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# EVEN11 - C H A P T E R 11 Vector-Valued Functions Section...

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Unformatted text preview: C H A P T E R 11 Vector-Valued Functions Section 11.1 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . 268 Section 11.2 Differentiation and Integration of Vector-Valued Functions . . . 273 Section 11.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . 278 Section 11.4 Tangent Vectors and Normal Vectors . . . . . . . . . . . . . . . 283 Section 11.5 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . 289 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 C H A P T E R 11 Vector-Valued Functions Section 11.1 Vector-Valued Functions Solutions to Even-Numbered Exercises 2. r t 4 t2 i t 2j 6t k 4 t2 6t Domain: ln t i ln t ln t Domain: 0, i 8. r t Ft Gt t3 3 4. r t t 2 sin t i 4 cos tj tk sin t 4 cos t t Component functions: f t gt ht Domain: 6. r t Ft 2, 2 Gt Component functions: f t gt ht , 5t j 1i 1i 3t 2k 5t tj 4t j i 4t j 3t 2 3t 2k 3t 2 k j t 1 t 1 t k t 2 tt 2 t t 1 i t3 t 2 t3 t j t3 t 1 t3 t k t 1, Domain: 10. r t cos t i i , 1, 2 sin t j (a) r 0 (b) r (c) r (d) r 4 2 i 2 cos t r 2j i cos 2 sin ti j 2 sin cos i tj 2 sin j cos i 2 sin j 6 ti k 6 e t4 6 6 6 6 12. r t (a) r 0 (b) r 4 (c) r c (d) r 9 t3 2j k 2i 2 t 8j c r9 e 1k 2i c 9 9 2 32 j 9 3i e c 24 k e t 32 9 t4 ti t t 9 32 j k e 3i 9 27j t4 e e 94 k 27 j 94 k 14. rt rt ti t t 3t j 2 4t k 3t 2 4t t1 2 9t2 16t2 25t 268 S ection 11.1 16. r t ut 3 cos t 4 sin t 2 sin t 6 cos t t 2 t2 t3 2t 2, a scalar. Vector-Valued Functions 269 The dot product is a scalar-valued function. 1≤t≤1 2t k, 0.1 ≤ t ≤ 5 3 2t 3 ln x. Matches (a) 18. r t x cos cos ti t,y y2 sin sin tj t,z t 2k, t2 20. r t x ti t, y ln t j ln t, z 2 3x Thus, x2 1. Matches (c) Thus, z and y 22. r t x ti t, y tj t, z 2k 2⇒x y (b) 10, 0, 0 z y (a) 0, 0, 20 (c) 5, 5, 5 z 1 2 3 3 1 −3 2 2 −2 1 −1 2 3 y −3 −2 −1 2 1 3 3 y −2 −3 x 1 2 3 x 24. x y 1 1 t, y x t 26. x t2 y 5 4 t, y t2 t 28. x y x2 y2 2 cos t 2 sin t 4 y Domain: t ≥ 0 y 6 5 4 3 2 x 1 2 3 4 −1 3 2 1 x −1 1 2 3 4 5 −1 −1 1 1 x − 4 −3 −2 −1 −2 30. x x 2 2 cos3 t, y 23 2 sin3 t cos2 t 1 sin2 t 32. x y y t 2t 3t 5 34. x x2 9 z 3 cos t, y y2 16 t 2 1 4 sin t, z t 2 y 2 23 Line passing through the points: 3 x2 3 y 3 2 y2 3 22 0, 5 ( 2 , 0, 15 ) 2 5, 0 , z 6 4 5 15 , 0, 2 2 Elliptic helix z 4 (0, −5, 0) −3 −2 −2 −3 x 6 x 2 3 −6 2 −2 2 4 −4 −6 2 4 −4 −6 y 4 x 4 y 270 Chapter 11 Vector-Valued Functions 