Business Calc Homework w answers_Part_88

Business Calc Homework w answers_Part_88 - 436 Chapter 10...

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Unformatted text preview: 436 Chapter 10 Review 25. 23. d 1 cos sin 2 d 1 sin sin cos cos cos 2 2 2 dx d 1 cos cos 2 d d 1 sin cos sin cos sin 2 2 2 dy Solve 0 for with a graphing calculator: the solutions d dy d [ 1.5, 1.5] by [ 1, 1] The tips have Cartesian coordinates 1 2 1 2 1 2 , 1 2 1 2 , . From the are 0, 2.243, 4.892, 7.675, 10.323, and 4 . Using r sin reveals 1.739. , 1 2 , 1 2 , 1 2 , and , the middle four solutions to y horizontal tangent lines at y Solve are 0, dx d curve's symmetries, it is evident that the tangent lines at those points have slopes of 1, 1, 1, and 1, respectively. 0.443 and y 0 for with a graphing calculator: the solutions 3.531, 2 , 9.035, 11.497, and 4 . So the equations of the tangent lines are y 1 2 1.070, x 2, x 2 1 2 1 2 or Using the middle five solutions to find x vertical tangent lines at x Where 2, x r cos reveals y 1.104. y y y y x 1 1 x 0.067, and x 0, 4 ), close dy dx and both equal zero ( dt dt or inspection of the plot shows that the tangent lines are horizontal, with equation y using L'Hpital's rule.) 24. dy d dx d d [2(1 d d [2(1 d 2, x x 1 2 1 2 0. (This can be confirmed or 2 2, and x 2 1 2 sin )sin ] sin )cos ] 2 sin 2 sin 2 4 sin cos 2 cos y y x or 2 cos2 4 sin2 dy Solve d 2 sin2 26. 0 for : 6 2 the solutions are , , 5 3 , and . 6 2 1 and 2 [ 3, 3] by [ 2, 2] Using the first, third, and fourth solutions to find y y Solve r sin 4. dx d reveals horizontal tangent lines at y 0 for (by first using the quadratic formula to 2 find sin ): the solutions are two solutions to find x lines at x zero 2 3 2 3 , 7 11 , and . Using the last 6 6 As the plot shows, the curve crosses the x-axis at (x, y)-coordinates ( 1, 0) and (1, 0), with slope 1 and 1, respectively. (This can be confirmed analytically.) So the equations of the tangent lines are y 0 (x 1) y x 1 and y 0 x 1 y x 1. 27. r cos r sin x y, a line 28. r r2 x2 3 cos 3r cos y2 3x 3 2 2 r cos reveals vertical tangent dy dx and both equal dt dt 2.598. Where , inspection of the plot shows that the tangent 0. (This can be confirmed line is vertical, with equation x using L'Hpital's rule.) 3x 9 4 x2 x y2 9 4 3 2 2 3 , 0 , radius 2 y2 a circle center 3 2 Chapter 10 Review 29. r 4 tan sec r sin 4 r cos dr d 437 37. sin , so 2 r cos x y 4 or x 2 x Length 0 2 ( 1 2 cos )2 2 cos 2 ( sin )2 d 4y, a parabola d 2 30. r cos r cos cos 3 3 3 2 2 3 r sin sin 2 3 0 2 0 4 sin2 d 4 cos 2 8. 