Exercicios RESOLVIDOS James Stewart vol. 2 7ª ed - ingles, Exercícios de Economia Agroindustrial

Baixe Exercicios RESOLVIDOS James Stewart vol. 2 7ª ed - ingles e outras Exercícios em PDF para Economia Agroindustrial, somente na Docsity! - STUDENT SOLUTIONS MANUAL for STEWART' S DAN CLEGG • BARBARA FRANK .Student Solutions Manual for . MULTIVARIABLE . CALCULUS SEVENTH EDITION DAN CLEGG Palomar College BARBARA FRANK· Cape Fear Community College ~- . · BROOKS/COLE taa. CENGAG ELearning· Australia · Brazil · j apan · Korea • Mexico · Singapore • Spain · United Kingdom · United States D ABBREVIATIONS AND SYMBOLS CD cu D PDT HA I 1/D IP R VA CAS H 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 Cengo:~gc LeMning. All RigblS Reserved. May not be scunncd, copied, or duplic:Hed, or posted to a publicly ucce:ssible websire, in whole or jn part. v D · CONTENTS 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 11 INFINITE SEQUENCES AND SERIES 45 11.1 Sequences 45 11.2 Series 51 11.3 The Integral Test and Estimates of Sums 59 11 .4 The Comparison Tests 62 11.5 Alternating Series 65 11 .6 Absolute Convergence and the Ratio and Root Tests 68 11 .7 Strategy for Testing Series 72 11.8 Power Series 74 11.9 Representations of Functions as Power Series 78 11.10 Taylor and Maclaurin Series 83 11.1 1 Applications of Taylor Polynomials 90 Review 97 Problems Plus 105 0 12 VECTQRS AND THE GEOMETRY OF SPACE 111 12.1 Three-Dimensional Coordinate Systems 111 12.2 Vectors 114 12.3 The Dot Product 119 0 201 2 Ccogagc Learning. AU Rights Rcscn-cd. May not be scanned, copi~.:d, or duplicouccJ, or posletl to a publicly acc.:es.!~ible website, in whole or in p!U1.. vii viii o CONTENTS 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 Q 2012 Ccngage Learning. All Righ ts Rcscn·ed. May not be scanned, copied, or dU!>licaLcd. or posted to a publicly accessible website, in whole or in part. 2 D CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 9. X = Vt, y = 1 - t (a) t 0 1 2 3 4 X 0 1 1.414 1.732 2 y 1 0 - 1 - 2 .:....3 (b) x = Vt =} t = x 2 =} y = 1- t = 1 - x 2 . Since t ~ 0, x ~ 0. So the curve is the right half of the parabola y = 1 - x 2 • 11. (a) x = sin ~0, y =cos ~(:1, -1r ~ (:1 ~ 1r. x2 + y2 = sin2 ~8 + cos2 ~8 = 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. 13. (a) x = sint, y = csct, 0 < t < ~- y = csct = ~ = !. . . smt x For 0 < t < ~.we haveO < x < 1 andy > 1. Thus, the curve is the portion of the hyperbola y = 1/x withy> 1. 15. (a) x = e2t =}. 2t = ln x =} t. = ~ ln x. y = 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 - x 2 = 1. y (0, I) ,f = O X (2, -3) t=4 (b) y -1 0 I X (b) '\,,, 0 X (b) y (b) y 0 X . 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=2 cost =} sint=~, cost=~· sin2 t+cos2 t =1 =} (~r + (~r = l.The motionofthe 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]. © 2012 Ccngogc Lcnming. AJI Rights RcsclVcd. May not bo scwmed, copied, or duplicated. or posted to a publicly accessible website, in whole or in part. 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 (-1,0) X increases to 0. As t increases from 0 to 1, x increases to 0 and y increases. to 1. As { increases beyo~d 1, both x and y increase. Fort < - 1, x is positive and decreasing and y is negative and increasing. We could achieve greater accuracy by estimating x- and y-values for selected values oft from the given graphs and plotting the correspo~ding 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 to~. 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 y X 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 of t 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 5 1. Clearly the curve passes through P1(X1, Yl) when t = 0 and through P2(x2 , y2) when t = 1. For 0 <.t < 1, x is strictly between x1 and x2 and y 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 - x 1 ; if we call that common value t , then the given . Y2 - Y1 X2 - X I , parametric equations yield the point (x, y); ancf any (x, y) on .th~ line between P1(x1, y1) and P2(x2, y2) yields a value of t in [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. 33. The circle x 2 + (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. ® 2012 Ccngoge Lc3ming. All Rights Reserved. Moy nol be SC4uncd, copied, or duplicated, or posted ton publicly ncc<:ssiblc website, in " i tolc or in par1. 4 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES (c) To start at (0, 3) us ing the original equations, we must have X1 = 0; that is, 2 cos t = 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 cos t , 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 + 2 sin t, 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.l cost, X = 3 + 0.1 COSt, y .= 3 + 0.1 sin t, y = 3 + 0.1 sin t,· 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 others." We 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 37. (a)x = t 3 :::? t = x 113 ,so y=t2 = x213 • (b) x = t6 :::?·. t = x 116 , so y= t4 = x416 = x213 • We get the entire curve y = x213 traversed in a left to right direction. Since x = t 6 2': 0, we only get the right half of the ' . curve y = x213 • y x= t~ y = 11 (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 and y are both larger than 1. !f t > 0, then x and y 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 + sin7T sin a· = -cos a ], so.P has coordinates (rB- rsin(1r - B),r(1- cos·B)) = (r(B :- sin B),r(l - cos B)) [sincesin(1r - a) = sin1rcosa - cos1r.sin a = sin a ]. Again we have the parametric equations x = r( (} - sin 8), y = r(l - cos B). y 0 X y © 2012 CC=nguge Learning. All Rights Res<:rvcd. May not be scrumcd. copied. or duplicated. or posted too publ icly accessible website. in wbolc or in port. X SECTION 10.2 CALCULUS WITH PARAMETRIC-CURVES 0 7 itself at the origin and there are loops above and b~low the x-axis. In general, the figures have n - 1 points of intersection, all of which are on the y-axis, and a total of n closed loops. 1.1 -1.1 - 1.1 a=b = l t1 =3 n =2 n=l - 2.1 \- ---+- (a,b) =( ~.2)' (a, b) = (2, I) -3.1 - 2.1 -3.1 n=2 n =3 10.2 Calculus with Parametric Curves 1. x = t sin t, y = t 2 + t ::} dy . dx . dy dy I dt 2t + 1 -d = 2t + 1, -d = t cost+ sm t , and -d = d l d = . . t t x x t t cos t + sm t z . 3 dy = - 3t2 dx = 4- 2t d dy = dyldt = -3tz -Wh - 1 3. x = 1 + 4t - t , y = 2 - t ; t = 1. dt ' dt , an dx dx I dt 4 - 2t . en t ' (a, b) = (2, 3) (a, b) = (3, 2) ( x, y) = ( 4, 1) and dy I dx = - ~, so an equa.tion of the tangent to the curve at the point corresponding to t = 1 is y- 1 = -~(x- 4), oq/ = -~x + 7. 5. x = t cos t, y = t sin t; t = 1r. dy . dx . ' dy dy I dt t cos t + sin t -d = tcost+smt, -d = t(-smt)+cost,and-d = d ld = . . t t x x t -tsm t+cost When t = 1r, (x, y) = ( -1r, 0) and dy I dx. = - 1r I( -1) = 1r, so an equation of the tangent to U1e curve at the point corresponding to t = 1r is y - 0 = 1r[x - ( -1r) ) , or y = 1rx + 1r2 • 7. (a) x = 1 + ln t, y = t2 + 2; (1, 3). dy dx 1 dy dyldt 2t 2 dt = 2t, dt = t' and dx = dxldt = 1l t = 2t · At(1•3), x=l+ ln t = 1 ::} ln t = O =? t =1 an~ : = 2,so an equationofthetangentis y-3=2(x - l), ory = 2x + 1. (b)x = 1 + ln t =? ln t=x -1 =? t=e"'-l,soy=t2.+2=(e"'-1 ) 2 + 2 = e2" - 2 +2,andy'=e2x-z . 2. At (1, 3), y' = e2(l)- z · 2 = 2, so an equation of the tangent is y - 3 = 2(x - 1), or y = 2x + 1. 9. x = 6sin t, y = t2+t; (0,0). dy dy 1 dt 2t + 1 . - = d ld = --. The potnt (0, 0) corresponds tot = 0, so the dx x t 6cost slope of the tangent at that point is· lJ. An equation of the tangent is therefore y - 0 = i(x- 0), or y = ix. © 2012 Ccnt;!lgc Learning. All Rights Rcscr\'Cd. May not be scnnncd, copied, orduplicotcd. or posted to n publicly ncccssiblc wcbsifc, in whole or in part. . 8 D CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 2 2 dy _ dy I dt _ 2t + 1 _ 1 .!_ 11. x = t + 1• Y = t + t '* dx - dx fdt - 2t - + 2t !!:._ (dy) dt dx dx/ dt The curve is CU when ~~ > 0, that is, when t < 0. dy dyfdt - te- t + e- t = e-t (1- t) = e- 2t( 1 _ t) 13. x = et, y = te-t ::::} dx = dxjdt = et et '* d(~) . d 2 y = dt (h; = e-2t(- 1) + (1 - t)( - 2e-2t) = e - 2t(-1 - 2 + 2t) = e-3t(2t- 3). The curve is CU when dx2 dx/dt et et d 2 y 0 th . h t 3 dx2 > , at ts, w en > 2 . 15. x = 2sint, y = 3cost, 0 < t < 21r. . i_OO(dy - - 3 2 dy _ dy/ dt _ -3sint _ -~ d2 y _ dt dx _ -2sec t _ -~ 3 dx- dxjdt - 2cost - 2 tant, so dx2 - dxjdt - 2cos t - 4 sec t. The curve is CU when sec3 t < 0 ::::} sect < 0 ::::} cost < 0 ::::} i < t < 3; . . d d 17. X = t3 - 3t, y = t 2 - 3. dy = 2t, SO dy = 0 ¢> t = 0 ¢> t ' t (x, y) = (0, - 3). dxd = 3t2 - 3 = 3(t + 1)(t - 1), so d:x = 0 <=> t . t t = - 1 or 1 <=> (x, y) = (2, -2) or ( -2, -2). The curve has a horizontal tangent at (0, - 3) and vertical tangents at (2, - 2) and ( -2, -2). 19. x = cos 9, y = cos 39. The whole curve is traced out for 0 ~ 9 ~ 1r. ~~ = - 3'sin 39, so ~~ = 0 <=> sin 39 = 0 <=> 38 = 0, 1r, 21r, or 37r <=> 'e =0, ~ . 23.,.,or1r <=> (x,y) = (1,1), (~ , -1), (-~ , 1) , or(-1,-1 ). dx . dx d9 = - sLD 8, so dO = 0 <=> sin 8 = 0 <=> 8 = 0 or 1r <=> dy d.x (x,y_) = (1, 1) or (-1,-1). Both dO and dO equal 0 when 8 = 0 and 1r. 2 (-t.1) 9=21T/3 (- 1,-1) (J= 1T -2 (1,1) 9 = 0 .., fi d th I I 9 0 fi d li dy lim - 3 sin 38 11 lim _g cos 38 9 hi h . th I h 0 10 n e s ope w ten = , we n m dx = . n = ,.. = , w c 1s e same s ope w en = 1r. o~o o~o - sm" O-•O -cos" Thus, the curve has horizontal tangents at ( ~ , -1) and (- t, 1), and there are no vertical tangents. © 2012 C..ngage Learning. All Righi!< Rcsen'Cd. Moy not be scnnnod, copiod, or duplicotod. or posted to a publicly accessible website, in whole or in port. SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES 0 9 21. From the graph, it appears that the rightmost point on the curve :J: = t - t 6 , y = et is about (0.6, 2). To find the exact coordinates, we find the value oft for which the graph has a vertical tangent, that is, 0 = dxl dt = 1 - 6t5 {::} t = 11 W. Hence, the rightmost point is 23. We graph the curve x = t 4 - 2t3 - 2e, y = t 3 - tin ~he viewing rectangle [-2, 1.1] by [- 0.5, 0.5]. This rectangle corresponds approximately tot E [ -1, 0.8] . o.s - 0.5 We estimate that the curve has horizontal tangents at about ( - 1, - 0.4) and ( - 0:11, 0.39) and vertica·L tangents at dy dyl dt· 3t2- 1 . about (0, 0) and ( -0.19, 0.37). We calculate -d = dxl d = 4 3 6 2 4 . The honzontal tangents occur when X t t - t - t · dy I dt = 3t2 - 1 = 0 1=} t = ± ~, so both horizontal tangents are shown in our graph. The vertical tange~ts occur when dxldt = 2t(2t2 - 3t- 2) = 0 {::} 2t(2t + 1)(t - . 2) = 0 {::} t = 0, -~or 2. It seems that we have missed one vertical tangent, and indeed if we plot the curve on the t-interval [-1.2, 2.2] we see that there is another vertical tangent at ( -8, 6). 25. x = cos t , y = sin tcos t . dxl dt = - sint, dyl dt = - sin2 t + cos2 t = cos 2t. (x,y )= (O, O) {::} cost = .O. {::} tisanoddmultipleoflf.Whent=~, dxl dt = - 1 and dyl dt = - 1, so dyl dx = 1. When t = 3; , dxl dt = 1 and dy I dt = - 1. So dy I dx = - 1. Thus, y = x andy = - x are both tangent to the curve at (0, 0). 27. x = rB - d sinB, y = r - dcosB. dx dy , . dy d sinB (a) dB = r- dcosll, dB = dsme, so dx = r _ dcosB ' (b) IfO < d < r, then jdcosllj:::; d <. r , so r- dcosfi ;::: r- d > 0. This shows that dxldll never vanishes, ·so the trochoid can have no vertical tangent if d < r . 29. X = 2t3' y = 1 + 4t - t2 ---'- dy - dy I dt = 4 - 2t N 1 . dy = 1 ~ 4 - 2t = 1. ......,._ _,.. ·dx - dxl dt 6t 2 • ow so ve dx .,_,. 6t2 .....,. 6t 2 + 2t - 4 = 0 {::} 2(3t - 2)(t + 1) = 0 {::} t = ~ or t := - 1. 1ft = ~ . the point is(~~ ' 2:) , and ift = -1, the point is ( -2, -4). © 2012 Ccngoge Learning. All Rights Reserved. Moy not be scllnnCd, copied, or duplicated, or pcstcd to a publicly occcssiblc website, in whole or in port. 12 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES (b) We use the CAS to find the derivatives dx/dt and dy/dt, and then use Theorem 6 to find the arc lengtll. Recent versions of Maple express the integratJ;" J(dxjdt) 2 + (dyjdt) 2 dt as 88E(2 V2 i), where E (x) is the elliptic i~tegral {1 v/1- x2t2 Jo v'f=t2 dt and i is the imaginary number yCI. I Some earlier versions of Maple (a5 well as Mathematica) cannot do the integral exactly, so we use the command evalf (Int (sqrt (diff (x, t) - 2+diff (y, t) -2), t=O .. 4*Pi)); to estimate the length, and find that the arc length is approximately 294.03. Derive's Para_ar c_length function in the utility fi le Int_apps simplifies the integral to 11 }~4,. J -4 cost cos(l~t ) - 4 sin t sin(l~t) + 5 dt. 57. x=tsint, y=tcost, O~t~Tr/2. dxjdt=tcost +sintanddyjdt= -tsint +cost,so (dx/dt? + (dyjdt)2 = t 2 cos2 t + 2tsin t cost+ sin2 t + e sin2 t- 2tsin t cos t+ cos2 t = t 2 (cos2 t + sin2 t) +.sin2 t + cos2 t = t2 + 1 59. x = 1 + tet, y = (t2 + l) et, 0 ~ t ~ 1. (~~ )2 + (!fltf = (te1 + et)2 + [(e + I )et + et(2tW = [et(t + 1}f + [et (e + 2t + 1W = e2t(t + 1? + e2t(t + 1)4 = e21 (t + 1)2[1 + (t + 1)2 ), so ( 3 (u 4) = 2Tr J 4 T Ju ( fg du) [ u = Dt2 + 4, t 2 = (u- 4)/9, J du = 18t dt, so t dt = · f.J tlu - 2!:. [~u5/2 - §.u3/2] 13 - 2!:. • .L [3u5/2 - 20u3/2] 13 - 81 5 3 4 - 81 15 4 = 1;~5 [(3. 132 v'13- 20. 13 VI3) - (3 ·.32 - 20. 8)] = 1;~5 (247 VI3 + 64) 63. x = acos3 8, y =a sin3 8, 0 ~ B ~ I· (~~)2 + (;¥tt)2 = ( - 3acos2 8 sinG?+ (3a sin 2 8 cos 0)2 = 9a2 sin2 (I cos2 0. S = .