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Complete Solutions Manual

for

MULTIVARIABLE CALCULUS

SEVENTH EDITION

DAN CLEGG

Palomar College

BARBARA FRANK

Cape Fear Community College

Australia. Brazil. Japan. Korea. Mexico. Singapore. Spain. United Kingdom. United States

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This Complete Solutions Manual contains detailed solutions to all exercises in the text Multivariable

Calculus, Seventh Edition (Chapters 10–17 of Calculus, Seventh Edition, and Calculus: Early

Transcendentals, Seventh Edition) by James Stewart. A Student Solutions Manual is also available,

which contains solutions to the odd-numbered exercises in each chapter section, review section,

True-False Quiz, and Problems Plus section as well as all solutions to the Concept Check questions.

(It does not, however, include solutions to any of the projects.)

Because of differences between the regular version and the Early Transcendentals version of the

text, some references are given in a dual format. In these cases, users of the Early Transcendentals

text should use the references denoted by “ET.”

While we have extended every effort to ensure the accuracy of the solutions presented, we would

appreciate correspondence regarding any errors that may exist. Other suggestions or comments are

also welcome, and can be sent to dan clegg at dclegg@palomar.edu or in care of the publisher:

Brooks/Cole, Cengage Learning, 20 Davis Drive, Belmont CA 94002-3098.

We would like to thank James Stewart for entrusting us with the writing of this manual and offer-

ing suggestions and Kathi Townes of TECH-arts for typesetting and producing this manual as well as

creating the illustrations. We also thank Richard Stratton, Liz Covello, and Elizabeth Neustaetter of

Brooks/Cole, Cengage Learning, for their trust, assistance, and patience.

DAN CLEGG

Palomar College

BARBARA FRANK

Cape Fear Community College

■ PREFACE

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x ■ CONTENTS

■ 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 711

17.1 Second-Order Linear Equations 711

17.2 Nonhomogeneous Linear Equations 715

17.3 Applications of Second-Order Differential Equations 720

17.4 Series Solutions 725

Review 729

■ APPENDIX 735

H Complex Numbers 735

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2 ¤^ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

5. { = 3 3 4 w, | = 2 3 3 w (a) w 31 0 1 2 { 7 3 31 35 | 5 2 31 34 (b) { = 3 3 4 w i 4 w = 3 { + 3 i w = 3 14 { + 34 , so | = 2 3 3 w = 2 3 3  3 14 { + 34 ^ = 2 + 34 { 3 94 i | = 34 { (^3 ) 6. { = 1 3 2 w, | = 12 w 3 1 , 32 $ w $ 4 (a) w 32 0 2 4 { 5 1 33 37 | 32 31 0 1 (b) { = 1 3 2 w i 2 w = 3 { + 1 i w = 3 12 { + 12 , so | = 12 w 3 1 = 12  3 12 { + 12 ^3 1 = 3 14 { + 14 3 1 i | = 3 14 { 3 34 , with 37 $ { $ 5 7. { = 1 3 w^2 , | = w 3 2 , 32 $ w $ 2

(a) w 32 31 0 1 2 { 33 0 1 0 33 | 34 33 32 31 0 (b) | = w 3 2 i w = | + 2, so { = 1 3 w^2 = 1 3 (| + 2)^2 i { = 3 (| + 2)^2 + 1, or { = 3 |^2 3 4 | 3 3 , with 34 $ | $ 0

8. { = w 3 1 , | = w^3 + 1, 32 $ w $ 2 (a) w 32 31 0 1 2 { 33 32 31 0 1 | 37 0 1 2 9 (b) { = w 3 1 i w = { + 1, so | = w^3 + 1 i | = ({ + 1)^3 + 1, or | = {^3 + 3{^2 + 3{ + 2, with 33 $ { $ 1

° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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°°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.duplicated, or posted to a publicly accessible website, in whole or in part.

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SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ¤^ 3

9. { = Iw, | = 1 3 w (a) w 0 1 2 3 4 { 0 1 1 = 414 1 = 732 2 | 1 0 31 32 33 (b) { = Iw i w = {^2 i | = 1 3 w = 1 3 {^2. Since w D 0 , { D 0. So the curve is the right half of the parabola | = 1 3 {^2. 10. { = w^2 , | = w^3 (a) w 32 31 0 1 2 { 4 1 0 1 4 | 38 31 0 1 8

(b) | = w^3 i w = s^3 | i { = w^2 =

 (^) s 3 |

= |^2 @^3. w M R, | M R, { D 0.

