27 n1 2 1 1 and n2 1 1 2 are normals to the given

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Unformatted text preview: 10 = a = (b) r = = 1 5d 1 5 cos θ = d , if θ = π : r0 = d/6, θ = 0 : r1 = d/4, 5 − cos θ 1 1 (r1 + r0 ) = d 2 2 11 + 46 = 5 144/5 144 d, d = 144/5, r = = 24 5 − cos θ 25 − 5 cos θ √ 3d 3 , if θ = 3π/2 : r0 = d, θ = π/2 : r1 = 3d, 4 = b = 3d/ 7, 4 − 3 sin θ 7 1− √ 4√ 47 d= 7, r = 3 4 − 3 sin θ (c) r = 3 4d 3 4 sin θ = (d) r = 4 5d 4 5 cos θ = 1+ c = 10 = 4 1 1 (r1 − r0 ) = d 4 − 2 2 9 15. (a) e = c/a = (b) e = 1 2 (r1 1 2 (r1 = 16 45 45/2 45 d, d = , r= = 9 8 5 + 4 cos θ 10 + 8 cos θ − r0 ) r1 − r0 = r 1 + r0 + r0 ) r1 /r0 − 1 r1 1+e , e(r1 /r0 + 1) = r1 /r0 − 1, = r1 /r0 + 1 r0 1−e 16. (a) e = c/a = (b) e = 4d 4 , if θ = 0 : r0 = d; θ = π : r1 = 4d, 5 + 4 cos θ 9 1 2 (r1 1 2 (r1 + r0 ) r1 + r0 = r 1 − r0 − r0 ) r1 /r0 + 1 r1 e+1 , e(r1 /r0 − 1) = r1 /r0 + 1, = r1 /r0 − 1 r0 e−1 17. (a) T = a3/2 = 39.51.5 ≈ 248 yr (b) r0 = a(−e) = 39.5(1 − 0.249) = 29.6645 AU ≈ 4,449,675,000 km r1 = a(1 + e) = 39.5(1 + 0.249) = 49.3355 AU ≈ 7,400,325,000 km (c) r = a(1 − e2 ) 39.5(1 − (0.249)2 ) 37.05 ≈ ≈ AU 1 + e cos θ 1 + 0.249 cos θ 1 + 0.249 cos θ p/ 2 (d) 50 -30 20 0 -50 18. (a) In yr and AU, T = a3/2 ; in days and km, so T = 365 × 10−9 a 150 3/2 days. T = 365 a 150 × 106 3/2 , 439 Chapter 12 (b) T = 365 × 10−9 57.95 × 106 150 3/2 ≈ 87.6 days (c) From (17) the polar equation of the orbit has the form r = (d) 37.82 a(1 − e2 ) = 1 + e cos θ 1 + 0.205 cos θ p/2 40 -50 0 20 -40 19. (a) a = T 2/3 = 23802/3 ≈ 178.26 AU (b) r0 = a(1 e) ≈ 0.8735 AU, r1 = a(1 + e) ≈ 355.64 AU (c) r = 1.74 a(1 − e2 ) ≈ AU 1 + e cos θ 1 + 0.9951 cos θ p/ 2 (d) 20 0 -400 -20 20. (a) By Exercise 15(a), e = r 1 − r0 ≈ 0.093 r1 + r0 (b) r = 1 (a0 + a1 ) = 225,400,000 km ≈ 1.503 AU, so T = a3/2 ≈ 1.84 yr 2 (c) r = a(1 − e2 ) 1.49 ≈ AU 1 + e cos θ 1 + 0.093 cos θ (d) p/ 2 1.49 1.6428 1.3632 0 1.49 21. r0 = a(1 − e) ≈ 7003 km, hmin ≈ 7003 − 6440 = 563 km, r1 = a(1 + e) ≈ 10,726 km, hmax ≈ 10,726 − 6440 = 4286 km Chapter 12 Supplementary Exercises 440 22. r0 = a(1 − e) ≈ 651,736 km, hmin ≈ 581,736 km; r1 = a(1 + e) ≈ 6,378,102 km, hmax ≈ 6,308,102 km a → +∞. e 1 − e and lim d = +∞; Let d be the distance between the directrix and the focus, then d = a e→0 e but the center and the focus come together, so distance between the directrix and the center also tends to +∞. 