EVNREV11 - 298 Chapter 11 Vector-Valued Functions GmM r r3...

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Unformatted text preview: 298 Chapter 11 Vector-Valued Functions GmM r r3 GM r r3 r is parallel to r, r r 0. Also, 86. F ma ⇒ m a a Since r is a constant multiple of a, they are parallel. Since a d r dt Thus, r r r r r r 0 0 0. r is a constant vector which we will denote by L. 88. dr dt GM L r r 1 r GM 1 0 GM r r3 1 r r3 r r 0 r GM r r3 r r L r 1 r r3 r r 1 r r3 r r r r r r 1 r r3 r r 0 Thus, r GM L r is a constant vector which we will denote by e. r 90. L Let: r r r r r cos i r sin i i sin j cos j d dt dr dt j r sin r cos d dt r r2 dr d d dt k 0 0 92. Let P denote the period. Then P A 0 dA dt dt 1 L P. 2 Also, the area of an ellipse is ab where 2a and 2b are the lengths of the major and minor axes. ab P 1 LP 2 2 ab L 4 4 4 2a 2 Then: r r r cos r sin d dt r2 d k and L dt r d . dt P2 L L 2 2 a2 ed a c2 4 4 2a 2 a2 1 L2 e2 2a 4 2 2ed L 2 a3 423 a GM Ka 3 L 2 GM 3 a L2 Review Exercises for Chapter 11 2. r t ti 1 t 4 j k 4. r t 2t 1i t 2j , tk (a) Domain: 0, 4 and 4, (b) Continuous except at t 4 (a) Domain: (b) Continuous for all t Review Exercises for Chapter 11 6. (a) r 0 (b) r 2 3i 2 j k i 3 cos 3 cos t t t 1 t 1 y 299 (c) r s (d) r t 3 cos s r 1 sin s ti 3i 1 j sin s k tj tk tk 3i j k sin t 8. r t xt y ti t, y t j 4 10. r t t 1 −2 −1 −2 3 2 1 x 1 2 3 4 2t i 1 2 x, tj t, z y2 1 2 1 1 t 2k t 2, 5 z x y t x y z 2t, y z 0 0 0 0 x x 1 2 1 1 2 4 2 4 4 x 3 y 12. r t x x2 t x y z 2 cos t i 2 cos t, y z2 0 2 0 0 4 tj t, z 2 sin t k 2 sin t 2 x 3 z 14. r t 1 2 ti z 6 5 4 tj 13 4t k 2 0 2 2 0 2 3 2 0 3 2 2 2π y −3 −2 1 −1 2 3 2 1 −1 3 1 2 −2 3 −3 x y 16. One possible answer is: r1 t r2 t r3 t 4t i, 4 cos t i 4 t j, 4 sin t j, 0≤t≤1 0≤t≤ 2 18. The x- and y-components are 2 cos t and 2 sin t. At t 3 , 2 0≤t≤4 the staircase has made 3 of a revolution and is 2 meters 4 high. Thus, one answer is rt 2 cos t i 2 sin t j 4 t k. 3 2 cos 2t i 1 j k 20. x2 x z2 t, y rt rt 4, x t, z ti ti tj tj y ± 0, t 4 4 4 t2 x z 22. lim t→0 sin 2t i t e tj et k lim t→0 j k 2i t2k t2k 4 x 5 y 3 300 Chapter 11 Vector-Valued Functions 1 k t (b) r t (d) u t 0 Dt u t 2r t cos t i sin t j sin t i 2r t cos t j sin t i cos t j 1 t 2t k 1 t2 2k 24. r t sin t i cos tj t k, u t k sin t i cos t j (a) r t (c) r t Dt r t cos t i ut ut 2 sin t j (e) r t Dt r t (f) r t Dt r t ut 1 t2 t 1 t2 t cos t i 1 cos t t2 1 sin t t t sin t t sin t j cos t i 1 cos t t 1 sin t t2 t cos t sin t j 1 cos t t ut 1 sin t t 26. The graph of u is parallel to the yz-plane. t2 4 28. ln t i t ln tj k dt t ln t ti 1 2 ln t j tk C 30. tj t 2k i tj t k dt t2 t3 i t 2j t k dt t3 3 t4 i 4 t3 j 3 t2 k 2 C 32. r t r0 rt C sec t i 3k tan t j t 2k dt ln sec t tan t i ln cos t j t3 k 3 C ln sec t tan t i ln cos t j t3 3 3k 1 34. 0 tj t sin t k dt 2 32 tj 3 1 sin t t cos t k 0 2 j 3 sin 1 cos 1 k 1 36. 