{0" 12 2Tr · a sin3 () · 3asin 0 cos() d() = 61ra2 J0rr/ 2 sin~ 0 cos 0 dO = ~1ra2 [sin5 0] ~12 = ~1ra2 65. X= 3t2, y = 2t3, 0 ~ t ~ 5 . ::} C~l + (~/} = (6t? + (6t2)2 = 36t2(1 + t2) ::} s = .r; 21rx J(dx/dt)2 + (dyJdt)2 dt = J; 27r(3e)6t vt + t 2 d.t = l81r J; t2J1 + t2 2tdt [ u = 1 +t2,] = 18Trj·2G(u3/2- u1/2) du = 18Tr ['du5/2- ~u3/2] 26 du = 2t dt 1 5 3 1 = 187r [ ( ~ . 676 J26 - ~ . 26 J26) - ( ~ - ~)] = ¥ 7r ( 949 J26 + 1) ® 201 1 2 Ccngngc Learning. All ltig.hts Rcscn·cd. M.3.y not be scarmcd. copied, orduplic~Jlcd+ or ~stcd to o publicly accessible website. in who le or in part. SECTION 10.3 POLAR COORDINATES 0 13 67. If f' is continuous and j' (t) i= 0 for a$ t $ b,. then either!' (t) > 0 for all tin (a, b) or j' (t) < 0 for all t in [a, bj. Thus, f . . ' is monotonic (in fact, strictly increasing or stri<:tly decreasing) on (a, b]. It follows that f has an inver;se. Set F =g o f- 1 , that is, define F by F(x) = g(r1 (x)). Th~n x = f(t) ~ r 1 (x) = t , soy= g(t) = g(f-1 (x)) = F(x). - 1 ( dy) d<P _ d -1 ( dy ) _ 1 [ d ( dy ) ] dy dy 1 dt i1 69· (a)¢> = tan dx ~ dt - -dt tan dx - 1 + (dyjdx)2 dt dx · But dx = dx/dt = ± ~ d ( dy ) d ( if ) fix - xif d¢> 1 ( fl± - xfl ) xy - xif ' . . · dt di = dt ;; = xz ~ dt = 1 + (iJ/x) 2 x2 = xz +iF. Usmg the Cham Rule, and tne "a·cttllat s- rot (d:r;)2+(!!11..)2dt d .• - /(d:r;)2+(!!11..)2 ('2+ •2)1/2 b h ! i J 0 dt dt ~ 'Jt - y dt dt = x y , we ave t at d¢> _ d¢>/dt _ (xii- xfl ) 1 _ xii - xiJ . _- I d¢> 1-1 xy- xy 1- l±ii - xiJI ds - ds/dt - :i;2 + y2 (±2 + y2)1/2 - (~2 + 1;2)3/2 · So"'- - ds - (±2 + y2)3/2 - (±2 + y2)3/2 · d !( ) . 1 .. 0 d . dy .. d2y (b) x = x_an y = x ~ x = 'x = an y = dx' y = dx2' - II· (d~y/dx2 ) - 0 . (dyjcb:) i - ld2 y/ dx2 1 So"' - (1 + (dyjdx)2 )312 - (1 + (dyjdx)2] 312 • 71. X = 8 - sin8 ~ :i; = 1----' cosB -~ X = sin8, and y = 1 - cos8 ~ iJ =sine ~ y = cos8. Therefore, icosB - cos2 e- sin2 el icosB - (cos2 8 + sin2 8)1 JcosB - II . "' = = . 2 = . The top of the arch 1s [(1 - cos B)2 + sin2 B)S/2 (1 - 2 cos e + cos2 e + sm 8)312 (2 - 2 cos 8)3/2 characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when B = (2n - l )1r, d b · B · ti . fi Jcos 1r -11 1-1- 11 1 sotaken=1 an su stttute =1rmto 1e ex~resston or ~~;: ~t= (2 _ 2 cos1r)3/ 2 = 12 _ 2 (- 1 ))3/ 2 =4· 73. The coordinates of T are .(r cosB, r sin B) . Since TP was unwound from arc T A, T P has length rB. Also LPTQ = LPT R - LQT R = t 1r - 8, soP has coordinates x = r cos B + r8 cos(~1r ~B) = r(cos B + B sin 8), y = rsin B - rBsin(~1r - 8) = r(sin8- 8cos9). 10.3 Polar Coordinates X 1. (a) (2, i) By adding 21r to i, we obtain the point ( 2, 7;). The direction opposite i is 43.,., so (- 2, ~) is a point that satisfies the r < 0 · requirement. © 2012 Cengage ~ing. All Rights Reserved. May not be scanned, copied, or duplicated, or posted too publicly accessible wcbsitc, in whole or in part. 14 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES (b) (1, - 3;) 0 (c) ( - 1, 2J) 0 3. (a) 'TT (l, 'TT) 0 (b) (c) _ 311' 4 ,. 2 r > 0: (1, - 3; + 21r) = (1, 5;) r < 0: (- 1,-3r+ 7r) = (-1,~) r > 0: (- (- 1) , 1J +1r) = (1, 3;) r < 0: (-1, ~ + 21r) = (- 1, 52,.) (-t,f) X= 1COS7r = 1(- 1) = -1 and y = 1 sin 1r = 1(0) = 0 give us the Cartesian coordin~tes ( -1, 0). x = 2cos(- 23"' ) = 2(-~) = - 1 and y = 2sin( - 2;) = 2( -~) ~ ~VS give us ( -1, -V3). x = -2cos 3; = -2( -4) = J2 and y = - 2sin 34" = -~( 4) = - J2 gives us ( J2, -J2). 5. (a) x = 2 andy = - 2 => r = J22 + (-2)2 = 2 J2 and 8 = tan- 1 ( :n := -~ . Since (2, -2) is in the fourth quadrant, the polar coordinates are (i) (2 J2, 7;) and (ii) (-2 J2, 3;). (b) x = - 1 andy = V3 => r = J< -1)2 + ( v'3)2 = 2 and 8 = tan~1 ( :1) = 2:;. Since_( -1, VS) is in the second quadrant, the polar coordinates are (i) (2, ~)and (ii) ( - 2, 6;). '· © 2012 Cengnge Learning. All Rights Reserved. May no1 be scnnned, copied, or dupll<otcd, or posted to a publicly accessible website, in whole or in pon. 39. r = 1--2sin8 41. r 2 = 9sin28 43. r = 2 + sin 38 45. r = 1 + 2cos28 3 -3 r 3 2 2..- 0 2..- ---r--~------~----+-t-~ 11 ..- 0 3..- 7..- 6 6 6 IJ SECTION 10.3 POLAR COORDINATES 0 17 0=!!. __________ ..• ----- 6 (3, ..-j6) 47. For(}=:= 0, 7T, and 27T, r has its minimum ;value of about 0.5. For 0 =~and 32", r attains its maximum va lue of2. We see that the graph has a similar shape for 0 ~ (} ~ 7T and 7T ~ 8 ~ 27T. . r 2 ® 2012 c~ngage Learning. All Rights 'Rescn· .. -d. May not be scanned. copic.•tl, or dupticntcd, or posted lo a publicly uccc:s.siblc website, ;n whole or in part. 18 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 49. x = r cos 8 = ( 4 + 2 sec 8) cos 8 = 4 cos 8 + 2. Now, r -+ oo => (4 + 2sec8)-+ 00 => 8-+ (~) - or 8 -+ e2"' )+ [since we need only consider 0 $ 8 < 271' ], so lim x = lim ( 4 cos 8 + 2) = 2. Also, . r -oo 8 -+-rr/2- T -+ -oo => ( 4 + 2 sec B) -+ -oo => 8 -+ ( ~) + or 8 -+ ( 3;)-, so lim x = lim ( 4 cos 8 + 2) = 2. Therefore, lim x = 2 => x = 2 is a vertical asymptote. r--oo 8--+rr/2+ T"--+±oo (6,0) 51 . To show that x = 1 is an asymptote we must prove lim x = 1. r - ±oo x= l x = (r) cosO= (sin 8 tan 8) cos 8 = sin2 ().Now; r -+ oo => sinO tan8-+ oo => 8 -+ (f) - ,so lim x= lim sin2 8 = l.Aiso,r-+-oo => sinBtanB-->-oo => r --+oo 0-+'Tr/2- . e-+ (~) +,so lim X = lim sin2 {;I,;, 1. Therefore, lim X= 1 . => X= 1 is . r ...... - oo 0--+1fj 2+ ,·-±oo a vertical asymptote. Also notice that X = sin2 e ~ 0 for a ll e, and X = sin2 {;I $ 1 for all e. And Xi= 1, since the curve is not defined at odd multiples of ~. Therefore, the curve lies entirely within the vertical strip 0 $ x < 1. 53. (a) We see that the curve r = 1 + csin 8 crosses itself at the origin, where r = 0 (in fact the inner loop corresponds to negative r-values,) so we solve the equation of the lima~on for r = 0 <=> c sin 8 = - 1 <=> sin 8 = - 1/ c. Now if lei < 1, then this equation has no solution and hence there is no inner loop. But if c < - 1, then on the interval (0, 27T) the equation has the two solutions 8 = sin-1 ( -1/c) and 8 = 1r- sin- 1 ( - 1/c), and if c > 1, the solutions are (} = 1r + sin - 1 (1/c) and ()= 21r- sin- 1 (1(c). Tn each case, r < 0 for (} between the two solutions, indicating a loop. (b) For 0 < c < 1, the dimple (if it exists) is characterized by the fact that y has a local maximum at 8 = 3; . So we determ.ine for what c-valu~s d 2 ~ is negative at(} = 32~, since by the Second Derivative Test tllis indicates a maximum: ~ . . dy d2 y y = rsin8 =sinO + c sin2 (} => dB = cos8 + 2csinfJ cos8 = cosfJ + csin 28 =>. - ., = - sin8 + 2ccos 28. dB~ At 8 = 32,.., this is equal to - ( - 1) + 2c( -1) = 1 - 2c, which is negative only for c > %. A similar argument shows that for - 1 < c < 0, y only has a local minimum at e = ~ (indicating a dimple) for c < -~ . . 55. r = 2 sin (} => x = r cos(} = 2 sin fJ cos 8 = sin 28, y = 1· sin 8 = 2 sin2 B => dy = dyjd(J = 2 · 2 sin fJ cosO= sin 28 = tan 2B dx dxjdB cos2fJ·2 cos28 7T dy ( 7r) 7T rr; · When 8 = 6, d.'t =tan 2 · 6 =tan 3 = v 3. [Another method: Use Equation 3.] 57. r= l / 8 => x=rcos8 = (cosfJ) j B, y=rsin8=(sinfJ) / 8 => dy dyjd8 sin.O( - 1/ 82 ) + (1/ 8) cos 8 82 - sin (}+(} cosO dx = dxj dO = cos8(- 1/ fJ2 ) - (1/ 8) sinB · (}2 = - cosO - Osin B .dy - 0 + 7r(- 1) - 'Tr When 8 = 1r, dx = -(-1) -1r(O) = l = -1r. ® 2012 Cengoge loaming. All Rights Resef'-.d. Moy not be: ~ned. copl<-d, ordupliuted, or po5led to a publicly accessible web•itc, in whole or in part. ' SECTION 10.~ POLAR COORDINATES 0 19 59. r = cos 2() => x = r cos() = cos 2() cos(}, y = r sin() = cos 2(} !lin(} => 1f dy When() = 4, dx dy dy / df) cos 2(} cos(}+ sin(} ( -2 sin 2()) dx = dx/ d() = cos 2() (- sin(})+ cos() ( -2 sin 2()) -,0,!-( J2---:2 /:---27) +_('-:-v-2-:::2 /,.--!2 )~( -_2;,_) - _- J2_2 - 1 O(-J2/ 2) + (J2/ 2}(-2) - -J2 - . 61 . r = 3cos0 => x = rcosO = 3cos0 cosO,_ y = rsinO = 3cos0 sinO => , ~ = -3sin2 9 +3cos2 ,() = 3cos20 = 0 => 2() = ~ or 3; <=> 0 = %or 3;. So the tangent is horizontal at ( ~, %) and ( - -32, 3;) [same as ( -32 , -:a:)] . ~~ = - 6 sin() cos~= -3 sin 20 = 0 => 29 = 0 or 1r <=> 0 = 0 or~· So the tangent is vertical at (3, 0) and (o, ~). 63. r = 1 + cos 0 · => x = r cos 9 = cos() (1 +cos 9), y = r sin 0 = sin 0 (1 + cos 9) => ·.~ = (1 +cosO) cosO- sin2 0 = 2cos2 0 + cos9 - 1 = (2cos9 - 1) (cos(J + 1) = 0 => cos()=~ or - 1 => () = 'i· 1r, or 5; => horizontal tangent at(~ , 'i), (0, tr), and(~ , 5; ) . ~~ = - ( 1 + cos 8) sin (} - cos ()sin() = - sin (} ( 1 + 2 cos 9) = 0 => sin 8 = 0 or cos 9 = - ~ => () = 0, 1r, 2;, or 4; => vertical tangent ~t (2, 0), (!, 2;), and (!, -t;). Note that the ~gent is horizontal, not vertical when 8 = 7r, since e : j :: = 0. 65. r = a sin() + b cos 9 => r 2 = ar sin() + br cos 8 '=> x2 + y2 = ay + bx => x 2 - bx + (~b? + y2 - ay + ( ~a)2 = (~b)2 + (ta? => (x - ~b)2 + (y - ~a? = Ha2 + b2 ) , and this is a circle with center (!b, ~a) and radius. ~.Ja2 + b2 • 67. r = 1 + 2 sin(8 / 2): The parameter intei-val is [0, 47r]. 69. r = e•inB - 2cos(40). The parameter interval is [0, 21l'j. 71. r = 1 + cos999 8. The parameter interval is [0, 2tr). 1.1 / - 1.1 © 2012 <:engage Learning. All Rights Reserved. May not be SCIInncd, copied, or duplicated, or posted too publicly accessible website, in whole or in port. 22 D CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 1 ( 3 9 1 . 9] 21r 1 ( ) 3 = 2 2 + 2o sm 10 o = 2 3'1T = 27T -1.4 17. The curve passes through the pole when r = 0 ==?- 4 cos 30 = 0 ==?- cos 39 = 0 ==?- 39 = ~ + 1rn =?- 9 = % +·in. The part of the shaded loop above the polar axis is traced out for 9 = 0 to 9 = 7T /6,' so we'll use -7T / 6 and 7T /6 as our limits of integration. r = 4 cos 30 19. r = 0 ==?- sin49 = 0 ==?- 48 = 1rn ==?- 9 = ~n. r=sin 40 r'4 . r'4 r '4 A = Jo Hsin48) 2 dfJ = t Jo sin 2 48d8 = t Jo t(1-cos89)d8 0 . . 0 0 1 (9 1 · 89] "' I 4 1 (,.) . 1 = 4 - B SlD 0 = 4 4 = Tii7T 21. This is a lima~on, with inner loop traced r = 1 + 2 sin 8 (rect.) 3 out between 8 = 7; and 1 ~" [found by solving r = 0]. 17T ------· 0=6 . ··--•• , ll l'l' 0= 6 [ 31r/2 . [37f/2 [ 37r/2 A = 2 !(1 + 2sin8? d8 = (1 + 4sin8 + 4sin2 8) d8 = [1 +4sin8 + 4· t{1- cos29)] d9 71r /6 77r /6 71r /6 . . = [9 - 4cos0 + 20- ~n28J~:~~ = (9;)- C; + 2 v'3 - 4) = 7T - ¥ 23. 2 cos B = 1 => cos 8 = ~ => 8 = :g. or 5; . A = 2 f0,.13 ~[(2 cos8)2 - 12) dB= J0" 13 (4cos2 8 - l) d8 = J; ; a { 4 [ ~ (1 + cos 28)] - 1} d8 = J; /S ( 1 + 2 cos 28) d8 = [8 + sin 28)~13 = i + 4 r = 2cos0 @) 2012 Cengogc Lcaming. All Rights Reserved. Moy not be oamned, copied, or duplicated, or posted to a publicly accessible website, in whole or in prut. SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES 0 23 25. To find the area inside the l~miniscate r2 = 8 cos 29 and outside the circle r = 2, we first note that the two curves intersect when r2 = 8 cos 29 and ~ = 2, . I that is, when cos2B = ~-For -1r < B ::; 1r, cos2B = ~ {::} 2B = ±7r/ 3 or ±511' /3 {::} B = ±1r / 6 or ±57r / 6. The figure shows that the desired area is 4 times the area between the curves from 0 to 1r /6. Thus, A = 4 J0" /. 6 [~(8 cos2B)- ~(2)2] dB = 8 J0" 16 (2 cos 2B - 1) dB [ ] rr~ . . =8 sin2B - B 0 .=8(VS/2-7r/6) =4\1'3-411'/3 27. 3cosB=1+cosB {::} cosB=~ => B=t or-i. A= 2 f0" 13 ~ [(3 cos 8?- (1 +cos 0)2 ) dB = J0"'13(8 cos2 8-'- 2 cos B- 1) dB = f0" 13[4(1 +cos 2B) - 2 cos B - 1) dB = J0.,.13 (3 + 4cos28- 2cosB) dB= [3B + 2sin2B- 2sinf!] ~13 = 1T +v'3-.J3=1T 29. v'3cosB = sinB => J3 = sinBB => tanB = J3 => B = i· . cos A - f "' 13 1 (sin B)2 dB + 1'-"' 12 1 (J3 cos B)2 d() -Jo 2 hr/ 3 . 2 - f"' / 3 1. 1(1 - cos28) dB+ f "/2 1 · 3 · 1(1 + cos2B) dB - Jo 2 2 rr/3 2 2 - 1 [B - 1 · 28]"13 +!! [B +.! · 28]1f12 - 4 2 sm o 4 2 sm 7f/3 = t [ ( i - 4) - o] + ~ [ (~ + o)- ( f + 4)] _ .l!.. _ _il + .!!: _ 1fl _ 5rr _ fl - 12 16 . 8 16 - 24 4 31. sin 2B = cos 2B => sin 2 B = 1 => tan 28 = 1 => 28 = ~ => cos2B B=. f => A= 8 · 2 f0"/8'~ sin 2 2B dB= 8 f0"'18 H1- cos 48) dB = 4[9- t sin48]~18 = 4(f - i ·1) = ~ - 1 33. sin2B = cos2B => tan2B = 1 => 2B = ~ => (} = f A= 4 f0"' 18 ~sin 2B dB [since r 2 = sin 2B) = f0"'18 2 sin2B dB= [-;- cos 28] ; 18 = -~ v'2- ( - 1) = 1- ~ Y2. r=2 8 =1!. I 3 ,l' r = J3cos8 .... ···B=i © 2012 Ccngage learning. All Riglus Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessibk website, in whole or in part. · 24 • 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 35. The darker shaded region (from 0 = 0 to 0 = 27T / 3) represents ~ of the desired area plus ~ of the area of the inner loop. From this area, we' ll subtract~ of the area of the inner loop (the lighter shaded region from 0 = 27r /3 to 0 = 7r), and then double that difference to obtain the desired ·area. A = 2 [J;.,.13 H~ +~os0) 2 dO- g.,.13 H~ +cos 0) 2 dB] = £0 2 .,. 13 (~+cosO+ cos2 0) dB- f2: 13 (~ + cos 0"+ cos2 0) dB . . = J0211'/S [~+cos 0 + ~(1 + cos28)) dO - I2:/3 a + cosO+ ~(1 + cos28)) d(J = [~ + sin 8 + ~ + sin 20] 2 " 13 _ [~ + sin 8 + ~ + sin 20 1.,. 4 2 4 0 4 2 4 211'/ 3 - (1!: + .il + 1!: - .:.::2) - (!!: + !!:) + ( 1!: + .il + 1!: - .:.::2) -6 2 3 8 4 2 6 2 3 8 = ~ + ~ J3 = H 7T+3v'3) 37. The pole is a point of intersection. 1 + sin 8 = 3 sin 0 => 1 = 2 sin (J => sin 0 = ~ => (J =~ or 5;. The other two ~oints of intersec~ion ~e ( ~, i) and ( f, 56"). 39. 2 sin 20 = 1 =? sin 20 = ~ =? 2o = 1!: &.,. 13.,. or 11" 6' 6 ' 6 ' 6 ° By symmetry, the eight points of intersection are given by (1 8) h (} _ 11' 5,. 