3 t 2 36. x x t x y z t2, y y2 ,z 4 2 4 4 3 2t, z 3 y 4 1 1 2 3 2 38. x y z cos t sin t t y2 t 1 t sin t t cos t 0 0 0 0 z 1 1 2 3 2 2 4 4 3 x2 z t2 1 z 2 or x2 y2 z2 1 Helix along a hyperboloid of one sheet z 4 3 2 −4 −3 −2 −1 1 1 −1 −2 5 x −3 2 3 4 y x 3 2 2 3 4 y 40. r t ti 32 tj 2 12 tk 2 z 42. r t Ellipse 2 sin t i 2 cos t j z 2 sin t k Parabola 2 1 1 −1 1 2 x 3 −2 −3 1 2 3 y x −2 −2 −3 −1 1 2 1 y −1 −1 44. r t ti z t2 j 13 th 2 (a) u t rt 2j is a translation 2 units to the left along the y-axis. z (b) u t 13 t k has the roles 2 of x and y interchanged. The graph is a reflection in the plane x y. t2i tj z 5 4 3 2 5 4 3 2 1 1 2 3 4 5 x x 5 x 5 4 4 3 12 34 5 −2 y 3 2 1 5 4 3 2 12 3 y −2 2 1 12 3 y rt 4k is an upward (c) u t shift 4 units. z 5 4 3 2 1 1 2 3 4 5 x 12 34 5 y (d) u t 13 t k shrinks the 8 z-value by a factor of 4. The curve rises more slowly. ti t2 j z 5 4 3 2 1 1 2 3 4 5 x 5 y (e) u t r t reverses the orientation. z 5 4 3 2 1 1 2 3 4 5 x 12 3 y 5 S ection 11.1 46. 2x Let x rt 3y 5 0 1 2t 3 5j 5. 48. y 4 x2 t, then y ti 4 4 Vector-Valued Functions 271 t, then y ti Let x rt t 2. t2 j 1 2t 3 50. x Let x 2 2 y2 2 2 4 2 sin t. 2 sin t j 52. x2 16 Let x y2 9 1 4 cos t, y 4 cos t i 3 sin t. 3 sin t j 2 cos t, y 2 cos t i rt rt 54. One possible answer is rt 1.5 cos t i 1.5i 1.5 sin t j 2 k. r1 0 0, r1 10 4 10i r2 0 10i, r2 r3 0 4 5 2i 5 2i 5 2j 0 1 t k, 0 ≤ t ≤ 2 Note that r 2 56. r1 t r2 t r3 t t i, 0 ≤ t ≤ 10 sin t j , ti 10 cos t i 5 21 0≤t≤ t j, 5 21 0≤t≤1 5 2 j, r3 1 (Other answers possible) 0≤t≤1 y 1 t j, 58. r1 t r2 t ti 1 t j, ti x x 0≤t≤1 y (Other answers possible) 60. z x2 y 2, z y2 4 4 or 2 sin t, z 2sin t j 4k 2 x 2 y 6 z Therefore, x2 x rt 2 cos t, y 2 cos t i 4. 62. 4x2 If z 4y2 z2 16, x t2 and y z2 1 2 0 0 2 0 t2j 4 z t, then x 16 4t4 t2. 2 2 4 x t x y z rt 1.3 1.69 0.85 1.3 t2i 1 2 1.2 1.44 1.25 1.2 16 1 1 1.66 1 4t4 1 1 1.66 1 tk 1.2 1.44 1.25 1.2 2 y 272 64. x2 Chapter 11 y2 2 z2 Vector-Valued Functions 10, x y 2 4 sin t and z 21 sin2 t 2 cos t. z Let x sin t, then y t x y z 2 1 3 0 6 3 2 5 2 6 2 sin t i 2 0 2 2 2 6 5 2 3 2 6 2 2 3 1 0 2 2 x 4 4 4 y 2 rt 66. x2 Let x y z 1 t y2 2 sin t j 2 cos t k z 4 z2 t, then 4 t t4 8 16, xy 4 (first octant) and 16t 2 x2 16 y2 z2 t2 16 t2 z2 16. 