0 2 3 38. 1 r cos 2 1 3 x y 2 2 r sin 3 3 dr d 0 0 2 cos /2 2 sin , so Length 2 cos )2 8 cos2 2. d (2 cos 2 sin )2 d (2 sin /2 /2 2 4 8 sin2 2 2d x 31. x r2 r 2 3y 2 3 or y 0 0 0 0 x 3 4, a line 0 y 5y 5r sin 5 sin 39. dr d 8 sin2 3 /4 cos 3 , so 2 32. x 2 y 2 2y r 2 2r sin r 2 sin 33. x 2 Length 0 /4 0 8 sin3 8 sin2 8 sin2 3 3 3 8 sin2 3 2 3 cos 3 3 d sin2 d cos2 d 4y 2 2 16 4(r sin ) 4r 2 sin2 (y 2)2 2 /4 (r cos ) r 2 cos2 34. (x 2)2 (r cos 35. 16 16, or r 2 cos2 16 4 sin2 0 8 1 2 3 2 sin 4 3 /4 3 0 5)2 16 (r sin 5)2 e t, so 2e2t 1 2t e 2 1 2 8 1 2 8 1 64 16 40. dr d /2 sin 2 1 /2 , so Length cos 2 )2 1 sin 2 cos 2 2 c os 2 dx = 2e2t dt 1 dy , 8 dt ln 2 0 ( 1 (et )2 dt e dt 2t /2 /2 /2 /2 d Length ln 2 0 ln 2 0 (1 1 1 cos 2 )2 cos 2 2 cos 2 1 sin2 2 d 1 cos 2 cos2 2 cos 2 sin2 2 4e 4t d 2e2t t 8 ln 2 8 ln 2 8 ln 2 0 dt /2 2d /2 2 dy dt e 4 3 2t 41. 1 ln 2 8 24 dx dt 2 sin t, /2 0 /2 2t, so (2t)2 dt Length 3.087. 0 ( 2 sin t)2 2 t2 sin2 t dt, 3.183. which using NINT evaluates to dx 36. dt dy 2t, dt 3 t 2 1, so 42. (2t) 2 Length 3 3 (t t4 2 2 1) dt 2t 2 1 dt 2 dx dt 3 cos t, 3 0 dy dt 3 t, so (3 t)2 dt 12.363. 3 Length (3 cos t)2 3 0 t cos2 t dt, 4t 2 3 3 which using NINT evaluates to 2 (t 3 2 3 1) dt 4 3 43. Area 0 t 3 3 t 3. 1 (4 4 cos 2 0 1 4 4 sin 2 1 (2 2 2 cos )2 d cos2 ) d 1 2 1 sin 2 4 2 0 9 2 438 Chapter 10 Review /3 0 /3 44. Area 0 1 sin2 3 d 2 1 1 2 2 1 sin (6 ) 12 50. 12 45. [ 1.5, 1.5] by [ 1, 1] [ 3, 3] by [ 2, 2] r 0, such as . Using Area sin 2 and /2 dr d cos 2 sin 2 , where 0 sin 2 cos2 2 d sin 2 2 , so The curves cross where cos 2 the curves' symmetry, /4 2 0 2 /2 0 sin 2 cos cos /2 4 4 cos 2 )2 1] d /4 d 4 Length 4 0 1 [(1 2 /4 4 51. (a) v(t) sin 0 2 0 (cos 2 1 sin 4 8 1 2 2 2 cos 2 ) d sin 2 0 2 4 d [(4 cos t)i dt ( 2 sin t)j] ( 2 cos t)j ( 2 cos t)j] ( 2 2 a(t) ( 4 sin t)i d [( 4 sin t)i dt 46. ( 4 cos t)i (b) v [ 4.5, 4.5] by [ 2, 4] 2 sin t)j 2 4 4 sin 8 1 4 2 cos 4 3 Since the two curves are covered over different -intervals, find the two areas separately. Then 2 (c) At t r2 4 ,v 1 v a 2 va 2i j, a 2 2i j, and Area 0 1 [2(1 2 2 sin )]2 d 2 sin sin 1 2 2 cos cos 2 2 0 (1 )d 5 0 2 dx dt dy dt 5 0 2 cos 1 sin 2 4 1 (3)(3) 7 cos 1 38.94 . 