13rr d 1771" d , , w ere - 12, 12, 12, an 12, an ( - 1, 8), where 0 = i;, 1t;, 1: 2"; and 2[;. [There are many ways to describe these points.] 41. The pole is a point of intersection. sin (J = sin 20 = 2 sin(} cos 0 <=> sinO (1 - 2 cos 9) = 0 <=> sinO= 0 or coslJ = t => 8 = 0, 7T, t. or - t => the other intersectio~ points" are ( ~. t ) and ( ~, 2; ) [by symmen:']. r=l + sinO r = 2sin20 r =sin20 © 2012 Ccngage Learning. All Rights Reserved. May not be scotuled, copied, or dupHcuted, or posted to a publicly accessible website, in whole or in part. 5. (x + 2)2 = 8 (y- 3). 4p = 8, sop= 2. The vertex is ( - 2, 3), the focus is ( -2, 5), and the directrix is y = 1. y -------------------------- __ ,y_~_!_ X SECTION 10.5 CONIC SECTIONS 0 27 7. y2 + 2y + 12x + 25 = 0 =l> y2 + 2y + 1 = - 12x- 24 =;.. (y + 1)2 = - 12(x + 2). 4p = - 12, sop= -3. The vertex is ( - 2, -1), the focus is ( -5, - 1), and the directrix is x = 1. 0 : ~ X I jx=l 9. The equation has the form y2 = 4px, where p < 0. Since the parabola passes through (- 1, 1), we have 12 = 4p( - 1), so 4p = - 1 and an equation is y 2 = - x or x = - y 2 • 4p = - 1, sop = -i and the focus is ( -;i , 0) while the directrix . 1 ISX = 4· y 2 ellipse is centered at {0, 0), with vertices at (0, ±2). The foci are (0, ±v'2). -..fi -2 x2 Y2 13. x 2 + 9y2 = 9 <=> 9 + l = 1 =l> a = v'9 = 3, 15. 9x2 - 18x + 4y2 = 27 <=> b = Vi.= 1, c = ~ = v'9=I = J8 = 2v'2. The ellipse is centered at {0, 0), with vertices (± 3, 0). The foci are (±2v'2, 0). - 3 y I 0 - I 3 X 9(x2 - 2x + 1) + 4y2 = 27 + 9 <=> 9(x- 1)2 + 4y2 = 36 <=> (x- 1)2 y2 4 + g- = 1 =} a = 3, b = 2, c = J5 =:- center (1, 0), vertices (1 , ± 3}, foci (1, ±J5) y (1, 3) 3 X (1,- 3) 2 2 17. The center is (0, 0), a= 3, and b = 2, so an equation is ~ + Y 9 = 1. c = ~ = J5, so the foci are (0, =;I=J5). © 2012 Ceng3ge LC3JT1ing. All Rights Reser\'ed. Muy not be scuMed. copied, or duplicated, or pos1cd to a publicly access ible wc:bsite, in whole or in pan. 28 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLARCOORDINATES yz xz 19. 25 - 9 = 1 ::::? a = 5, b = 3, c = J25 + 9 = J34 ::::? center (0, 0), vertices (0, ±5), foci (0, ±J34), asymptotes y = ±~x. Note: It is helpful to draw a 2a-by-2b rectangle whose center is the center of the hyperbola. The asymptotes are the extended diagonals of the rectangle. x2 y2 21 . x2 - y2 = 100 <=> 100 - 100 = 1 ::::? a = b = 10, c = JlOO + 100 = 10J2 => center (0, 0), vertices (±10, 0), foci (±10 J2, 0), asymptotes y = .±f§x = ±x 23. 4x2 - y2 - 24x - 4y + 28 = 0 <=> 4(x2 - 6x + 9) - (y2 + 4y + 4) = -28 + 36- 4 <=> 4(x '_ 3)2 - (y + 2)2 = 4 <=> (x - 3)2 - (y + 2)2 = 1 => 1 4 a = v'f = 1, b = v'4 = 2, c = Vf+4 = J5 ::::? center (3, - 2), vertices (4, - 2) and (2, -2), foci (3 ± J5, -2), asymptotes y + 2 = ±2(x- 3). y ,.._--!-----",, ,~_; ....... X 25. x2 = y + 1 <=> x2 = 1(y + 1 ). This is an equation of a parabola with 4p = 1, so p = i. The vertex is (0, -1) and the focus is (0, - ~). 27. x 2 = 4y- 2y2 <=> x 2 + 2y2 - 4y = 0 <=> x 2 + 2(y2 - 2y + 1) = 2 <=> x 2 + 2(y - 1)2 = 2 <=> x; + (y ~ 1)2 = l..This is an equation of an ellipse with vertices a~ (±J2, 1) . The foci ~eat (±.,J2=1, 1) = (±1, 1). 29. y2 + iy = 4x2 + 3 <=> y2 + 2y + 1 = 4x2 + 4 <=> (y + 1? - 4x 2 = 4 <=> (y: 1) 2 - x2 = 1. This is an equation of a hyperbola with vertices (0, - 1 ± 2) = {0, 1) and (0, -3). The foci are at (0, - 1 ± J4+T) = (0, - 1 ± J5). 31. The parabola wi~ vertex {0, 0) and focus (1 , 0) ?pens to the right and hasp= 1, so its equation is y 2 = 4px, or y 2 = 4x. ' 33. The distance from. U1e focus ( -4, 0) to the directrix x = 2,is 2 - ( - 4) = 6, so the distance from the focus to the vertex is ~(6) = 3 and the vertex is ( - 1, 0). Since the focus is to the left of the vertex, p = - 3. An equation is y 2 .= 4p(x + 1) => y2 = -12(x + 1). © 2012 CcnGogc: Lcmning. Al l Rjglns Rest.!rvcd. Mny not he scanned. copied, or duplicutccl, or posted to a publicly occcssiblc website. in whole or in p..1rt. SECTION 10.5 CONIC SECTIONS 0 29 35. A parabola with vertical axis and vertex (2, 3) has equation y- 3 = a(x- 2?. Since it passes through (I; 5), we have 5-3=a(I-2)2 '* a =2,soanequationisy -3 =2(x-2?. 37. The ellipse with foci (±2, 0) and vertices (± 5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 2, x2 1 2 so b2 = a2 - c2 = 25- 4 = 21. An equation is 25 + ; 1 = 1. 39. Since the vertices are (0, 0) and (0, 8), the ellipse has center (0, 4) with a vertical axis and a = 4. The foci at (0, 2) and (0, 6) are 2 units from the center, soc= 2 and b = .,Ja2 - c2 = .,j42 - 22 = v'f2. An equation is (x - 2 0) 2 + (y- 4) 2 = 1 '* b a2 _41. An equation of an ellipse with center (-1,4) and vertex (- 1,0) is (x !21)2 + (y ~24)2 = 1. The focus (-1,6) is 2 units from the center, soc= 2. Thus, b2 + 22 = 42 '* b2 = 12 and the equation is (x + 1? + (y - 4)2 = 1 . , 12 16 . 2 2 43. An equation of a hyperbola with vertices (±3, 0) is ~2 ....:. ; 2 = 1. Foci (±5, 0) '* c = 5 and 32 + b2 =52 '* . 2 2 b2 = 25 - 9 = 16, so the equation is ~ - r6 = 1. 45. The center of a hyperbola with vertices ( -3, - 4) and ( -3, 6) is ( -3, 1), so a = 5 and a.n equation is (x + 3)2 . · · b2 = l.Focl(- 3, - 7)and(- 3,9)-* c =8,so5 2 +b2 =82 '* b2 =64-25=39andthe ( - 1}2 {x + 3)2 equation is y - = I 25 39 . . . 2 • 2 47. The center of a hyperbola with vertices (±3, 0) is (0, 0) , so a = 3 and an equation is ~2 - t2 = 1. b . 2 2 Asymptotes y = ±2x '* - = 2 '* b = 2(3) = 6 and the equation is ~ - JL = 1. a 9 36 49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance a - c from it) while the farthest point is the other vertex (at a distance of a+ c). So for this lunar orbit, (a- c)+ (a+ c) = 2a = (1728 + 110) + (1728 + 314), or a = 1940; and (a+ c)- .(a - c) = 2c = 314 - 110, x2 y2 or c = 102. Thus, b2 = a2 - c2 = 3,753,196, and the equation is 3 , 763 , 600 + 3 , 753 , 196 =I. 51. (a) Set up the coordinate system so that A is ( -200, 0) and B is (200, 0) . IPAI-\PBI = (1200)(980) = 1,176,000 ft = 21~ mi = 2a -* a= 1~5 , and c ~ 200 so b2 - 2 - 2 - 3,339,375 - c a - 121 121x 2 1,500,625 121y 2 - 1 3,339,375 - . © 20 12 Ccngoge Leoming. All Rights Rescn'Cil. May not be scanned, copied. or duplicated. or posted to a publi<:ly uccusible website, in whole or in part. 32 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 10.6 Conic Sections in Polar Coordinates 1. The. directrix x = 4 is to the right of the focus at the origin, so we use the form with "+ e cos(}" in the denominator. ed l. 4 4 (See Theorem 6 and Figure 2.) An equation is r = (} = 21 (} = 2 (}. ' 1+ ecos 1+ 2 cos +cos 3. The directrix y = 2 is above the focus at the origin, so Wf! use the form w ith "+ e sin(}" in the denominator. An equation is _ ed _ 1.5(2) _ 6 r - 1 + e sin () - 1 + 1.5 sin(} - 2 + 3 sin(}· 5. The vertex ( 4, 3Jr /2) is 4 units below the focus at the origin, so the directrix is 8 units below the focus ( d = 8), and we use the form with "- e sin(}" in the denominator. e = 1 for a parabola, so an equation is ed 1(8) 8 r = 1 - e sin () = 1 - 1 sin (} = 1 - sin (} · 7. The directrix r = 4 sec (} (equ ivalent to r cos(} = 4 or x· = 4) is to the right of the focus at the origin, so we w ill use the form with"+ e cos()" in the denominator. The distance from the focus to the directrix is d = 4, so an equation is ed ~{4) 2 4 r = 1 + ecosB = 1 + ~cos(}· 2 = 2 +cos(( 9. r = 4 . l / 5 = 4 / 5 where e = .1 and ed = .1 => d = 1. 5 - 4sinB 1/5 . 1 - ~sinB' 5 & (a) Eccentricity = e = ~ (b) Since e = ~ < 1, the conic is an ellipse. (c) Since"- e sin (}" appears in the denominator, the directrix is below the focus at the origin, d = IFl l = 1, so an equat'ion of the directrix is y = - 1. (d) The vertices are (4, ¥) and(~ , 3;). 2 1/3 2/ 3 ' 2 11. r = 3 + 3 sinB · . 1/~ = l + 1 sinB ' wheree=1and ed= 3 => d= ~. (a) Eccentricity= e = 1 (b) Since e = .1, Lhe conic is a parabola. (c) Since"+ e sin(}" appears in the denom inator, the directrix is above the focus at the origin. d = IFll = j , so an equation oflhe directrix is y = i· (d) T~e vertex is at ( ~, ¥), midway between .the focus and directrix. y (4 , 7r/2) X -------1 ---------· (4 3'1T) y=-1 F2 y y = 2/ 3 @) 2012 CcngJige Leon~ing. All Rights Rc.c:n-..1. May not be scanned. copied, or duplieoted. or posted to a publicly accessible website, in whole or in P"fL SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES 0 33 13. r = 6 : 9 · 1 1 1 1 6 6 = 3 ( 2 , where e = ~ and ed = ~ :::} d = ~ - + cos 1+ 3 cos0 (a) Eccentricity= e = t (b) Since e = t < 1, the conic is an ellipse. (c) Since "+ e cos e" appears in the denominator, the directrix is to the right of the focus at the origin. d = !Fll = ~.so an equation of the directrix is x = ~- (d) The vertices are ( ~, 0) and ( t, 1r), so the center is midway between them~ that is, ( fn, 1r) . 3 1/ 4 3/ 4 3 15. r = 4 - 8 cos 9 . 1 I 4 = 1 - 2 cos 9' where e = 2 and ed = 4 (a) Eccentricity = e = 2 (b) Since e = 2 > 1, the conic is a hyperbola. d - 1 - s · (c) Since "- e cos 0" appears in the denominator, the directrix is to the left of the focus at the origin. d'= IFll = ~.so an equation of the directrix is X - 3 . - - s· (d) The vertices are ( -~. 0) and (i, 1r), so the center is midway between them, that is, (~ , 1r). 17. (a) r = ; . 9 , where e = 2 and ed = 1 :::} d = 1 2 . The eccentricity 1 - sm e = 2 > 1, so the conic is a hyperbola. Since" -e sin B" appears in the denominator, the directrix is below the focus at the origin. d = IFl l = ~. so an equation of the directrix is y = - ~ . The vertices are ( - 1, ~) and ( 1 3'~~" ) h . "d b th h . ( 2 3" ) 3 , 2 , sot e center IS m1 way etween l'!m, t at 1s, 3 , 2 . (b) By the discussion that precedes Example 4, the equation . 1 IS T = ( 3 ) • 1 - 2sin e- ; I (f,1T) 9 : x=21 : ! -3 -2 ® 2012 Cengagc Learning. All Rights Resen·<."<<. May not be scanned. COJ1ied. or duplicated. or posled to a publicly accessible website. in whole or in ~'U't. 34 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 19. Fore < 1 the curve is an ellipse. It is nearly circular when e is close to 0. As e increases, the graph is stretched out to the right, and grows larger (that is, its right-hand focus moves to the right while its left-hand focus remains at the origin.) At e = 1, the curve becomes a parabola with focus at the origin. 21.IPFI= e iPLI => r =e[d-rcos(7r-9)J = e(d+rcos8) =* r(1 - ecos~) = ed ed => r=---....,. 1- ecosO 23. IPFI = eiP ll => r = e[d- rsin(9 -1r)] = e(d+r sinO) => r(1 - esin 9) = ed => ed r = ---""7 .. 1- esin9 x=-d y l • y= - d 25. We are given e = 0.093 and a = 2.28 x 108 . By (7), we have a(1 - e2 ) 2.28 x 108 (1- (0.093)2] . 2.26 x 108 r = = ~ -::-----::-=::------:: 1 + ecos 8 1 + 0.093 cosO 1 + 0.093cos8 X X 27. Here 2a = length of major axis = 36.18 AU => a = 18.09 AU and e = 0.97. By (7), the equation of the orbit is 18.09(1 - (0.97) 2 ] 1.07 B (&) th · d" fr th h · r = ~ 8 . y , e maxtmum 1stance om e comet to t e sun 1s 1 + 0.97 cos 9 1 + 0.97 cos 18.09(1 + 0.97) ~ 35.64 AU or about 3.314 billion miles. 29. The minimum distance is at perihelion, where 4.6 x 107 = r = a(1 - e) = a(1 - 0.206) = a(0.794} => a = 4.6 x 107/ 0.794. So the maximum distance, which is at aphelion, is r = a(1 + e) = ( 4.6 x 107/ 0.794) (1.206} ~ 7.0 x 107 km. 31. From Exercise 29, we have e = 0.206 and a(l- e)= 4.6 x 107 km. Thus, a= 4.6 x 107/ 0.794. From (7), we can write the . . 1 - e2 • equation of Mercury's orbit as r =a 1 8 . So smce + ecos dr a(1- e2 )esin8 -= => d9 (1 + ecos8)2 © 2012 Ccn!lllg< looming. All Rights Rcsen ·cd. May not be scanned, copied, orduplicotcd, or posted to a publicly accessible website , in whole or in part. CHAPTER 10 REVIEW 0 37 EXERCISES 1. X = t 2 + 4t, y = 2 - t, - 4 ~ t ~ 1. ~ = 2 - y, SO X = {2 - y)2 + 4{2- y) = 4- 4y + y2 + B- 4y = y2 - By+ 12 {::} x + 4 = y 2 - By+ 16 = (y - 4)2 . This is part of a parabola with vertex ( -4, 4), opening to the right. 3. y = sec() = ~() = .! . Since 0 ~ () ~ 1r /2, 0 < x ~ 1 and y ;::: 1. COS X This is part of the hyperbola y = 1/x. 5. Three different sets of parametric equations for the curve y = -IX are (i) X = t, y = Vt (ii) X = t 4 , y = t 2 (iii) x = tan2 t, y = tan t, 0 ~ t < 1r /2 There are many other sets of equations that also give this curve. y y \ ,1,1),0-0 0 7. (a) The Cartesian coordinates are x = 4 cos 2; = 4 (- ~) = -2 and y = 4sin ;,. = 4( 4) = 2 J3, that is, the point ( - 2,2 J3). X (b) Give~ x = - 3 and y= 3, we haver = J( -3)2 + 32 = JIB = 3 v'2. Also, tan() = ~ '* tan() = ~3 , and since ( -3, 3) is in the second quadrant,(}= ' 34". Thus, one set of polar coordinates for ( -3, 3) is (3 V2, 3,;'"), and two others are (3 '2 1111") and (- 3 '2 7,.) y ~, 4 v. .L.i, 4 . 9. r = 1 - cosO. This cardioid is symmetric about the polar axis. r 2 (2, 7T) 7T 27T 0 ® 2012 Censagcl..eaming. All Rights Rcscn"<:d. May not be sconncd, copied, or duplicated. or posted to a publicly acc..'SSiblc website, in whole or in Jl'lrL 38 0 CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 11. r = cos 30. This is a three-leaved rose. The curve is traced twice. 13. r = 1 +cos 2e. The curve is sym·metric about the pole and both the horizontal and vertical axes. r 1T 2 1T 31T 2 3 15. r = =} e = 2 > 1, so the conic is a hyperbola. ·de = 3 => t + 2sin£1 d = ~ and the form "+2 s in(}" imply that the directrix is above the focus at the origin and bas equation y = ~-The vertices are (1, ~) and ( - 3, 3;) . (2, 1r) 2 . 