4 x 4 y 4 3≤t≤ 8 43 t x y z rt 8 43 1.0 3.9 0 1.5 1.5 2.7 2.6 1 t 2 2 2 2 2.8 16t 2 2.5 2.5 1.6 2.7 16 k 3.0 3.0 1.3 2.3 3.5 3.5 1.1 1.6 8 43 3.9 1.0 0 ti 4 j t t4 68. x2 y2 160 120 80 40 40 120 80 e t cos t z e t sin t 2 e 2t z2 70. lim et i t→0 sin t j t e tk i j k since lim t→0 40 sin t t lim t→0 cos t 1 1 (L’Hôpital’s Rule) x 80 120 y 72. lim t→1 ti ln t j t2 1 2t 2 k i 1 j 2 2k 74. lim e t i t→ 1 j t t t2 1 k 0 since lim ln t 1 t→1 t 2 since lim t→1 1t 2t 1 . (L’Hôpital’s Rule) 2 t→ lim e t 0, lim t→ 1 t 0, and lim t→ t t2 1 0. S ection 11.2 76. r t ti t 1j Differentiation and Integration of Vector-Valued Functions 78. r t 2e t, e t, ln t 1 . 273 Continuous on 1, 80. r t 8, t, 3 Continuous on t 1 > 0 or t > 1: 1, t Continuous on 0, ut 2. 82. No. The graph is the same because r t For example, if r 0 is on the graph of r, then u 2 is the same point. a if 84. A vector-valued function r is continuous at t the limit of r t exists as t → a and lim r t t→a ra. i i j t≥0 is not continuous at t j t<0 0. The function r t 86. Let r t t→c x1 t i ut y1 t j t→c t→c z1 t k and u t x2 t i y2 t j z2 t k. Then: lim z1 t lim z2 t t→c t→c lim r t lim x1 t x2 t lim x1 t lim x2 t t→c y1 t y2 t t→c z1 t z 2 t t→c lim y1 t lim y2 t lim z1 t k t→c lim x1 t i t→c lim y1 t j t→c lim x2 t i t→c lim y2 t j t→c lim z2 t k t→c lim r t t→c lim u t t→c 88. Let ft 1, 1, if t ≥ 0 if t < 0 0, 90. False. The graph of x y z t 3 represents a line. and r t f t i. Then r is not continuous at c whereas, r 1 is continuous for all t. Section 11.2 2. r t xt y x3 r1 rt r1 i i i j 3t 2j 3j ti t 3 j, t0 t3 Differentiation and Integration of Vector-Valued Functions 1 4. r t xt x 1 y2 r2 rt r2 4i 2t i 4i 1 j 2 1 j t2 1 j 4 t 2i 1 j, t0 t 1 t 2 t, y t t 2, y t r t0 is tangent to the curve. y 4 3 2 r' (1, 1) x 1 2 3 4 r −4 −3 −2 r t0 is tangent to the curve. y 1 −2 −3 −4 2 r 1 r' x 8 −1 274 Chapter 11 Vector-Valued Functions 6. r t (a) ti 4 y 5 t2 j 8. r t y ti x 2, z rt t2j 3 2 i 2i i z 3 k, t0 2 2 r (1) 3 2 1 −3 −1 −1 2t j 4j 4j 3 k 2 r (1.25) r (1.25) − r (1) x 3 r2 r2 (b) r1 r 1.25 r 1.25 r1 rt i 3j 2.4375 j −2 2 1.25 i 0.25 i i r1 2t j i −4 −2 0.5625 j 2 x 4 (c) r −4 −6 y 2j 0.5625j 0.25 i 2.25 j r' −6 r 1.25 1.25 r1 1 0.25i This vector approximates r 1 . 10. r t rt 1 i t 1 i t2 16t j 16 j t2 k 2 tk 12. r t rt 4 ti 2 t i t2 t j 2t t ln t2 k t2 2t j 2 k t 14. r t rt sin t t cos t, cos t t sin t, t 2 16. r t rt arcsin t, arccos t, 0 1 1 t2 , 1 1 t2 ,0 t sin t, t cos t, 2t 18. r t (a) r t rt (b) r t t2 ti 2t 2i rt t2 1i 2j 2t tj 2t 1j 20. r t 8 cos t i 3 sin tj 3 cos t j 3 sin t j 8 cos t 3 cos t 3 sin t (a) r t rt 8 sin t i 8 cos t i rt 12 2t 12 8t (b) r t 8 sin t 55 sin t cos t 22. r t ti i 2t 2j 3j 3k 3t 5k 24. r t (a) r t rt e t, t2, tan t e t, 2t, sec2 t e t, 2, 2 