A i 4j 3 4 1 3i 4j 5 so the required vector is

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Unformatted text preview: e solutions, t = 0, ±π/2; the last two correspond to the crossing point. dy 2 2 For t = ±π/2, m = = ; the tangent lines are given by y = ± (x − 2). dx ±π π 17. If y = 4 then t2 = 4, t = ±2, x = 0 for t = ±2 so (0, 4) is reached when t = ±2. dy/dx = 2t/(3t2 − 4). For t = 2, dy/dx = 1/2 and for t = −2, dy/dx = −1/2. The tangent lines are y = ±x/2 + 4. 18. If x = 3 then t2 − 3t + 5 = 3, t2 − 3t + 2 = 0, (t − 1)(t − 2) = 0, t = 1 or 2. If t = 1 or 2 then y = 1 so (3, 1) is reached when t = 1 or 2. dy/dx = (3t2 + 2t − 10)/(2t − 3). For t = 1, dy/dx = 5, the tangent line is y − 1 = 5(x − 3), y = 5x − 14. For t = 2, dy/dx = 6, the tangent line is y − 1 = 6(x − 3), y = 6x − 17. 19. (a) 1 -1 1 -1 (b) dx dy = −3 cos2 t sin t and = 3 sin2 t cos t are both zero when t = 0, π/2, π, 3π/2, 2π , dt dt so singular points occur at these values of t. 20. (a) when y = 0 (b) dy a sin θ = = 0 when θ = 2nπ + π/2, n = 0, 1, . . . (which is when y = 0). dx a − a cos θ √ √ 21. Substitute θ = π/3, r = 1, and dr/dθ = − 3 in equation (7) gives slope m = 1/ 3. 22. As in Exercise 21, θ = π/4, dr/dθ = √ 2/2, r = 1 + √ √ 2/2, m = −1 − 2 23. As in Exercise 21, θ = 2, dr/dθ = −1/4, r = 1/2, m = tan 2 − 2 2 tan 2 + 1 √ √ 24. As in Exercise 21, θ = π/6, dr/dθ = 4 3a, r = 2a, m = 3 3/5 √ √ 25. As in Exercise 21, θ = 3π/4, dr/dθ = −3 2/2, r = 2/2, m = −2 26. As in Exercise 21, θ = π , dr/dθ = 3, r = 4, m = 4/3 r cos θ + (sin θ)(dr/dθ) cos θ + 2 sin θ cos θ dy ; if θ = 0, π/2, π , = = dx −r sin θ + (cos θ)(dr/dθ) − sin θ + cos2 θ − sin2 θ then m = 1, 0, −1. 27. m = 28. m = dy cos θ(4 sin θ − 1) = ; if θ = 0, π/2, π then m = −1/2, 0, 1/2. dx 4 cos2 θ + sin θ − 2 Exercise Set 12.2 416 29. dx/dθ = −a sin θ(1 + 2 cos θ), dy/dθ = a(2 cos θ − 1)(cos θ + 1) (a) horizontal if dy/dθ = 0 and dx/dθ = 0. dy/dθ = 0 when cos θ = 1/2 or cos θ = −1 so θ = π/3, 5π/3, or π ; dx/dθ = 0 for θ = π/3 and 5π/3. For the singular point θ = π we find that lim dy/dx = 0. There is a horizontal tangent line at (3a/2, π/3), (0, π ), and (3a/2, 5π/3). θ →π (b) vertical if dy/dθ = 0 and dx/dθ = 0. dx/dθ = 0 when sin θ = 0 or cos θ = −1/2 so θ = 0, π , 2π/3, or 4π/3; dy/dθ = 0 for θ = 0, 2π/3, and 4π/3. The singular point θ = π was discussed in part (a). There is a vertical tangent line at (2a, 0), (a/2, 2π/3), and (a/2, 4π/3). 