17. x+y=2 ¢:> rcosll+rsinll = 2 ¢:> r(cos0+sin 0)=2 ¢:> r= - - - -- cos(J + s in£1 19. r = (sin £1)/0. As B-. ±oo, r __, 0. As (} __, 0, r --.. 1. In the first figure, there are an infinite number of x -intercepts at x = 1rn, n a nonzero integer. These correspond to pole points in the second figure. -0.25 2 dy dx 1 dy dy / dt 2t 2 21. x = lnt, y = 1 + t ; t = 1. dt = 2t and dt = t' so dx = dxjdt = 1/t = 2t · When t = 1, (x , y) = (0, 2) and dyjdx = 2. 23. r = e-0 => y = rsin£1 = e-0 s in(} and x = T"cos O = e- 0 cosO => dy dyjd£1 *sin(} +rcosO _ -e- 0 sin(}+ e-0 cos(} - e0 sine- cosO dx = dx j d(} = ~;cos(} - rsin(} - -e-0 cosB - e-0 sinO · - e0 = cos B+sine · dy 0 - (- 1) 1 When (} = 1r, - = = - = - 1. dx -1 + 0 -1 (2,0) -1 - 0.75 ® 2012 Ccngage Lcnming.. All Rights Reserved. "Mo.y not be scanned. copied, or duplicated, or po!:ticd lou publicly accessible Wt!bSile., in whole or in p~rt. 25. x = t + sin t, y = t - cost ~ dy dyjdt 1 + sint dx = d.xjdt = 1 +cos t {1+cost) cost- {1 +sint)(- sint) CHAPTER 10 REVIEW 0 39 d (dy) dt dx dxjdt ___ _ ___:(,_1,...-+.:..__c_os_t-'-)-2 _____ = cost+ cos2 t +sin t + sin2 t = l +cost (l+cost)3 1 + cost + sin t (1 + cost)3 3 2 . 27. We graph the curve x = t - 3t, y = t + t + 1 for -2.2 :=::; t :=::; 1.2. By zooming in or using a cursor, we find that the lowest point is about (1.4, 0. 75). To find the exact values, we find the t-value at which dyjdt=2t+1 = 0 ¢=> t=~~ <=> , (x,y),;,(lj,~). 29. x·= 2acost- acos2t ~ dx = -2asin t + 2asin2t = 2asint(2cost - 1) = 0 ¢:> dt sint = 0 or cost = ~ ~ t = O,·t, 1r, orr;. y = 2asin t - a sin 2t ~ dy = 2a cost - 2a cos 2t = 2a(l +cost- 2 cos2 t) = 2a(1 -cos t)(1 + 2 cost) = 0 =} dt t 0 271" 4" = '3,or3. Thus the graph has vertical tangents where t = t. 1r and ~">3" , and horizontal tangents where t = 2; and 4;. To determine what the slope is where t = 0, we use !'Hospital's Ru;e to evaluate lim ddy/jddt = 0, so there is a. horizontal tangent there. t->0 X t t X y y 0 a 0 " ~a :.::]:a 3 2 2rr -~a ~a 3 2 (-3a,O) (a,O) )C 7r - 3a 0 4 7r -~a -~a 3 2 51r ~a _ :.::]:a 3 2 31. The curve r 2 = 9 cos 50 has 10 "petals." For instance, for - -fu :=::; 8 :=::; {'0 , there are two petals, one with r > 0 and one with r < 0. 33. The curves intersect when 4 cos 8 = 2 ~ cos 8 = ~ ~ 8 = ± i for -1r :=::; 8 :=::; 1r. The points of intersection are (2 , i) and (2,- ~). r = 4cos 0 (f) 2012 Ce:ngoge Learning. All Rights Reserved. M.o.y nol be scanm-d. copiL-d, or duplica1ed, or posted toll publicly accessible: website; in whole or In part. 42. D . CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES (b) dy = (1 +t 3 )(Gt) - 3t2(3t 2 ) = Gt - 3t 4 =Dwhen6t - 3t4 =3t(.2-t3 )=D ·=> t=Dort= 3'2,so thereare dt (1+t3)2 (1+t3)2 y.r. horizontal tangents at (0, D) and ( V2', ?-'4). Using the symmetry from part (a), we see that there are vertical tangents at (0, D) and ( ?-'4, .v2). (c) Notice that as t --> -1 +,we have x--> -oo andy --> oo. As t--> -1-, we have x --. oo andy --> - oo. Also . 3t+3t2 +{1+t3) {t+1)3 {t+1)2 y - { -x - 1) = y + x + 1 = 3 = 3 = 2 -> 0 as t -> -1. Soy = -x - 1 is a 1+ t 1 +t t -t+1 slant asymptote. ( .d) dx = (I-+ t3 )(3)- 3t(3t2) = 3 - 6t3 dy 6t- 3t4 dy dyjdt t(2 - t 3) dt (1 + t3)2 (1 + t3)2 and from part (b) we have dt = (1 + t3)2. So dx = dxjdt = 1- 2t3 . So the curve is concave upward there and has a minimum point at (D, D) and a maximum point at ( .v2, W). Using this together with the information from parts (a), (b), and (c), we sketch the curve. y X ( 3t ) 3 ( 3t2 ) 3 27t3 + 27t6 .(1 + t3)3 27t3 (1 +t3 ) (1 + t3)3 'y =-x - l (e) x? + y3 = 1 + t 3 + 1 + t3 . 2 3 3xy = 3C!\3)C~t3 ) = (1~~3) 2 ,sox3 +y3 =3xy. (f) We start with the equation from part (e) and substitute x = r cos 0, y = r sin 0. Then x 3 + y 3 = 3xy => r 3 cos3 0 + 1'3 s in3 e ~ 3r2 cos e sin e. For r =1- 0, this gives r = 3 ~OS e s~: . Dividing numerator and denominator cos O+ sm 0 3(_1 ) sinO . cosO · cosO by cos3 0, we obtam r = . 3 1 sm e +--cos3 0 3sec0 tan B 1 + tan3 e . (g) The loop corresponds to 0 E (D, ~).so its area is - {"/ 2 r 2 -.! ("/2 (3sec0 tan0)2 - ~ r /2 sec2 e tan2 e - ~ {""' u 2 du A-} 0 2d0-2} 0 1+tan3 0 d0-2} 0 (l+ tan3 0)2d0 -2} 0 {1+u3 ) 2 = lim l!.[-~(l+u3)-l]b =;! b--+oo 2 3 a 2 [let u = tan OJ (h) By symmetry, the area between the foUum and the line y = -x - 1 is equal to the enclosed area in the third quadrant, plus twice the enclosed area in the fourth quadrant. The area in the third quadrant is~ . and since y = -x - 1 => r sine = - r cos 0 - 1 => r = - . e 1 e' the area in the fourth quadrant is !;JU +cos 1 /_-"'/ 4 [( 1 ) 2 (3sec0 tan0) 2 ] CAS 1. . 1 (1) 3 - 2 - . () () - 1 3 () d() = - 2 . Therefore, the total area IS 2 + 2 2 = 2 . - rr/2 sm + cos +tan © 2012 Ccngage Learning. Al l Rights Reserved. M1ty not be scanned, copied. ~r duplicated, or posted too. publicly accessible wcb~itc, in whole or in part. 0 PROBLEMS PLUS l t cos u lt sin u dx cost dy sin t 1. x = -- du, y = . -- du, so by FTCl, we have dt = -t- and dt = - t - . Vertical tangent lines occur when 1 tL 1 tL dx = 0 {:} cost = 0. The parameter value corresponding to (x, y) = {0, 0) is t = 1, so tl1e nearest vertical tangent dt . occurs when t = ~. Therefore, tl1e arc length between these points is 1 7r/2 L= 1 (dx) 2 (dy) 2 1 1r/Z - + - dt= dt dt 1 2 t . 2 t j '1f/Z dt cos ~dt = - ~ [ lnt]7f12 = ln :!!. t2 + t2 1 t 1 2 3. ln. terms ofx andy, we have x =?'cosO= (1 + csin B)cosB =cos O+ csinBcosB =cosO + ~csin2B and y = rsinB = (1 + csinO)sinB =sinO+ csin2 B. Now - 1 $sinO $ 1 => -1 $sinO -t csin2 0 $ 1 + c $2, so -1 $ y $ 2. Furthermore, y = 2 when c = 1 and B = ~, while y = -1 for c = 0 and 0 = 3; . Therefore, we need a viewing. rectangle with - 1 $ y $ 2. To find the x-values, iook at the equation x =cos B + ~csin 20 and use the fact that sin 20 ~ 0 for 0 $ B $ ~ and sin 28 $ 0 for - ~ $ B $ 0. [Because r =::, 1 + csin 0 is symmetric about the y-axis, we only need to consider - ~ S 8. $ ~ .) So for - ~ $ B .$ 0, x has a maximum value when c = 0 and then x = cos 0 has a maximum value of 1 at B = 0. Thus, the maximum value of x must occur on [ 0, ~] with c = 1. Then x = cos 0 + ~ sin 20 => ~~ = -sinB +cos2B=- sinB + 1-2 sin2 9 =? j~ = -(2sin9-1)(sin9+1) = 0when sin9 = -1or~ [but sin B # - 1 for 0 $ 9 $ H If sin 9 = ~, then B = * and x = cos {f + ~sin 3 = ~V3. Thus, the maximum value of xis ~V3, and, by symmetry, the minimum value is - ~ V3. Therefore, the smallest viewing rectangle that contains every member of the family of polar curves r = 1 + csinB, where 0 $ c $ 1, is [-~V3, ~V3] x [-1 , 2]. 2.1 c=t 5. Without Joss of generality, assume the hyperbola has equation :: - ~: = 1. Use implicit differentiation to get 2 : - 2 Yb 2 y' = 0, soy' = b:x. The tangent line at the point (c, d) on the hyperbola has equation y- d = b:dc (x - c). a a y . a ' 2 The tangent line intersects the asymptote y = !!. x when .!!.x - d = bzdc (x- c) => abdx- a 2d2 = b2 cx - b2 c2 => a a a © 2012 Cengnge le>ming. All Rigllls Reserved. Moy not be: seonncd, copied, or duplicated, or posted to n publicly a=ssiblc website. in whole or in pan. 43 44 D CHAPTER 10 PROBLEMS PLUS a 2 d 2 -b2c2 ad + bc d h 1 . bad +be ad '-t-be => X = b(ad _be) = - -b- an t e y-va Ue IS a --b- = --a- · S. .1 1 h 1. . b (be - ad ad - be ) Th .d . ftl . . . . 1m1 ar y, t e tangent me mtersects y = -ax at --b-, --a- . e m1 pomt o 1ese mtersect1on pomts 1s ( 1 (ad+bc be-ad) 1 (ad+ be ad -be)) (12bc 12ad) . 2 --b- + --b- • 2 --a- + --a- = 2 b , 2--;;:- = ( c, d), the pomt of tangency. Note: Jf y = ·o, then at (±a, 0), the tangent line is x = ±a, and the points of intetsection are clearly equidistant from the point of tangency. ® 2012 Cengnge learning. All Rights Reserwf. Mny not be scanned. eopictl. or duplicated, or posted ton publicly accessible website, in whole or in pilrt. SECTION 11.1 SEQUENCES 0 47 35. a,. = cos( n /2). This sequence diverges s ince the tenns don 't approach any particular real number as n---> oo. The terms take on val4es between - 1 and I. _ (2n - 1)! = (2n - 1}! · = 1 · ___. 0 as n---> oo. C 37· an - (2n + 1)! (2n + 1}(2n}(2n- 1)! (2n + 1}(2n) onverges 1 + e - 2n --'------ ---> 0 as n---> oo because 1 + e- 2n ---> 1 and e" - e-n -+ oo. en- e n Converges 2 -n n 2 • • x2 H 2x 11 2 41. an = n e = -. Smce lim - = lim - = lim - = 0, it follows from Theorem 3 that lim an= 0. Converges en X--+00 e X X-tCO eX X--tOO e:t n-oo 43. 0 < cos 2 n < _..!:.._ · [since 0 :s; cos2 n :s; 1], so since lim 2 1 = 0, { 'c0 2 82 n } converges to 0 by the Squeeze Theorem. - 2n - 2" n->oe> " n · . ( / ) sin(1/ n) s· li sin(1/x) 1. sin t [ h 1/ ] . ., II fr 45. an = n sm 1 n = 1 / n . mce m / = tm w ere t = x = 1, tt •O ows · om Theorem 3 x ..... oo 1 X t-o+ t that {an } converges to I . = lim 2 = 2 ::::} x ->oo 1 + 2/x lim 1 + - = lim e1n Y = e2 , so by Theorem 3, lim 1 + - = e2 • Converges ( 2)"' . ( 2)" !t-oo X :r-oo n -+oo n ( 2n 2 + 1) (2 + 1/n 2 ) 49. an = In(2n2 + 1)- ln(n2 + 1) =In n 2 + 1 = In . 1 + 1/ n 2 -+ In 2 as n ---> oo. Converges 51. a.,, = arctan(ln n). Let f(x) = arctan(ln x). Then lim f(x) = ~since In x -+ oo as x---> oo and arctan is continuous. x - oo Thus, lim a,= lim f(n) = ~- Converges n-+oo n -+oo 53. {0, 1, 0 , 0, 1, 0, 0, 0, 1, ... } diverges since the sequence takes on only two values, 0 and I , and never stays arbitrarily close to either one (or any other value) for n sufficiently large. n ! 1 2 3 (n- 1) n 1 n n 55. an= 2n = 2 · 2 · 2 · · · · · - -2- · 2 2': 2 · 2 (for n > 1] = 4--> oo as n--+ oo, so {an} diverges. 57. 2~------------~ From the graph, it appears .that the sequence converges to 1. { ( -2/e)n} converges to 0 by (7), and hence {1 + ( - 2/e)" } converges to 1 + 0 = 1. .. . . .. . . 0'---~--~--~---~ 21 ® 2012 Ccng;,gc Lc011T1ing. AU RighiS R.escn·l.-d. May not be scanned, copied. or duplicated. or posted to a publicly accessible website, in whole or in part. I 48 D CHAPTER 11 INFINITE SEQUENCES AND SERIES 59. .. ... 0 '----~--~--"--'----./ 21 From the graph, it appears that the sequence converges to t. As n--+ oo, so lim an =~ · n -+oo 3/ n 2 +2 8 + 1/ n => 61. 2.~-------------.. From the. graph, it appears 'that the sequence {an} ,;_ { ~ 2 ~o~: } is divergent, since it osci llates between 1 and -1 (approximately). To n2 .. .. - 2 63. 0~-~-~~-----~-J IO prove this, suppose that {a,.} converges to L. If bn = -- 2 , then 1+n {bn} converges to 1, and lim abn =!::. = L. But an = cosn, so n-+oo n 1 bn lim abn does not exist. This contradiction shows that {an} diverges. -oon • From the graph, it appears that the sequence approaches 0. 1 · 3 · 5 · · · · · (2n- 1) 1 3 5 2n- 1 O < a,.= n = - · - · - · · .. ·-- (2n) 2n 2n 2n 2n 1 1 :<::; 2 n · (1) · (1) · · · · · (1) = 2n --+ 0 as n --+ oo { 1 · 3 · 5 · · · · · (2n - 1)} So by the Squeeze Theorem, ( 2 n)" converges to 0. 65. (a) a,. = 1000(1.06)." => at = 1060, a2 = 1123.60, a3 = 1191.02, a4 = 1262.48, and as = 1338.23. (b) lim a,. = 1000 lim (1.06)", so the sequence diverges by (9) with r = 1.06 > 1. n - oo n-+oo 67. (a) We are given that the init!al population is 5000, so Po = 5000. The number of catfish increases by 8% per month and is decreased by 300 per month, so Pt = Po + 8%Po - 300 = 1.08Po - 300, H = l.08P1 - 300, and so on. Thus, P,. = 1.08Pn-t - 300. (b) Using the recursive formula with Po = 5000, we get P1 = 5100, P2 = 520.8, P3 = 5325 (rounding any portion of a catfish), P,. = 5451, P5 = 5587, and P6 = 5734, which is lhe number of catfish in the pond after six monlhs. 69. If lrl ~ 1, then {r"} diverges by (9), so {nrn} diverges also, since lnr'' l = n lr"l ~ lr'' l· If lrl < 1 then lim xr"' = lim -=..._ :lb ~ ( In\ = lim ___!__I"'. :::: 0, so lim nr" = 0, and hence { nr"} converges. :C-tOO :c-oo r-:Z: X-+00 - T T-% %-+00 - n T n-oo . whenever lrl < 1. ® 2012 Ccngogc Lcoming. All Rights Rc:sen'Ctl. May not be S<llllllCd. copied, orduplic:oted, or posted ton publicly accessible website, in whole or in p.1rt. ' SECTION 11.1 SEQUENCES D 49 71. Since {a,.} is a decreasing sequence, an > an+l for all n ~ 1. Because all of its terms lie between 5 and 8, {an} is a bounded sequence. By the Monotonic Sequence Theorem,'{ an} is convergent; that is, {an} has a limit L. L must be less than 8 since {an} is decreasing, so 5 $ L < 8. 1 ' d . . 1 1 1 " 73. a,. = 2 n + 3 IS ecreasmg smce an+1 = 2(n + 1) + 3 = 2n + 5 < 2n + 3 = a1, tOr each n ~ 1. The sequence is bounded since 0 <an $ i fo r all n ~ 1. Note that a1 = t· 75. The tenns of an = n( -1)n alternate in sign, so the sequen~e is not monotonic. The first five terms are - 1, 2, - 3, 4, and -5. Since l im Ia .. I = lim n = oo, the sequence is not bounded. n.-+oo n-+oo n . . ( ) x '( ) (x2 + 1)(1)- x(2x) 1 - x2 77. an= - 2 -- defines a decreasmg sequence smce for f x = - 2 - -1 , f x = ( 2 )2 = < 0 n + 1 . x +. x + 1 (x2 + 1)2 - for x ~ 1. The sequence is bounded since 0 < a" $ ~ for all n ~ 1. { tn rn-Fn . I rn-Fn } · _ 112 _ 3/4 _ 7/B _ (2"-1)/2" = 21-(1/2">. 79. For v 2, y 2 y 2, y 2 y 2 v 2, .. . , a1 - 2 , a2 - 2 , aa - 2 , ... , so an - 2 lim a, = lim 21- <112") = 21 = 2. n-oo n-+oo Alternate solution: Let L = lim an. (We could show the limit exists by showing that {an} is bounded and increasing.) n--+oo . . Then L must satisfy L = ~ => L2 = 2L => L(L - 2) = 0. L =/= 0 since the sequence increases, soL = 2. 1 81 . a 1 = 1, an.