sec2 t tan t rt e 2t (a) r t rt (b) r t 0 rt 0 (b) r t 4t 2 sec4 t tan t S ection 11.2 1 4 Differentiation and Integration of Vector-Valued Functions 275 26. rt rt r 1 r 4 r14 r 14 rt r r 1 4 1 4 1 4 ti i i t 2j 2 tj 1 j 2 12 1 20 e0.75t k, t0 0.75e0.75t k 0.75e0.1875 k 1 2 9e3 8 2 ( 1) 1 r' ( 4 ) r' 4 −2 −1 z 2 ( 1) 1 r'' ( 4 ) r'' 4 −2 −1 i 2 16 1 j 2 33 e 4 5 4 16k 2 x 1 −1 1 2 y 33 e 4 2j 93 e 16 8 20 4 9e 38 −2 4i 3e3 16k 2i 2i 9 0.75t e k 16 93 e 16 22 1 1024 81e3 8 16k 93 e 16 2 16 4 9e3 16k 81 3 e 256 8 1024 16 81 e3 8 r 14 r14 1 t t 32j 28. rt rt 1 i 1 1 3t j i 3j 1 30. r r 1 sin cos i i 1 cos j sin j 2 r 2n 1 0, n any integer 1 , 2n 1 Not continuous when t Smooth on , 1 , 1, 2t2 8 t3 32t t3 Smooth on 2n 32. r t rt rt 2t 8 16 t3 i t3 4t3 i 82 j 2t 4 j 82 34. r t rt et i et i e tj e tj 3t k 3k 0 , 36. r t rt ti 1 i 2t t2 2t j 1j 1 k 4 1 tk 4 0 r is smooth for all t: 0 for any value of t. 2. 2, . r is smooth for all t > 0: 0, r is not continuous when t Smooth on 38. r t ut ti 1 i t i ut ut , 2 sin tj 2 sin tj 2, 2 cos tk 2 cos t k 2 sin t k 4 cos2 t 5 (b) r t (d) 3r t Dt 3r t 2 sin tj ut ut 3t 3 2 cos tk 1 i t 4 sin t j 1 i t2 4 cos tk 4 sin tk (a) r t (c) r t Dt r t 2 cos t j 1 4 sin2 t 0, t 0 4 cos tj —CONTINUED— 276 Chapter 11 Vector-Valued Functions 38. —CONTINUED— (e) r t ut i t 1t 2 cos t Dt r t ut j k 2 sin t 2 cos t 2 sin t 2 cos t 1 t tj 1 t 2 sin t t t 1 t 1 k t 1 t2 1 t2 1j k 2 sin t 2 cos t 2 sin t 1 2 cos t t (f) r t Dt r t t2 4 12 t 2 t2i 2t i 2t 3 t4 tj j t t 2, r t 2t 3 t4 arccos 0.340 t t 2 4t 2 2t3 t4 4 12 2t t t2 4 1.0 40. rt rt rt rt rt cos 0 4t 2 1 − 0.5 1 1 0.707 2 . 2 t t2 4t 2 19.47 maximum at t 2 for any t. 42. r t lim rt t t t ti rt 3 t t 3 j t t→0 j 2t t 3 tk ti 2tk lim t→0 lim t t→0 t t t t i t t t 3 t j 2k lim t→0 t t 1 t t i t i t 3 t tt 3t j tt t j 2k 2k lim t→0 t 3 j t2 t 2k 1 i 2t 44. 4t3i 6tj 4 t k dt t 4i 3t 2 j 832 tk 3 C 46. ln t i 1 j t k dt t ln t ti ln t j tk C (Integration by parts) 48. et i sin t j cos t k dt et i cos t j sin t k C S ection 11.2 Differentiation and Integration of Vector-Valued Functions 277 50. e t sin t i e t cos t j dt et 2 sin t cos t i et 2 cos t sin t j C 1 52. 1 ti t 3j 3 t k dt t2 i 2 1 1 t4 j 4 1 1 343 tk 4 1 0 1 2 54. 0 ti etj tetk dt t2 i 2 2i 2 2 2 etj 0 0 t e2 1 et k 0 56. r t r0 rt C i 3t 2 j i 2 6 t k dt 2j t3 j 4t3 2k t3j 4t3 2k C e2 1j 1k 58. r t rt r0 rt r0 rt 4 cos t i 4 sin t i 3k 3k 3 sin t k 3 cos t k C1 ⇒ C1 3 sin t k C2 4j 3 sin t k 1 k dt t 1 j t j 4i 4 j ⇒ C2 4i 1 j t2 C 4j C1 0 4 cos t i 4i C2 4 cos t 1 1 4 2 i j t2 60. r t r1 rt i arctan t i 2 1 j t i ln t k ln t k C 2i ⇒ C 1 4 4 arctan t i 62. To find the integral of a vector-valued function, you integrate each component function separately. The constant of integration C is a constant vector. 