Exercícios Resolvidos - James Stewart - Vol 2 - 7 ed - Ingles

Exercícios Resolvidos - James Stewart - Vol 2 - 7 ed - Ingles

(Parte 1 de 6)

-STUDENT SOLUTIONS MANUAL for STEWART'S

.Student Solutions Manual for.

DAN CLEGG Palomar College

BARBARA FRANK· Cape Fear Community College

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This Student Solutions Manual contains detailed solutions to selected exercises in the text Multivariable Calculus, Seventh Edition (Chapters 10-17 of Calculus, Seventh. Edition, and

Calculus: Early Transcendentals, Sevent,h Edition) by James Stewart. Specifically, it includes solutions to the odd-numbered exercises in each chapter section, review section, True-False Quiz, and Problems Plus section. Also included are all solutions to the Concept Check questions.

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, readers of the Early Transcendentals text should use the references denoted by "ET."

Each solution is presented in the context of the corresponding section of the text. In general, solutions to the initial exercises involving a new concept illustrate that concept in more detail; this knowledge is then utilized in subsequent solutions. Thus, while the intermediate steps of a solution are given, you may need to refer back to earlier exercises in the section or prior sections for additional explanation of the concepts involved. Note that, in many cases, different routes to an answer may exist which are equally valid; also, answers can be expressed in different but equivalent forms. Thus, the goaJ of this manual is not to give the definitive solution to each exercise, but rather to assist you as a student in understanc#ng the concepts of the text and learning how to apply them to the. chal- lenge of solving a problem.

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

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CD cu

I 1/D IP concave downward concave upward the domain off First Derivative Test horizontal asymptote(s) interval of convergence

Increasing/Decreasing Test inflection point(s) radius of convergence vertical asymptote(s)

· indicates the use of a computer algebra system. indicates the use of ljHospital's Rule.

'indicates the use ofFormu'Ja j in the Table of Integrals in the back endpapers. indicates the use of the substitution { u = sin x, if.u = cos x dx}.

indicates the use of the substitution { u = cos x, du = -sin x dx}. 2012 LeMning. All RigbReserved. May not be scunncd, copied, or duplic:Hed, or posted to a publicly ucce:ssible websire, in whole or jn part. v

0 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 1

10.1 Curves Defined by Parametric Equations

10.2 Calculus with Parametric Curves 7 1 0.3 Polar Coordinates 13

10.4 Areas and Lengths in Polar Coordinates 20 10.5 Conic Sections 26

1 0.6 Conic Sections in Polar Coordinates 32 Review 35

Problems Plus 43

0 1 INFINITE SEQUENCES AND SERIES 45

1.1 Sequences 45

1.2 Series 51 1.3 The Integral Test and Estimates of Sums 59 1 .4 The Comparison Tests 62

1.5 Alternating Series 65 1.6 Absolute Convergence and the Ratio and Root Tests 68 1.7 Strategy for Testing Series 72

1.8 Power Series 74

1.9 Representations of Functions as Power Series 78 1.10 Taylor and Maclaurin Series 83 1.1 Applications of Taylor Polynomials 90

Review 97

Problems Plus 105

0 12 VECTQRS AND THE GEOMETRY OF SPACE 1

12.1 Three-Dimensional Coordinate Systems 1

12.2 Vectors 114 12.3 The Dot Product 119

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12.4 The Cross Product 123

12.5 Equations of Lines and Planes 128 . 12.6 C,ylinders and Quadric Surfaces 135

Review 140

Problems Plus 147

0 13 VECTOR FUNCTIONS 151

13.1 Vector Functions and Space Curves 151

13.2 Derivatives and Integrals of Vector Functions 157

13.3 Arc Length and Curvature 161 13.4 Motion in Space: Velocity and Acceleration 168 Review 173

Problems Plus 179

0 14 PARTIAL DERIVATIVES 183 14.1 Functions of Several Variables 183

14.2 14.3

Limits and Continuity Partial Derivatives 192

195 14.4 Tangent Planes and Linear. Approximations 203 14.5 The Chain Rule 207

14.6 Directional Derivatives and the Gradient Vector 213 14.7 Maximum and Minimum Values · 220 14.8 Lagrange Multipliers 229

Review 234

Problems Plus 245

0 15 MULTIPLE INTEGRALS 247

15.1 Double Integrals over Rectangles 247

15.2 Iterated Integrals 249 15.3 Double Integrals over General Regions 251 15.4 Double Integrals in Polar Coordinates 258

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15.6 15.7

15.8 15.9 15.10

Applications of Double Integrals 261'

Surface Area 267 Triple Integrals 269

Triple Integrals in Cylindrical Coordinates

Triple Integrals in Spherical Coordinates Change ofVariables in Multiple Integrals Review 289

