + 6 + C1 e−x + C2 xe−x .

16.

(r − 1)2 + 4 = 0 ⇒ r = 1 ± 2i y = C1 t cos(2 ln |t|) + C2 t sin(2 ln |t|)

12.

dy d2 y −5 + 6y = xe2x dx2 dx Same complementary function as in Exercise 13: C1 e2x + C2 e3x . For a particular solution we try y = (Ax 2 + Bx)e2x . Substituting this into the given DE leads to

2

−t/2

d3 y d2 y dy =0 + 8 + 16 3 2 dt dt dt Aux: r 3 + 8r 2 + 16r = 0

r (r + 4)2 = 0 ⇒ r = 0, −4, −4

d2 y − 2y = x 3 . dx2 The corresponding homogeneous equation has auxiliary equation r (r − 1) − 2 = 0, with roots r = 2 and r = −1, so the complementary function is y = C1 x 2 + C2 /x. A particular solution of the nonhomogeneous equation can have the form y = Ax 3 . Substituting this into the DE gives x2

y = C1 + C2 e−4t + C3 te−4t

13.

6 Ax 3 − 2 Ax 3 = x 3 ,

d2 y dy + 6y = e x + e3x −5 dx2 dx Aux: r 2 − 5r + 6 = 0 ⇒ r = 2, 3. Complementary function: y = C1 e2x + C2 e3x . Particular solution: y = Ae x + Bxe3x y ′ = Ae x + B(1 + 3x)e3x x

′′

y = Ae + B(6 + 9x)e e x + e3x = Ae x (1 − 5 + 6) 3x

3x

+ Be (6 + 9x − 5 − 15x + 6x)

= 2 Ae x + Be3x .

1 x e + xe3x + C1 e2x + C2 e3x . 2

so that A = 1/4. The general solution is y=

17.

C2 1 3 x + C1 x 2 + . 4 x

dy x2 = 2 , y(2) = 1 dx y y2 d y = x 2 d x

y3 = x 3 + C 1 = 8 + C ⇒ C = −7

y 3 = x 3 − 7 ⇒ y = (x 3 − 7)1/3

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REVIEW EXERCISES 18 (PAGE 1038)

18.

19.

ADAMS and ESSEX: CALCULUS 8

dy y2 = 2 , y(2) = 1 dx x dy dx 1 1 = 2 ⇒ − = − −C 2 y x y x 1 1 1= +C ⇒ C = 2 2   2x 1 −1 1 = + y= x 2 x +2

23.

dy xy = 2 , y(0) = 1. Let y = xv(x). Then dx x + y2

24.

y ′′ + 10y ′ + 25y = 0, y(1) = e−5 , y ′ (1) = 0 Aux: r 2 + 10r + 25 = 0 ⇒ r = −5, −5. y = Ae−5x + Bxe−5x

y ′ = −5 Ae−5x + B(1 − 5x)e−5x . We require e−5 = (A + B)e−5 and 0 = e−5 (−5 A − 4B). Thus A + B = 1 and −5 A = 4B, so that B = 5 and A = −4. The solution is y = −4e−5x + 5xe−5x .

dv v = dx 1 + v2 dv v v3 x = − v = − dx 1 + v2 1 + v2 dx 1 + v2 dv = − v3 x 1 − ln |v| = ln |x| + ln C 2v 2 x2 1 = 2 = ln(Cv x)2 = ln(C 2 y 2 ) y2 v

x 2 y ′′ − 3x y ′ + 4y = 0, y(e) = e2 , y ′ (e) = 0 Aux: r (r − 1) − 3r + 4 = 0, or (r − 2)2 = 0, so that r = 2, 2.

v+x

C 2 y2 = ex y2 = e

2 /y 2

x 2 /y 2

,

,

y(0) = 1 ⇒ C 2 = 1

or y = e x

2 /(2y 2 )

y = Ax 2 + Bx 2 ln x y ′ = 2 Ax + 2Bx ln x + Bx. We require e2 = Ae2 + Be2 and 0 = 2 Ae + 3Be. Thus A + B = 1 and 2 A = −3B, so that A = 3 and B = −2. The solution is y = 3x 2 − 2x 2 ln x, valid for x > 0.

25.

d2 y + 4y = 8e2t , y(0) = 1, y ′ (0) = −2 dt 2 Complementary function: y = C1 cos(2t) + C2 sin(2t). Particular solution: y = Ae2t , provided 4 A + 4 A = 8, that is, A = 1. Thus

.

y = e2t + C1 cos(2t) + C2 sin(2t)

y ′ = 2e2t − 2C1 sin(2t) + 2C2 cos(2t).

20.

dy + (cos x)y = 2 cos x, y(π ) = 1 dx     d sin x sin x d y e y =e + (cos x)y = 2 cos xesin x dx dx esin x y = 2esin x + C y = 2 + Ce− sin x 1 = 2 + Ce0 ⇒ C = −1 y = 2 − e− sin x

21.

y ′′ + 3y ′ + 2y = 0, y(0) = 1, y ′ (0) = 2 Aux: r 2 + 3r + 2 = 0 ⇒ r = −1, −2. y = Ae−x + Be−2x

y ′ = − Ae−x − 2Be−2x

⇒

1= A+ B

⇒

2 = − A − 2B.

Thus B = −3, A = 4. The solution is y = 4e−x − 3e−2x .

22.

We require 1 = y(0) = 1 + C1 and −2 = y ′ (0) = 2 + 2C2 . Thus C1 = 0 and C2 = −2. The solution is y = e2t − 2 sin(2t). d2 y dy − 3y = 6 + 7e x/2 , y(0) = 0, y ′ (0) = 1 +5 dx2 dx Aux: 2r 2 + 5r − 3 = 0 ⇒ r = 1/2, −3. Complementary function: y = C1 e x/2 + C2 e−3x . Particular solution: y = A + Bxe x/2  x y ′ = Be x/2 1 + 2  x y ′′ = Be x/2 1 + . 4 We need   x 5x Be x/2 2 + + 5 + − 3x − 3 A = 6 + 7e x/2 . 2 2

26. 2

This is satisfied if A = −2 and B = 1. The general solution of the DE is

y ′′ + 2y ′ + (1 + π 2 )y = 0, y(1) = 0, y ′ (1) = π Aux: r 2 + 2r + 1 + π 2 = 0 ⇒ r = −1 ± πi . y = Ae−x cos(π x) + Be−x sin(π x) y ′ = e−x cos(π x)(− A + Bπ ) + e−x sin(π x)(−B − Aπ ).

y = −2 + xe x/2 + C1 e x/2 + C2 e−3x . Now the initial conditions imply that

Thus − Ae−1 = 0 and (A − Bπ )e−1 = π , so that A = 0 and B = −e. The solution is y = −e1−x sin(π x).