66. Let r t x1 t i rt ±ut Dt r t ± u t y1 t j z1 t k and u t x2 t i y2 t j 64. The graph of u t does not change position relative to the xy-plane. z 2 t k. z1 t ± z2 t k z1 t ± z2 t k y2 t j z2 t k x1 t ± x2 t i x1 t ± x2 t i x1 t i y1 t j r t ±u t y1 t ± y2 t j y1 t ± y2 t j z1 t k ± x2 t i 68. Let r t rt Dt r t x1 t i ut ut y1 t j z1 t k and u t z1 t y2 t i y1 t z2 t x2 t i y2 t j z 2 t k. z1 t x2 t j z1 t y2 t i y1 t x2 t k z1 t x2 t j z1 t x2 t j x1 t y2 t x1 t y2 t y1 t x2 t k y1 t x2 t k x1 t y2 t x1 t z2 t y1 t x2 t k x1 t z2 t z1 t x2 t z1 t x2 t j y1 t z2 t y1 t z2 t x1 t y2 t y1 t z2 t x1 t z2 t z1 t y2 t y1 t x2 t x1 t z2 t x1 t y2 t z1 t y2 t i z1 t y2 t i rt ut y1 t z2 t rt ut x1 t z2 t 278 Chapter 11 xti rt Dt r t rt rt Vector-Valued Functions ytj z t k. Then r t ytz t ytz t xty t ytz t zty t i y tz t x ty t zty t i x ti y tj z t k. xty t xtz t ytx t k x tz t ztx t z tx t j 70. Let r t xtz t zty t ytx t xtz t ztx t j z ty t i y tx t k ztx t j xty t ytx t k rt rt 72. Let r t then: xti ytj x2 t z t k. If r t y2 t y2 t z2 t z2 t 2z t z t ztz t rt r t is constant, C Dt C 0 0 0 74. False Dt r t ut rt ut rt ut (See Theorem 11.2, part 4) Dt x2 t 2x t x t 2xtx t 2y t y t yty t 2rt Therefore, r t rt 0. Section 11.3 2. r t vt at x 6 6 rt rt t, y 0 t, y ti i tj Velocity and Acceleration y 4. r t vt t2i rt rt t2, y t3j 2t i 2i t3 1. 2i 2i 3j 6j x 3t2j 6tj y2 3 8 7 6 5 4 3 2 1 −1 y j 4 a at v (3, 3) v 6 x 2 x x At 1, 1 , t v1 a1 (1, 1) x 234567 8 2 4 6. r t vt at x 3 cos ti 3 sin ti 3 cos ti 3 cos t, y 0. 2j 3i 2 sin tj 2 cos tj 2 sin tj 2 sin t, x2 9 y2 4 1 Ellipse y 3 8. r t vt at x e e t, et rt rt t y e t, et e 1 ,y et 0. i i j j t, 2 et 1 v (1, 1) a et, y 1 x 1 2 x At 3, 0 , t v0 a0 At 1, 1 , t v0 v a x 2 1 1, 1 1, 1 1 −2 −1 a0 −1 (3, 0) −3 10. r t vt st at 4t i 4i vt 0 4t j 4j 2t k 2k 16 16 4 6 Section 11.3 12 tk 4 1 tk 2 1 12 t 4 10 12 t 4 Velocity and Acceleration 279 12. r t vt st at 3ti 3i tj j 9 14. r t vt st at t2 i 2t i 4t 2i 2 tj j 1 2t 3 2k 3 tk 9t 4t 2 9t 1 3 k 2t 1 k 2 et cos t, et sin t, et et cos t e2t cos t et 3 et sin t i sin t 2 16. r t vt st 18. (a) et sin t e2t cos t et cos t j sin t 2 rt rt r3 t, 1, 1, 3 t, y 0.1 25 t 25 3 , 4 z 3 t 2, t2 3 4 4 , 25 t 25 t 2 , t0 t2 3 et k e2t at 2et sin t i 2et cos t j et k x (b) r 3 3 t 4 3 0.1 , 4 4 3 0.1 4 0.1, 4 3.100, 3.925, 3.925 20. a t vt v0 rt r0 r2 C 4i C 2i 2i 3k 3k dt 4j ⇒ v t 2ti 4j 2ti 2ti 3tk 4j t2i 4tj C 3tk 4tj 32 tk 2 32 tk 2 C 22. a t vt v0 vt rt j cos t i cos t i C sin t i sin t i cos t i r0 rt r2 24. (a) The speed is increasing. (b) The speed is decreasing. 