1 t3i arcsin tj t2k dt t4 i 4 2 k 3 t arcsin t 1 t2 j t3 k 3 1 1 38. rt rt vt rt at t, vt 1 0, tan t, et 1, sec t 2 sec t 2 4 40. sec2 t, et e 2t t rt rt r0 Since r 0 y t, z r t0 3 cosh t, sinh t, 3 sinh t, cosh t, 0, 1, 2t , t0 2 0 2 direction numbers 3, tan t, e 3, 0, 0 , the parametric equations are x 2t. 0.1 r 0.1 3, 0.1, 0.2 Review Exercises for Chapter 11 v02 sin cos 16 6 2 13 4 2 13 3v02 104 2 13 6 301 42. Range 4 v02 16 416 3 v02 ⇒ v0 11.776 ft sec 4 44. r t (a) r t 20 v0 cos ti v0 sin t 1 2 9.8 t2 j 4.9t 2 j (b) r t 20 20 cos 30 t i 20 sin 30 t 20 cos 45 t i 20 sin 45 t 4.9t 2 j 0 0 45 0 0 45 Maximum height (c) r t 20 5.1 m; Range 35.3 m 4.9t 2 j Maximum height 10.2 m; Range 40.8 m 20 cos 60 t i 20 sin 60 t 0 0 45 Maximum height 15.3 m; Range 35.3 m (Note that 45 gives the longest range) 2 t 1 14 12 3j 2j 46. rt vt v at Tt 1 4i 5 0 1 4i 5 4t i 3j 2 3t j 48. rt vt vt 2t 2i 2 t t 4 t t i t t t t 1 1 1 1 1i 2 t 1 j 1 3j at Tt Nt a a T N N t does not exist a a T 0 1 2i t 14 t 4 j 1 N does not exist 1 2j 1 1 4 3 t 1 t 4 t 1 2 4 1 1 1 4t 3 1 1 4 4 302 50. Chapter 11 rt vt vt at Tt Nt at at Tt Nt t t cos t i rt speed rt vt vt Vector-Valued Functions t sin t j t sin t cos t i cos t 2 sin t i cos t i t2 t sin t 1 2 t cos t sin t j sin t 2 t sin t t cos t t sin t sin t i t2 t cos t t sin t t cos t 1 cos t j t2 1 2 cos t j sin t j t cos t t t2 2 1 2 1 1 k t t 2 52. rt vt vt at Tt Nt a a T N i t 1i j 2t 4 t2 tj 1 k t2 1 54. r t ti t2j 2, x i 23 3 t k, x t, y 4, z t 2, z 16 3. 23 3t When t rt 2, y 2t j 2t 2k 2, a 4, z 8t 1, b 16 3 Direction numbers when t x t 2, y 4t 4, c 8 2 k t3 t 2i i t2j k 2t 4 1 2t 2k 2 2t 4 1 j 1 1 2 t 3 2t 4 4 t 2 2t 4 56. Factor of 4 t2i 2t i b 58. r t rt s 2tk, 0 ≤ t ≤ 3 2k 3 6 5 4 z r t dt a 0 4t2 1 t t t2 4 dt 3 3 2 ln ln t2 10 1 0 −2 (0, 0, 0) 1 2 1 2 y (9, 0, 6) 3 4 5 6 7 9 x 3 3 10 11.3053 Review Exercises for Chapter 11 60. rt rt rt s 0 y z 8 6 4 2 − 8 − 6 − 4 −2 −4 −6 −8 2 x x 2468 2 1 1 1 4 3 303 10 cos t i 10 sin t i 10 2 10 sin tj 10 cos tj 62. r t rt s ti i b t2j 2tj 2tk, 0 ≤ t ≤ 2 2k, r t 2 5 5 4t2 dt 105 4t2 r t dt a 0 10 dt 20 21 5 ln 5 4 5 ln 4 45 6.2638 (2, 4, 4) 2 3 4 y 64. r t xt s 2 sin t t cos t , 2 cos t t sin t , t , 0 ≤ t ≤ 4t2 1 dt 4.6468 1 2 66. rt rt rt et sin t i et cos t et cos t 2et et cos t k, 0 ≤ t ≤ et sin t i et sin t 2 2t sin t, 2t cos t, 1 , r t b 2 et sin t et sin t et cos t k et cos t 2 r t dt a 0 4t2 17 4 17 1 ln 4 s 0 r t dt 2 0 et dt 2et 0 2e 1 68. rt rt rt 2 ti 1 t i 3tj 3j, r t 3 2i 1 t 9 1 t 9t 1 t 2 i 1 t 1 t 2 j 3 32 k 0 0 3 2t3 2 1 9t 3 2 t3 5 sin tk 5 cos tk, r t 29 3 2 r r 3 t 2 3 2k; r r 3 2t 3 2 0 K rt rt rt3 2ti 2i 5 cos t j 5 sin tj 21 9t 32 70. rt rt rt r r r r K 5 cos tj i 2 0 725 r r r 3 5 sin tk k 5 cos t 5 sin t 25i 10 sin tj 10 cos tk j 5 sin t 5 cos t 725 29 3 2 25 29 29 29 5 29 304 Chapter 11 x2 Vector-Valued Functions 72. y y e 74. y x 2, tan x sec2 x 2 sec2 x tan x 1 4 y y ,K 232 1 e 2 y 1 e 4 x2 y y K 1 y y 232 1x e 4 1 e 1 4 2 2 K x 32 2 sec2 x tan x 1 sec4 x]3 2 4 55 45 and r 25 55 . 4 At x 2 55 25 ,r 25 55 . 2 At x 0, K 14 5 43 2 53 2 4 53 2 Problem Solving for Chapter 11 2. 2 x 3 13 x2 3 y2 1 3y 3 a2 0 3 4. Bomb: r1 t Projectile: r2 t 5000 400t, 3200 t, v0 sin 16t2 t 16t2 2 y 3 v0 cos At 1600 feet: Bomb: y y1 x1 sin3 tj 3 sin2 t cos tj 3 3 Slope at P x, y . 3200 16t2 1600 ⇒ t 10 rt rt r ti Tt Tt cos3 ti Projectile will travel 5 seconds: 5 v0 sin 16 25 v0 sin Horizontal position: sin tj 1600 400. 