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0 = y(0) = −2 + C1 + C2 C1 1 = y ′ (0) = 1 + − 3C2 , 2

INSTRUCTOR’S SOLUTIONS MANUAL

REVIEW EXERCISES 18 (PAGE 1038)

which give C1 = 12/7, C2 = 2/7. Thus the IVP has solution

31.

1 y = −2 + xe x/2 + (12e x/2 + 2e−3x ). 7

27.

2x 2 v ′ + x 3 v ′′ − (xv + x 2 v ′ )(2 + x cot x) + xv(2 + x cot x) = 0

Z

dy 3y dy dx = ⇒ =3 dx x −1 y x −1 ⇒ ln |y| = ln |x − 1|3 + ln |C|

x 3 v ′′ − x 3 v ′ cot x = 0,

or, putting w = v ′ , w′ = (cot x)w, that is,

⇒ y = C(x − 1)3 . Since y = 4 when x = 2, we have 4 = C(2 − 1)3 = C, so the equation of the curve is y = 4(x − 1)3 .

dw cos x d x = w sin x ln w = ln sin x + ln C2 v ′ = w = C2 sin x ⇒ v = C1 − C2 cos x.

28. The ellipses 3x 2 + 4y 2 = C all satisfy the differential equation

6x + 8y

dy = 0, dx

A family of curves that intersect these ellipses at right 4y dy = . angles must therefore have slopes given by dx 3x Thus Z Z dy dx 3 =4 y x 3 ln |y| = 4 ln |x| + ln |C|. The family is given by y 3 = C x 4 .

29. [(x +

A)e x

sin y + cos y] d x is M d x + N d y. We have

+ x[e x

cos y + B sin y] d y = 0

∂M = (x + A)e x cos y − sin y ∂y ∂N = e x cos y + B sin y + xe x cos y. ∂x These expressions are equal (and the DE is exact) if A = 1 and B = −1. If so, the left side of the DE is dφ(x, y), where φ(x, y) = xe x sin y + x cos y. The general solution is xe x sin y + x cos y = C.

30. (x 2 + 3y 2 ) d x + x y d y = 0. Multiply by x n : n

2

2

x (x + 3y ) d x + x

n+1

y dy = 0

is exact provided 6x n y = (n + 1)x n y, that is, provided n = 5. In this case the left side is dφ, where φ(x, y) =

A second solution of the DE is x cos x, and the general solution is y = C1 x + C2 x cos x.

dy 3x =− . dx 4y

or

1 6 2 1 8 x y + x . 2 8

The general solution of the given DE is 4x 6 y 2 + x 8 = C.

x 2 y ′′ − x(2 + x cot x)y ′ + (2 + x cot x)y = 0 If y = x, then y ′ = 1 and y ′′ = 0, so the DE is clearly satisfied by y. To find a second, independent solution, try y = xv(x). Then y ′ = v + xv ′ , and y ′′ = 2v ′ + xv ′′ . Substituting these expressions into the given DE, we obtain

32.

x 2 y ′′ − x(2 + x cot x)y ′ + (2 + x cot x)y = x 3 sin x Look for a particular solution of the form y = xu 1 (x) + x cos xu 2 (x), where u ′1

xu ′1 + x cos xu ′2 = 0 + (cos x − x sin x)u ′2 = x sin x.

Divide the first equation by x and subtract from the second equation to get −x sin xu ′2 = x sin x. Thus u ′2 = −1 and u 2 = −x. The first equation now gives u ′1 = cos x, so that u 1 = sin x. The general solution of the DE is y = x sin x − x 2 cos x + C1 x + C2 x cos x.

33. Suppose y ′ = f (x, y) and y(x0 ) = y0 , where f (x, y) is continuous on the whole x y-plane and satisfies | f (x, y)| ≤ K there. By the Fundamental Theorem of Calculus, we have y(x) − y0 = y(x) − y(x0 ) Z x Z = y ′ (t) dt = x0

x x0

  f t, y(t) dt.

Therefore, |y(x) − y0 | ≤ K |x − x0 |. Thus y(x) is bounded above and below by the lines y = y0 ± K (x − x0 ), and cannot have a vertical asymptote anywhere.

Remark: we don’t seem to have needed the continuity of ∂ f /∂ y, only the continuity of f (to enable the use of the Fundamental Theorem). 697 Copyright © 2014 Pearson Canada Inc.

Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

1.

APPENDIX I. (PAGE A-10)

APPENDICES

15.

Appendix I. Complex Numbers (page A-10)

16.

z = −5 + 2i,

Re(z) = −5, z = −5 + 2i

Im(z) = 2 y z-plane

17.

z = −6

x z = 4−i z = −πi

18. |z| = 2, arg (z) = π ⇒ z = 2(cos π + i sin π ) = −2 19.

Fig. .1

2.

z = 4 − i,

Re(z) = 4,

Im(z) = −1

3.

z = −πi,

Re(z) = 0,

Im(z) = −π

4.

z = −6,

5.

6.

20.

Re(z) = −6, Im(z) = 0 √ z = −1 + i, |z| = 2, Arg (z) = 3π/4 √ z = 2 (cos(3π/4) + i sin(3π/4)) z = −2, |z| = 2, Arg (z) = π z = 2(cos π + i sin π ) z = 3i, |z| = 3, Arg (z) = π/2 z = 3(cos(π/2) + i sin(π/2))

8.

z = −5i, |z| = 5, Arg (z) = −π/2 z = 5(cos(−π/2) + i sin(−π/2)) √ z = 1 + 2i, |z| = 5, θ = Arg (z) = tan−1 2 √ z = 5(cos θ + i sin θ ) √ z = −2 + i, |z| = 5, θ = Arg (z) = π − tan−1 (1/2) √ z = 5(cos θ + i sin θ )

10.