26. r t 900 cos 45 t i 450 2t i 3 3 900 sin 45 t 16t 2 j 32t 50,649 8 0, which implies that t 225 2 16. 16t 2 j i C j sin t j, v 0 sin t j dt k⇒C cos t j cos t j tk 0 tk 2k k k dt C k j k, r 0 sin t i i cos t j C 3tk dt t2i 0 ⇒ rt 8j 6k sin t j i⇒C sin t j cos t i cos 2 i sin 2 j 450 2t 450 2 The maximum height occurs when y t The maximum height reached by the projectile is y 3 450 2 225 2 16 16 225 2 16 3 2 6331.125 feet. 16t 2 0 which implies that The range is determined by setting y t t Range: x 450 2 32 450 2 450 2 32 405,192 450 2t 39.779 seconds. 405,192 25,315.500 feet 280 Chapter 11 Vector-Valued Functions 28. 50 mph rt 220 ft/sec 3 220 cos 15 t i 3 5 220 sin 15 t 3 16t 2 j The ball is 90 feet from where it is thrown when x 220 cos 15 t 3 90 ⇒ t 27 22 cos 15 1.2706 seconds. The height of the ball at this time is y 5 220 sin 15 3 27 22 cos 15 16 27 22 cos 15 2 3.286 feet. 30. y x 0.005x 2 is the coefficient of x. Therefore, tan 1, 4 rad 45 . Also From Exercise 34 we know that tan 16 sec2 v02 16 2 v02 rt negative of coefficient of x 2 0.005 or v0 40 2 t i 60, 32 4 40 2 40 2 32t j 24 2 j 8 2 5i 2j direction 80 ft/sec 16t 2 j. Position function. 40 2t When 40 2 t t vt v 32 4 Speed 60 40 2 40 2 i 40 2 i 32 4 v 8 2 25 4 8 58 ft sec 32. Wind: 8 mph rt 50 176 ft sec 15 176 i 15 2.5 140 sin 22 t 16t2 j 140 cos 22 t 0 0 450 When x 375, t 2.98 and y 16.7 feet. Thus, the ball clears the 10-foot fence. Section 11.3 34. h 7 feet, rt 35 , 30 yards v0 cos 35 t i 7 90 feet v0 sin 35 t v0 sin 35 t 16t 2 j 16t 2 t 7 v0 sin 35 90 v0 cos 35 16 90 v0 cos 35 2 Velocity and Acceleration 281 (a) v0 cos 35 t 90 when 7 4 90 v0 cos 35 4 3 v02 v0 129,600 v02 cos2 35 129,600 cos2 35 90 tan 35 54.088 feet per second (c) x t t 90 ⇒ v0 cos 35 t 90 54.088 cos 35 90 3 90 tan 35 (b) The maximum height occurs when yt t v0 sin 35 v0 sin 35 32 32t 0. 2.0 seconds 0.969 second 22.0 feet. 0, v0 16t2 j 30,000 At this time, the height is y 0.969 36. Place the origin directly below the plane. Then rt v0 cos 792 t i vt 792i ti 30,000 16t2 j 0⇒ v0 sin t 792 and α 30,000 32t j. α At time of impact, 30,000 r 43.3 v 43.3 v 43.3 tan 34,294.6i 792i 16t2 t2 1875 ⇒ t 43.3 seconds. (0, 0) 34,295 1385.6j 1088 mph 0.8748 ⇒ 0.7187 41.18 1596 ft sec 30,000 34,294.6 38. From Exercise 37, the range is x Hence, x v02 sin 2 . 