ODDREV11 - 68 Chapter 11 xti dr dt Vector-Valued Functions...

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Unformatted text preview: 68 Chapter 11 xti dr dt Vector-Valued Functions ytj xt 2 85. Let r r z t k. Then r yt 2 r 2 xt 1 xt 2 r 2 2 yt yt 2 2 zt zt 2 2 and r x ti y tj 2y t y t z t k. Then, 2z t z t zt 12 2x t x t xtx t 87. Let r xi yj rr x2 yty t ztz t r. r. rr (using Exercise 77) z k where x, y, and z are functions of t, and r r dr dt r2 y2 z2 x i rr yj rr r2 zk r3 r r xx r 2r yy r r3 dr dt r zz xi yj zk 1 x y2 r3 1 yz r3 i x z2 xyy j xzz i x 2y k z 2y xx y 1 r r3 zz y j x 2z y 2z xx z yy z k yz x xz y xz xy z xy r r 89. From Exercise 86, we have concluded that planetary motion is planar. Assume that the planet moves in the xy-plane with the sun at the origin. From Exercise 88, we have r L GM r r e. y Planet Sun r Since r L and r are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate the coordinate system so that e lies along the positive x-axis and is the angle between e and r. Let e e . Then r e r e cos re cos . Also, L 2 θ e x L r L r r L r r L GM e r r GM r e r r r GM re cos r Thus, L 1 2 GM e cos r and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse. 91. A 1 2 rd 2 Thus, dA dt dA d d dt 1 2d r 2 dt 1 L 2 and r sweeps out area at a constant rate. Review Exercises for Chapter 11 1. r t ti csc t k n , n an integer n , n an integer 3. r t ln t i tj tk (a) Domain: t (a) Domain: 0, (b) Continuous for all t > 0 (b) Continuous except at t Review Exercises for Chapter 11 5. (a) r 0 (b) r (c) r c 2 1 i 3i 2c 2c (d) r 1 t r1 4j 1 1i 8 3k 1 3 69 1i c 21 2 ti c 1 2j t t t 1 2j 1 3 c 1 3k 1 3 1 3k c 1 3k 1 1 3 1i 2j 9. r t t 2j t3 i 1 t t2 ⇒ z z 1 t 3k 3 tk 3i j 3 t2 tj t 2k 7. r t xt x2 cos t i 2 sin2 t j 2 sin2 t 11. r t x t x y z i 1, y 0 1 0 1 sin t j sin t, z k 1 3 2 1 0 1 1 1 1 cos t, y t y 2 y 1 21 x y z y2 x2 2 1 1 1 z 1≤x≤1 y 2 1 1 2 1 y 3 2 1 −2 2 3 1 1 2 −1 x 1 x y x 13. r t ti z 3 2 1 1 ln t j 12 2t k 15. One possible answer is: r1 t r2 t r3 t 4t i 4i 4 3t j, 3 t i, t j, 0≤t≤1 0≤t≤3 0≤t≤4 1 2 3 2 y x 17. The vector joining the points is 7, 4, rt 2 7t, 3 4t, 8 10 . One path is 19. z x x2 t, y rt z 5 y 2, x t, z ti tj y 0, t 2t 2 2t 2k x 10t . −3 2 3 1 2 3 y x 21. lim t 2 i t→2 4 t2j k 4i k 70 23. r t Chapter 11 3t i 3i ut ut t j Vector-Valued Functions 1 j, u t ti t 2j 23 tk 3 (b) r t t2 t 3t 2 4t 1 t3 2t 2 (d) u t Dt u t (f ) r t 1 Dt r t ut ut 0 2r t 2r t 24 t 3 83 t 3 5t i 5i t3 i t2 2t 2t 4j 8t 3j 2t 2j 2j 3t 