11. 12. 13. 14.

z = −3 − 4i, |z| = 5, z = 5(cos θ + i sin θ )

21.

|z| = 5, θ = arg (z) = π ⇒ sin θ = 3/5, cos θ = 4/5 z = 4 + 3i   3π 3π 3π |z| = 1, arg (z) = ⇒ z = cos + i sin 4 4 4 1 1 ⇒ z = −√ + √ i 2 2  π π π ⇒ z = π cos + i sin |z| = π, arg (z) = 6 6 6 √ π 3 π ⇒ z= + i 2 2

22. |z| = 0 ⇒ z = 0 for any value of arg (z)

7.

9.

4π 4π + 3i sin 5 5 4π |z| = 3, Arg (z) = 5 3π π If Arg (z) = and Arg (w) = , then 4 2 3π π 5π arg (zw) = + = , so 4 2 4 5π −3π Arg (zw) = − 2π = . 4 4 5π π If Arg (z) = − and Arg (w) = , then 6 4 5π π 13π arg (z/w) = − − =− , so 6 4 12 13π 11π Arg (z/w) = − + 2π = . 12 12 z = 3 cos

23.

θ = Arg (z) = −π + tan−1 (4/3)

1 π 1 π π , arg (z) = − ⇒ z= cos − i sin 2 3 2 3 3 √ 1 3 ⇒ z= − i 4 4

|z| =

24. 5 + 3i = 5 − 3i 25. −3 − 5i = −3 + 5i 26. 4i = −4i 27. 2 − i = 2 + i 28. |z| = 2 represents all points on the circle of radius 2 centred at the origin.

29. |z| ≤ 2 represents all points in the closed disk of radius 2 centred at the origin.

z = 3 − 4i, |z| = 5, θ = Arg (z) = −tan−1 (4/3) z = 5(cos θ + i sin θ ) √ z = 3 − i, |z| = 2, Arg (z) = −π/6 z = 2(cos(−π/6) + i sin(−π/6)) √ √ z = − 3 − 3i, |z| = 2 3, Arg (z) = −2π/3 √ z = 2 3(cos(−2π/3) + i sin(−2π/3))

30. |z − 2i | ≤ 3 represents all points in the closed disk of radius 3 centred at the point 2i .

31. |z − 3 + 4i | ≤ 5 represents all points in the closed disk of radius 5 centred at the point 3 − 4i .

32. arg (z) = π/3 represents all points on the ray from the origin in the first quadrant, making angle 60◦ with the positive direction of the real axis.

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APPENDIX I. (PAGE A-10)

ADAMS and ESSEX: CALCULUS 8

33. π ≤ arg (z) ≤ 7π/4 represents the closed wedge-shaped

48.

region in the third and fourth quadrants bounded by the ray from the origin to −∞ on the real axis and the ray from the origin making angle −45◦ with the positive direction of the real axis.

cos(3θ ) + i sin(3θ ) = (cos θ + i sin θ )3

= cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ Thus cos(3θ ) = cos3 θ − 3 cos θ sin2 θ = 4 cos3 θ − 3 cos θ sin(3θ ) = 3 cos2 θ sin θ − sin3 θ = 3 sin θ − 4 sin3 θ.

34. (2 + 5i ) + (3 − i ) = 5 + 4i 35. i − (3 − 2i ) + (7 − 3i ) = −3 + 7 + i + 2i − 3i = 4 36. (4 + i )(4 − i ) = 16 − i 2 = 17

49.

37. (1 + i )(2 − 3i ) = 2 + 2i − 3i − 3i 2 = 5 − i 38.

(a + bi )(2a − bi) = (a + bi )(2a + bi ) = 2a 2 − b2 + 3abi 3

2

3

39. (2 + i ) = 8 + 12i + 6i + i = 2 + 11i 40. 41. 42. 43.

2−i (2 − i )2 3 − 4i = = 2+i 4 − i2 5

50.

1 + 3i (1 + 3i )(2 + i ) −1 + 7i = = 2 2−i 4−i 5

1+i (1 + i )(−3 − 2i ) −1 − 5i 1+i = = = i (2 + 3i ) −3 + 2i 9+4 13

a) z = 2/z can be rewritten |z|2 = zz = 2, so is √ satisfied by all numbers z on the circle of radius 2 centred at the origin. b) z = −2/z can be rewritten |z|2 = zz = −2, which has no solutions since the square of |z| is nonnegative for all complex z. √ If z = w =√−1, then zw = 1, so zw = 1. √But if we √ use and the same value for w, then √ √ z = 2 −1 = i √ z w = i = −1 6= zw.

51. The three cube roots of −1 = cos π + i sin π are of the

form cos θ +i sin θ where θ = π/3, θ = π , and θ = 5π/3. Thus they are

(1 + 2i )(2 − 3i ) 8+i = =1 (2 − i )(3 + 2i ) 8+i

√ 1 3 +i , 2 2

44. If z = x + yi and w = u + vi , where x, y, u, and v are

−1,

√ 1 3 −i . 2 2

real, then

z + w = x + u + (y + v)i = x + u − (y + v)i = x − yi + u − vi = z + w.



3π 3π + i sin 2 2 are of the form 2(cos θ + i sin θ ) where θ = π/2, θ = 7π/6, and θ = 11π/6. Thus they are

52. The three cube roots of −8i = 8 cos

45. Using the fact that |zw| = |z||w|, we have z w

46.

47.

=



zw |w|2



2i, zw zw z = = = . 2 ww w |w|

√ √  π π z = 3 + i 3 = 2 3 cos + i sin 6 6   √ 2π 2π w = −1 + i 3 = 2 cos + i sin 3 3   √ 5π 5π zw = 4 3 cos + i sin 6 6   √ √ −π −π z = 3 cos + i sin = −i 3 w 2 2   √ 3π 3π z = −1 + i = 2 cos + i sin 4 4  π π w = 3i = 3 cos + i sin 2 2   √ 5π 5π + i sin = −3 − 3i zw = 3 2 cos 4 4 √  z 2 π π 1 1 = cos + i sin = + i w 3 4 4 3 3

√ − 3 − i,

√



3 − i.