+l = 3 - - . We show by induction that {an} is increasing and bounded above by 3. Let Pn be the proposition an that an+l > an and 0 < an < 3. Clearly P1 is true. Assume that Pn is true. Then an+l > an => 1 1 -- < - => an+l an 1 1 1 1 - - ·- >--.Now an+2 = 3 - - - > 3-- = an+l ~ Pn+l· This proves that {an} is increasing and bounded an+l a,., an+l a., above by 3, so 1 = a1 < an < 3, that is, {an} is bound~d , and hence convergent by the Monotonic Seque11ce Theorem. If L = lim an, then lim an+l = L also, soL must satisfy L = 3- 1/ L => £ 2 - 3L + 1 = 0 => n---+oo n---+oo £ _ 3±.,/5 - 2 • But L > 1, soL=¥. 83. (a) Let a,. be the number of rabbit pairs in the nth montl1. Clearly a1 = 1 = a2. In the nth month, each pair that is 2 or more months old (that is, a,.-2 pairs) will produce a new pair to add to the an-1 pairs already present. Thus, an= qn- 1 + an-2, so that {an}= {fn}, the Fibonacci sequence. (b) an= f,.+1 f ,. fn fn-1 + fn - 2 = 1 + .f,.- 2 = 1 + 1 1 an-1 = - - = / ·= 1 + --. If L = lim an, fn - 1 fn- 1 f n-1 f n- 1 fn-2 an-2 n-oo then L = . lim an- 1 and L = lim a,.- 2, soL must satisfy L = 1 + -L1 => L 2 - L - l = 0 => L = ¥ n-+oo n -+oo · [s ince L must be positive]. ® 2012 Ccngagc Lcoming. /\11 Riglns Rcsc~rvcd. May no1 be scnnncd. copied, or duplicated, or poi!.cd loa publicly occ<..-ssiblc wcbsilc, in whole or in parL 52 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES 00 3. I: an = lim s, = lim [2- 3(0.8)n) = lim 2 - 3 lim (0.8)" = 2 - 3(0) = 2 n=l n-oo n-.oo n--too n-+oo 00 1 1 1 1 5. For I: 3 • a,,= 3· 81 = a1 = 13 = 1, 82 = s1 + a2 = 1 + 23 = 1.125, sa = 82 + aa ~ 1.1620, n=l n n s~ = 83 +a~ ~ 1.1777, ss = 84 +as~ 1.1857, so = ss + ao ~ 1.1903, s 7 = 8G + a 7 ~ 1.1932, and ss = sr + as ~ 1.1952. It appears that the series is convergent. ' oo n n 1 2 7. For I: ~·a,, = ----r.:; · St = a1 i;= - -- = 0.5 82 = St + a2 = 0.5 + ~ ~ 1.3284, n=l 1 + yn 1 + yn 1 + J1 ' 1 + v2 9. 11. s3 = s2 +as~ 2.4265, S4 = sa+ a4 :::::: 3.7598, 8::, = 84 + a 5 :::::: 5.3049, 8 6 = s5 + a 6 :::::: 7.0443, sr = so+ ar:::::: 8.9644, ss = 87 +as~ 11.0540. It appears that the series is divergent. n 8n 1 - 2.40000 Ot----+--;---+---+----..----+--i ll 2 - 1.92000 3 - 2.01600 4 - 1.99680 5 - 2.00064 6 - 1.99987 - 3 7 -2.00003 From the graph and the table, it seems that the series converges to - 2. In fact, it is a geometric 8 -1.99999 9 - 2.00000 10 -2.00000 · ·th 2 d I · · . ~ 12 -2.4 -2.4 sen es w1 a= - .4 an r =-&·so tis sum IS 6 ( - 5 )n = ( 1) = -- = - 2. . n=l 1 - - & 1.2 Note that the dot corresponding to n = 1 is part of both {an} and { Sn}. TI-86 Note: To graph {an} and { sn}, set your calculator to Param mode and Draw Dot mode. (Draw Dot is under GRAPH, MORE, FORMT (F3).) Now under E (t) = make the assignments: xt1=t, yt1=12/ ( -5) -t, xt2=t , yt2=surn seq ( ytl, t, 1, t , 1) . (sum and seq are under LIST, OPS (F5), MORE.) Under WIND use 1 , 10, 1, 0, 1 0, 1, - 3, 1, 1 toobtain a graph s im ilar to the one above. Then use TRACE (F4) to see the values. n 1 2 3 4 5 6 7 8 9 10 Sn. 0.44721 1.15432 1.98637 2.88080 3.80927 4.75796 5.71948 6.68962 7.66581 8.64639 10/------------~ 0....___....__ _ __.. __ .._ _ __.._ _ ..__, 11 The series f: ~ diverges, since its terms do not approach 0. n=l n 2 + 4 ® 2012 Cengagc Lcaming. All Rights Rcscr.-cd. Mny not be scanned, cor>icd. or duplicated, or posted to a publicly accessible website, in whole or in p.nrt. SECTION 11.2 SERIES 0 53 13. n s., 1 0.29289 2 0.42265 3 0.50000 4 0.55279 5 0.59175 6 0.62204 o'-~~~~--~~~~~L-~~ u 7 0.64645 8 0.66667 From the graph and the table, it seems that the series converges. 9 0.68377 10 0.69849 00 (1 1) ( 1) so I: - - --- = lim 1 - - -- = 1 n=l Vn v'n + 1 k -oo y'kTI · 15. (a) lim an = lim ~ = ~.so the seqitence {an} is convergent by (11.1.1). n-oo n-oo 3n + 1 3 (b) Since lim an = ~ =I 0, the series f: an is divergent by the Test for Divergence. n -+oo n=l 17. 3 - 4 + 1; - 694 + · · · is a geometric series with ratio r = -l Since lrl = ~ > 1, the series diverges. 19. 10 - 2 + 0.4 - 0.08 + · · · is a geometric series with ratio - 120 = -i· Since lr\ = k < 1, the series converges to 21. f: 6(0.9)n-l is a geometric series with first term a = 6 and ratio r = 0.9. Since lrl = 0.9 < 1, the series converges to n=l a 6 6 -- = --=- = 60. 1- 7· 1-0.9 0.1 23. f: ( ~3~n-l = ~ f:· (-~) n-1 . The latter series is geometric w.ith a= 1 and ratio r =-~.Since lrl = ~ < 1, it n = l 4 4 n=l 4 converges to ( 1 / ) = ~. Thus, tbe given series converges to (-!) ( ~) = t. 1 - - 3 4 oo ?Tn 1 oo (?T) n· . . . . ?T . . . 25. I: n+l = - I: - IS a geometnc senes w1th rat1o r = - 3 . Smce \r l > 1, the senes dtverges. n=O 3 3 n=O 3 1 1 1 1 1 ~ 1 1 ~ 1 Th' . I . I f h d' h . . 27. - + - + - + - + - + · · · = L- -- = - 3 L- - . IS ts a constant mu t1p e o t e tvergent armoruc senes, so 3 6 9 12 15 n = l 3n n=l n it diverges. ~ n - 1 . b h "' " o· . lim li n - 1 1 ....£ 0 29. L., ----- dtverges y t e .est .or 1vergence smce . an = m - 3 1 = - 3 r . n=l 3n - 1 n-oo n-oo n - © 2012 Ccnll"lle l earning. All RighiS Reserved. May not be scanned. copied. or duplicotcd. or poste-d loa publicly a<:<:essiblc website, in wbolc or in part. 54 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES 31. Converges. 00 1 + 2" 00 ( 1 2") 00 [ ( 1 ) " ( 2 ) " ] L:--=L: - + - =L: - +- n=1 3" n=1 3n 3" . n=1 3 3 [sum of two convergent geometric series] -~ ~- ! 2-~ - 1 - 1/ 3 + 1 - 2/3 - 2 + - 2 00 33. L: o/2 = 2 + v'2 + ?12 + ~ + · · · diverges by the Test for Divergence since n=l lim a,. = lim o/2 = lim 21/ n = 2° = 1 =/= 0. n.-oo n-oo n-oo 2 ) . 35. E In ( 2 n 2 + 1 diverges by the Test for Divergence since n=l n + 1 37. E (~)" is a geometric series with ratio r = -I ~ 1.047. It diverges because lrl ~ 1. k=O 00 39. L: arctan n diverges by the Test for Divergence since lim an = lim arctan n = % =/= 0. n=l n.-+oo n-+oo 41 ~ 1 ~ ( 1 )" · · · · h fi 1 d · 1 s· I I 1 th · . 6 ~ = 6 - 1s a geometric senes w1t rst term a = - an ratio r = - . mce r = - < 1, e senes converges n = l e n=1 e e e e 1/e 1/e e 1 oo 1 · to -- 1 - = -- 1 - · - = --.By Example 7, L: ( 1 ) = 1. Thus, by Theorem 8(ii), 1 - 1 e 1 - 1 e e e - 1 n=1 n n + oo (1 1 · ) 00 1 "" · 1 · 1 1 e - 1 e n~1 en + n(n + 1) = n~l en + n~l n(n + 1) = e- 1 + 1 = e - 1 + e- 1 = e - 1· 43. Using partial fractions, the partial sums of the series E +are ' n =2 n -1 s,. = L: . . = L: -. -- -. -n 2 n ( 1 1 ) i = 2 (t - 1)(t + 1) i=2 t- 1 t + 1 = (1- !) + (! -!) + (! -!) + ... + ( '-1 - _1 ) + (-1 - .!.) 3 2 4 3 5 n-3 n- 1 n-2 n Th. . I . . d 1 1 1 1 1s sum 1s a te escopmg senes an sn = + - 2 ~ -- 1 - - · n - n Thus, E --i:-- 1 = l~ Sn = lim (1 +! - - 1-- .!.) = ~- n = 2 n - n-+oo n-+00 2 n- 1 n 2 45. For the series :E , s,. = :E -. -.-- = L: -:- - -. -oo 3 n 3 n (1 1 ) n=1 n(n+3) i =1 t(1 + 3) i=1 \ t+3 [using partial fractions). The latter sum is (1 - l) + (l ....: l) + (1 - l) + (l - l) + ... + (-1 - l) + (-1 - _1 ) + (-1 - _1 ) + (.!. - --4) 4 2 5 3 6 4 7 n-3 n n -2 n + 1 n-1 n+2 n n+3 1 + 1 + 1 1 1 1 [tl. ' ] = 2 3- n+1 - n+ 2 - n+3 e escopmg senes TI ~ 3 l' l' (1 1 1 1 1 1 ) 1 1 11 c ms, 6 ( 3) = tm Bn = 1m + 2 + 3 - - + 1 - - + 2 - ..,----+3 = 1 + 2 + 3 = 6 . onverges n = l n n + n-oo n -oo n 1l n © 2012 Cengagc Learning. All Righ!S Reserved. Muy not be scanned. copied. or duplicotcd, or po>led lo n publicly occcs:Jiblc wobsitc. In whole or ip part. SECTION 11.2 SERIES D 57 . . (1+ c)- 2 scn es and set 1t equal to 2: ) 1 = 2 <:::} 1- (1 + c ( 1 ) 2 ( 1 ) -- -2 - 2 -- 1+ c - 1 +c <:::} 1 = 2(1 + c)2 - 2(1 +c) <:::} 2c2 + 2c - 1 = 0 <:::} c = - 2 ~ m = ±'1- 1 . However, the negative root is inadmissible because -2 < --q- 1 < 0. Soc= :0 2 - 1 . 1 1 1 . . 75. e"" = e1+2+3+··-+;; = e1e112e113 · · · e1/ n > (1 + 1) (1 + ~) (1 + ~) · · · (1 + ~) [e"' > 1 + x] = ~~~ ... n + 1 =n + 1 123 n Thus, e·'" > n + 1 and lim e"" = oo. Since { sn} is increasing, lim s,. = oo, implying that the harmonic series is n-+oo n -+oo divergent. 77. Let dn be the diameter of Cn. We draw lines from the centers of the C; to the center of D (or C), and using the Pythagorean Theorem, we can write 12 + (1 - ~d1) 2 = (1 + ~d1)2 ¢:} 1 = (1 + ~d1 ) 2 - (1- ~d1 )2 = 2d1 [differenceofsquares] ==> d1 = ~ · Similarly, 1 = (l + ~d2) 2 - (1 - d1 - ~d2) 2 = 2d2 + 2d1- di - d1d2 = (2 - dl)(d1 + d2) ¢:} T 1 (1- dd ( 1 )2 ( 1 )2 - [1- (dl + d2))2 . d2 = 2 _ d1 - d1 = 2 _ d1 , 1 = 1 + 2d3 - 1 - d1 - d2 - 2ds <:::} d3 - 2 _ (dt + d2 ) , and m general, d,.+1 = ( 1 -=-¥ 1 ~)2 . lfwe actually calculate d2 and d~ from the formulas above, we find that they are.!. = - 1- and 2 i=1 ' 6 2 . 3 1 1 · I h . al d 1 .,., th' . d . - = - - respect1ve y, so we suspect t at m gener , n = ( 1 ) . •0 prove IS, we use m uct1on: Assume that for all 12 3 · 4 n n+ 1 1 1 n 1 n ly < n dk = = - - --.Then 2:= d; = 1 - - - = - - [telescoping sum). Substituting this into our - ' k (k + 1) k k + 1 . ;=1 n + 1 n + 1 [1 _ _ n ] 2 . n+1 formula for dn+l. we get dn+l = ( n ) 2- - - n+1 1 (n+1)2 1 n + 2 = (n + 1)(n + 2), and the induction is complete. n+1 Now, we observe that the partial s ums 2:=~=1 d; of the diameters of the circles approach 1 as n--+ oo; that is, 00 00 1 :2:= an = :2:= ( ) = 1, which is what we wanted to PJOVe. n=1 n =1 n n + 1 79. The series 1 - 1 + 1- 1 + 1 - 1 + · · · diverges (geometric series with r = -1) so we cannot say that 0=1- 1 +1 - 1 + 1-1+· · · . 81. 2:=""=1 can = lim l:=~=l ca; = lim c 2:='~1 a; = c lim 2:=~~1 a; = c 2:::,'=1 a,. , which exists by hypothesis.\ n n-+ oo n-oo ' n -+oo ® 201 2 Cc:ngagc l camins- All RighLtt Resen:~.-d. ~~fay no1 be scunrx.-d. copied. or duplicated. or posted to a publicly accessible website, in whole or in part. 58 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES 83. Suppose on the contrary that l:(an + bn) converges. Then l:(an + bn) and I: an are convergent series. So by Theorem 8(iii), I: [(a,. + bn) - a,.] would also b~ convergent. But I: [(an + bn) -an] = I: bn, a contradiction, since I: bn is given to be divergent. 85. The partial sums {sn} form an increasing sequence, since s,. - Sn- 1 =an > 0 for all n. Also, the sequence {sn} is bounded since Sn ::::; 1000 for all n. So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series I: an is convergent. 87. (a) At the first step, only the interval ( ~ , ~) (length ~) is removed. At the second step, we remove the intervals ( i, ~) and ( ~ ' ~).which have a total length of2 · Ut At the third step1 we remove 22 intervals, each of length (~ ) 3 • In general, at the nth step we remove 2n-1 intervals, each oflength ( ~) n, for a length of 2"-1 · ( ~)"' = ~ ( ~) n-1 • Thus, the total length of all removed intervals is f ~ ( t) n-1 = 1 :./;13 = 1 [geometric se~ies with a = t and 7' = t] . Notice. that at n =l the n th step, the leftmost interval that is removed is ( ( 1 )" , ( j )"),so we never remov~ 0, and 0 is in the Cantor set. Also, the rightmost interval removed is (1 - (j)", 1 - (tf'), so 1 is never removed. Some other numbers in the Cantor set 12127 d B are 3• 3•·9• 9• 9• an 9· (b) The area removed at the first step is ~; at the second step, 8 · ( ~) 2 ; at tbe third step, {8)2 · ( i) 3 • In general, the area removed at t.he nth step is {8)"-1 a r = H ~r-l. so the totat area of all removed squares is 00 1 (8)"-1 1 / 9 n>;1 9 9 = 1 - 8/ 9 = 1. oo n 1 1 1 2 5 5 3 23 89. (a) For n>;1 (n + 1)!' 81 = 1. 2 = 2' S2 = 2 + ~ = 6' sa = 6 + 1 . 2. 3. 4 24, 23 4 119 . (n + 1}!-1 s4 = - + = -.The denommators are (n + 1}!, so a guess would be sn = ( 1) 1 24 1 · 2 · 3 · 4 · 5 120 n + . 1 21 - 1 (k + 1)! - 1 (b) For n = 1, 8 1 = 2 = ~·so the formula holds for n = 1. Assume 8k = (k + 1)! . Then (k + 1}! - 1 k + 1 (k + 1}! - 1 k + 1 (k + 2)! - (k + 2) + k + 1 Sk+l = (k + 1)! + (k + 2)! = (k + 1)! + (k + l)!(k + 2) = (k + 2)! - (k + 2)! -1 - (It+ 2)! Thus, the formula is true for n = k + 1. So by induction, the guess is correct. . · . (n + 1)! - 1 . [ 1 ] oo n (c) lim 8n = lim ( )I = hm 1 - ( ) ' = 1 and so L: ( 1)! = 1. n-oo ,._00 n + 1 . n-oo n + 1 . "=1 n + ® 2012 Ccngagc LearninH,. All Rights Reserved. Muy not be scanned, copied, or duplicated, or poste<.llo a publicly accessible websile, in whole or In part. SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS 0 59 11.3 The Integral :rest and Estimates of Sums 1 r2 ·1 1. The picture sh?WS that a 2 = 2 1.3 < 1 1 x 1.3 dx, y 1 131 00 1 [ 00 1 a 3 = "1.3 < "1.3 dx, and so on, so 2:: 1':3 < "1.3 dx. The . 3 . 2 X n = 2 n . 1 X integral converges by (7.8.2) withp = 1.3 > 1, so the series converges. 0 2 3 4 X 3. The function f(x) = 1/ ~ = x- 1/ 5 is continuous, positive, and decreasing on [1 , oo ), so the Integral Test applies. f oo x - 115 dx = lim 1 J' x- 1/ 5 dx = lim [2x415] t = lim (~t415 - .§.) = oo, so E. 1/ ifrf diverges. 1 . . t-.oo 1 t-.oo 4 1 t--.oo ·. 4 n = 1 5. The function f(x) = ( 2 x ~ 1 )3 is continuous, positi~e, and decreasing on [1 , oo), so the lf1tegral Test applies . . roo 1 dx ~ lim r 1 dx = lim [-.! 1 ] t = lim ( - 1 + _.!_) = _.!__ 11 (2x + 1)3 t-+oo 11 (2x + 1)3 t-.oo. 4 (2x + 1)2 1 t- oo 4(2t + 1)2 36 36 Since this improper integral is-convergent, the series E ( 2 1 1 )3 is also convergent by the Integral Test. . n=l n + 7. The function f(x) =-/-- is continuous, positive, and decreasing on [1, oo), so the futegral Test applies. X + 1 . . roo ~ 'dx = lim r -/-- 1 dx = lim [- 2 1 ln( x2 + 1 )] t = - 2 1 lim [ln( e + 1) - ln 2] = oo. Sine~ this improper } 1 X + 1 t -.oo } 1 X + t-->oo . 