30. dx/dθ = a(cos2 θ − sin2 θ) = a cos 2θ, dy/dθ = 2a sin θ cos θ = a sin 2θ (a) horizontal if dy/dθ = 0 and dx/dθ = 0. dy/dθ = 0 when θ = 0, π/2, π, 3π/2, 2π ; dx/dθ = 0 for (0, 0), (a, π/2), (0, π ), (−a, 3π/2), (0, 2π ); in reality only two distinct points (b) vertical if dy/dθ = 0 and dx/dθ = 0. √ dx/dθ = 0 when θ = π/4, 3π/4, 5π/4, 7π/4; √ dy/dθ = 0 √ √ there, so vertical tangent line at (a/ 2, π/4), (a/ 2, 3π/4), (−a/ 2, 5π/4), (−a/ 2, 7π/4), only two distinct points 31. dy/dθ = (d/dθ)(sin2 θ cos2 θ) = (sin 4θ)/2 = 0 at θ = 0, π/4, π/2, 3π/4, π ; at the same points, dx = 0 at θ = π/2, a singular point; and dx/dθ = (d/dθ)(sin θ cos3 θ) = cos2 θ(4 cos2 θ − 3). Next, dθ θ = 0, π both give the same point, so there are just three points with a horizontal tangent. 32. dx/dθ = 4 sin2 θ − sin θ − 2, dy/dθ = cos θ(1 − 4 sin θ). dy/dθ = 0 when cos θ = 0 or sin θ = 1/4 so θ = π/2, 3π/2, sin−1 (1/4), or π − sin−1 (1/4); dx/dθ = 0 at these points, so there is a horizontal tangent at each one. 33. p/2 34. p/2 p/2 35. 0 0 4 2 36. θ0 = ±π/4 θ0 = π/2 θ0 = π/6, π/2, 5π/6 p/2 p/2 37. 0 θ0 = 0, π/2 θ0 = 2π/3, 4π/3 2π adθ = 2πa 0 p/2 38. 3 39. r2 + (dr/dθ)2 = a2 + 02 = a2 , L = 0 0 0 θ0 = 0 417 Chapter 12 π /2 40. r2 + (dr/dθ)2 = (2a cos θ)2 + (−2a sin θ)2 = 4a2 , L = 2adθ = 2πa −π/2 π 41. r2 + (dr/dθ)2 = [a(1 − cos θ)]2 + [a sin θ]2 = 4a2 sin2 (θ/2), L = 2 2a sin(θ/2)dθ = 8a 0 π 42. r2 + (dr/dθ)2 = [sin2 (θ/2)]2 + [sin(θ/2) cos(θ/2)]2 = sin2 (θ/2), L = sin(θ/2)dθ = 2 0 2 43. r2 + (dr/dθ)2 = (e3θ )2 + (3e3θ )2 = 10e6θ , L = √ 10e3θ dθ = √ 10(e6 − 1)/3 0 44. r2 + (dr/dθ)2 = [sin3 (θ/3)]2 + [sin2 (θ/3) cos(θ/3)]2 = sin4 (θ/3), π /2 L= √ sin2 (θ/3)dθ = (2π − 3 3)/8 0 3 sin t dy = dx 1 − 3 cos t dy 3 sin 10 (b) At t = 10, = ≈ −0.4640, θ ≈ tan−1 (−0.4640) = −0.4344 dx 1 − 3 cos 10 45. (a) From (3), dy dy = 0 when = 2 sin t = 0, t = 0, π, 2π, 3π dx dt dx = 0 when 1 − 2 cos t = 0, cos t = 1/2, t = π/3, 5π/3, 7π/3 (b) dt 46. (a) 47. (a) r2 + (dr/dθ)2 = (cos nθ)2 + (−n sin nθ)2 = cos2 nθ + n2 sin2 nθ = (1 − sin2 nθ) + n2 sin2 nθ = 1 + (n2 − 1) sin2 nθ, π /(2n) 1 + (n2 − 1) sin2 nθdθ L=2 0 π /4 1 + 3 sin2 2θdθ ≈ 2.42 (b) L = 2 0 (c) n L 2 3 4 5 6 7 8 9 10 11 2.42211 2.22748 2.14461 2.10100 2.07501 2.05816 2.04656 2.03821 2.03199 2.02721 n L 13 14 15 16 17 18 19 20 12 2.02346 2.02046...
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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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