53. The three cube

  √ 3π 3π 2 cos + i sin are of the 4 4 form 21/6 (cos θ + i sin θ ) where θ = π/4, θ = 11π/12, and θ = 19π/12.

roots of −1 + i =

54. The four √ fourth roots of 4 = 4(cos 0 + i sin 0) are of the 55.

form 2(cos θ + i sin θ ) where θ = 0,√θ = π/2, π√ , and √ √ θ = 3π/2. Thus they are 2, i 2, − 2, and −i 2. √ The equation z 4 + 1 − i 3 = 0 hassolutions that are the  √ 2π 2π four fourth roots of −1 + i 3 = 2 cos + i sin . 3 3 Thus they are of the form 21/4 (cos θ + i sin θ ), where θ = π/6, 2π/3, 7π/6, and 5π/3. They are the complex numbers ! √ √ ! 3 1 3 1/4 1/4 1 ±2 + i , ±2 − i . 2 2 2 2

700 Copyright © 2014 Pearson Canada Inc.

INSTRUCTOR’S SOLUTIONS MANUAL

APPENDIX II. (PAGE A-19)

56. The equation z 5 + a 5 = 0 (a > 0) has solutions that are

8. The function w = z 2 = x 2 − y 2 + 2x yi transforms the

57. The n nth roots of unity are

9. The function w = z 2 = x 2 − y 2 + 2x yi transforms the

−a 5

the five fifth roots of = a (cos π + i sin π ); they are of the form a(cos θ + i sin θ ), where θ = π/5, 3π/5, π , 7π/5, and 9π/5.

line x = 1 to u = 1 − y 2 , v = 2y, which is the parabola v 2 = 4 − 4u with vertex at w = 1, opening to the left. line y = 1 to u = x 2 − 1, v = 2x, which is the parabola v 2 = 4u + 4 with vertex at w = −1 and opening to the right.

ω1 = 1

2π 2π + i sin n n 4π 4π ω3 = cos + i sin = ω22 n n 6π 6π ω4 = cos + i sin = ω23 n n .. . 2(n − 1)π 2(n − 1)π ωn = cos + i sin = ω2n−1 . n n ω2 = cos

10. The function w = 1/z = (x − yi )/(x 2 + y 2 ) transforms the line x = 1 to the curve given parametrically by u=

Appendix II. Complex Functions (page A-19)

u2 + v 2 =

−y . 1 + y2

1 + y2 = u, (1 + y 2 )2

with centre w = 1/2 and radius 1/2.

11. The function w = e z = e x cos y + i e x sin y transforms

the horizontal strip −∞ < x < ∞, π/4 ≤ y ≤ π/2 to the wedge π/4 ≤ arg (w) ≤ π/2, or, equivalently, u ≥ 0, v ≥ u.

12. The function w = ei z = e−y (cos x + i sin x) transforms

the vertical half-strip 0 < x < π/2, 0 < y < ∞ to the first-quadrant part of the unit open disk |w| = e−y < 1, 0 < arg (w) = x < π/2, that is u > 0, v > 0, u 2 + v 2 < 1.

In Solutions 1–12, z = x + yi and w = u + vi , where x, y, u, and v are real.

1. The function w = z transforms the closed rectangle

13.

2. The function w = z transforms the line x + y = 1 to the line u − v = 1.

3. The function w = z 2 transforms the closed annular sector 1 ≤ |z| ≤ 2, π/2 ≤ arg (z) ≤ 3π/4 to the closed annular sector 1 ≤ |w| ≤ 4, π ≤ arg (w) ≤ 3π/2.

14.

circular disk 0 ≤ |z| ≤ 2, 0 ≤ arg (z) ≤ π/2 to the closed three-quarter disk 0 ≤ |w| ≤ 8, 0 ≤ arg (w) ≤ 3π/2.

5. The function w = 1/z = z/|z|2 transforms the closed

6. The function w = −i z rotates the z-plane −90◦ , so transforms the wedge π/4 ≤ arg (z) ≤ π/3 to the wedge −π/4 ≤ arg (z) ≤ −π/6. √ The function w = z transforms the ray arg (z) = −π/3 (that is, Arg (z) = 5π/3) to the ray arg (w) = 5π/6.

f (z) = z 3 = (x + yi )3 = x 3 − 3x y 2 + (3x 2 y − y 3 )i

u = x 3 − 3x y 2 , v = 3x 2 y − y 3 ∂u ∂v ∂u ∂v = 3(x 2 − y 2 ) = , = −6x y = − ∂x ∂y ∂y ∂x ∂u ∂v ′ 2 2 f (z) = +i = 3(x − y + 2x yi ) = 3z 2 . ∂x ∂x

4. The function w = z 3 transforms the closed quarter-

quarter-circular disk 0 ≤ |z| ≤ 2, 0 ≤ arg (z) ≤ π/2 to the closed region lying on or outside the circle |w| = 1/2 and in the fourth quadrant, that is, having −π/2 ≤ arg (w) ≤ 0.

f (z) = z 2 = (x + yi )2 = x 2 − y 2 + 2x yi

u = x 2 − y 2, v = 2x y ∂u ∂v ∂u ∂v = 2x = , = −2y = − ∂x ∂y ∂y ∂x ∂u ∂v ′ f (z) = +i = 2x + 2yi = 2z. ∂x ∂x

0 ≤ x ≤ 1, 0 ≤ y ≤ 2 to the closed rectangle 0 ≤ u ≤ 1, −2 ≤ v ≤ 0.

7.

v=

This curve is, in fact, a circle,

Hence ω1 + ω2 + ω3 + · · · + ωn = 1 + ω2 + ω22 + · · · + ω2n−1 1 − ω2n 0 = = 0. = 1 − ω2 1 − ω2

1 , 1 + y2

15.

1 x − yi = 2 z x + y2 −y x u= 2 , v= 2 x + y2 x + y2 ∂u y2 − x 2 ∂v ∂u −2x y ∂v = 2 = , = 2 =− ∂x (x + y 2 )2 ∂y ∂y (x + y 2 )2 ∂x ∂u ∂v −(x 2 − y 2 ) + 2x yi −(z)2 1 f ′ (z) = +i = = = − 2. 2 2 2 ∂x ∂x (x + y ) (zz)2 z f (z) =

701 Copyright © 2014 Pearson Canada Inc.

APPENDIX II. (PAGE A-19)

16.