32 150 v02 sin 24 32 i ⇒ v02 4800 ⇒ v0 sin 24 108.6 ft sec. 40. (a) r t t v0 cos 16t x t v0 sin 0 when t v0 sin 32 16t 2 j v0 sin . 16 v02 sin 2 32 (b) y t dy dt tv0 sin v0 sin 16t 2 32t 0 when t v0 sin . 32 t v0 sin Range: v0 cos Maximum height: y v0 sin 32 v02 sin2 32 16 v02 sin2 322 2 . v02 sin2 64 The range will be maximum when dx dt or 2 2 , 4 rad. v02 2 cos 2 32 0 Minimum height when sin 1, or 282 42. r t Chapter 11 v0 cos ti Vector-Valued Functions h v0 sin v0 sin 8 t 50 ⇒ t 4.9 t 4.9t 2 j 4.9t 2 j 50 . For this value of t, y v0 cos 8 2 v0 cos 8 t i x 50 when v0 cos 8 t v0 sin 8 50 v0 cos 8 0: 50 v0 cos 8 0 4.9 2500 ⇒ v02 v02 cos2 8 ⇒ v0 4.9 50 tan 8 cos2 8 42.2 m sec 1777.698 50 tan 8 44. r t vt Speed bt b 1 vt sin t i cos t i 2b b1 cos t j sin t j 1 cos t and has a maximum value of 2b when t 80.67 rad sec since since b 1 ,3 ,... . 55 mph 80.67 ft sec Therefore, the maximum speed of a point on the tire is twice the speed of the car: 2 80.67 ft sec 46. (a) Speed v 110 mph b2 b2 2 2 sin2 t sin2 t b2 2 cos2 t b (b) − 10 10 cos2 t 10 − 10 The graphing utility draws the circle faster for greater values of . 48. a t 50. v t vt b at F mb b 2 2 b 2 cos t i sin t j b 2 30 mph 44 ft sec 605 44 rad sec 300 3000 n θ 44 3000 300 32 300 2 605 lb Let n be normal to the road. n cos n sin 3000 605 Dividing the second equation by the first: tan 605 3000 arctan 605 3000 11.4 . 52. h t 6 feet, v0 45 feet per second, 45 sin 42.5 32 69.02 feet. 2 2 42.5 . From Exercise 47, 2 32 6 2.08 seconds. 45 sin 42.5 At this time, x t S ection 11.4 54. r t yt rt vt st Thus, x t xt at 0 x ti mx t j 0. xti mxt xti x ti xt ytj b, m and b are constants. mxt mx t j 2 Tangent Vectors and Normal Vectors xti r1 2t 2r1 2t 4r1 2t r1 t , then: r1 t 2 283 56. r1 t r2 t bj ytj ztk Velocity: r2 t C, C is a constant. In general, if r3 t Velocity: r3 t Acceleration: r2 t 2 mx t C 1 m2 Acceleration: r3 t r1 t Section 11.4 2. rt rt rt Tt T1 t3i 3t2i 9t rt rt 1 9 16 4 Tangent Vectors and Normal Vectors 2t2j 4tj 16t 2 4. rt rt rt 6 cos ti 6 sin ti 36 sin t rt rt 2 2 sin tj 2 cos tj 4 cos2 t 6 sin ti 36 sin2 t j 14 2 cos tj 4 cos2 t 1 28 3 3i j 1 9t4 3i 16t2 4j 3t2i 3 i 5 4tj 4 j 5 T Tt 3 3 3i 36 3 4 6. r t rt When t T1 t2i 2ti tj j 4 k 3 8. r t rt t, t, 1, 1, 4 4 t2 t t2 1, 1, 1 ,t 3 1 3 1 3 t 3 1, 1 at 1, 1, 3. 