11. (a) { = sin 12 , | = cos 12 , 3  $  $ . {^2 + |^2 = sin2 1 2  + cos2 1 2  = 1. For 3  $  $ 0 , we have 31 $ { $ 0 and 0 $ | $ 1. For 0?  $ , we have 0? { $ 1 and 1 A | D 0. The graph is a semicircle.

(b)

12. (a) { = 12 cos , | = 2 sin , 0 $  $ .

(2{)^2 + ^12 |^2 = cos^2  + sin^2  = 1 i 4 {^2 + 14 |^2 = 1 i {^2 (1@2)^2 +^

|^2

22 = 1, which is an equation of an ellipse with {-intercepts ± 12 and |-intercepts ± 2. For 0 $  $ @ 2 , we have (^12) D { D 0 and 0 $ | $ 2. For @ 2?  $ , we have 0 A { D 3 (^12) and 2 A | D 0. So the graph is the top half of the ellipse.

(b)

13. (a) { = sin w> | = csc w, 0? w?  2. | = csc w = (^) sin^1 w =^1 {.

For 0? w?  2 , we have 0? {? 1 and | A 1. Thus, the curve is the portion of the hyperbola | = 1@{ with | A 1.

(b)

° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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°°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.licated, or posted to a publicly accessible website, in whole or in par

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SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ¤^ 5

21. { = 5 sin w, | = 2 cos w i sin w = { 5 , cos w = | 2. sin^2 w + cos^2 w = 1 i

= 1. The motion of the particle takes place on an ellipse centered at (0> 0). As w goes from 3  to 5 , the particle starts at the point (0> 3 2) and moves clockwise around the ellipse 3 times.

22. | = cos^2 w = 1 3 sin^2 w = 1 3 {^2. The motion of the particle takes place on the parabola | = 1 3 {^2. As w goes from 32  to 3 , the particle starts at the point (0> 1), moves to (1> 0), and goes back to (0> 1). As w goes from 3  to 0 , the particle moves to ( 31 > 0) and goes back to (0> 1). The particle repeats this motion as w goes from 0 to 2 . 23. We must have 1 $ { $ 4 and 2 $ | $ 3. So the graph of the curve must be contained in the rectangle [1> 4] by [2> 3]. 24. (a) From the first graph, we have 1 $ { $ 2. From the second graph, we have 31 $ | $ 1 = The only choice that satisfies either of those conditions is III. (b) From the first graph, the values of { cycle through the values from 32 to 2 four times. From the second graph, the values of | cycle through the values from 32 to 2 six times. Choice I satisfies these conditions. (c) From the first graph, the values of { cycle through the values from 32 to 2 three times. From the second graph, we have 0 $ | $ 2. Choice IV satisfies these conditions. (d) From the first graph, the values of { cycle through the values from 32 to 2 two times. From the second graph, the values of | do the same thing. Choice II satisfies these conditions. 25. When w = 31 , ({> |) = (0> 3 1). As w increases to 0 , { decreases to 31 and | increases to 0. As w increases from 0 to 1 , { increases to 0 and | increases to 1. As w increases beyond 1 , both { and | increase. For w? 31 , { is positive and decreasing and | is negative and increasing. We could achieve greater accuracy by estimating {- and |-values for selected values of w from the given graphs and plotting the corresponding points. 26. For w? 31 , { is positive and decreasing, while | is negative and increasing (these points are in Quadrant IV). When w = 31 , ({> |) = (0> 0) and, as w increases from 31 to 0 , { becomes negative and | increases from 0 to 1. At w = 0, ({> |) = (0> 1) and, as w increases from 0 to 1 , | decreases from 1 to 0 and { is positive. At w = 1> ({> |) = (0> 0) again, so the loop is completed. For w A 1 , { and | both become large negative. This enables us to draw a rough sketch. We could achieve greater accuracy by estimating {- and |-values for selected values of w from the given graphs and plotting the corresponding points. 27. When w = 0 we see that { = 0 and | = 0, so the curve starts at the origin. As w increases from 0 to 12 , the graphs show that | increases from 0 to 1 while { increases from 0 to 1 , decreases to 0 and to 31 , then increases back to 0 , so we arrive at the point (0> 1). Similarly, as w increases from 12 to 1 , | decreases from 1 to 0 while { repeats its pattern, and we arrive back at the origin. We could achieve greater accuracy by estimating {- and |-values for selected values of w from the given graphs and plotting the corresponding points.

° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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°°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.licated, or posted to a publicly accessible website, in whole or in par

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6 ¤^ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

28. (a) { = w^4 3 w + 1 = (w^4 + 1) 3 w A 0 [think of the graphs of | = w^4 + 1 and | = w] and | = w^2 D 0 , so these equations are matched with graph V. (b) | = Iw D 0. { = w^2 3 2 w = w(w 3 2) is negative for 0? w? 2 , so these equations are matched with graph I. (c) { = sin 2w has period 2 @2 = . Note that |(w + 2) = sin[w + 2 + sin 2(w + 2)] = sin(w + 2 + sin 2w) = sin(w + sin 2w) = |(w), so | has period 2 . These equations match graph II since { cycles through the values 31 to 1 twice as | cycles through those values once. (d) { = cos 5w has period 2 @ 5 and | = sin 2w has period , so { will take on the values 31 to 1 , and then 1 to 31 , before | takes on the values 31 to 1. Note that when w = 0, ({> |) = (1> 0). These equations are matched with graph VI= (e) { = w + sin 4w, | = w^2 + cos 3w. As w becomes large, w and w^2 become the dominant terms in the expressions for { and |, so the graph will look like the graph of | = {^2 , but with oscillations. These equations are matched with graph IV. (f) { = 4 +sin 2 ww 2 , | = 4 +cos 2 ww 2. As w < ", { and | both approach 0. These equations are matched with graph III. 29. Use | = w and { = w 3 2 sin w with a w-interval of [ 3 > ]. 30. Use { 1 = w, | 1 = w^3 3 4 w and { 2 = w^3 3 4 w, | 2 = w with a w-interval of [ 33 > 3]. There are 9 points of intersection; (0> 0) is fairly obvious. The point in quadrant I is approximately (2= 2 > 2 =2), and by symmetry, the point in quadrant III is approximately ( 32 = 2 > 32 =2). The other six points are approximately (~ 1 = 9 > ± 0 =5), (~ 1 = 7 > ± 1 =7), and (~ 0 = 5 > ± 1 =9). 31. (a) { = { 1 + ({ 2 3 { 1 )w, | = | 1 + (| 2 3 | 1 )w, 0 $ w $ 1. Clearly the curve passes through S 1 ({ 1 > | 1 ) when w = 0 and through S 2 ({ 2 > | 2 ) when w = 1. For 0? w? 1 , { is strictly between { 1 and { 2 and | is strictly between | 1 and | 2. For every value of w, { and | satisfy the relation | 3 | 1 = | {^22 33 |{^11 ({ 3 { 1 ), which is the equation of the line through S 1 ({ 1 > | 1 ) and S 2 ({ 2 > | 2 ). Finally, any point ({> |) on that line satisfies (^) || 2 33 | |^11 = (^) {{ 2 33 { {^11 ; if we call that common value w, then the given parametric equations yield the point ({> |); and any ({> |) on the line between S 1 ({ 1 > | 1 ) and S 2 ({ 2 > | 2 ) yields a value of w in [0> 1]. So the given parametric equations exactly specify the line segment from S 1 ({ 1 > | 1 ) to S 2 ({ 2 > | 2 ). (b) { = 3 2 + [3 3 ( 3 2)]w = 3 2 + 5w and | = 7 + ( 31 3 7)w = 7 3 8 w for 0 $ w $ 1.