2

f (z) = e z = e x 2

2 −y 2

2

ADAMS and ESSEX: CALCULUS 8

2

2 −y 2

cos z = 0 ⇔ e zi = −e−zi ⇔ e2zi = −1

⇔ e−2y [cos(2x) + i sin(2x)] = −1

2

u = e x −y cos(2x y), v = e x −y sin(2x y) ∂u 2 2 ∂v = e x −y (2x cos(2x y) − 2y sin(2x y)) = ∂x ∂y ∂u ∂v x 2 −y 2 = −e (2y cos(2x y) + 2x sin(2x y)) = − ∂y ∂x ∂u ∂v ′ f (z) = +i ∂x ∂x 2 2 x −y =e [2x cos(2x y) − 2y sin(2x y) + i (2y cos(2x y) + 2x sin(2x y))] = (2x + 2yi )e x

21.

(cos(2x y) + i sin(2x y))

⇔ sin(2x) = 0, e−2y cos(2x) = −1 ⇔ y = 0, cos(2x) = −1 3π π =⇔ y = 0, x = ± , ± , . . . 2 2 Thus the only complex zeros of cos z are its real zeros at z = (2n + 1)π/2 for integers n.

22.

sin z = 0 ⇔ e zi = e−zi ⇔ e2zi = 1

⇔ e−2y [cos(2x) + i sin(2x)] = 1

2

(cos(2x y) + i sin(2x y)) = 2ze z .

⇔ sin(2x) = 0, e−2y cos(2x) = 1 ⇔ y = 0, cos(2x) = 1 =⇔ y = 0, x = 0, ±π, ±2π, . . .

17. e yi = cos y + i sin y (for real y). Replacing y by −y, we get e−yi = cos y − i sin y (since cos is even and sin is odd). Adding and subtracting these two formulas gives e yi + e−yi = 2 cos y,

e yi − e−yi = 2i sin y.

e yi + e−yi e yi − e−yi Thus cos y = and sin y = . 2 2i

18.

e z+2π i = e x (cos(y + 2π ) + i sin(y + 2π )) = e x (cos y + i sin y) = e z . z Thus e is periodic with period 2πi . So is e−z = 1/e z . Since ei(z+2π ) = e zi+2π i = e zi , therefore e zi and also e−zi are periodic with period 2π . Hence cos z =

cosh z =

19.

20.

= = = =

24.

e z = e x+yi = e x cos y + i e x sin y

e−z = e−x−yi = e−x cos y − e−x sin y e z + e−z e x + e−x e x − e−x cosh z = = cos y + i sin y 2 2 2 = cosh x cos y + i sinh x sin y Re(cosh z) = cosh x cos y, Im(cosh z) = sinh x sin y.

e z − e−z e z + e−z and sinh z = 2 2

26.

e z − e−z e x − e−x e x + e−x = cos y + i sin y 2 2 2 = sinh x cos y + i cosh x sin y Re(sinh z) = sinh x cos y, Im(cosh z) = cosh x sin y.

sinh z =

ei z = e−y+xi = e−y cos x + i e−y sin x

e−i z = e y−xi = e y cos x − i e y sin x

i e zi − e−zi e zi + e−zi = = − sin z 2 2 zi −zi zi −zi ie + e e −e = = cos z 2i 2i z −z z −z e +e e −e = = sinh z 2 2 z −z z −z e −e e +e = = cosh z 2 2

ei z + e−i z = cosh z 2 i z −i 1 e −e z −i sinh(i z) = = sin z i 2 −z z e +e cos(i z) = = cosh z 2 −z z e −e −e−z + e z sin(i z) = =i = i sinh z 2i 2 cosh(i z) =

cos(i z) = 0, that is, if and only if z = (2n + 1)πi /2 for integer n. Similarly, sinh z = 0 if and only if sin(i z) = 0, that is, if and only if z = nπi for integer n.

25.

are periodic with period 2πi . d dz d dz d dz d dz

23. By Exercises 20 and 21, cosh z = 0 if and only if

e zi − e−zi e zi + e−zi and sin z = 2 2i

are periodic with period 2π , and

d cos z dz d sin z dz d cosh z dz d sinh z dz

Thus the only complex zeros of sin z are its real zeros at z = nπ for integers n.

ei z + e−i z e−y + e y e−y − e y = cos x + i sin x 2 2 2 = cos x cosh y − i sin x sinh y Re(cos z) = cos x cosh y, Im(cos z) = − sin x sinh y

cos z =

ei z − e−i z e−y − e y e−y + e y = cos x + i sin x 2i 2i 2i = sin x cosh y + i cos x sinh y Re(sin z) = sin x cosh y, Im(sin z) = cos x sinh y.

sin z =

27. 28.

z 2 + 2i z = 0 ⇒ z = 0 or z = −2i

z 2 − 2z + i = 0 ⇒ (z − 1)2 = 1 − i   √ 7π 7π + i sin = 2 cos 4 4   7π 7π ⇒ z = 1 ± 21/4 cos + i sin 8 8

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INSTRUCTOR’S SOLUTIONS MANUAL

29.

30.

31.

APPENDIX II. (PAGE A-19)

z 2 + 2z + 5 = 0 ⇒ (z + 1)2 = −4 ⇒ z = −1 ± 2i

35. Since P(z) = z 5 + 3z 4 + 4z 3 + 4z 2 + 3z + 1 has real coefficients, if z 1 = i is a zero of P(z), then so is z 2 = −i . Now

z 2 − 2i z − 1 = 0 ⇒ (z − i )2 = 0 ⇒ z = i (double root)

(z − z 1 )(z − z 2 ) = (z − i )(z + i ) = z 2 + 1.

z 3 − 3i z 2 − 2z = z(z 2 − 3i z − 2) = 0

By long division (details omitted) we discover that

2

⇒ z = 0 or z − 3i z − 2 = 0   1 3 2 =− ⇒ z = 0 or z − i 2 4   3 1 ⇒ z = 0 or z = ± i 2 2 ⇒ z = 0 or z = i or z = 2i

32.