1 . t--+oo · integral is divergent, the series E ~ is also divergent by the Integral Test. n=1 .n + 1 9. E ~ is a p-series with p = -/2 > 1, so it converges by (1 ). n=1 n . . 1 1 1 1 ~ 1 Thi . . . ' th 3 1 . b ( ) 11.1 +-+-+ - + - + ···= 6 3 · s tsap-seneswL p = > ,so1tconverges y I. 8 27 64 125 n=l n 1 1 1 1 00 1 1 13. 1 +- +- + - + - + · · · = L: --. The function f(x) = 2x _- 1· is 3 5 7 9 n =l 2n - 1 continuous, positive, and decreasing on [1, oo ), so the· Integral Test applies. r oo - 1-·- dx = lim ft -2 1 dx = lim a 1n 12x - 111 ~ = ~ lim (1n(2t - 1) - 0) = oo: so the series r: -2 1 } 1 2x- 1 t -+oo 1 X- 1 t -+ oo t-->oo n=l n - 1 diverges. oo vfn+ 4 'oo ( vfn 4 ) oo 1 oo 4 15. L: --2- = L: - 2 +2 = L: 3 / 2 + L: 2• n =1 n n = l n n n =l n n =1 n ~ 1 . . 'th 3 . 6 3!2 IS a convergent p-senes w1 p = 2 > 1. n = l n E 4 2 = 4 E ~ is a constant multiple of a convergent p-series with p = 2 > 1, so it converges. The sum of two n=l n . n = l n convergent series_ is-convergent, so the original series is conv~rgent. © 2012 Ccognse Learning. All Righls RCSCJV<d. Mny not be SC41111cd, copied, or duplicalcd, or posted lo o publicly accessible wclnile, in whole or in pnrt. 62 0 CHAPTER 11 INFINITE SEQUENCES 1,\ND SERIES (c) The estimate in part (b) iss ~ 1.64522 with error ~ 0.005. The exact value given in Exercise ~4 is 1r2 / 6 ~ 1.644934. The difference is less than 0.0003. (d) Rn ~ /.~ --;. dx = .!.. So Rn < 0.001 if.!. < - 1- {::} n > 1000. ,. x n n 1000 39. f(x) = 1/{2x + 1)6 is continuous, pos itive, and decreasing on [1, oo), so the Integral Test applies. Using (3), R,. ~ 1 oo (2x + 1) - 6 dx = t~ [ 10(2; ~ 1 )5 r l0(2n \ 1)5 . To be correct to five decimal places, we want 10( 2 n 1 + 1 )5 ~ 1~6 ¢:} (2n + 1)5 ~ 20,000 {::} n ~ ~ ( .e-'20,000- 1) ~ 3.12, ;muse n = 4. 4 1 1 1 1 1 S4 = n~1 (2n + 1)6 = 36 + S6 + 76 + ge ~ 0.001446 ~ 0.00145. 41. f: n-1.oo1 = f: 1~01 is a convergent p-series with p = 1.001 > 1. Using (2), we get n = 1 n=1 n Rn < !.oo x-l.001 dx = lim [ x - o.oo1 ] t = - 1000 lim [-1-]t = - 1000(--1-) = 1000. - n t->oo - 0.001 " t->oo x0.001 n n0.001 n0.001 1000 - 1000 We want Rn < 0.000 000 005 {::} n0.001 < 5 x 10 9 {::} n°·001 > 5 x 10 9 {::} n > (2 X 10ll) 1000 = 21000 X 1011,000 ~ l.07 X 10301 X 1011.000 = l.07 X 1011.301. 43. (a) From the figure, a2 +as+···+ an ~ Jt f(x) dx, so with 1 1 1 1 1 1" 1 f(x)=-,-+-+-+ .. ·+ - ~ -dx= ln n. x2 3 4 n 1 x 1 1 1 1 Thus, Bn = 1 + z + '3 + 4 + .. · + ;:;: ~ 1 + ln n. (b) By part (a), s10o ~ 1 + ln 106 ~ 14.82 < 15 and 8109 ~ 1 + ln 109 ~ 21.72 < 22. y 0 2 3 4 ·" n 45. bin" = ( e1" b) Inn = ( e1" ") In b = nln b = n _1 1 n b. This is a p-series, which converges for all b such that - ln b > 1 ¢:} ln b < - 1 ¢:} b < e-1 {::} b < 1/e [with b > 0]. 11.4 The Comparison Tests 1. (a) We cannot say anything about I: an. If an > bn for all nand :L b,. is convergent, then :L an could be convergent or divergent. (See the note after Example 2.) (b) If an < bn for all n, then I; an is convergent. [This is part (i) of the Comparison.Test.) n n 1 1 " II ~ n b . 'th ~ 1 h' I 3. 2 n 3 + 1 < -2 3 = - 2 < 2 .or a n ~ 1, so L..- -2- 3-- converges y companson WI L..- 2 , w 1c 1 converges n 2n n n=l n + 1 n = l n because it is a p-series with p = 2 > 1. ® 2012 Cengage Learning. A ll Ri£bts Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. SECTION 11.4 THE COMPARISON TESTS D 63 n + 1 n 1 ~ n + 1 d' b · ··h ~ 1 h' h d' b · · 5. -- > - - = - for all n 2: 1, soL- 1 1verges y companson w1t L- 1 , w 1c Jverges ecause 1t 1s a nVn nVn Jn n=l n vn n=l vn p-series with p = ! ~ 1. 7. < - = - for all n > 1. L:: (for is a convergent geometric series (lrl'= fo < 1 ), so L:: n 9" 9" ( 9 )" 00 oo '9n 3 + 10n 10" 10 - . n=l n=l 3 + 10 converges by the Comparison Test. lnk 1 [' l k "k .>3] ~ lnkd. b · 'h~ 1 h'h d' b · 9. - > - for all k 2: 3 smce n • > 1 .or _ , so LJ - k 1verges y com pan son w1t LJ -k, w 1c 1verges ecause 1t k k k = 3 k=3 • is a p-series with p =: 1 ~ 1 (the harmonic series). Thus, f lnkk diverges since a finite number of terms doesn't affect the I k=l convergence or divergence of a series. {fk ifk kl/3 1 00 • {fk . . 00 1 11 . < ~ = k 312 = k7 16 for all k 2: 1, so L:: converges by compar1son w1th L:: 716 , Jk3 + 4k + 3 .v k3 k=l ,jk3 + 4k + 3 k=l k · which converges because it is a p-series with p = ~ > 1. arctann 7r/2 fi II ~ arctann b . . h 1r ~ 1 hi h . 13. 1.2 < ""1.2 or a n 2: 1, so LJ ~,2 converges y comparison w1t -2 LJ ""1.2• w c converges because 1t is a n . n n =l n n=l n constant times ap-series with p = _1.2 > 1. 4'' + 1 4 . 4" (4)" 00 (4)" 00 (4)" 15. 3 n _ 2 > 3n"" = 4 3 for all n 2: 1. n~l 4 3 = 4 n~l 3 is a divergent geometric series (lrl = ~ > 1), so 00 4n+l L:: - -- diverges by the Comparison Test. n=l 3"- 2 1 1 17. Use the Lirni~ Comparison Test with an = JTi2'+T and bn = n: lim an = lim n = lim ' 1 = 1 > 0. Since the harmonic series f .!. diverges, so does n-oo bn n-oo Jn2 + 1 n-<oo )1 + (1 fn2) n=l n 00 1 L:: Jn2+1' n=l n + 1 th L. . C . Ti . I 1 + 4" db 4" 19. Use e 1m1t ompanson estwJtlan·= - 1 3 an n = - 3 : . + n n 1 +4" . . a.. . 1 + 3" . 1 + 4" 3" . 1 + 4" 3" . ( 1 ) 1 lim-= lim--,-, -= hm --·- = hm --·---= hm - +1 ·---=1 > 0 n__.ob bn n__.oo !._ · n-oo 1 + 3" 4" n__.oo 4" 1 + 3" n-.oo 4" .I_ + 1 ~ ' ~ Since the geometric series 2::: bn = 2::: ( 1 r diverges, so does f 1 1 + 4 3 : . Alternatively, use the Comparison Test with n=l + 1+4" 1+4" 4" 1(4)" . - - - > --- > - ( - ) = - - or use the Test for Divergence. 1 + 3" 3n + 3" 2 3" 2 3 © 2012 Cengoge Le:uning. All R;ghL< Resel'\'ed, May not be SCIUUled, copied, or duplicated, or posted to a publicly acc'cssihlc website, in whole or in pM. 64 D CHAPTER 11 INFINITE SEQUENCES AND SERIES 21 . Use the Limit Comparison Test with an = 2 f+"2 and bn = ;12 : n +n+ 1 n . an . n312 Jn+2 . (n312 Jn+2)l(n312 vn) . ) 1 +2ln /1 1 hm - = hm = lim = lim = - = - > 0 n--.oo bn n--.oo 2n2 + n + 1 n--.oo (2n2 + n + 1)ln2 n--.oo 2 + 1ln + 1l n2 2 2 · Since f: ;12 is a convergent p-series [P = ~ > 1], the series f: 2 f+"2 also converges. n=l n . n= l n + n + 1 5+ 2n 1 23. Use the Limit Compa'rison Test with a,. = ( 1 + n2 )2 and bn = n 3 : lim an li n 3 (5 + 2n) l' 5n 3 + 2n 4 1ln 4 l' * + 2 . 2 0 s· ~ 1 . n --.oo bn. = n--.~ (1 + n2)2 = n~ (1 + n2)2 · l l (n2 )2 = n!...~ (.l;,. + l )2 = > . mce n'S::l n3 IS a convergent ... . [ 3 ] h . ~ 5 + 2n l p-senes p = > 1 , t e senes L.J ( 2 ) 2 a so converges. n.=l 1 + n n2 1 oo ../n4 + 1 2 ( ) = -- for all n ~ 1, so I: 3 2 diverges by comparison with n n + 1 n + 1 · n=l n + n f: - 1- = f: ~.which diverges because it is a p-series with p = 1 ~ 1. n.=ln+1 n=2n 27. Use the Limit Comparison Test with an = (1 + ~) 2 e- n and bn = e-": lim abn = lim .(1 + ~) 2 = 1 > 0. Since n n-oo n n-oo n f: e-n = f: · :. is a convergent geometric series [lrl = ~ < 1], the series f: (1 + ~)2 e-n also converges. n=l n=l e n=l n 29. Clearly n! = n(n - 1)(n - 2) · · · (3)(2) > 2 · 2 · 2 · · .. · 2 · 2 = 2''-1 so..!_ < - 1-. E - 1- is a convergent geometric . - ' n! - 2n-l n =1 2 n -1 series [lr I = ~ < 1], so E --\ converges by the Comparison Test. n=l n . 31. Use the Limit Comparison Test with an == sin (;) and bn = ; . Then I: an and 2: bn are series with positive terms and I . an 1 . sin(1ln) . sinO . ~ . . . . 1m - = rm _ = hm - 0 - = 1 > 0. Smce L.J bn IS the divergent harmontc senes, n--.oo bn n--.oo 1ln 0--+D n=l 00 2: sin (1 l n) also diverges. [Note that we could also use !'Hospital's Rule to ev~luate the limit: n = 1 · . sin(ll x) H • cos(l/x) · ( -l/x2 ) . 1 hm I = lim I 2 = hm cos - =cos O= 1.) x - oo 1 X %-oo -1 X x_____. oo X lO 1 1 1 1 1 1 1 1 . 33. n~l ~ = .J2 + ..ff7 + v'82 + · · · + .JI0,05I ~ 1.24856. Now ~ < ..JTiJ = n 2 , so the error IS 1 00 1 [ 1 ] t ( 1 1 ) 1 Rw ~ Tw ~ 2 dx = lim -- = lim -- +- = - = 0.1. 10 X t --.oo X 10 t --.oo t 10 10 ©. 2012 Ccngage Leaming. All Ri(;hts Reserved. Mny not be scanned, copied, or duplicated, or posted ton publicly accessible website, in whole or ln pan. SECTION 11.5 ALTERNATING SERIES D 67 ~ ( -0.8t :.._ - ,;.., ( - 0.8t LJ ' ~57- LJ ' n=1 n. n=1 n. ~ -0.8 + o.32 - o.oss3 + o.oi 706- o.oo2 731 + o.ooo 364- o.ooo 042 ~ -o.sso7 Adding b8 to s7 does t:~ot change the fourth decimal place of 57, so the sum_ of the series, correct to four decimal places, is - 0.5507. 23. The series f: ( -1);+1 satisfies (i) of the Alternating Series Test because ( 1 )6 < -i and (ii)' lim -i = 0, so the n=l n · n + 1 n n -+oo n 1 1 series is convergent. Now bs = 56 = 0.000064 > 0.00005 and b6' = 66 ~ 0.00002 < 0.00005, so by the Alternating Series Estimation Theorem, n = 5. (That is, since the 6th term is less than the desired error, we need to afld the first 5 terms to get the sum to the desired accuracy.) 25. The series f: ( -1)",· satisfies (i) of the Alternating Series Test because 10 . +lt 1 )' < - 0 1 1 and (ii) lim - 1 - = 0, n =O 10" n. " n + . 1 " n n-ioo 10" n ! so the series !s convergent. Now bs = 10 ! 31 ~ 0.000 167 > 0.000 005 and b4 = 10 ! 41 = 0.000 004 < 0.000 005, so by the Alternating Series Estimation Theorem, n = 4 (since the series starts with n = 0, -not n = 1). (That is, since the 5th tenn is less than the desired error, we need to add the first 4 terms to get the sum to the desired accuracy.) 1 1 27. b4 = I = -- :::; o.ooo 025, so 8. 40,320 00 (- 1)" 3 (-1)" ' 1 1 1 n~1 (2n)! ~ 53 = n~l (2n)! = -2 + 24 - . 720 ~ ~0.459722 Adding b4 to S3 does not change the fourth decinlal place of 5s, so by th~ Alternating Series Estimation Theorem, the sum of . the series, correct to four decimal places, is - 0.4597. 72 - 29. b7 = 107 = 0.000 004 9, so 00 (-1)n- ln2 6 (-1)n- ln2 "' "' 1 4 9 16 2s 36 ·0 067 n~1 10" :::; 56 = n-7;:1 10" = 10 - 100 + 1000 - 10,000 + 100,000 - 1,000,000 = · 614 Adding b7 to 56 does not change the fourth decimal place of 56, so by the Alternating Series Estimation Theorem, the sum of the series, correct to four decimal places, is 0.0676. I 00 ( -1)n-l 1 1 1 1 1 1 1 31. I: = 1- - + - - - + · · · + - - - + --- + · · · . The 50th partial sum of this series is an n = 1 n 2 3 4 49 50 51 52 underestimate, since "~1 ( - 1~n-l = 5so + ( 511 - 512) + ( ; 3 - 5 1 4 ) + ; · ·, and the te~s in parentheses are all positive. ' The result can be seen geometrically in Figure 1. 33. Clearly bn = - 1 - is decreasing and eventually positive and lim bn = 0 f~r any p. So the Series converges (by the n+p n -+ oo Alternating Series Test) for any P. for which every bn is defined, that is, n + p :/:- 0 for n 2 1, or p is not a negative integer. ® 2012 Ccngnge Learning. All Rights Reserved. May not be scanned, copied, or dupJicatcd, or posted to a publicly actcss ible website, in whole or in part. 68 D CHAPTER 11 INFINITE SEQUENCES AND SERIES 35. 2: b2n = 2: 1/(2n)2 clearly converges (by comparison with ~e p-series for p = 2). So suppose that 2: ( -1r-I bn . . converges. Then by Th~orem 11.2.8(ii), so does 2: [ ( -1 t-1 bn + bn] ::: 2 ( 1 + l + i + · · · ) = 22: - 2 1 . But thi~ . n -1 diverges by comparison with the harmonic series, a contradiction. Therefore, 2: ( ....:1)"- 1 b,. must diverge. The Alternating Series Test does not apply since {bn} is not. decreasing. 11.6 Absolute Convergence and the Ratio and Root Tests 1. (a) Siilce lim I an+l I = 8 > 1, part (b) of the Ratio Test tells us that the series 2: an is divergent. n-oo an. . . (b) Since lim I an+l I = 0.8 < 1: part (a) of the Ratio Test tells us that the series 2: an is absolutely convergent (and n.-CX> an therefore convergent). (c) Since lim I an+l I = 1, the Ratio Test fails and the series 2: an might converge or it might diverge. -~ an . . 3. lim lan+l l = lim In+ 1 · 5" 1 = lim 1.!. · n+ 1 1 =.!.lim 1+1/n = .!. (1) = .!. < 1,sotheseries E ~is n-too an n-oo 5n+l n n-too 5 n 5 n-too 1 5 5 n=l 5n absolutely co.nvergent by the Ratio Test. 5. bn = - 5 1 1 > 0 for n 2::' 0, { bn} is decreasing for n ~ 0, and lim bn = 0, so E ( -1 t converges by the Alternating n + n-->oo . n=O 5n + 1 Series Test. To determine absolute convergence, choose an = ~ to get - n li an 1' 1/n l' ' 5n + 1 5 0 ~ 1 d. . b .h L ' . C . ,.. . h th m -b = tm 1 / ( 5 1 ) = rm -- = > , so ~ -- 1 1verges y t e mut ompar1son . est w1t e n~~ n n-~ n + n-+oo n n=l 5n + ' . hannonic series. Thus, the series E 5 ( - 1 )" 1 is conditionally convergent. n=O n+ 7. lim lak+Il = lim [(k-+: 1) (~_)'"+ 1 ] = lim k + 1 (~)1 = ~ lim (1 + .!..) = ~'( 1) = 1 < ·1,sotheseries . k~oo a k k-oo k(~)k k -too k 3 3 k~oo ' k 3 ' 3 E k ( ~) k is apsolutely convergent by the Ratio Test. since the terms of this series are positive, absolute convergence is the n=1 same as convergence. 9. lim 1 an+ I I = lim [ (l.l)n+l . ~J = lim (1.1)n 4 ·= (1.1) lim 1 · = (1.1) lim ( ~~ )4 n --<oo an n-too (n + 1)4 (1.1)n n-+oo (n + 1)4 ~-oo (n + 1)4 n -o oo 1 + n = (1.1)(1) = 1.1 > 1, so the series f: ( - 1)n (l.~)n diverges by the Ratio Test. n = l n el/n e ( 1 ) oo 1 · ' 00 e l /n 11 . Since 0 ~ - 3- ~ 3 = e 3 and 2: 3 is a convergent p-series [p = 3 > 1 ], L: - 3- converges, and so n n n n = l n n = l n oo (- 1)"el/n. 2:: 3 is absolutely convergent. n=l n ® 2012 Cengagc: Learning. All JUghts Reserved. May not be scan~. copied. or duplicated, or posled 10 a publicly accessible website, in whole or in p3l1. SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS D 69 . lan+l l . [ w n+l {n+1)4 2 "+ 1 ] . (10 n + 1) 5 . . 