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8 ¤^ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

36. If you are using a calculator or computer that can overlay graphs (using multiple w-intervals), the following is appropriate. Left side: { = 1 and | goes from 1 = 5 to 4 , so use { = 1> | = w> 1 = 5 $ w $ 4 Right side: { = 10 and | goes from 1 = 5 to 4 , so use { = 10> | = w> 1 = 5 $ w $ 4 Bottom: { goes from 1 to 10 and | = 1= 5 , so use { = w> | = 1= 5 > 1 $ w $ 10 Handle: It starts at (10> 4) and ends at (13> 7), so use { = 10 + w> | = 4 + w> 0 $ w $ 3 Left wheel: It’s centered at (3> 1), has a radius of 1 , and appears to go about 30 ^ above the horizontal, so use { = 3 + 1 cos w> | = 1 + 1 sin w> 56  $ w $ 136  Right wheel: Similar to the left wheel with center (8> 1), so use { = 8 + 1 cos w> | = 1 + 1 sin w> 56  $ w $ 136  If you are using a calculator or computer that cannot overlay graphs (using one w-interval), the following is appropriate. We’ll start by picking the w-interval [0> 2 =5] since it easily matches the w-values for the two sides. We now need to find parametric equations for all graphs with 0 $ w $ 2 = 5. Left side: { = 1 and | goes from 1 = 5 to 4 , so use { = 1> | = 1=5 + w> 0 $ w $ 2 = 5 Right side: { = 10 and | goes from 1 = 5 to 4 , so use { = 10> | = 1=5 + w> 0 $ w $ 2 = 5 Bottom: { goes from 1 to 10 and | = 1= 5 , so use { = 1 + 3= 6 w> | = 1= 5 > 0 $ w $ 2 = 5 To get the x -assignment, think of creating a linear function such that when w = 0, { = 1 and when w = 2= 5 , { = 10. We can use the point-slope form of a line with (w 1 > { 1 ) = (0> 1) and (w 2 > { 2 ) = (2= 5 > 10). { 3 1 = 210 = 5 33 1 0 (w 3 0) i { = 1 + 3= 6 w. Handle: It starts at (10> 4) and ends at (13> 7), so use { = 10 + 1= 2 w> | = 4 + 1= 2 w> 0 $ w $ 2 = 5 (w 1 > { 1 ) = (0> 10) and (w 2 > { 2 ) = (2= 5 > 13) gives us { 3 10 =^132 = 5 3 3 10 0 (w 3 0) i { = 10 + 1= 2 w. (w 1 > | 1 ) = (0> 4) and (w 2 > | 2 ) = (2= 5 > 7) gives us | 3 4 = (^27) = 5 3 3 4 0 (w 3 0) i | = 4 + 1= 2 w. ° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ¤^ 9 Left wheel: It’s centered at (3> 1), has a radius of 1 , and appears to go about 30 ^ above the horizontal, so use { = 3 + 1 cos^815  w + 56 ^ > | = 1 + 1 sin^815  w + 56 ^ > 0 $ w $ 2 = 5

(w 1 >  1 ) =  0 > 56 ^ ^ and (w 2 >  2 ) = ^52 > 136 ^ ^ gives us  3 56  =

52 3 0 (w^3 0)^ i^ ^ =^56  +^815  w. Right wheel: Similar to the left wheel with center (8> 1), so use { = 8 + 1 cos^815  w + 56 ^ > | = 1 + 1 sin^815  w + 56 ^ > 0 $ w $ 2 = 5

37. (a) { = w^3 i w = {^1 @^3 , so | = w^2 = {^2 @^3. We get the entire curve | = {^2 @^3 traversed in a left to right direction.

(b) { = w^6 i w = {^1 @^6 , so | = w^4 = {^4 @^6 = {^2 @^3. Since { = w^6 D 0 , we only get the right half of the curve | = {^2 @^3.

(c) { = h^33 w^ = (h^3 w)^3 [so h^3 w^ = {^1 @^3 ], | = h^32 w^ = (h^3 w)^2 = ({^1 @^3 )^2 = {^2 @^3. If w? 0 , then { and | are both larger than 1. If w A 0 , then { and | are between 0 and 1. Since { A 0 and | A 0 , the curve never quite reaches the origin.

38. (a) { = w, so | = w^32 = {^32. We get the entire curve | = 1@{^2 traversed in a left-to-right direction.

(b) { = cos w, | = sec^2 w = (^) cos^12 w = (^) {^12. Since sec w D 1 , we only get the parts of the curve | = 1@{^2 with | D 1. We get the first quadrant portion of the curve when { A 0 , that is, cos w A 0 , and we get the second quadrant portion of the curve when {? 0 , that is, cos w? 0.

(c) { = hw, | = h^32 w^ = (hw)^32 = {^32. Since hw^ and h^32 w^ are both positive, we only get the first quadrant portion of the curve | = 1@{^2.

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