33.

z 5 + 3z 4 + 4z 3 + 4z 2 + 3z + 1 = z 3 + 3z 2 + 3z + 1 z2 + 1 = (z + 1)3 . Thus P(z) has the five zeros: i , −i , −1, −1, and −1.

z 4 − 2z 2 + 4 = 0 ⇒ (z 2 − 1)2 = −3 √ √ or z2 = 1 + i 3 z2 = 1 − i 3    5π 5π π π z 2 = 2 cos + i sin , z 2 = 2 cos + i sin 3 3 3 3   √ 5π 5π z = ± 2 cos + i sin , or 6 6  √  π π z = ± 2 cos + i sin 6 6 ! ! r r i i 3 3 −√ , z=± +√ z=± 2 2 2 2 z 4 + 1 = 0 ⇒ z 2 = i or z 2 = −i 1−i 1+i ⇒ z=± √ , z=± √ 2   2  1+i 1−i 4 z +1 = z− √ z− √  2   2 1−i 1+i z+ √ × z+ √ 2 ! 2 !  2   1 1 1 2 1 z−√ = + z+ √ + 2 2 2 2 √ √ 2 2 = (z − 2z + 1)(z + 2z + 1)

34. Since P(z) =

z4

4z 3

12z 2

− + − 16z + 16 has real √ coefficients, if z 1 = 1 − 3i is a zero of P(z), then so is z 1 . Now (z − z 1 )(z − z 1 ) = (z − 1)2 + 3 = z 2 − 2z + 4.

By long division (details omitted) we discover that z 4 − 4z 3 + 12z 2 − 16z + 16 = z 2 − 2z + 4. z 2 − 2z + 4 Thus z 1 and z 1 are both double zeros of P(z). These are the only zeros.

36. Since P(z) = z 5 − 2z 4 − 8z 3 + 8z 2 + 31z − 30 has real coefficients, if z 1 = −2 + i is a zero of P(z), then so is z 2 = −2 − i . Now (z − z 1 )(z − z 2 ) = z 2 + 4z + 5. By long division (details omitted) we discover that z 5 − 2z 4 − 8z 3 + 8z 2 + 31z − 30 z 2 + 4z + 5 3 2 = z − 6z + 11z − 6. Observe that z 3 = 1 is a zero of z 3 − 6z 2 + 11z − 6. By long division again: z 3 − 6z 2 + 11z − 6 = z 2 − 5z + 6 = (z − 2)(z − 3). z−1 Hence P(z) has the five zeros −2 + i , −2 − i , 1, 2, and 3.

37. If w = z 4 + z 3 − 2i z − 3 and |z| = 2, then |z 4 | = 16 and |w − z 4 | = |z 3 − 2i z − 3| ≤ 8 + 4 + 3 = 15 < 16. By the mapping principle described in the proof of Theorem 2, the image in the w-plane of the circle |z| = 2 is a closed curve that winds around the origin the same number of times that the image of z 4 does, namely 4 times.

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APPENDIX II. (PAGE A-19)

ADAMS and ESSEX: CALCULUS 8

Appendix III. Continuous Functions (page A-25) 1. To be proved: If a < b < c, f (x) ≤ g(x) for a ≤ x ≤ c, lim x→b f (x) = L, and limx→b g(x) = M, then L ≤ M.

Proof: Suppose, to the contrary, that L > M. Let ǫ = (L − M)/3, so ǫ > 0. There exist numbers δ1 > 0 and δ2 > 0 such that if a ≤ x ≤ b, then

7. Suppose n is a positive integer and a > 0. Let ǫ > 0 be given. Let b = a 1/n , and let δ = min{a(1 − 2−n ), bn−1 ǫ}. If |x − a| < δ, then x > a/2n , and if y = x 1/n , then y > b/2. Thus 1/n − a 1/n = |y − b| x |y n − bn | y n−1 + y n−2 b + · · · + bn−1 |x − a| bn−1 ǫ < n−1 < n−1 = ǫ. b b =

|x − b| < δ1 ⇒ | f (x) − L| < ǫ |x − b| < δ2 ⇒ |g(x) − M| < ǫ. Thus if |x − b| < δ = min{δ1 , δ2 , b − a, c − b}, then f (x)− g(x) > L −ǫ − M −ǫ = L − M −2ǫ =

L−M > 0. 3

This contradicts the fact that f (x) ≤ g(x) on [a, b]. Therefore L ≤ M.

2. To be proved: If f (x) ≤ K on [a, b) and (b, c], and if lim x→b f (x) = L, then L ≤ K .

Proof: If L > K , then let ǫ = (L − K )/2; thus ǫ > 0. There exists δ > 0 such that δ < b − a and δ < c − b, and such that if 0 < |x − b| < δ, then | f (x) − L| < ǫ. In this case L−K f (x) > L − ǫ = L − > K, 2 which contradicts the fact that f (x) ≤ K on [a, b) and (b, c]. Therefore L ≤ K .

3. Let ǫ > 0 be given. Let δ = ǫ 1/r , (r > 0). Then 0 0 we have m  m  = lim x 1/n lim x m/n = lim x 1/n x→a

x→a

x→a

= (a 1/n )m = a m/n ,

and x m/n is continuous at each positive number.

9. If m and n are integers and n is odd, then (−x)m/n = cx m/n , where c = (−1)m/n is either −1 or 1 depending on the parity of m. Since x m/n is continuous at each positive number a, so is cx m/n . Thus (−x)m/n is continuous at each positive number, and x m/n is continuous at each negative number. If r = m/n > 0, then lim x→0+ x r = 0 by Exercise 3. Hence lim x→0− x r = (−1)r lim x→0+ x r = 0, also. Therefore lim x→0 x r = 0, and x r is continuous at x = 0.

10. Let ǫ > 0 be given. Let δ = ǫ. If a is any real number

a) Let f (x) = C, g(x) = x. Let ǫ > 0 be given and let δ = ǫ. For any real number x, if |x − a| < δ, then | f (x) − f (a)| = |C − C| = 0 < ǫ, |g(x) − g(a)| = |x − a| < δ = ǫ. Thus lim x→a f (x) = f (a) and lim x→a g(x) = g(a), and f and g are both continuous at every real number a.

5. A polynomial is constructed by adding and multiplying finite numbers of functions of the type of f and g in Exercise 4. By Theorem 1(a), such sums and products are continuous everywhere, since their components have been shown to be continuous everywhere.

6. If P and Q are polynomials, they are continuous everywhere by Exercise 5. If Q(a) 6= 0, then P(x) P(a) lim x→a = by Theorem 1(a). Hence P/Q is Q(x) Q(a) continuous everywhere except at the zeros of Q.

then

|x| − |a| ≤ |x − a| < ǫ

if

|x − a| < δ.