9 1 8 47. t, 2, so 2 (2t) t 2 22 dt 5 0 52. (a) v(t) d [( dt 3 sec t)i ( 3 tan t)j] ( 3 sec2 t)j ( 3 sec2 t)j] (2 3 sec2 t tan t)j 0 3 3 Area ( a(t) 3 sec t tan t)i 3 sec t tan t)i 4 (t 2 3 4) 3/2 76 3 d [( dt 3(sec t tan2 t (b) v(0) 4 dt, (c) At t 0, v v a cos 1 va 2 sec3 t)i 48. dx dt 2t 1 1/ 1 dy , 2t 2 dt 4, so 2 Area 2 2 t 1 2t 2t 1 2 2t 2 3 sec2 0 tan2 0 3j, a 0 ( 3)( 0 3) 3 sec4 0 3i cos 1 which using NINT evaluates to dr 49. d sin 2 cos 2 /4 10.110. 0 90 , so cos 2 sin d /4 Area 0 0 2 /4 cos 2 sin2 2 d cos 2 53. v(t) dr dt (1 t i t 2)3/2 2 sin cos 0 v(t) 1 j (1 t 2)3/2 2 t (1 t 2)3/2 (1 2 1 t 2)3/2 1 1 t2 , 2 (2 2) 1.840 which is at a maximum of 1 when t 0. Chapter 10 Review dr = (e t cos t dt dx (e t cos t dt 439 54. v(t) a(t) e t sin t)i e t sin t e cos t t t (e t sin t e t sin t e cos t t e t cos t )j e t cos t)i 61. (a) v(t) v(3) v(3) dx dy , dt dt 3 4 2 9 2 32 3 5 sin t, cos t , 4 4 4 4 , 5 4 2 25 2 32 , and 34 4 2 4 17 (e sin t ( 2e t sin t)i r(t) a(t) t e sin t)j (2e t cos t)j (e t sin t)(2e t cos t) 0 3.238 2 (e t cos t)( 2e t sin t) (b) x-component: for all t. The angle between r and a is always 90 . 1 1 d 2x dt 2 t 3 3 2 cos 16 4 5 16 2 3 3 y 2 5 3 16 2 5 2 d 2y y-component: 2 dt t 3 1 sin x 2 3 4 16 2 55. 0 (3 3t 2 e 6t) dt i 0 6 cos t dt j (c) 6i 6 sin t j 0 2 e e 3t 2 1 i 0 x 3 x2 9 dx dt cos y 25 2 4 t and 1. y 5 sin 4 t, so 1 or 56. e 2 ln t dt i t e 2 1 dt j t ln t e 2 62. (a) (ln 2)j C 1 dy and 2 dt 10 0 5 1 2 2 t so (5 t)2 dt, which using NINT ln2 t 57. r(t) r(0) r(t) 58. r(t) r(0) r(t) j i i e ln (ln t) e j 3i (sin t)j i Length evaluates to 10 dr dt dt (cos t)i j, so C 1)i (tan i t 1)i (sin t 1 25.874. y2 dx dt dt t(10 t) 2 dt 2 C j, and (b) Volume 1)j t2 i and 1j 10 (cos t dr dt dt 0 10 2 t)i 1j C 8 8 0 10 0 (100t 2 20t 3 5t 4 1 2 2 t 4) dt 1250 3 C 1 j, so C t2 (tan 100 3 t 3 t(10 t) 2 1 5 10 t 5 0 dr d 2r 59. dt 2tj C1, r(t) dt dt 2 dr C1 0, so r(t) t 2j dt t 0 dr dt dt (c) Area t 2j C1 t C2 C2 i, so 63. (a) dy dx dy dx t dy/dt dx/dt 0 2 (5 t)2 dt, which using NINT evaluates to e t sin t e t cos t 1 1 1040.728. cos t cos t sin t sin t C2. And r(0) r(t) 60. dr dt i t 2j ( 2t)i t 2i 2j 6t)i j C2 6t ( 2t)j t 2j C1 ( t2 3i 2)i C1t 4i, so C1 2t)j C2 2i 2)j 2j, and C1 , C2 6i 2j and e t cos t e t sin t r(t) dr dt t 1 d 2r dt dt 2 dr dt dt 1 (b) 2i ( t2 5i ( t2 r(t) r(1) r(t) dy dx e t(sin t cos t), e t(cos t sin t) dt dt dy 2 e 2t(sin2 t 2 sin t cos t cos2 t) dt e 2t(1 dx 2 dt 2 sin t cos t) 2 cos t sin t