1, r t r1 r1 2i r1 j 5 2, b 2t 2i j t j 0 t 4 1 at 1, 1, 3 . When t T1 1, r 1 r1 r1 5 2i 5 1, c 1, y 21 1, 1, 7 1, b t 1, c 1, y 1 t 3 Direction numbers: a Parametric equations: x 1, z 4 3 Direction numbers: a Parametric equations: x z 10. r t rt When t T 2 sin t, 2 cos t, 4 sin2 t 2 cos t, ,r r r 2 sin t, 8 sin t cos t 3, 6 6 1 4 3, b 3t 1, 2 3 , 3, 1, 2 3 1, c 1, y 23 t 3, z 2 3t 1 t at 1, 3, 1 . 6 6 6 6 Direction numbers: a Parametric equations: x 284 12. r t rt Chapter 11 3 cos t i 3 sin t i ,r r r Vector-Valued Functions 4 sin t j 4 cos t j 3i 2 2 1 k 2 5 z 1 k 2 1 k, 2 2 37 3i 0, c 4, z t 2 1 k 2 1 t 4 at 0, 4, 1 37 4 . x 54 4 3 2 When t T 1 1 4 5 y 2 2 2 6i k Direction numbers: a Parametric equations: x 6, b 6t, y 14. r t rt r0 e ti e i T0 ti 2 cos t j 2 sin t j 2 sin t k, t0 2 cos t k i i 5 1 s, y s 1 2k, r 0 2k 0 2j, r 0 r0 r0 5 Parametric equations: x s r t0 0.1 r0 0.1 2, z s 2s 0.1, 2, 2 0.1 0.9, 2, 0.2 16. r 0 u0 0, 1, 0 0, 1, 0 Hence the curves intersect. rt us u0 cos r0 r0 1, sin t, cos t , r 0 sin s cos s 1, 0, 1 u0 u0 0⇒ cos s, 1, 0, 1 sin s cos s cos s, 1 cos 2s 2 1 2 2 18. rt rt Tt ti i 6 j, t t 6 j t2 3 20. rt rt Tt cos ti sin t i rt rt 2 sin t j 2 cos t j sin t i sin2 t k, t 4 rt rt t t4 2 1 1 36 72t 36 i i 36 t4 6 j t2 12t3 t4 36 2j i 6 j t2 2 cos t j 4 cos2 t The unit normal vector is perpendicular to this vector and points toward the z-axis: Nt 2 cos ti sin tj . sin2 t 4 cos2 t Tt N2 t4 32 32 j T2 T2 1 3i 13 S ection 11.4 22. r t vt at Tt Tt Nt 4t i 4i O vt vt O Tt Tt is undefined. 1 2i 5 j 2t j 2j 24. r t vt at Tt Tt Nt t2j 2t j 2j Tangent Vectors and Normal Vectors k 285 vt vt O Tt Tt 2t j 2t j is undefined. The path is a line and the speed is constant. 26. rt vt at Tt T1 t2i 2ti 2tj, t 2j, v 1 2i 1 4t2 j 4 2 i 2 1 Nt Tt Tt 1 t2 N1 aT aN 30. r t0 v t0 a t0 T 2 The path is a line and the speed is variable. 28. r t a cos t i a sin t i bj a a v0 v0 2 1 2i 2j b sin t j b cos t j vt v0 2i, a 1 vt vt 1 i 2 2ti 2j 2 j 2 t2 1 1 t 1 1 t2 1 ti j at a0 T0 cos t i b 2 sin t j 2i j t2 1 3 2i 3 2j Motion along r t is counterclockwise. Therefore, N0 aT aN a a i. T N 0 a 2 t2 1 i 2 j 2 2 2 sin t0 i cos t0 i 1 tj 2 i 2 a a t0 1 T N cos t0 j sin t0 j cos t0 j sin t0 i 1 v v cos t0 i sin t0 j 2 1 cos t0 sin t0 i 1 cos t0 j . 2 1 cos t0 2 Motion along r is clockwise. Therefore, N aT aN a a T N sin t0 2 1 cos t0 2 2 2 1 cos t0 2 1 cos t0 32. T t points in the direction that r is moving. N t points in the direction that r is turning, toward the concave side of ...
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