3 cos2 t sin ti 3 cos t sin t rt rt sin ti cos ti cos tj At t At t Thus, 10, bomb is at 5000 400 10 5. 9000. 5, projectile is at v0 cos Q 0, 0, 0 origin P \ cos3 T t, sin3 t, 0 on curve. j sin3 t sin t k 0 0 5v0 cos v0 cos Combining, v0 v0 sin v0 cos 9000 1800. 400 ⇒ tan 1800 1843.9 ft sec 2 ⇒ 9 12.5 . PQ i cos3 t cos t cos3 t sin t \ sin3 t cos t k D K PQ T T Tt rt 3 cos t sin t 1 3 cos t sin t 1800 cos 1 Thus, the radius of curvature, , is three times the K distance from the origin to the tangent line. Problem Solving for Chapter 11 305 6. r r st 1 sin t cos t 1 t cos d 2 sin2 d t 2 4 cos t 2 2 cos d 2 sin K 2r r 2 sin2 2 2 2 4 cos r2 32 2 rr r2 1 cos cos 2 1 cos 2 8 sin3 3 3 cos 8 sin3 sin2 sin3 2 3 4 sin 2 3 4 2 2 1 K s2 9 2 4 sin 3 2 16 cos2 2 16 sin2 2 16 8. (a) r r dr dt a xi yj position vector r sin j r sin d i dt dr sin dt dr cos dt dr sin dt d dt d dt dr sin dt dr cos dt r cos d dt d dt d j dt r cos r sin d dt d dt 2 r cos i dr cos dt d 2r dt2 d 2r cos dt2 d 2r sin dt2 r sin 2 d2 i dt2 d2 dt2 r cos ar a ur a dt2 cos i d 2r cos2 d 2r 2 sin dt2 sin j 2 dr sin cos dt 2 2 d dt d dt r cos2 r sin2 d dt 2 r cos sin 2 d2 dt2 d2 dt2 dr sin dt cos d dt r cos sin d 2r dt2 a a a a 2 r d dt u ur ur r a a d dt 2 sin i uu ur 2 cos j 2 dr d dr dt r d2 dt2 dr dt2 dr d dt dt r d2 u dt2 —CONTINUED— 306 Chapter 11 Vector-Valued Functions 8. —CONTINUED— (b) r r d dt t 42,000 cos i 12 42,000, dr dt 0 2 10. r t t 42,000 sin j 12 0 rt T T N B T 875 3 cos ti sin ti sin ti cos ti cos ti N ,T N B 4 4 4 sin tj k, t 4 1 −2 −1 2 z cos tj, r t cos tj sin tj sin tj k 2 i 2 2 i 2 k −2 0, d 2r dt2 1 −1 2 −1 2 x −2 y d2 , 12 d t 2 Therefore, a 42000 12 2 ur 2 ur . At t 4 2 j 2 2 j 2 Radial component: 875 3 Angular component: 0 12. y y y 15 x 32 53 x 64 2 2 15 1 x 128 2 K 15 1 2 x 128 25 3 x 1 4096 32 At the point 4, 1 , K 120 ⇒r 89 3 2 1 K 89 3 2 120 7. 14. (a) Eliminate the parameter to see that the Ferris wheel has a radius of 15 meters and is centered at 16 j. j, which is the low point on the Ferris wheel. At t 0, the friend is located at r1 0 (b) If a revolution takes t seconds, then t 10 and so t t t 10 2 20 seconds. The Ferris wheel makes three revolutions per minute. 8.032 11.472 14 m sec. The angle of 20 8.03i 11.47j. The speed is (c) The initial velocity is r 2 t 0 0.96 radians or 55 . inclination is arctan 11.47 8 03 (d) Although you may start with other values, t0 0 is a fine choice. The graph at the right shows two points of intersection. At t 3.15 sec the friend is near the vertex of the parabola, which the object reaches when t t0 2 11.47 4.9 1.17 sec. 0 30 0 Thus, after the friend reaches the low point on the Ferris wheel, wait t0 order to allow it to be within reach. 2 sec before throwing the object in (e) The approximate time is 3.15 seconds after starting to rise from the low point on the Ferris wheel. The friend has a constant speed of r 1 t 15 m sec. The speed of the object at that time is r 2 3.15 8.032 11.47 9.8 3.15 2 2 8.03 m sec. ...
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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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