Thus lim x→a |x| = |a|, and the absolute value function is continuous at every real number.

11. By the definition of sin, Pt = (cos t, sin t), and

Pa = (cos a, sin a) are two points on the unit circle x 2 + y 2 = 1. Therefore |t − a| = length of the arc from Pt to Pa > length of the chord from Pt to Pa p = (cos t − cos a)2 + (sin t − sin a)2 .

If ǫ > 0 is given, and |t − a| < δ = ǫ, then the above inequality implies that

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| cos t − cos a| ≤ |t − a| < ǫ, | sin t − sin a| ≤ |t − a| < ǫ.

INSTRUCTOR’S SOLUTIONS MANUAL

APPENDIX IV. (PAGE A-30)

Thus g is continuous and increasing on R. If f is continuous on [a, b], then   f (x) h(x) = g f (x) = 1 + | f (x)|

Thus sin is continuous everywhere.

12. The proof that cos is continuous everywhere is almost 13.

identical to that for sin in Exercise 11. n a ǫa o Let a > 0 and ǫ > 0. Let δ = min , . 2 2 a 1 2 If |x − a| < δ, then x > , so < whenever t is 2 t a between a and x. Thus

is also continuous there, being the composition of continuous functions. Also, h(x) is bounded on [a, b], since   | f (x)| ≤ 1. g f (x) ≤ 1 + | f (x)|

| ln x − ln a|

By assumption in this problem, h(x) must assume maximum and minimum values; there exist c and d in [a, b] such that       g f (c) ≤ g f (x) ≤ g f (d)

1 = area under y = between t = a and t = x t 2 2 ǫa < |x − a| < = ǫ. a a 2

for all x in [a, b]. Since g is increasing, so is its inverse g −1 . Therefore

Thus lim x→a ln x = ln a, and ln is continuous at each point a in its domain (0, ∞).

f (c) ≤ f (x) ≤ f (d)

14. Let a be any real number, and let ǫ > 0 be given. Assume (making ǫ smaller if necessary) that ǫ <

ea .

for all x in [a, b], and f is bounded on that interval.

Since

    ǫ  ǫ  ǫ2 ln 1 − a + ln 1 + a = ln 1 − 2a < 0, e e e   ǫ  ǫ  we have ln 1 + a < − ln 1 − a . e e  ǫ  Let δ = ln 1 + a . If |x − a| < δ, then e   ǫ  ǫ  ln 1 − a < x − a < ln 1 + a e e ǫ ǫ 1 − a < e x−a < 1 + a e e x−a ǫ e − 1 < a e |e x − ea | = ea |e x−a − 1| < ǫ. Thus lim x→a e x = ea and e x is continuous at every point a in its domain.

15. Suppose a ≤ xn ≤ b for each n, and lim xn = L. Then

16.

a ≤ L ≤ b by Theorem 3. Let ǫ > 0 be given. Since f is continuous on [a, b], there exists δ > 0 such that if a ≤ x ≤ b and |x − L| < δ then | f (x) − f (L)| < ǫ. Since lim xn = L, there exists an integer N such that if n ≥ N then |xn − L| < δ. Hence | f (xn ) − f (L)| < ǫ for such n. Therefore lim( f (xn ) = f (L). t . For t 6= 0 we have Let g(t) = 1 + |t| g ′ (t) =

1 + |t| − t sgn t 1 + |t| − |t| 1 = = > 0. (1 + |t|)2 (1 + |t|)2 (1 + |t|)2

If t = 0, g is also differentiable, and has derivative 1: g ′ (0) = lim

h→0

1 g(h) − g(0) = lim = 1. h→0 1 + |h| h

Appendix IV. The Riemann Integral (page A-30) 1.



1 if 0 ≤ x ≤ 1 0 if 1 < x ≤ 2 Let 0 < ǫ < 1. Let P = {0, 1 − 3ǫ , 1 + ǫ3 , 2}. Then  ǫ ǫ L( f, P) = 1 1 − +0+0=1− 3 3    ǫ ǫ 2ǫ +1 +0 =1+ . U ( f, P) = 1 1 − 3 3 3 f (x) =

Since U ( f, P) − L( f, P) < ǫ, f is integrable on [0, 2]. Since L( f, P) < 1 < U ( f, P) for every ǫ, therefore Z 2 f (x) d x = 1. 0

2.

n 1 if x = 1/n (n = 1, 2, 3, . . .) f (x) = 0 otherwise If P is any partition of [0, 1] then L( f, P) = 0. Let 0 < ǫ ≤ 2. Let N be an integer such that 2 N + 1 > ≥ N . A partition P of [0, 1] ǫ can be constructed so that the first two points of P ǫ 1 are 0 and , and such that each of the N points 2 n (n = 1, 2, 3, . . . , n) lies in a subinterval of P having ǫ 1 . Since every number with n a poslength at most 2N n h ǫi itive integer lies either in 0, or one of these other 2 N subintervals of P, and since max f (x) = 1 for these subintervals and max f (x) = 0 for all other subintervals ǫ ǫ of P, therefore U ( f, P) ≤ + N = ǫ. By Theorem 2 2N 3, f is integrable on [0, 1]. Evidently Z 1 f (x) d x = least upper bound L( f, P) = 0. 0

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APPENDIX IV. (PAGE A-30)

3.

4.

ADAMS and ESSEX: CALCULUS 8

1/n if x = m/n in lowest terms 0 otherwise Clearly L( f, P) = 0 for every partition P of [0, 1]. Let ǫ > 0 be given. To show that f is integrable we must exhibit a partition P for which U ( f, P) < ǫ. We can assume ǫ < 1. Choose a positive integer N such that 2/N < ǫ. There are only finitely many integers n such that 1 ≤ n ≤ N . For each such n, there are only finitely many integers m such that 0 ≤ m/n ≤ 1. Therefore there are only finitely many points x in [0, 1] where f (x) > ǫ/2. Let P be a partition of [0, 1] such that all these points are contained in subintervals of the partition having total length less than ǫ/2. Since f (x) ≤ 1 on these subintervals, and f (x) < ǫ/2 on all other subintervals P, therefore U ( f, P) ≤ 1 × (ǫ/2) + (ǫ/2) × 1 = ǫ, R1 and f is integrable on [0, 1]. Evidently 0 f (x) d x = 0, since all lower sums are 0. I∗ − I ∗ Suppose, to the contrary, that I∗ > I ∗ . Let ǫ = , 3 so ǫ > 0. By the definition of I∗ and I ∗ , there exist partitions P1 and P2 of [a, b], such that L( f, P1 ) ≥ I∗ − ǫ and U ( f, P2 ) ≤ I ∗ + ǫ. By Theorem 2, L( f, P1 ) ≤ U ( f, P2 ), so f (x) =

n

3ǫ = I∗ − I ∗ ≤ L( f, P1 ) + ǫ − U ( f, P2 ) + ǫ ≤ 2ǫ. Since ǫ > 0, it follows that 3 ≤ 2. This contradiction shows that we must have I∗ ≤ I ∗ .