23. Since the foci are fixed, a is constant; since e → 0, the distance x2 y2 x2 y2 c x−a 24. (a) From Figure 12.4.22, 2 − 2 = 1, 2 − 2 = 1, (x − c)2 + y 2 = a b a c − a2 a c (x − c)2 + y 2 = x − a for x > 0. a (b) From Part (a), P F = 2 , PF c P D, = c/a. a PD CHAPTER 12 SUPPLEMENTARY EXERCISES √ 2. (a) ( 2, 3π/4) √ (b) (− 2, 7π/4) √ (c) ( 2, 3π/4) √ (d) (− 2, −π/4) 3. (a) circle (e) lima¸on c (b) rose (f ) none (c) line (g) none (d) lima¸on c (h) spiral 4. (a) r = 1/3 , ellipse, right of pole, distance = 1 1 + 1 cos θ 3 (b) hyperbola, left of pole, distance = 1/3 (c) r = 1/3 , parabola, above pole, distance = 1/3 1 + sin θ (d) parabola, below pole, distance = 3 5. (a) (b) p/ 2 p/ 2 (1, 9 ) 0 0 (c) p/ 2 (d) (e) p/ 2 0 p/ 2 (2, 3 ) 0 (1, 3 ) (1, # ) (–1, 3 ) 0 441 Chapter 12 6. Family I: x2 + (y − b)2 = b2 , b < 0, or r = 2b sin θ; Family II: (x − a)2 + y 2 = a2 , a < 0, or r = 2a cos θ 7. (a) r = 2a/(1 + cos θ), r + x = 2a, x2 + y 2 = (2a − x)2 , y 2 = −4ax + 4a2 , parabola (b) r2 (cos2 θ − sin2 θ) = x2 − y 2 = a2 , hyperbola √ √ (c) r sin(θ − π/4) = ( 2/2)r(sin θ − cos θ) = 4, y − x = 4 2, line (d) r2 = 4r cos θ + 8r sin θ, x2 + y 2 = 4x + 8y, (x − 2)2 + (y − 4)2 = 20, circle 9. (a) 2 4 45 7 5 2 12 c = e = and 2b = 6, b = 3, a2 = b2 + c2 = 9 + a2 , a2 = 9, a = √ , x + y =1 a 7 49 49 49 9 5 (b) x2 = −4py , directrix y = 4, focus (−4, 0), 2p = 8, x2 = −16y √ √ (c) For the ellipse, a = 4, b = 3, c2 = a2 − b2 = 16 − 3 = 13, foci (± 13, 0); √ 4 13 2 a, for the hyperbola, c = 13, b/a = 2/3, b = 2a/3, 13 = c2 = a2 + b2 = a2 + a2 = 9 9 x2 y2 a = 3, b = 2, − =1 9 4 10. (a) e = 4/5 = c/a, c = 4a/5, but a = 5 so c = 4, b = 3, (x + 3)2 (y − 2)2 + =1 25 9 (b) directrix y = 2, p = 2, (x + 2)2 = −8y (c) center (−1, 5), vertices (−1, 7) and (−1, 3), a = 2, a/b = 8, b = 1/4, y 11. (a) (b) 2 y (y − 5)2 − 16(x +1)2 = 1 4 y (c) 4 4 x -4 x 8 x 2 -8 8 10 -3 -12 -10 470 13. (a) The equation of the parabola is y = ax2 and it passes through (2100, 470), thus a = , 21002 470 2 x. y= 21002 2100 1+ 2 (b) L = 2 0 470 x 21002 2 dx ≈ 4336.3 ft 14. (a) As t runs from 0 to π , the upper portion of the curve is traced out from right to left; as t runs from π to 2π the bottom portion of the curve is traced out from right to left. The loop 1 1 occurs for π + sin−1 < t < 2π − sin−1 . 4 4 (b) lim x = +∞, lim y = 1; lim x = −∞, lim y = 1; lim x = +∞, lim y = 1; t→0+ t→0+ t→π − t→π − t→π + t→π + lim x = −∞, lim y = 1; the horizontal asymptote is y = 1. t→2π − t→2π − (c...
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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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