sin2 t) 3j, so C2 ( t2 2t e 2t(cos2 t e 2(1 2 cos t sin t) v(t) v(3) et 2 e3 2 3 (c) Distance 0 3 0 v(t) dt et 2 2 et (e3 3 0 1) 2 440 Chapter 10 Review 69. (a) 64. (a) v(t) v(4) dx dy , dt dt 2t, t 2 , v(4) 96 2 5 104 5 6 2 2 t dt 5 6 5 8, 96 , and 5 82 4 (b) Distance 0 4 0 (2t)2 [ 2, 10] by [ 2, 6] 2 t 25 9t 2 dt 5 2 4 (25 9t 2)3/2 135 0 dy dx dy/dt dx/dt 6t 2/5 2t (b) v(t) 4144 135 3 t 5 3 5 a(t) v(0) a(0) v(2) a(2) dx dy , dt dt d 2x d 2y , dt 2 dt 2 cos t, 2 sin t 2 sin t, v(1) a(1) v(3) a(3) cos t 2 ,0 0, 2 ,0 0, 2 (c) t x 2, so x 2 0, 0 0, 2 0, 0 0, 2 2 65. x degrees east of north is (90 Add the vectors: 540 cos 10 , 540 sin 10 595 cos 10 , 485 sin 10 585.961, 84.219 . Speed Direction 585.9612 84.2192 x) degrees north of east. 55 cos ( 10 ), 55 sin ( 10 ) (c) Topmost point: 2 ft/sec center of wheel: ft/sec Reasons: Since the wheel rolls half a circumference, or feet every second, the center of the wheel will move feet every second. Since the rim of the wheel is turning at a rate of ft/sec about the center, the velocity of the topmost point relative to the center is ft/sec, giving it a total velocity of 2 ft/sec. 70. v0 R Rg , where sin 2 591.982 mph. 81.821 east of north 585.961 tan 1 84.219 45 , g 32, and range 12,975 ft: v0 14,256 ft: v0 644.360 ft/sec 675.420 ft/sec 59.195 ft/sec 66. Add the vectors: 120 cos 20 , 120 sin 20 411.622, 14.896 . Direction Length tan 1 for 4325 yds 300 cos ( 5 ), 300 sin ( 5 ) for 4752 yds 71. (a) v0 2.073 411.891 lbs Rg sin 2 (109.5)(32) 14.896 411.622 411.6222 14.8962 (b) The cork lands at y Solve y 45 : v0 4, x 177.75. 67. Taking the launch point as the origin, y (44 sin 45 )t 16t 2 equals 6.5 when t 2.135 sec (as can be determined graphically or using the quadratic formula). Then x (44 cos 45 )(2.135) 66.421 horizontal feet from where it left the thrower's hand. Assuming it doesn't bounce or roll, it will still be there 3 seconds after it was thrown. 68. ymax (80 sin 45 ) 2(32) 2 g x 2 (tan )x for v0, with 2v02 cos2 gx 2 74.584 ft/sec y x 72. (a) The javelin lands at y Solve y 40 : v0 (b) ymax g 2v02 cos2 6.5, x x2 262 5 . 12 (tan )x for v0, with 7 57 feet gx 2 (2 cos 40 )( y x tan 40 ) 2 91.008 ft/sec (v0 sin )2 2g (91.008 sin 40 )2 64 6.5 6.5 59.970 ft 73. We have x y x 2 (v0t) cos gt 2 2 2 and gt 2 2 (v0t) sin . Squaring and adding gives (v0t)2(cos2 sin2 ) v02t 2. y ...
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