Problems Plus 297

0 16. VECTOR CALCULUS 303

16.1 Vector Fields 303 16.2 Line Integrals 305

16.3 The Fundamental Theorem for Line Integrals 310 16.4 Green's Theorem 313

16.5 Curl and Divergence -316 16.6 Parametric Surfaces and Their Areas 321

16.7 Surface Integrals 328 16.8 Stokes' Theorem 3

16.9 The Divergence Theorem 335 . Review 337

Problems Plus 343

0 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 345

17.1 Second-Order Linear Equations 345 ·

17.2 Nonhomogeneous Linear Equations 347

17.3 Applications Of Second-Order Differential Equations 350 17.4 Series Solutions 352 Review 354

. 0 A.PPENDIX 359 H Complex Numbers 359

CONTENTS D ix

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10 D PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.1 Curves Defined by Parametric Equations

1. X= t2 + t, y = t2 -t, -2 S t S 2 t -2 -1 0 l 2 X 2 0 0 2 6 y 6 2 0 0 2

3. x = cos2 t, y = 1 -sin t, 0 S t S 7r /2 t 0 7r/6 7r/3 7r/2 X 1 3/4 1/4 0 y 1 1/2 1-1 ~ 0.13 0

5. X = 3 -4t, y = 2 -3t

(a) t -1 0 1 2

X 7 3 -1 -5 y 5 2 -1 -4

(b) X= 3-4t =} 4t =-X+ 3 =} t =-~+~SO y = 2-3t = 2--~x + = 2 +~X-=} y =~X-

1. x = 1-e, v = t-2, -2 s t s 2

(a) t -2 -1 0 1 2

X -3 0 1 0 -3 y -4 -3 -2 -1 0

(b) y = t-2 =} t = y + 2, SO X= 1-t2 = 1-(y + 2? =} x = -(y + 2)2 + 1, or x -y2-4y -e 3, with -4 S y S 0 y y

(-3, 0) t=2 r=JI. 6

(-5, -4) 1=2

(7, S) t=-1

(3, 2) t=O

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(1, -2) t=O

2 D CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

9. X = Vt, y = 1 -t

(a) t 0 1 2 3 4

y 1 0 -1 -2 .:3

X 0 1 1.414 1.732 2

(b) x = Vt =} t = x2 =} y = 1-t = 1-x2. Since t ~ 0, x ~ 0.

So the curve is the right half of the parabola y = 1 -x2 •

1. (a) x =sin y =cos -1r ~ ~ 1r.

x2 + y2 = sin2 + cos2 = 1. For -1r ~ 0 ~ 0, we have

-1 ~ x ~ 0 and 0 ~ y ~ 1. For 0 <. 8 ~ 1r, we have 0 < x ~ 1 and 1 > y ~ 0. The graph is a semicircle.

smt x

13. (a) x = sint, y = csct, 0 < t < y = csct = ~ = !. For 0 < t < ~.whaveO < x < 1 andy > 1. Thus, the curve is the

portion of the hyperbola y = 1/x withy> 1.

15. (a) x = e2t =}. 2t = lnx =} t. = lnx. =t+1=~lnx+1.

17. (a) x =sinh t, y =cosh t '=} y2 -x2 = cosh2 t-sinh2 t = 1. Since y = cosh t ~ 1, we have the upper branch of the hyperbola y2 -x2 = 1.

y (0, I) ,f=O

(2, -3) t=4 (b) y

-1 0 I (b)

0 (b) y

(b) y

. 19. x = 3 + 2 cost, y = 1 + 2 sin t, 1r /2 ~ t ~ 31r /2. By Example 4 with r = 2, h = 3, and k = 1, the motion of the particle takes place on a circle centered at (3, 1) with a radius of2. As t goes from to 3;, the particle starts at the point (3, 3) and moves counterclockwise along the circle (x-3)2 + (y-1)2 = 4 to (3, -1) [one-half of a circle].

21.x=5sint,y=2cost =} t=~cost=~· sin2t+cos2t=1 =} (~r+(~r= particle takes place on an ellipse centered at (0, 0). As t goes from -1r to 511', the particle starts at the point (0, -2) and moves clockwise around the ellipse 3 times.

23. We must have 1 ~ x ~ 4 and 2 ~ y ~ 3. So the graph of the curve must be contained in the rectangle (1, 4] by [2, 3].

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SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS D 3 25. When t = -1,-(x,y) = (0, -1). As t increases to 0, x decreases to -1 andy increases to 0. As t increases from 0 to 1, x increases to 0 and y increases. to 1. As {increases 1, both x andy increase. Fort < -1, x is positive and decreasing and y is negative and increasing. We could achieve greater accuracy by estimating x-andy-values for selected values oft from the given graphs and plotting the points.

t=O t=-1

27. When t = 0 we see that x = 0 and y = 0, so the curve starts at the origin. As t increases from 0 the graphs show that y increases from 0 to 1 while x increases from 0 to 1, decreases to 0 and to -1, then increas.es back to 0, so we arrive at the point (0, 1). Sinlilarly, as t increases from to 1, y decreases from 1 to 0 while x repeats its pattern, and we arrive back at the origin. We could achieve greater accuracy by estimating x-and y-values for selected values oft from the given graphs and plotting the corresponding points.