3 9t 2 23 tk 3 2t 2k t2 2t tk 1k (a) r t (c) r t Dt r t 3t 2 (e) r t Dt r t 10t 2 2t 1 1 2t 10t 10t 2 2t 2 i 25. x t and y t are increasing functions at t decreasing function at t t 0. t0, and z t is a 27. cos t i t cos t j dt sin t i t sin t cos t j C 29. cos t i sin t j t k dt 1 t 2 dt 1 t1 2 t2 ln t 1 t2 C 2 31. r t r0 rt j t2 2t i k 1i etj C i et e t k dt 3j 2j t 2i etj i e tk 2j C 4k 33. 2 3t i 2t 2j t 3k dt 3t 2 i 2 2t 3 j 3 t4 k 4 2 2 32 j 3 5k ⇒ C e t 4k 2 2 35. 0 et 2i 3t2j k dt 2et 2i t3j tk 0 2e 2i 8j 2k 37. r t vt cos3 t, sin3 t, 3t rt vt 3 3 cos2 t sin t, 3 sin2 t cos t, 3 9 cos4 t sin2 t cos2 t sin2 t 9 sin4 t cos2 t cos2 1 sin2 t 2 2 9 1 t sin2 t 3 cos2 t sin2 t at vt 6 cos t 3 cos2 t cos t, 6 sin t cos2 t cos t , 3 sin t 2 cos t 2 3 sin2 t sin t , 0 3 cos t 2 sin t 1 3 , t 2, t , t 0 2 3 , 2t, 1 2 sin t , 0 v02 sin 2 32 75 2 sin 60 32 2 39. rt rt r4 ln t 1 t 4 41. Range x 152 feet 1, 8, 1 2 direction numbers 0, 16, 2 , the parametric equations are Since r 4 1 x t, y 16 8t, z 2 2 t. r t0 0.1 r 4.1 v02 sin 2 9.8 0.1, 16.8, 2.05 43. Range x 80 ⇒ v0 80 9.8 sin 40 34.9 m sec Review Exercises for Chapter 11 45. rt vt vt at Tt 5t i 5i 5 0 i at Tt a N does not exist Nt a a T N i (The curve is a line.) i vt 47. rt vt ti i tj 1 j 2t 4t 1 2t 1 j 4t t 12 tj 4t 1 2 t 2 tj 4t 1 1 1 12 tk 2 tk 5t 2 k 2t j t k 1 5t2 5t i 2j k 5 1 5t2 5t 1 5t 2 5 5t 2 1 5t 2 5 51 2 ti 4t j 1 71 N t does not exist a T 0 1 4t t 4t 1 2t 4t ti i 1 2j i t 2j 2t j 49. rt vt vt at Tt Nt a a T N et i et i e2t et i e tj e tj e e tj 2t 51. rt vt v at Tt Nt a a T N et i e t j e2t e 2t e ti e2t etj e 2t e2t e 2t e2t e 2t 2 e2t e 2t 53. r t 2 cos t i 3 ,x 4 2 sin t j t k, x 2, y 2 cos t, y 2, z k 2, b 2, z t 3 . 4 2 sin t, z t When t rt 2 sin t i 2 cos t j 3 ,a 4 2t Direction numbers when t x 2t 2, y 2, c 3 4 1 55. v 9.56 104 4600 2t i 2i b 4.56 mi sec 57. r t rt s 3tj, 0 ≤ t ≤ 5 3j 5 2 −4 −2 y (0, 0) x 246 8 10 12 14 r t dt a 5 0 4 9 dt −4 −6 −8 − 10 − 12 − 14 − 16 13t 0 5 13 (10, − 15) 72 59. Chapter 11 rt rt rt 10 cos3 t i Vector-Valued Functions 10 sin3 tj 30 sin2 t cos tj sin4 t cos2 t − 10 10 y 30 cos2 t sin ti 30 cos4 t sin2 t 30 cos t sin t 2 2 −2 x 2 10 s 4 0 30 cos t sin t dt 120 sin2 t 2 2 60 0 − 10 61. r t rt s a 3ti 3i b 2tj 2j 4tk, 0 ≤ t ≤ 3 4k 3 3 12 10 8 6 4 2 2 x z (−9, 6, 12) r t dt 0 9 4 16 dt 0 29 dt 3 29 (0, 0, 0) 24 68 10 y 63. r t rt s 8 cos t, 8 sin t, t , 0 ≤ t ≤ < b 2 65 65 2 65. r t rt s 1 ti 2 1 i 2 sin t j cos t j r t dt cos t k, 0 ≤ t ≤ sin t k 8 