00 10" 13. Ji_~ ~ = J2...~ (n + 2) 42n+3 . . 10" = n~ 42 . n + 2 = 8 < 1, so the senes n~l (n + 1)42~+1 , is absolutely convergent by the Ratio Test. Since the. terms of this series are positive, absolute convergence is the same as convergence. \ (-1)" arctann\ rr/ 2 . oo rr/ 2 rr 00 1 . . oo (-1)" arctann 15. 2 < - 2 , so smce 2::: - 2 = -2 2::: 2 converges (p = 2 > 1), the giVen senes 2::: 2 n n n = l n n =l n n=l n converges absolutely by the Comparison Test. 17. f: ( -1)" converges by the Alte~nating Series Test s.ince lim - 1 1 = 0 and {-ln1 } is decreasing. Now Inn < n, so .. = 2 Inn n -oo nn n - 1- > .!, and since f: .! is the divergent (partial) harmonic series, f: - 1 1 diverges by the Comparison Test. Thus, Inn n n=2.n n = 2 nn ~ (-1)". di' II w -In IS con twna y convergent. n=2 n icos{nrr/ 3)1 1 00 1 . . 00 cos{nrr/ 3) 19. 1 :5 1 and 2::: 1 converges (use the Ratto Test), so the senes 2::: 1 converges absolutely by the n. n. n=l n. n=l n. Comparison Test. . . n 2 + 1 . 1 + 1/n 2 1 . 00 ( n 2 + 1 ) " . 21 . lim y/i(lJ = lim - 2 2 1 = hm 2 + 1 / 2 = -2 < 1, so the senes 2::: 2 2 1 IS absolutely convergent by the n-oo n-+oo n + n -oo n n =l n + Root Test. 23. lim y'jaJ = lim " (1 + .!)" 2 = lim (1 + .!)" = e > 1 [by Equation 7.4.9 (or 7.4* .9) [ ET 3.6.6] ], n--..oo n-... oo n n~oo n 00 ( 1)"2 so the series L;: 1 + - diverges by the Root Test. n =l n . . . I an+ll . I (n + 1)10C!1QO"+l n! I . 100 (n + 1)100 . 100 ( 1 )100 25. hm -- - Inn · . - lim -- -- - lim -- 1 + - n-oo a,. -n- co (n + 1)! n 100100" .-n-oo n+ 1 n - n-co n + 1 n = 0 · 1 = 0 < 1 oo nl00100" . so the series 2::: 1 is absolutely convergent by the Ratio Test. n=l n 27. Use the Ratio Test with the series 1 · 3 1 · 3 · 5 1 · 3 · 5 · 7 n - 11 · 3 · 5 · · · · · (2n - 1) _ co ,._11 · 3 · 5 · · · · · (2n - 1) 1 -3!+ - 5-!- - 7! + .. · +(-1) {2n - 1)! +···-n~l(- 1) · (2n - 1)! lim lan+ll = lim ~ (- 1)"~1·3·5 .... ·(2n -1)[2(n +1)-l]_ (2n-1)! I n-oo a,. n-oo (2(n+1) - 1]! (- 1)"-1 ·1·3·5·····(2n-1) =lim ~ (-1)(2n+l)(2n- 1)! 1 = lim _.!_=0 < 1, n -oo (2n + 1)(2n)(2n - 1)! n-oo 2n.. so the given series is ahsolutely convergent and therefore convergent. ® 2012 Ccngagc Lc.lm..ing. All Rights Resen·cd. May not be scanned. copiL-d. or dupllcaled. or posted to a publicly ncccssiblc: \\'Cbsilc. in whole or in pan. 72 D CHAPTER 11 INFINITE SEQUENCES AND SERIES 11.7 Strategy for Testing Series 1. -- < - = - for all n > 1. 1 1 (1)n n+3n 3n 3 - f (~)n is a convergent geometric series [lrl = t < 1), so f ~ n = l . n=l n + 3 converges by the Comparison Test. 3. lim la..l = lim _2:_ 2 = 1, so lim a n= lim (-1}"_2:_ does not exist. Thus, the series f (- 1)n __.!:_diverges by n - oo n-oo n + n--<00 n -<oo n + 2 n=l n + .2 the Test for Divergence. 5. lim Jan+II = lim I (n + 1? 2n • (-5)" I = lim 2(n+ 1)2 = ~ llm (1 + .!)2 = ~(1) = ~ < 1, so the series n-oo an . n-<oo ( -5)n+l n2 2 n-l n - oo 5n2 5 n-oo n 5 · 5 00 n2 2n-l n~l ( _ 5)" converges by the Ratio Test. 7. Let f(x) = ~-Then f is positive, continuous, and decreasing on [2, oo), so we can apply the In..tegral Test. xvlnx Since J ~ dx [d..,:' 11°/x, ] = j u- 112 du = 2u112 + C = 2 JinX+ C, we find xvlnx u- ex x f oo ~ = lim r d~ = lim [2J!nX] t = lim ·(2 villt- 2 v1Ii2) = oo. Since the integral diverges, the J2 X y lnx t .... oo J2 X y lnx t_.oo 2 t-oo given series f ~ diverges. n = 2 n vlnn 00 • 00 k2 . . 9. 2: k 2e-k = 2: ""'"'k· Usmg the Rat1o Test, we get k=l k=I e = 12 . .! = .! < 1, so the series converges. e e 11. f: (~ + 3~, ) ~ E ~ + E (.!)n· The first series converges since it is a p-series with p = 3 > 1 and the second n = l n n=l n n=l 3 series converges since it is geometric with lrl = ~ < 1. The sum of two convergent series is convergent. 13. lim la"+1 1= lim J 3n+l (n+ 1) 2 -~1= lim 3(n + 1)2 =3lim n~ 1 =0 < 1,sotheseries f 3n~2 n-oo an . . n--<00 (n + 1)! 3nn2 n--<00 (n + 1)n2 n --<00 n . ' n=l n. converges by the Ratio Test. 00 (6)k 00 2k-13k+l 00 3 (6)k k~l k converges. It follows from Theorem 8(i) in Section 11.2 that the given series, k~l kk = k~l 2 k , also converges. © 2012 <:engage Lcnming. All Rights Rese"·ed. Moy not be scanned, eopied, or dupuented, or posted to • publicly accessible website. in whole oc in p:lrt. SECTION 11.7 STRATEGY FOR TESTING SERIES 0 73 17 lim I an+l I= lim 11 · 3 · 5 .. · · · (2n- 1)(2n + 1) . 2 · 5 · 8 · .... (3n - 1) I = lim 2n + 1 · n-oo a n n--+oo 2 · 5 · 8 · · · · · (3n- 1)(3n + 2) 1 · 3 · 5 · · · · · (2n- 1) n--+oo 3n + 2 = lim 2 + 1/ n = ~ < 1, n--+oo 3 + 2/ n 3 oo 1 . 3. 5 .. · .. (2n - 1) so the series 2::: ( 3 ) converges by the Ratio Test. n=l 2 · 5 · 8 · · · · · n - 1 . ln x 1 2 - ln x ? lrin. . 2 19. Let f(x) = r . Then f (x) = 2 x 312 < 0 when lnx > 2 or x > e-, so r ts decreasmg for n > e . vx . . vn . I R I lim In n I' 1/n 1' 2 0 h . ~ ( )"Inn b By I'Hosptta 's u e, r = tm ( ) = tm r = , sot e senes n6- 1 - 1 r;;, converges y the n--+oo vn n--+oo 1/ 2 Vn n - oo vn - vn Alternating Series Test. 21 . lim lan. l = lim IC- l )ncos(l/n2)1 = lim icos(1/n2).1 =cos O= 1, so the series E (-l)'' cos(l/ n 2 ) diverges by the n-oo n-oo n -'oo . n =l Test for Divergence. 23. Using the Limit Comparison Test with a n = tan(.!.) and b,. = .!., we have n . n lim an = lim tan(1/ n) = lim tan(l/ x) M: lim sec2(1/x). (- 1/x2) = lim sec2(1/ x) = 12 = 1 > 0. Since n-oo bn n --+oo 1/ n :z:--+oo 1/x :r-oo -1/x2 :z:--+oo 00 00 2::: bn is the divergent harmonic series, 2::: a,.. is also divergent. n = l n.=l U I · ~ li I an+l I . 1' I (n + 1)! c" 2 1 1. (n + 1)n! · e" 2 1. n + 1 oo n! 25. se t 1e Ratto . est. m - - = un < +t)~ · - 1 = rm 2 2 = 1m ~+t = 0 < 1, so :L ---:r f"L-H>O an .,._.00 e n n. ft. - 00 en + n+ln! r&.-oc e n n= l en converges. 27. roo ln: dx = lim [- ln X - .!.] t . [using integration by parts] M: 1. So E ln: converges by the Integral Test, and since ./2 X t-oo X X 1 n=l n k ln k k ln k ln k . . oo k In A: • -----,-3 < - k3 = -k? , the gtven senes 2::: 3 converges by the Companson Test. (k + l) .. ·- k=t(k+ l) 29. f a n = f ( -1)" ___!_h = f ( - 1)" b,,. Now b,. = - 1- 1 - > 0, {b,..} is decreasing, and lim b, = 0, so the series n =l n=l cos n "·=1 cos 1 n n -oo converges by the Alternating Series Test. 0 W . 1 2 2 d ~ 1 '. . . ~ 1 . b h r: rtte co~lt n = < - an 6 - ts a convergent geometrtc senes, so 6 --1 - ts convergent y t e ~ e" + e-" e" n = l e" n = l cos 1 n Comparison Test. So E ( - 1)" ___!_h is absolutely convergent and therefore convergent. n=l cos n k 31. lim a~.; = lim k 5 4 k = (divide by 4") k--+oo k-oo 3 + oo s" Thus, 2::: ~ diverges by the T<;st for Divergence. k = 1, 3 +4 ' <D 2012 Ccn~:tgc Lc:tmin{l- All Rights Rescn·cd. May no1 be scanned, copied, or dup1icau:d, or ('K)stl-d h) n publicly ncccs..'lihl~ wcbsih!, in whole or in rart. 74 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES "2 jn 1 1 1 oo n2 33. lim ~ = lim (~1) = lim [( 1) I t = li (1· 11 r = - < 1, so the series I: (·-n-) n-+oo n-+00 n + n-+00 n + n m + n e . n=1 n + 1 n-->oo converges by the Root Test. 35. an = n1; 1/n = n . ~1/n, so let bn = ; and use the Limit Comparison Test. l. a, lim 1 1 0 un - = . --= > n-+oo bn n-+oo n1/n [see Exercise 4.4.61 ], so th~ series f: 1 11,. diverges by comparison with the divergent harmonic series. . n=l n . 00 37. lim ~ = lim (21/" - 1) = 1 - 1 = 0 < 1, so the series I: ( y/2 - 1)"' converges by the Root Test. n-~oo n-too n-=1 . 11.8 Power Series 1. A power series is a series of the form :L:::'=o enx" = co + c1x + c2x2 + c3x3 + · ··,where x is a variable and the en's are constants called the coefficients of the series. More generally, a series of the form :L:::'=o c,,(x - a)" =eo+ c1 (x- a)+ c2(x- a)2 + · · · is called a power series in ( x - a) or a power series centered at a or a power series about a, where a is a constant. I a I ~ (-1)"+1 (n+1)x"-H I I n+1 I [( 1) ] lim n+1 = lim ( ) = lim ( -1)--x = lim 1 + - JxJ = JxJ. By the Ratio Test, the n--+00 Un n-tOO - 1 71. nxn n-H)O n n-00 n ' 00 . series I: ( -1)"nxn converges when !x i < 1, so the radius of convergence R = 1. Now we' II check !he endpoints, that is, n=l x = ± 1. Both series f: ( - 1)"n(±1)" = f: (=F1)"n diverge by the Test for Divergence since lim J(=F1)"nl = oo. Thus, n=l · n=l TI.--+OO the interval of convergence is I = ( -1 , 1). x" ' . I a,.+1 1 . I x"+l 2n - 11 . (2n - 1 ) . (2- 1/n ) 5. If an = - 2 --, then lim -- = hm - 2 1 · - = !1m - 2 1 Jx l = hm 2 1 / lxJ = JxJ. By n- 1 n-+oo an n-+oo n + x" n-+oo n + n-+oo + n ·oo ~ oo 1 · the Ratio Test, the series I: - 2 -- converges when Jx! < 1, so R = 1. When x = 1, the series I: - 2 1 diverges by n=l n - 1 n=l n - . . I ~ 1 . 1 1 d 1 ~ 1 d' . . . I . I fth h . . companson Wit 1 6 - srnce --- > - an - 6 - 1verges smce I t ts a constant mu t ip e o e armoruc senes. n=1 2n 2n - 1 . 2n 2 n=l n When x = - 1, the series f: ( -1) n converges by t~e Alternating Series Test. Thus, the interval of converge~ce is [ -1, 1) . n=1 2n -1 7. If an= x",, then lim lan+l l = lim I ( x"+\; ·. n ! I= lim 1- x -1 = !xi lim - 1- = lxl · 0 = 0 < 1 for all real x. n. n -+ oo an n-+oo n + 1 . X" n -+oo n + 1 n-+oo n + 1 So, by the Ratio Test, R = oo and I= ( - oo, oo). lim I an+1 I = lim J (n + 1)2 x"+l . ~~ = lim I x(n + 1)2 1 = lim [JxJ (1 + .!. )2] = EJ.(1)2 = ~ JxJ. By the n - oo . an n-CXJ 2n+l n 2 x"· n.-oo 2n2 n-oo 2 n 2 . . © 2012 Cengagc Learning. All Rishts Reserved. May not be scanned. copied. or duplicated. or posted to u publicly occcssiblc website:, in whole or in part. SECTION 11.8 POWER SERIES 0 n (b) It does not follow that I:::'=o en( - 4)" is necessarily convergent. [See the comments after Theorem 3 about convergence at the endpoint of an interval. An example is Cn = ( - 1 t / ( n4").] (n!)k n 31. If a,.=-( ) ' x , then kn. lim I an+ll = lim [(n + 1)!]k (kn)! lxl = lim - (n + 1)k lxl ....... oo a., n ..... ex> (n!)k [k(n + 1)]! ,.....,CX> (kn + k)(kn + k- 1) · · · (kn + 2)(kn + 1) . [ (n+1) (n+1) (n+1)] = J~ (kn + 1) (kn + 2) ... (kn + k) lxl l. [ n + 1 ] lim [ n + 1 ] 1. [ n + 1 ] I I - t.m --- --- · · · liD --- X - n-+ex> kn + 1 ,. ..... ex> kn + 2 n-oo kn + k = ( ~ J lxl < 1 ¢:} lxl < kk for convergence, and the radius of convergence is R = kk. 33. No. If a power series is centered at a, its interval of convergence is symmetric about a. lf a power series has an infinite radius of convergence, then its interval of convergence must be ( - oo, oo), not [0, oo). (-1t X2n+l 35. (a) If a,. = I( 1)1 22n+l, then n. n+ . r n+l li . . lim I a I I x2n+3 . nl(n + 1)'22n+l l (x)2 1 ,.:.,~ ---a,: = n ..... ~ (n + 1)!(n + 2)! 22n+3 . x2n+l = 2 ,..._ ..... oo (n + 1)(n + 2) = 0 for all x. So J 1 ( x) converges for all x and its domain is ( - oo, oo). (b), (c) The initial terms of J1(x) upton = 5 are ao = ~· xu and a(j = - 176 , 94 7 , 200 . The partial sums seem to approximate J1 ( x) well near the origin, but as lx l increases, we need to take a large number of terms to get a good approximation. 37. S2n-l = 1 + 2x + x2 + 2x3 + x 4 + 2x5 + · · · + x2"-2 + 2x2"- 1 = 1(1 + 2x) + x2 (1 + 2x) + x 4 (1 + 2x) + · · · + x 2" - 2(1 + 2x) = (1 + 2x)(1 + x 2 + x 4 + ... + x2~-2 ) 1 - x 2" 1 + 2x = (1 + 2:c)- 1 -2 [by (11.2.3) with r = x 2] --+ - 1 --2 as n--+ oo by (11.2.4), when lx l < 1. - X - X I 2n 1 +2x. 2n 0" I I 1Th &: 1+2x. . A so S2n = S2n-1 + x --+ 1 _ x2 smce x --+ •Or x < . ere,ore, Sn --+ 1 _ x 2 smce S2n and S2n-1 both 1 + 2x . . 1 + 2x approach ---2 as n --+ oo. Thus, the mterval of convergence IS ( - 1, 1) and f ( x) = - - 2 . . 1 -x 1 - x © 2012 Cengagc Lcoming. All Rights Reserved. May not be scnnncd, copied, or dupliCD.tcd: or posted to a publicly acccs!tiblc website. in whole or in part. 78 D CHAPTER 11. INFINITE SEQUENCES AND SERIES 39o We use the Root Test on the series I:; c..x" 0 We need lim \/lc,x"l =I xi lim y'jCnT = c !x! < 1 for convergence, or n.- oo n--too lxl < 1/c, so R = 1/c. 41 . For 2 < x < 3, I:; cnx" diverges and I:; dnx" .convergeso By Exercise 11.2.69, I.:;( en+ dn) x" diverges. Since both series converge for I xi < 2, the radius of convergence ofi.:;(en + dn) x" is 2. 11.9 Representations of Functions as Power Series 00 00 10 If f(x) = I:; c..xn has radius of convergence 10, then!' (x) = I:; nc..x"-1 also has radius of convergence 10 by n = O n = l · Theorem 20 3. Our goal is to write the function in the form - 1 · 1 , and then use Equation ( I) to represent the function as a sum of a power - r 1 1 00 00 0 series. f(x) = -- = ( ) = I:; ( -xr' = I:; ( -1)"x" with 1-xl < 1 ¢:? lxl < 1, so R = 1 and I= ( - 1, 1). 1 +X 1 - -X n =O n= O 2 2( 1) 200 (x)" oo 1 lxl 5o f(x) = 3 _ x = 3 1 _ x/ 3 = 3 n~o 3 or, equivalently, 2 fo 3,.+1 x" 0 The series converges when 3 < 1, that is, when I xi < 3, so R = 3 and I = ( -3, 3). The geometric series ... ~o [- (~) 2] " conver~es when 1- (~) 21 < 1 ¢:? ~ 2 1 < 1 ¢:? lxl2 < 9 ¢:? lxl < 3, so R = 3and I= (-3,3). The series converges when lxl < 1, so R = 1 and I = ( - 1, 1). · ( ) 1 +X -(1 - x) + 2 ( 1 ) ~ n 2 ~ n A second approach: f x :::; -- = = - 1 + 2 -- = -1 + 2 LJ x = 1 + LJ x . 1 - X 1 - X 1 - X n = O n=l A third approach: f(x)= 1 +x =(l +x)(- 1 -) = (l +x)(l +x+x2 +x3 + ·o·) 1 - x 1 -x = (1 + x + x2 + x3 + .. o) + (x + x2 + x 3 + x4 + o o o) = I+ 2x + 2x2 + 2x3 + o .. = 1 + 2 f: Xno n=l © 2012 Cengoge Learning. All Rights Reserved. Mo.y not be scanned. copied, or duplicated, or posted toR publicly accessible website. in whole or in p:ut. SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES D 79 3 11. f (x) = 2 2 x - x- 3 A B ..,....