# 94 with the positive x axis be 47 let f1 f2 the

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Unformatted text preview: 2.01802 2.01600 2.01431 2.01288 2.01167 2.01062 2.00971 48. (a) p/2 0 +∞ (b) r2 + (dr/dθ)2 = (e−θ )2 + (−e−θ )2 = 2e−2θ , L = θ0 (c) L = lim θ0 →+∞ 0 √ 2e−θ dθ = lim θ0 →+∞ √ 0 2(1 − e−θ0 ) = √ 2 √ 2e−θ dθ Exercise Set 12.2 418 49. x = 2t, y = 2, (x )2 + (y )2 = 4t2 + 4 4 4 (2t) 4t2 + 4dt = 8π S = 2π 0 8π 2 (t + 1)7/2 3 t2 + 1dt = t 0 4 = 0 √ 8π (17 17 − 1) 3 50. x = et (cos t − sin t), y = et (cos t + sin t), (x )2 + (y )2 = 2e2t π /2 π /2 √ √ S = 2π (et sin t) 2e2t dt = 2 2π e2t sin t dt 0 0 √ 1 2t e (2 sin t − cos t) 5 = 2 2π π /2 0 √ 22 π (2eπ + 1) = 5 51. x = −2 sin t cos t, y = 2 sin t cos t, (x )2 + (y )2 = 8 sin2 t cos2 t √ 8 sin2 t cos2 t dt = 4 2π π /2 cos2 t S = 2π 0 √ cos3 t sin t dt = − 2π cos4 t π /2 0 1 1 + 16t2 dt = t 0 π 53. x = −r sin t, y = r cos t, (x )2 + (y )2 = r2 , S = 2π 2π S = 2π dx dφ 2 + dy dφ √ 2π √ π (17 17 − 1) 24 √ r sin t r2 dt = 2πr2 0 dy dx = a(1 − cos φ), = a sin φ, dφ dφ = 0 52. x = 1, y = 4t, (x )2 + (y )2 = 1 + 16t2 , S = 2π 54. π /2 π sin t dt = 4πr2 0 2 = 2a2 (1 − cos φ) √ a(1 − cos φ) 2a2 (1 − cos φ) dφ = 2 2πa2 0 2π (1 − cos φ)3/2 dφ, 0 √ φ φ so (1 − cos φ)3/2 = 2 2 sin3 for 0 ≤ φ ≤ π and, taking advantage of the but 1 − cos φ = 2 sin 2 2 π 3φ 2 symmetry of the cycloid, S = 16πa sin dφ = 64πa2 /3 2 0 2 55. (a) dθ dr dr/dt 2 dr = 2 and = 1 so = = = 2, r = 2θ + C , r = 10 when θ = 0 so dt dt dθ dθ/dt 1 10 = C, r = 2θ + 10. (b) r2 + (dr/dθ)2 = (2θ + 10)2 + 4, during the ﬁrst 5 seconds the rod rotates through an angle 5 (2θ + 10)2 + 4dθ, let u = 2θ + 10 to get of (1)(5) = 5 radians so L = 0 L= = 1 2 20 u2 + 4du = 10 2 + dy dθ β r2 + L= α u2 + 4 + 2 ln |u + u2 + 4| 20 10 √ √ √ 1 20 + 404 √ 10 404 − 5 104 + 2 ln ≈ 75.7 mm 2 10 + 104 dr dy dr dx = cos θ − r sin θ, = r cos θ + sin θ, dθ dθ dθ dθ 2 dr = r2 + , and Formula (6) of Section 8.4 becomes dθ 56. x = r cos θ, y = r sin θ, dx dθ 1u 22 2 dr dθ 2 dθ 419 Chapter 12 EXERCISE SET 12.3 π 1. (a) π/2 2π (d) 0 1 (1 − cos θ)2 dθ 2 π /2 (b) 0 (c) 1 (1 − sin θ)2 dθ 2 (f ) 2 π /2 12 θ dθ 2 (e) −π/2 2. (a) 3π/8 + 1 2π 3. (a) A = 0 (f ) π/8 π (b) A = 0 −π/2 4. (a) r2 = r sin θ + r cos θ, x2 + y 2 − y − x = 0, 3π/4 (b) A = −π/4 π /6 0 + y− 0 1 2 2 = 1 2 1 (1 + sin θ)2 dθ = 3π/8 + 1 2 1 (16 cos2 3θ)dθ = 4π 2 1 4 sin2 2θ dθ = π/2, so total area = 2π . 