sin t, 8 cos t, 1 , r t 2 r t dt a z 65 dt 0 0 π 2 (0, 8, π ) 2 4 6 8 y 0 1 4 dt 0 cos2 t 5 t 2 sin2 t dt 5 2 4 86 x (8, 0, 0) 5 2 0 67. r t Line k 0 3ti 2tj 69. rt rt rt r r 2ti 2i j i 2 0 r r 12 tj 2 tj 2k j t 1 r 3 t2k 2tk, r 5t2 4 k 2t 2 5t2 4j 20 4 2k, r r 25 5t2 20 K 32 4 32 71. y y y K At x 12 x 2 x 1 1 2 73. y y K ln x 1 ,y x 1 y y 1, K 1 x2 232 y y 4, K 232 1 1 x2 1 1 22 1 x2 1 x 2]3 2 32 At x 32 1 and r 173 2 1 23 2 2 and r 4 2 2. 17 17 17. 75. The curvature changes abruptly from zero to a nonzero constant at the points B and C. P roblem Solving for Chapter 11 73 Problem Solving for Chapter 11 t 1. x t 0 cos u2 du, y t 2 t sin 0 u2 du 2 3. Bomb: r1 t Projectile: r2 t 5000 400t, 3200 t, v0 sin 16t2 t 16t2 xt (a) s t2 cos ,y t 2 a t2 sin 2 a v0 cos At 1600 feet: Bomb: dt 0 xt 0 2 y t 2 dt t2 2 t2 2 1 a. ,y t a t2 2 3200 16t2 1600 ⇒ t 10 seconds. Projectile will travel 5 seconds: t cos 5 v0 sin 16 25 v0 sin t Horizontal position: At t At t 10, bomb is at 5000 400 10 1000. 1600 400. (b) x t t sin t cos2 t sin2 K At t (c) K a, K a t2 2 5, projectile is at 5v0 cos . 200. 400 ⇒ tan 200 447.2 ft sec 2⇒ 63.4 . length Thus, v0 cos Combining, v0 sin , 0 ≤ ≤2 v0 sin v0 cos 200 cos 5. x x 2 1 y cos , y 2 1 2 cos 2 cos 2 sin2 4 sin2 2 t t st x K 2 sin 2 d 4 cos cos cos 3 2 4 cos t 2 sin , y 11 cos sin sin 2 1 cos 8 sin3 1 4 sin 2 2 1 2 sin Thus, s2 1 K 2 t 4 sin and 2 16 cos2 t 2 16 sin2 t 2 16. 7. r2 t d rt dt rt 2 rt 2rt rt d rt dt rt rt rt ⇒ d rt dt rt rt rt 74 Chapter 11 Vector-Valued Functions dB ds N 9. r t rt rt T T N B At t T 4 cos ti 4 sin ti 4 cos ti 4 sin ti 5 4 cos ti 5 cos ti N ,T N B z 6π 4 sin tj 4 cos tj 4 sin tj 4 cos tj 5 4 sin tj 5 sin tj 3 sin ti 5 3tk, t 2 5 11. (a) B T dB ds N d T ds T T T 1 constant length ⇒ N N N T T T dB ds N T T T N T T N B 3k, r t 3 k 5 T dB ds T 0 3 cos tj 5 3 k 5 4 k 5 dB dB B and ds ds for some scalar . Hence, (b) B B T N T⇒ 2 2 2 2 3 i 5 4 i 5 j N. Using Exercise 10.3, number 64, T N N N N T T NT N N TN 4 k 5 T B T T N T T T T NT N T TN B N B N T N. Now, KN Finally, dT ds Ts Ts Ts dT . ds 3π 4 x 3 2 1 4 y Ns d B ds T B B KT T KN B. B T N T 13. r t (a) −3 t cos t, t sin t , 0 ≤ t ≤ 2 2 2 (b) Length 0 2 3 r t dt 2t 2 0 1 dt 6.766 (graphing utility) −2 22 (c) K K0 K1 K2 2 t 2 1 32 22 t (d) 5 2 2 2 1 32 1.04 0 0 5 0.51 0 (f ) As t → , the graph spirals outward and the curvature decreases. (e) lim K t→ ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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