32 f1 x y z 2x2 3y 2 z 2 f2 x y z x2

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Unformatted text preview: C2 = k, r(t) = 3ti − ln(t + 1)j + 4 4 9 1 −2t 1 −e − t k. 44 2 19. If a = 0 then x (t) = y (t) = z (t) = 0, so x(t) = x1 t + x0 , y (t) = y1 t + y0 , z (t) = z1 t + z0 , the motion is along a straight line and has constant speed. 20. (a) If r is constant then so is r 2 , but then x2 + y 2 = c2 (2-space) or x2 + y 2 + z 2 = c2 (3-space), so the motion is along a circle or a sphere of radius c centered at the origin, and the velocity vector is always perpendicular to the locating vector. (b) If v is constant then by the Theorem, v(t) · a(t) = 0, so the velocity is always perpendicular to the acceleration. 21. v = 3t2 i + 2tj, a = 6ti + 2j; v = 3i + 2j and a = 6i + 2j when t = 1 so √ cos θ = (v · a)/( v a ) = 11/ 130, θ ≈ 15◦ . 22. v = et (cos t − sin t)i + et (cos t + sin t)j, a = −2et sin ti + 2et cos tj, v · a = 2e2t , v = √ a = 2et , cos θ = (v · a)/( v a ) = 1/ 2, θ = 45◦ . 23. (a) displacement = r1 − r0 = 0.7i + 2.7j − 3.4k (b) ∆r = r1 − r0 , so r0 = r1 − ∆r = −0.7i − 2.9j + 4.8k. y 24. (a) (b) distance = 50(1 − cos 2πt) 4 2 x -4 -2 2 4 -2 -4 3 25. ∆r = r(3) − r(1) = 8i + 26/3j; v = 2ti + t2 j, s = t √ √ 4 + t2 dt = (13 13 − 5 5)/3. 1 3π/2 26. ∆r = r(3π/2) − r(0) = 3i − 3j; v = 3 cos ti − 3 sin tj, s = 3dt = 9π/2. 0 √t 2e , 515 Chapter 14 27. ∆r = r(ln 3) − r(0) = 2i − 2/3j + √ √ 2 ln 3k; v = et i − e−t j + 2 k, s = ln 3 (et + e−t )dt = 8/3. 0 28. ∆r = r(π ) − r(0) = 0; v = −2 sin 2ti + 2 sin 2tj − sin 2tk, π π /2 3| sin 2t|dt = 6 v = 3| sin 2t|, s = 0 sin 2t dt = 6. 0 29. In both cases, the equation of the path in rectangular coordinates is x2 + y 2 = 4, the particles move counterclockwise around this circle; v1 = −6 sin 3ti + 6 cos 3tj and v2 = −4t sin(t2 )i + 4t cos(t2 )j so v1 = 6 and v2 = 4t. 30. Let u = 1 − t3 in r2 to get r2 (u) = (3 + 2u)i + uj + (1 − u)k so both particles move along the same √ √ line; v1 = 2i + j − k and v2 = −6t2 i − 3t2 j + 3t2 k so v1 = 6 and v2 = 3 6t2 . 31. (a) v = −e−t i + et j, a = e−t i + et j; when t = 0, v = −i + j, a = i + j, v = √ v × a = −2k so aT = 0, aN = 2. √ (c) κ = 1/ 2 (b) aT T = 0, aN N = a − aT T = i + j √ 2, v · a = 0, 32. (a) v = −2t sin(t2 )i + 2t cos(t2 )j, a = [−4t2 cos(t2 ) − 2 sin(t2 )]i + [−4t2 sin(t2 ) + 2 cos(t2 )]j; when √ √ √ √ √ √ t = π/2, v = − π /2i + π /2j, a = (−π/ 2 − 2)i + (−π/ 2 + 2)j, v = π , √ v · a = 2 π , v × a = π 3/2 k so aT = 2, aN = π √ √ (b) aT T = − 2(i − j), aN N = a − aT T = −(π/ 2)(i + j) (c) κ = 1 33. (a) v = (3t2 − 2)i + 2tj, a = 6ti + 2j; when t = 1, v = i + 2j, a = 6i + 2j, v = √ √ v × a = −10k so aT = 2 5, aN = 2 5 √ 25 (b) aT T = √ (i + 2j) = 2i + 4j, aN N = a − aT T = 4i − 2j 5 √ (c) κ = 2/ 5 √ 5, v · a = 10, √ 34. (a) v = et (− sin t+cos t)i+et (cos t+sin t)j, a = −2et sin ti+2et cos tj; when t = π/4, v = 2eπ/4 j, √ √ √ √ a = − 2eπ/4 i + 2eπ/4 j, v = 2eπ/4 , v · a = 2eπ/2 , v × a = 2eπ/2 k so aT = 2eπ/4 , √ aN = 2eπ/4 √ √ (b) aT T = 2eπ/4 j, aN N = a − aT T = − 2eπ/4 i (c) κ = √ 1 2eπ/4 35. (a) v = i + 2tj + 3t2 k, a = 2j + 6tk; when t = 1, v = i + 2j + 3k, a = 2j + 6k, v = √ √√ v · a = 22, v × a = 6i − 6j + 2k so aT = 22/ 14, aN = 76/ 14 = 38/7 (b) aT T = √ 14, 22 33 8 9 11 11 i + j + k, aN N = a − aT T = − i − j + k 7 7 7 7 7 7 √ 19 (c) κ = √ 7 14 36. (a) v = et i − 2e−2t j + k, a = et i + 4e−2t j; when t = 0, v = i − 2j + k, a = i + 4j, v = √ v · a = −7, v × a = −4i + j + 6k so aT = −7/ 6, aN = 53/6 √ 6, Exercise Set 14.6 516 19 7 7 13 i+ j+ k (b) aT T = − (i − 2j + k), aN N = a − aT T = 6 6 3 6 √ 53 (c) κ = √ 66 37. (a) v = 3 cos ti − 2 sin tj − 2 cos 2tk, a = −3 sin ti − 2 cos tj +4 sin 2tk; when t = π/2, v = −2j +2k, √ a = −3i, v = 2 2, v · a = 0, v × a = −6j − 6k so aT = 0, aN = 3 (b) aT T = 0, aN N = a = −3i (c) κ = 3 8 38. (a) v = 3t2 j − (16/t)k, a = 6tj + (16/t2 )k; when t = 1, v = 3j − 16k, a = 6j + 16k, v = √ √ v · a = −238, v × a = 144i so aT = −238/ 265, aN = 144/ 265 (b) aT T = − (c) κ = 39. 40. √ 265, 3808 432 714 2304 j+ k, aN N = a − aT T = j+ k 265 265 265 265 144 2653/2 v = 4, v · a = −12, v × a = 8k so aT = −3, aN = 2, T = −j, N = (a − aT T)/aN = i √ √ √ 5, v · a = 3, v × a = −6k so aT = 3/ 5, aN = 6/ 5, T = (1/ 5)(i + 2j), √ N = (a − aT T)/aN = (1/ 5)(2i − j) v= √ √ 41. v = 3, v · a = 4, v × a = 4i − 3j − 2k so aT = 4/3, aN = √ N = (a − aT T)/aN = (i − 8j + 14k)/(3 29) 42. v = 5, v · a = −5, v × a = −4i − 10j − 3k so aT = −1, aN = √ N = (a − aT T)/aN = (8i − 5j + 6k)/(5 5) 29/3, T = (1/3)(2i + 2j + k), √ 5, T = (1/5)(3i − 4k), 43. aT = d2 s d = dt2 dt 3t2 + 4 = 3t/ 44. aT = d2 s d = dt2 dt t2 + e−3t = (2t − 3e−3t )/[2 t2 + e−3t ] so when t = 0, aT = −3/2. 3t2 + 4 so when t = 2, aT = 3/2. d2 s d = (4t − 1)2 + cos2 πt = [4t − 1 − π cos πt sin πt]/ (4t − 1)2 + cos2 πt so when 2 dt dt √ t = 1/4, aT = −π/ 2. 45. aT = 46. aT = d2 s d = dt2 dt t4 + 5t2 + 3 = (2t3 + 5t...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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