------,-,-,---"7 = -- + -- => 3 = A(x + 1) + B(x - 2). Let x = 2 to get A = 1 and (x- 2)(x + 1) x - 2 x + 1 · x = - 1 to get B = - 1. Thus 3 __ 1 _ _ _ 1_=~( 1 )- 1 =-~ E (=)" _ E(-x)n x2- x - 2 - X- 2 X+ 1 - 2 1- (x/2) 1 - (- x) 2 n=O 2 n=O = f: [-~ (1:)" -1(-1)"']xn = E [(- 1t+l- n~l]x" n=O 2 2 n=O 2 We represented f as the sum of two geometric series; the first converges for x E (-2, 2) and the second converges for ( -1, ~) . Thus, the sum converges for x E ( - 1, 1) =I. ) 1 · d ( -1 ) , d [ ~ ( 1)n " ] 13. (a) f(x = = - -- = -- L..... - x (1+x)2 dx 1+x dx n=O [from Exercise 3] ~ ~ = L ( -1)"+1nx"- 1 [from Theorem 2(i)] = L ( -1)"(n + l)x" with R = ~ . n=l n=O In the last step, note that we decreased the initial value of the summation variable n by 1, and then increased each occurrence ofn in the term by 1 [also note that (-1)"+2 = (- 1)" ]. (b)f(x)= 1 =-1:j_[ 1 ] =-1:j_[E(-1)"(n+ 1)x"] [frompart(a)] (1+x)3 2dx (1+x)2 2dx n=O ~ ~ = -~ L ( -1)"(n + 1)nxn-1 = ~ L ( - 1)n(n + 2)(n + 1)x" with R = 1. n=l n=O x2 1 1 ~ (c)f(x) = (1 +x)3 =x 2 · (l+x)3 = x 2 · 2nJ; 0 {- 1)''(n+2)(n+1)x" [frompart(b)] = 1: E (-1)"{n+2)(n+ l)x"+~ 2 n=O To write the power series with x " rather ~an ·x"+2 , we will decrease each occurrence of n in the term by 2 and increase 1 . CX> the initial value of the summation variable by 2. This gives us 2 .. ~2 ( -1)"(n)(n- 1)x,. with R = 1. 15. f(x) = ln(5 - x) = -~~ = _1: ~~ = -1:1 [f: (::\"] dx = C- ~ E xn+l = ·C..:... f: x" 5-x 5 1 - x/5 5 n=O 5} 5n=o5"(n+ 1) n=tn5n Putting x = 0, we get C = ln 5. The series converges for Jx/51 < 1 <=> Jx J < 5, so R = 5. 1 1 00 17. We know that - 1 4 = ( 4 ) = L ( -4x)". Differentiating, we get + X 1- - X n=O f(x) = {1+x4x)2 = · -4x. -4 2 =-X f;{ - 4)n+1 (n+1)xn = f;(-1)n4n(n+1)xn+l (1 + 4x) 4 n=O n=O for J-4xJ < 1 <=> Jxl < ;}, so R = ;} . ® 2012 Ccngage lc.aming. All Rights Rescrvct.l. Moy not be scanned, copic.-d. or duplicntcd. or posted to u pllblic!y ucccssiblc: website. in whole: or in pan. 82 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES x 3 x 5 x1 (0.2? (0.2)5 (0.2)7 33 By Example 7 arctan x = x - - + - - - + · · · so arctan 0 2 = 0 2 - -- + -- - --+ ... . ' 3 57' .. 3 5 7 . Th . . I . 'f t1 1 · (0.2) 7 e senes 1s a ternatmg, so 1 we use 1ree terms, t 1e error ts at most - 7 - ~ 0.000 002. (0 2? (0 2? Thus, to five decimal places, arctan0.2 ~ 0.2- - · 3 - + -· 5 - ~ 0.197 40. 2 , 1 2 00 (- 1)"2n(2n- 1)x 2 " 00 (-1t2nx2"' 00 (- 1)"x2"+2 X Jo (x) + XJo(X) +X Jo(x) = n~l 22n(n!)2 + 1~1 22"(n!)2 + n~O 22" (n!)2 00 (- 1)"2n(2n-1)x2" oo (- 1)"2nx2'' oo (- 1t-1 x2" = n~l 22"(n!)2 + ,~1 22"(n!J2 + ,~1 22n-2 [(n- 1)!]2 _ 00 (- 1)"2n(2n -1)x2" 00 (- 1t2nx2" 00 (-1)"(-1)-122n2x2" - r~l 22"(nl)2 + n~l 22"(n!)2 + ,~1 22"(n!)2 = ~ (-1)" [2n(2n - 1) + 2n - 2 2 n 2 ] x2" n~l 22"(n!)2 , = ~ (-1)" [4n 2 - 2n + 2n- 4n2 ] 2n = 0 L- 22" ( 1)2 X n=l n. Since 16 .i28 ~ 0.000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places, J01 Jo (x) dx ~ 1 - 1; + 3~0 ~ 0.920. 00 Xn I 00 nxn- l 00 Xn-1 37. (a) f(x) = n~O nf ::} f (x) = •~1 ~ = •~1 (n- 1)! oo x.,." I: - 1 = f(x) n = O n (b) By Theorem 9.4.2, the only solution to the differential equation df(x)/d.'C = f(:c) is f(x) = Ke" , but f(O) = 1, so K = 1 and f(x) = e". Or: We could solve the equation df(x) / dx = f(x.) as a separable differential equation. n I I I n+l 21 . ( ) 2 . 39. If a,.. = X 2' then by the Ratio Test, lim an+ I = lim (X )2 . ~ = lxl lim ....2::_1 = lx l < 1 for n - n.-oo an n-oo n + 1 xn n----+oo n + convergence, so R = 1. When x = ± 1, f I x: I = f ~which is a convergentp-series (p = 2 > 1), so the interval of n = l n " =1 n convergence for f is [-1, 1]. By Theorem 2, the radii of convergence off' and f" are both 1, so we need only check the 00 x" 00 nxn- 1 00 x" endpoints. f(x) = I; 2 => . f'(x ) = I: - -2- = I; --, and this series diverges for x = 1 (harmonic series) n=l n n=l n n=O n + 1 @ 2012 Cengage Lc:tminiJ. All Rights Reserved. May not be sc:~.Mcd, copied, or duplicated, or posted w n publicly accessible wcbsile, in whole or in part. SECTION 11.10 TAYLOR AND MACLAURIN SERIES D 83 00 n.-1 and converges for x = - 1 (Alternating Series Test), so the interval of convergence is [-1, 1). !" (x) = I: ~ diverges n=l n+ 1 at both 1 and - 1 (Test for Divergence) since lim ~ = 1 =/= 0, so its interval ofc~nvergence is ( - 1, 1). n--.oo n+ 1 • 00 ·X2n+ l 1 41. By Example 7, tan-1 x = L: ( -1)"-- for lxl < 1. In particular, for x = r.;• we n=O 2n+ 1 y3 7r - 1 ( 1 ) 00 n (1/v'3) 2 n +l ~ ( )" (1)" 1 1 have 6 = tan y3 = n~O ( - 1) 2n + 1 = n~O - 1 3 y3 2n + 1' SO 7r=~f (-1)" =2v'3f (-1)" . y'3 n = O (2n + 1)3" n=O (2n + 1)3" 11.10 Taylor and Maclaurin Series oo J<">(a) j <B>(5) 1. Using Theorem 5 with n~o b .. (x - 5)", bn = -----n!' so ba = - 8 -1 - . 3. Since j <">(o) = (n + 1)!, Equation 7 gives the Maclaurin series oo J (n) (O) oo (n + 1) 1· oo L: --1-x" = L .1 • x" = L: (n + 1)xn. Applying the Ratio Test with an= (n + 1)xn gives us n=O n. n=O n. n=O . I an+l I . I (n + 2)x"+ll I I . n + 2 I I I I hm - - = lim ( . ) = x hm -- = x · 1 = x . For convergence, we must have lxl < 1, so the n-oo an n--.oo n + 1 xn n-+oo n + 1 radius of convergence R = 1. 5. n f(n) (:c) j<">(o) {1 - x)-2 = f(O) + f' (O)x + !"(O) x2 + f"'(O) x3 + j< 4 >(o) x4 + ... 2! 3! 41 0 {1- x)-2 1 = 1 + 2x + ¥x2 + ¥x3 + l224ox4 + ... 1 2(1- x)-3 2 2 6(1 - x)-4 6 = 1 + 2x + 3x2 + 4x3 + 5x4 + · · · = f: (n + 1)xn n=O 3 24(1 - x)-5 24 4 120(1- x)-6 120 lim I a,.+l l :;= lim I (n + 2)xn+ll = lxl lim n + 2 = lxl (1) = lxl < 1 n-+00 a,. n-oo (n + 1)xn n --.oo n + 1 for convergence, so R = 1. sin 7rX = f(O) + !' (O)x + !" (O) x2 + !"' (O) x3 2! 3! 7. n J<n>(x ) / {n ) (0) / (4 ) (0) ~ !(5) (0) 5 + -4-! -x + - 5-! -x +·· · 0 sin 1rx 0 1 7rCOS 7rX 7r 7ra 7ro = 0 + 1rx + 0 - -x3 + 0 + - x5 + ... 3! 5!. 7r3 3 7r/j 5 7r 7 7 = 1rx - -x + - x - - x + · · · 31 5! 71 2 -1r2 sin 1rx 0 3 - 7r3 COS 7rX -7r3 4 1r4 s in 1rx 0 oo · 2n+l = L ( - 1)" 7r x2n+l n = O (2n+ 1)! 5 7r° COS7rX 5 7f lim n+l = lim · = lim = 0 < 1 I a I 1 7r2n+3 x2n+3 (2n + 1)1 I 7r2 x2 n--.oo a,. n --.oo (2n + 3)! 7r2n+l x2n+l .,._,00 (2n + 3)(2n + 2) for all x, so R = oo. .. © 2012 Cenga;~e Learning. All RighiS Rese:n'\.'1.1. May not be scanned. copil.!d, or dupl icaled, or poslct.llo 3 publicly accessible wcbsih:, in whole or in part I 84 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES 9. n 0 1 2 3 4 11. n 0 1 2 3 4 13. n 0 1 2 3 4 5 6 15. n 0 1 2 3 <! 5 f (nl (x) t<">(o) 2"' 1 2"'(ln 2) ln2 2"'(1n 2)2 (In 2)2 2"'(1n 2)3 (In2? 2"' (In 2)4 (In 2)4 f (n)(x) J (n) (0) sinh x 0 cosh x 1 sinhx 0 cosh x 1 sinhx 0 J Cnl(x) J(nl (l) x 4 - 3x2 + 1 - 1 4x3 - 6x -2 12x2 - 6 6 24x 24 24 24 0 0 0 0 J(nl(x) J(n)(2) In x ln2 1/x 1/2 -1/x2 - 1/22 2/x3 2/23 - 6/x4 - 6/24 24/x5 24/25 lim I a .. +! I= lim I (ln2)n+lxn+l . n ! I n-oo a,. n- oo (n + 1)! (In 2)"x" = lim (In 2) !xi = 0 < 1 for all x , so R = oo. n-oo n -1-1 { 0 if n is even J<~> (o) = 1 ifn is odd 00 x2n+l so sinhx= I: (2 _ )1• n=o n+ 1. x2n+l Use the Ratio Test to find R. If an = ( )I, then 2n + 1 . r I an+ II r ·I x2n+3 (2n + 1)! I 2 r 1 "~ ~ = ,..:.,~ (2n + 3)! · x2n+l = x · n~ (2n + 3)(2n + 2) = 0 < 1 for all x, so R = oo. f (n) (x) = 0 for n ;:: 5, so f has a finite series expansion about a = 1. 4 t<"l (1) f(x) = x4 - 3x2 + 1 = 2::::: --(x - 1t n=O n! - 1· 0 - 2 1 6 . 2 = 0! (x - 1) + lT (x - 1) + 2f (x - 1) 24 (' 3 24 4 + 3f x - 1) + 41 (x- 1) = - 1 - 2(x- 1) + 3(x - t? -1- 4(x- 1)3 + (x- 1)4 A finite series converges for all x , so R = oo. f(x) = In x = f: t <n>?) (x- 2)" n=O n. ln 2 0 1 1 -1 2 2 )3 = 0! (x- 2) + 1! 21 (x- 2) + 2! 22 (x- 2) + 3! 23 (x- 2 + ~ (x - 2)4 + ~ (x - 2)5 + · · · 4J2d 5!25 = ln2 + f:(-1)"+dn ~:)! (x -2)'' n=l n.2 c:E> :!012 Cengnge Lc:unin~. All Rights Rcscn1:d. ~ tay not be SCnMCd. copk-d. or duplicated, or postct.l loa publicly ncccs.~iblc wcbsilt:, in whole or in part. SECTION 11.10 TAYLOR AND MACLAURIN SERIES D 87 1.5 - 1.5 t--r.'-----+---~ .. :--11.5 = 1- ~x'' + fixs- 7~oxl2 + ... The series for cos x converges for all x , so the same is true of the series for f(x), that is, R = c;o· Notice that, as n increases, Tn(x) becomes a better approximation to f(x). oo n oo ( )" oo n 41 :r; (11) "" X - x "" ....=::___1 ·= ""{-1)" .:._,SO . e = L- I, so e = L- L- 1 n =O n. n =O n. n =O n f (x) = xe- "' = f: ( -1)" ~ x"+l · n=O n. The series for e"' converges for all x, so the same is true of the series for f(x); that is, R = oo. From the graphs off and the fi rst few Taylor polynomials, we see that T,. ( x) provides a closer fit to f ( x) near 0 a!; n increases. - 1.5 ( 7T ) 7r . oo x2" x2 x4 x6 43. 5° = 5° 1800 = 36 radians and cos x = .. ~o ( -1t (2n) ! = 1 - 2f + 41 - 6f + ·· ·,so \ . T8 = T9 = T10 = T11 • \ 't \ . 7T (7r/ 36)2 (7r/ 36)4 (7r/ 36)6 (7r/ 36)2 . (7r/36)~ cos - = 1 - --- + - + ... Now 1 - ~ 0 99619 and addmg ~ 2 4 X w-G 36 2! 4! 6! . 21 . 4! ~ . does not affect the fifth decimal place, so cos 5° ~ 0.99619 by the Alternating Series Estimation Theorem. (b)sin- lx= l 1 dx=C + x + f: 1·3 · 5 ·· .. · (2n-1)x2n+l V1- x2 n = l (2n + 1)2n · n ! = x + f; 1 · 3 · 5 · · · · · {2n - 1) x 2 n+l since 0 = s in- 1 0 = C. n=l (2n + 1)2". nl (16) 00 x 2n 47. COSX = "~0(-1)" {2n) ! I oo X Gn+2 x cos(x 3 ) dx = C + .~o ( - 1)" (6n + 2)(2n)! , with R = oo. (16) oo n x2" oo x2" 49. cosx = .~o ( -1) (2n)! => cosx - 1 = .~1 ( - 1)" {2n)! . 1 oo 2n - 1 cos x - = 2: {-1)" .::_____ => X n=l (2n) ! I COS X - 1 oo n X 2n • dx = C + 2:::{-1) 2 · (2 )1, wtth R = oo. X n =l n n . @ 20 12 Ccngagc Lcunling. All Rights Ri:st:rvc:J. May not be scwmct.l. cupicd, or duplicated, or pO!'Itr.:d to a publicly IU:cc~ siblc website, in whole or in pan. 88 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES • oo x2n+l 00 x2n+ 4 51 . arctanx = 2: (- 1)" - 2 1 for lxl < 1, so x3 arctanx = 2: ( -1)" - - for Jxl < 1 and n = O n + n =O 2n+ 1 I oo x2n+5 x3 arctanx dx = C + ,..;;:o ( -1)"' (2n + 1)(2n + 5)' Since ~ < 1, we have (1 /2)5 (1/ 2r (1/ 2)9 · · {1/ 2)11 1.5 -~ + ~ ~ 0.0059 and subtracting Til ~ 6.3 x 10- 6 does not affect the fourth decimal place, so J0112 x3 arctan x dx ~ 0.0059 by the Alternating Series Estimation Theorem.' 53. v'1 + x4 = (1 +.x4 ) 112 = 2: (x4)", so }1 + x4 dx = C + L: -- and hence, since 0.4 < 1, 00 (1/ 2) I oo ( 1/ 2) x4n+1 n =O n , n =O n 4n + .1 we have I= } 1 +x4 dx = E 10.4 00 (1/ 2) (0.4)4n+l o n = O n 4n + 1 (o.4)1 ~ (o.4)5 H-4) (o.4)9 H-t)(-~) (o.4)13 H-4)(-~)(-~) (o.4)17 =(1)or+l!-5- +-2-,---9-+ 3! ~+ 4! ---rr- +· ·· (0.4)5 (0.4)0 (0.4)13 5(0.4) 17 = 0.4 + ----w- - ----;:;2 + 208 - 2176 + ... Now (O~~)o ~ 3.6 x 10- 6 < 5 x 10- 6 , so by the Alternating Series Esti~ation Theorem, I ~ 0.4 + (0~~) 5 ~ 0.40102 (correct to five decimal places). x -ln(1 + x) x- (x - l x2 + l x3 - lx4 + lx5 - • • ·) lx2 - lx3 + l x4 - l x5 + · · · 55. l im = lim 2 3 ., 4 fi = lim 2 :l 4 5 x-0 x2 x -.0 x· .,_,o X 2 = lim(! - l x + lx2 - ix3 +: .. ) = ! m-+0 2 3 4 5 2 since power series are continuous functions. sinx- x + l x3 57. lim 6 x-o x5 since power series are continuous functions. . 2 ~ ~ ~ ~ ~ _59. From Equation 1 I, we have e-x = 1 - I + -21 - I + · · · and we know that cos x . = 1 - -21 + -41 - · · · from 1. . 3. . . Equation 16. Therefore, e-"' 2 cosx = (1 - x 2 + ~x4 - ·- ·) (1 - ~x2 + i4 x4 - - - • ). Writing only the tenns with - degree < 4 we get e-"'2 cos x = 1 - lx2 + .l.x4 - x 2 + lx" + lx4 + . · . = 1 - !!.x2 + ~x4 +- ·. - ' 2 24 2 2 2 24 . © 2012 Cc.ngace Learning. All Rights Rcscn·cd. May not be SCOJ:mcd, oopicd~ or duplicated, or po.stcd too publiCly accessible website, in whole or in part. SECTION 11 .10 TAYLOR AND MACLAURIN SERIES 0 89 61 _ x_ <~> x · sin x - x- tx3 + 1~0 x5 - · · · · 1 I 2 7 4 + 6x +aGo X + · · · x- *x3 + l~o xs - .. . j x x - ~ x3 + rlo xs - .. . l x3 _ ....Lxs + .. . 6 120 1 3 1 5 oX - 36X + .. . From the long division above, ~ = 1 + ~x2 + 3~x4 + · · ·. · I SlllX · I oo 4n oo (-x~~) n a 63. L; (-1)"~ = L; - - 1 - = e- x· , by (I I ). n=O n. n =O n. 65. ~(-1)"-J 3nn = ~(-1)"_ 1 (3/5)" = ln(1 + ~) [fromTable l ] = ln~5 n=l n5 n = 1 n 5 n. ( i ) ( ) 71. lfp is an n th-degree polynomial, then p<'l (x) = 0 for i> n, so its Taylor series at a is p(x) = L; ~(x - a)i. i =O t. n. p(i) (a) Put x- a.= 1, so that x = a+ 1. Then p(a + 1) = 2: - .- 1 -. i=O t. n p(i)(x) This is true for any a, so replace a by x : p(x + 1) = L: -.-1 -i = O t. 73. Assume that lf"(x)l ~ M , so /"1 (x) ~ M for a ~ x ~a + d. Now fa"' !'11 (t) dt ~ J: M dt => f"(x) - J"(a) ~ M (x - a) => f"(x) ~ J"(a) + M(x - a). Thus, I: J"(t) dt ~ I: [!"(a)+ M(t- a)] dt => J'(x) - f '(a) ~ J"(a)(x - a)+ ~M(x- a )2 => J'(x) :::; /'(a) + J"(a)(x - a)+ ~M(x- a) 2 => .r: f(t )dt ~ .r: [!' (a) + J"(a) (t - a)+ tM(t - a?] dt => f(x) - f(a) :::; · f' (a)(x- a) + tf"(a)(x - a? + iM(x- a) 3 . So f(x) - f(a) - J'(a)(x- a)- ~f"(a)(x - a) 2 ~ ~M(x - a)3 . But R2(x) = f (x)- T2(x) = f (x) - f(a) - J'(a)(x- a)- ~f"(a)(x- a) 2 , so R2(x) :::; !M(x - a) 3 • A similar argument using f"'(x ) ~ -M shows that R2(x) ~ - t M(x- a)3 . So IR2 (x2)l ::; tM lx - a j3 . Although we have assumed that x > a, a similar calculation shows that. this inequality is a lso true if x < a. © 2012 Ccngngc Lc:uning. All Righls Rcscr'\'cd. May nol be scanned, copied, or duplicated, or posted lo • publicly accessible wcbsilc. in whole or in pan.