Divide and solve dt dt 4 8t 2t to get y 256x one

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Unformatted text preview: r × r = (ac/w3 ) sin(s/w)i − (ac/w3 ) cos(s/w)j + (a2 /w3 )k, (r × r ) · r = a2 c/w6 , r (s) = a/w2 , so τ = c/w2 and B = (c/w) sin(s/w)i − (c/w) cos(s/w)j + (a/w)k dT ds dT = = (κN)s = κs N, dt ds dt dN ds dN = = (−κT + τ B)s = −κs T + τ s B. N= dt ds dt 66. (a) T = (b) r (t) = s so r (t) = s T and r (t) = s T + s T = s T + s (κs N) = s T + κ(s )2 N. (c) r (t) = s T + s T + κ(s )2 N + [2κs s + κ (s )2 ]N = s (κs N) + s T + κ(s )2 (−κs T + τ s B) + [2κs s + κ (s )2 ]N = [s − κ2 (s )3 ]T + [3κs s + κ (s )2 ]N + κτ (s )3 B. (d) r (t) × r (t) = s s T × T + κ(s )3 T × N = κ(s )3 B, [r (t) × r (t)] · r (t) = κ2 τ (s )6 so τ= [r (t) × r (t)] · r (t) [r (t) × r (t)] · r (t) = 2 (s )6 κ r (t) × r (t) 2 67. r = 2i + 2tj + t2 k, r = 2j + 2tk, r = 2k, r × r = 2t2 i − 4tj + 4k, r × r = 2(t2 + 2), τ = 8/[2(t2 + 2)]2 = 2/(t2 + 2)2 68. r = −a sin ti + a cos tj + bk, r = −a cos ti − a sin tj, r r × r = ab sin ti − ab cos tj + a k, r × r 2 = = a sin ti − a cos tj, a2 (a2 + b2 ), τ = a2 b/[a2 (a2 + b2 )] = b/(a2 + b2 ) √ √ √ 69. r = et i − e−t j + 2k, r = et i + e−t j, r = et i − e−t j, r × r = − 2e−t i + 2et j + 2k, √ √ √ r × r = 2(et + e−t ), τ = (−2 2)/[2(et + e−t )2 ] = − 2/(et + e−t )2 70. r = (1 − cos t)i + sin tj + k, r = sin ti + cos tj, r = cos ti − sin tj, r × r = − cos ti + sin tj + (cos t − 1)k, r ×r = cos2 t + sin2 t + (cos t − 1)2 = 1 + 4 sin4 (t/2), τ = −1/[1 + 4 sin4 (t/2)] Exercise Set 14.6 512 EXERCISE SET 14.6 1. v(t) = −3 sin ti + 3 cos tj a(t) = −3 cos ti − 3 sin tj 9 sin2 t + 9 cos2 t = 3 √ r(π/3) = (3/2)i + (3 3/2)j √ v(π/3) = −(3 3/2)i + (3/2)j √ a(π/3) = −(3/2)i − (3 3/2)j 2. v(t) = i + 2tj a(t) = 2j v(t) = v(t) = √ 1 + 4t2 r(2) = 2i + 4j v(2) = i + 4j a(2) = 2j y y v = − 3√3i + 3 j 2 2 v = i + 4j 8 a = 2j (3 , 3 √ 3) 22 (2, 4) x 3 a=− 3 3√3 j i− 2 2 3. v(t) = et i − e−t j 4. v(t) = 4i − j −t t x 4 a(t) = e i + e j √ v(t) = e2t + e−2t a(t) = 0 r(0) = i + j r(1) = 6i v(0) = i − j v(1) = 4i − j a(0) = i + j a(1) = 0 v(t) = √ 17 y y a=i+j (6, 0) (1, 1) x x v = 4i − j a=0 v=i−j 5. v = i + tj + t2 k, a = j + 2tk; at t = 1, v = i + j + k, v = √ 3, a = j + 2k 6. r = (1 + 3t)i + (2 − 4t)j + (7 + t)k, v = 3i − 4j + k, √ a = 0; at t = 2, v = 3i − 4j + k, v = 26, a = 0 7. v = −2 sin ti + 2 cos tj + k, a = −2 cos ti − 2 sin tj; √ √ √ √ √ at t = π/4, v = − 2i + 2j + k, v = 5, a = − 2i − 2j 8. v = et (cos t + sin