29. Use y = t and x = t-2sin 7ft with at-interval of [-7r, 7r].

31. (a) x = X1 + (x2-x1)t, y = YI + (y2-Y1)t, 0 5 t 51. Clearly the curve passes through P1(X1, Yl) when t = 0 and through P2(x2, y2) when t = 1. For 0 <.t < 1, xis strictly betweenx1 and x2 andy is strictly between y1 and y2. For.

every value oft, x andy satisfy the relation y-Y1 = y2-YI (x-x1), which is the equation of the Line through X2 -Xl

H (x1, Yl) and P2(;,;2, y2).

Finally, any point (x, y) on that line; satisfies y-Yl = x-x1 ; if we call that common value t, then the given . Y2 -Y1 X2 -XI , parametric equations yield the point (x, y); ancf any (x, y) line between P1(x1, y1) and P2(x2, y2) yields a value of tin [0, 1]. So the given parametric equations exactly specify the line segment from P1 (x1, y1) to P2(x2, y2).

(b) x = -2 + (3-( -2)]t = -2 + 5t andy= 7 + ( -1-7)t = 7-Bt for 0 5 t 5 1.

3. The circle x2 + (y -1 )2 = 4 has center .(0, 1) and radius 2, so by Example_ 4 it can be represented by x = 2 cost, y = 1 + 2 sin t, 0 5 t 5 27r. This representation gives us the circle with a counterclocifwise orientation starting at (2, 1).

(a) To get a clockwise orientation, we could change the equations to x = 2 cost, y =. 1-2 sin t, 0 5 t 5 27r.

(b) To get three times around in the counterclockwise direction, we use the original equations ~ = 2 cost, y = 1 + 2 sin t with the domain expanded to 0 5_t 5 67r.

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4 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

(c) To start at (0, 3) using the original equations, we must have X1 = 0; that is, 2 cost= 0. Hence, t = So we use x = 2cost, y = 1 + 2sint, ~ ~ t ~ 3 ;.

Alternatively, if we want t to start at 0, we could change the equations of the curve. For example, we could use x = -2 sin t, y = 1 + 2 cost, 0 ~ t ~ 1r.

35. Big circle: It's centered at (2, 2) 'with a radius of 2, so by Example 4, parametric equations are x = 2 + 2cost, y = 2 + 2sint,

Small circles: They are centered at (1, 3) and (3, 3) with a radius ofO.l. By Example 4, parametric equations are and (left) (right) x;, 1 + O.lcost,

X = 3 + 0.1 COSt, y.= 3 + 0.1 sint, y = 3 + 0.1 sint,· 0 ~ t ~ 27T 0 ~ t ~ 27T

Semicircle: It's the lower half of a circle centered at (2, 2) with radius 1. By Example 4, parametric equations are

x = 2 + 1 cost, . y = 2 + 1 sin t, To get all four graphs on the same screen with a typical graphing calculator, we need to change the last t-interval to[O, 21r] in order to match the othersWe can do this by changing t to 0.5t. This change gives us the upper half. There are several ways to get the lower half-one is to change the"+" to a"-" in the y-assigr1ment, giving us x = 2 + 1 cos(0.5t), y = 2-1 sin(0.5t), · 0 ~ t ~ 27T

We get the entire curve y = x213 traversed in a left to right direction. Since x = t6 2': 0, we only get the right half of the .

x=y = 1

(c) x =e-st = (e-t)3 [so e-t = xl/3], y = e-2t = (e-t)2 = .(xl/3)2 = x2/3.

If t < 0, then x andy are both larger than 1. !ft > 0, then x andy are between 0 and 1. Since x > 0 and y > 0, the curve never quite reaches the origin.

39. The case ¥ < (} < 7T is illustrated. 0 has coordinates ( rO, r) as in Example 7, and Q has coordinates (rB, r + r cos(1r-B)) = (rB, 7"(1-cos B)) [sincecos(1r-a)= cos7rc;osa+sin7Tsina· =-cos a], so.P has coordinates (rB-rsin(1r-B),r(1-cos·B)) = (r(B :-sinB),r(l-cos B))

[sincesin(1r-a)= sin1rcosa-cos1r.sina =sin a]. Again we have the parametric equations x = r( (} -sin 8), y = r(l -cos B).

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(Parte 1 de 6)

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