2 0 2π/3 2 π /2 π /2 π 1 2 6. A = 8. The petal in the ﬁrst quadrant has area 9. A = 2 x− 1 (sin θ + cos θ)2 dθ = π/2 2 1 (2 + 2 cos θ)2 dθ = 6π 2 7. A = 6 122 4a sin θ dθ = πa2 2 12 4a cos2 θ dθ = πa2 2 (c) A = 0 0 1 cos2 2θ dθ 2 (c) π/8 12 a dθ = πa2 2 π /2 1 sin2 2θ dθ 2 π /4 (e) 3π/4 (d) 4π /3 π 0 (b) π/2 3 5. A = 2 π /2 1 4 cos2 θ dθ 2 √ 1 (1 + 2 cos θ)2 dθ = π − 3 3/2 2 π /2 11. area = A1 − A2 = 0 π 12. area = A1 − A2 = 0 π /6 A = A1 + A2 = 0 0 0 2 dθ = 4/3 θ2 1 1 cos 2θ dθ = π/2 − 2 4 π /2 1 (1 + cos θ)2 dθ − 2 √ 1 π /4 1 4 cos2 θ dθ − 2 13. The circles intersect when cos t = 3 10. A = 1 cos2 θ dθ = 5π/8 2 √ 3 sin t, tan t = 1/ 3, t = π/6, so 1√ (4 3 sin t)2 dt + 2 π /2 π/6 √ √ √ 1 (4 cos t)2 dt = 2π − 3 3+4π/3 − 3 = 10π/3 − 4 3. 2 14. The curves intersect when 1 + cos t = 3 cos t, cos t = 1/2, t = ±π/3, and hence π /3 total area = 2 0 1 (1+cos t)2 dt+2 2 π /2 π/3 √ √ 1 9 cos2 t dt = 2(π/4+9 3/16+3π/8−9 3/16) = 5π/4. 2 Exercise Set 12.3 420 π /2 15. A = 2 π/6 π √ 1 [25 sin2 θ − (2 + sin θ)2 ]dθ = 8π/3 + 3 2 1 [16 − (2 − 2 cos θ)2 ]dθ = 10π 2 16. A = 2 0 π /3 √ 1 [(2 + 2 cos θ)2 − 9]dθ = 9 3/2 − π 2 π /4 1 (16 sin2 θ)dθ = 2π − 4 2 17. A = 2 0 18. A = 2 0 2π/3 19. A = 2 0 π /3 20. A = 2 0 1 (1/2 + cos θ)2 dθ − 2 cos−1 (3/5) 0 π /8 0 2π/3 √ 1 (1/2 + cos θ)2 dθ = (π + 3 3)/4 2 9 9√ 1 (2 + 2 cos θ)2 − sec2 θ dθ = 2π + 3 2 4 4 21. A = 2 22. A = 8 π 1 (100 − 36 sec2 θ)dθ = 100 cos−1 (3/5) 48 2 1 (4a2 cos2 2θ − 2a2 )dθ = 2a2 2 23. (a) r is not real for π/4 < θ < 3π/4 and 5π/4 < θ < 7π/4 π /4 12 a cos 2θ dθ = a2 2 π /6 √ 2π 1 4 cos 2θ − 2 dθ = 2 3 − 2 3 (b) A = 4 0 (c) A = 4 0 π /2 24. A = 2 0 4π 1 sin 2θ dθ = 1 2 25. A = 2π 122 a θ dθ − 2 2π 0 122 a θ dθ = 8π 3 a2 2 26. (a) x = r cos θ, y = r sin θ, (dx/dθ)2 + (dy/dθ)2 = (f (θ) cos θ − f (θ) sin θ)2 + (f (θ) sin θ + f (θ) cos θ)2 = f (θ)2 + f (θ)2 ; β 2πf (θ) sin θ S= f (θ)2 + f (θ)2 dθ if about θ = 0; similarly for θ = π/2 α (b) f , g are continuous and no segment of the curve is traced more than once. 27. r2 + dr dθ 2 = cos2 θ + sin2 θ = 1, π /2 2π cos2 θ dθ = π 2 . so S = −π/2 421 Chapter 12 π /2 28. S = √ 2πeθ cos θ 2e2θ dθ 0 √ = 2 2π π /2 e2θ cos θ dθ = 0 √ 2 2π π (e − 2) 5 π 2π (1 − cos θ) sin θ 29. S = 1 − 2 cos θ + cos2 θ + sin2 θ dθ 0...
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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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