t)i + et (cos t − sin t)j + k, a = 2et cos ti − 2et sin tj; at t = π/2, v = eπ/2 i − eπ/2 j + k, v = (1 + 2eπ )1/2 , a = −2eπ/2 j 513 Chapter 14 9. (a) v = −aω sin ωti + bω cos ωtj, a = −aω 2 cos ωti − bω 2 sin ωtj = −ω 2 r (b) From Part (a), a = ω 2 r 10. (a) v = 16π cos πti − 8π sin 2πtj, a = −16π 2 sin πti − 16π 2 cos 2πtj; at t = 1, v = −16π i, v = 16π , a = −16π 2 j (b) x = 16 sin πt, y = 4 cos 2πt = 4 cos2 πt − 4 sin2 πt = 4 − 8 sin2 πt, y = 4 − x2 /32 (c) Both x(t) and y (t) are periodic and have period 2, so after 2 s the particle retraces its path. √ 11. v = (6/ t)i + (3/2)t1/2 j, v = 36/t + 9t/4, d v /dt = (−36/t2 + 9/4)/(2 36/t + 9t/4) = 0 √ if t = 4 which yields a minimum by the first derivative test. The minimum speed is 3 2 when r = 24i + 8j. 12. v = (1 − 2t)i − 2tj, v = (1 − 2t)2 + 4t2 = √ 8t2 − 4t + 1, 8t − 2 d 1 v =√ which yields a minimum by the first derivative test. The = 0 if t = dt 4 8t2 − 4t + 1 √ 1 3 i − j. minimum speed is 1/ 2 when the particle is at r = 16 16 13. (a) 6 0 3 8 (b) v = 3 cos 3ti + 6 sin 3tj, v = 9 cos2 3t + 36 sin2 3t = 3 1 + 3 sin2 3t; by inspection, maximum speed is 6 and minimum speed is 4 (d) d 9 sin 6t v= = 0 when t = 0, π/6, π/3, π/2, 2π/3; the maximum speed is 6 which dt 2 1 + 3 sin2 3t occurs first when sin 3t = 1, t = π/6. 8 14. (a) 0 0 c (d) v = −6 sin 2ti + 2 cos 2tj + 4k, v = 36 sin2 2t + 4 cos2 2t + 16 = 2 8 sin2 t + 5; by inspec√ √ tion the maximum speed is 2 13 when t = π , the minimum speed is 2 5 when t = π/2. 15. v(t) = − sin ti + cos tj + C1 , v(0) = j + C1 = i, C1 = i − j, v(t) = (1 − sin t)i + (cos t − 1)j; r(t) = (t + cos t)i + (sin t − t)j + C2 , r(0) = i + C2 = j, C2 = −i + j so r(t) = (t + cos t − 1)i + (sin t − t + 1)j Exercise Set 14.6 514 16. v(t) = ti − e−t j + C1 , v(0) = −j + C1 = 2i + j; C1 = 2i + 2j so v(t) = (t + 2)i + (2 − e−t )j; r(t) = (t2 /2 + 2t)i + (2t + e−t )j + C2 r(0) = j + C2 = i − j, C2 = i − 2j so r(t) = (t2 /2 + 2t + 1)i + (2t + e−t − 2)j 17. v(t) = − cos ti + sin tj + et k + C1 , v(0) = −i + k + C1 = k so C1 = i, v(t) = (1 − cos t)i + sin tj + et k; r(t) = (t − sin t)i − cos tj + et k + C2 , r(0) = −j + k + C2 = −i + k so C2 = −i + j, r(t) = (t − sin t − 1)i + (1 − cos t)j + et k. 1 1 1 j + e−2t k + C1 , v(0) = −j + k + C1 = 3i − j so t+1 2 2 1 1 1 −2t 1 C1 = 3i − k, v(t) = 3i − j+ e k; − 2 t+1 2 2 18. v(t) = − r(t) = 3ti − ln(t + 1)j − 1 −2t 1 e + t k + C2 , 4 2 9 1 r(0) = − k + C2 = 2k so...
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