5. Multiplying a function by a constant multiplies all its Riemann sums by the same constant. If the constant is positive, upper and lower sums remain upper and lower; if the constant is negative upper sums become lower and vice versa. Therefore Z b Z b A f (x) d x = A f (x) d x. a

U ( f + g, P) − ǫ ≤ I + J ≤ L( f + g, P) + ǫ. Hence

Z b a

a

If ǫ > 0, then there exist partitions P1 and P2 of [a, b] such that ǫ ǫ U ( f, P1 ) − ≤ I < L( f, P1 ) + 2 2 ǫ ǫ U (g, P2 ) − ≤ J < L(g, P2 ) + . 2 2 Let P be the common refinement of P1 and P2 . Then the above inequalities hold with P replacing P1 and P2 . If m 1 ≤ f (x) ≤ M1 and m 2 ≤ g(x) ≤ M2 on any interval, then m 1 + m 2 ≤ f (x) + g(x) ≤ M1 + M2 there. It follows that U ( f + g, P) ≤ U ( f, P) + U (g, P), L( f, P) + L(g, P) ≤ L( f + g, P).

 f (x) + g(x) d x = I + J .

6. Assume a < b < c; the other cases are similar. Let ǫ > 0. If Z

b

f (x) d x = I,

a

and

c

Z

f (x) d x = J,

b

then there exist partitions P1 of [a, b], and P2 of [b, c] such that ǫ L( f, P1 ) ≤ I < L( f, P1 ) + 2 ǫ L( f, P2 ) ≤ J < L( f, P2 ) + 2 (with similar inequalities for upper sums). Let P be the partition of [a, c] formed by combining all the subdivision points of P1 and P2 . Then L( f, P) = L( f, P1 ) + L( f, P2 ) ≤ I + J < L( f, P) + ǫ. Similarly, U ( f, P) − ǫ < I + J ≤ U ( f, P). Therefore Z

c a

f (x) d x = I + J.

7. Let

a

It therefore remains to be proved only that the integral of a sum of functions is the sum of the integrals. Suppose that Z b Z b f (x) d x = I, and g(x) d x = J. a

Therefore

Z

b

f (x) d x = I,

a

and

Z

a

b

g(x) d x = J,

where f (x) ≤ g(x) on [a, b]. We want to show that I ≤ J . Suppose, to the contrary, that I > J . Then there would exist a partition P of [a, b] for which I < L( f, P) +

I−J , 2

and U (g, P) −

I−J < J. 2

I+J > U (g, P) ≥ L(g, P). However, 2 f (x) ≤ g(x) on [a, b] implies that L( f, P) ≤ L(g, P) for any partition. Thus we have a contradiction, and so I ≤ J.

Thus L( f, P) >

Since −| f (x)| ≤ f (x) ≤ | f (x)| for any x, we can apply the above result to obtain −

Z

b

a

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| f (x)| d x ≤

Z

b a

f (x) d x ≤

Z

a

b

| f (x)| d x.

INSTRUCTOR’S SOLUTIONS MANUAL

Z Therefore

a

8. We have Z

a

−a

b

Z f (x) d x ≤

b

a

f (x) d x = = =

Z

Z0 a 0

By repeated applications of the triangle inequality,

| f (x)| d x.

0

Z−a a

APPENDIX IV. (PAGE A-30)

f (x) d x +

Z

f (−x) d x +

| f (xk−1 ) − f (a)| = | f (xk−1 ) − f (x0 )| < k − 1.

a

f (x) d x

0

Z

a

f (x) d x

If x is any point in [a, b], then x belongs to one of the intervals [xk−1 , xk ], so, by the triangle inequality again,

0

[ f (−x) + f (x)] d x.

If f is odd,Z the last integral is 0. If f is even, the last a integral is 2 f (x) d x. 0

| f (x)− f (a)| ≤ | f (x)− f (xk−1 )|+| f (xk−1 )− f (a)| < k ≤ N. Thus | f (x)| < | f (a)| + N , and f is bounded on [a, b].

9. Let ǫ > 0 be given. Let δ = ǫ 2 /2. Let 0 ≤ x ≤ 1 and 0√≤ y ≤ 1. If √ x < ǫ 2 /4 and y < ǫ 2 /4 then √ √ | x − y| ≤ x + y < ǫ. If |x − y| < δ and either x ≥ ǫ 2 /4 or y ≥ ǫ 2 /4 then √ |x − y| 2 ǫ2 √ | x − y| = √ = ǫ. √ < × ǫ 2 x+ y √ Thus f (x) = x is uniformly continuous on [0, 1].

11. Suppose that | f (x)| ≤ K on [a, b] (where K > 0), and

that f is integrable on [a, b]. Let ǫ > 0 be given, and let δ = ǫ/K . If x and y belong to [a, b] and |x − y| < δ, then

10. Suppose f is uniformly continuous on [a, b]. Taking ǫ = 1 in the definition of uniform continuity, we can find a positive number δ such that | f (x)− f (y)| < 1 whenever x and y are in [a, b] and |x − y| < δ. Let N be a positive integer such that h = (b − a)/N satisfies h < δ. If xk = a + kh, (0 ≤ k ≤ N ), then each of the subintervals of the partition P = {x0 , x1 , . . . , x N } has length less than δ. Thus | f (xk ) − f (xk−1 )| < 1

for

1 ≤ k ≤ N.

Z x Z y f (t) dt − f (t) dt |F(x) − F(y)| = Za x a ǫ = f (t) dt ≤ K |x − y| < K = ǫ. K y

(See Theorem 3(f) of Section 6.4.) Thus F is uniformly continuous on [a, b].

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