Drd abeb b chapter 12 horizon module 446 chapter 12

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Unformatted text preview: 6, – p) √ 3. (a) (3 3, 3) (2, g) 0 (2, $) √ (b) (−7/2, 7 3/2) √ (e) (−7 3/2, 7/2) √ (c) (3 3, 3) √ √ 4. (a) (−4 2, −4 2) (d) (5, 0) √ √ (b) (7 2/2, −7 2/2) (e) (0, −2) √√ (c) (4 2, 4 2) (f ) (0, 0) 5. (a) both (5, π ) √ √ (d) (8 2, 5π/4), (8 2, −3π/4) (b) (4, 11π/6), (4, −π/6) (c) (2, 3π/2), (2, −π/2) √ (f ) both ( 2, π/4) (d) (0, 0) 6. (a) (2, 5π/6) (f ) (−5, 0) (e) both (6, 2π/3) (b) (−2, 11π/6) (c) (2, −7π/6) (d) (−2, −π/6) 7. (a) (5, 0.6435) √ (b) ( 29, 5.0929) (c) (1.2716, 0.6658) 8. (a) (5, 2.2143) (b) (3.4482, 2.6260) (c) ( 4 + π 2 /36, 0.2561) 9. (a) r2 = x2 + y 2 = 4; circle (b) y = 4; horizontal line (c) r2 = 3r cos θ, x2 + y 2 = 3x, (x − 3/2)2 + y 2 = 9/4; circle (d) 3r cos θ + 2r sin θ = 6, 3x + 2y = 6; line 10. (a) r cos θ = 5, x = 5; vertical line (b) r2 = 2r sin θ, x2 + y 2 = 2y , x2 + (y − 1)2 = 1; circle (c) r2 = 4r cos θ + 4r sin θ, x2 + y 2 = 4x + 4y, (x − 2)2 + (y − 2)2 = 8; circle 1 sin θ , r cos2 θ = sin θ, r2 cos2 θ = r sin θ, x2 = y ; parabola (d) r = cos θ cos θ 11. (a) r cos θ = 7 (b) r = 3 (c) r − 6r sin θ = 0, r = 6 sin θ 2 (d) 4(r cos θ)(r sin θ) = 9, 4r2 sin θ cos θ = 9, r2 sin 2θ = 9/2 12. (a) r sin θ = −3 (b) r = (c) r2 + 4r cos θ = 0, r = −4 cos θ (d) r4 cos2 θ = r2 sin2 θ, r2 = tan2 θ, r = tan θ 408 √ 5 409 Chapter 12 p/ 2 13. 2 14. 3 -2 -3 2 0 3 -2 -3 r = 2 cos 3θ r = 3 sin 2θ 15. 16. p/ 2 p/2 0 -4 -1 4 0 r = 3 − 4 sin 3θ r = 2 + 2 sin θ 17. (a) r = 5 (b) (x − 3)2 + y 2 = 9, r = 6 cos θ (c) Example 6, r = 1 − cos θ 18. (a) From (8-9), r = a ± b sin θ or r = a ± b cos θ. The curve is not symmetric about the y -axis, so Theorem 12.2.1(a) eliminates the sine function, thus r = a ± b cos θ. The cartesian point (−3, 0) is either the polar point (3, π ) or (−3, 0), and the cartesian point (−1, 0) is either the polar point (1, π ) or (−1, 0). A solution is a = 1, b = −2; we may take the equation as r = 1 − 2 cos θ. (b) x2 + (y + 3/2)2 = 9/4, r = −3 sin θ (c) Figure 12.1.18, a = 1, n = 3, r = sin 3θ 19. (a) Figure 12.1.18, a = 3, n = 2, r = 3 sin 2θ (b) From (8-9), symmetry about the y -axis and Theorem 12.1.1(b), the equation is of the form r = a ± b sin θ. The cartesian points (3, 0) and (0, 5) give a = 3 and 5 = a + b, so b = 2 and r = 3 + 2 sin θ. (c) Example 8, r2 = 9 cos 2θ 20. (a) Example 6 rotated through π/2 radian: a = 3, r = 3 − 3 sin θ (b) Figure 12.1.18, a = 1, r = cos 5θ (c) x2 + (y − 2)2 = 4, r = 4 sin θ Exercise Set 12.1 410 21. 22. 23. 2 3 ( Line Line 24. Circle 25. 26. 4 2 6 Circle 1 Circle 27. Cardioid 28. 29. 2 1 2 3 4 6 30. Cardioid Cardioid Circle 31. 10 4 32. 3 1 5 8 1 Cardioid Limaçon Cardioid 33. 34. 35. 4 1 3 1 2 7 2 1 Cardioid 36. Limaçon Limaçon 37. 3 38. 8 3 4 2 1 7 5 2 Limaçon Limaçon Limaçon 411 Chapter 12 39. 40. 5 41. 1 3 3 3 7 Lemniscate 7 Limaçon Limaçon 42. 43. 1 44. 4 2p 4p 8p 6p 45. Spiral Lemniscate Lemniscate 46. 2p 47. 2p 4p 8p 4p 1 6p 6p Spiral Four-petal rose Spiral 48. 49. 50. 3 9 2 Eight-petal rose Four-petal rose 1 52. -1 Three-petal rose 3 53. 1 -3 -1 -3 2 54. -2 1 55. 2 -2 3 -1 1 -1 Exercise Set 12.1 412 56. 0 ≤ θ ≤ 8π 57. (a) −4π < θ < 4π 58. In I, along the x-axis, x = r grows ever slower with θ. In II x = r grows linearly with θ. √ Hence I: r = θ; II: r = θ. 59. (a) r = a/ cos θ, x = r cos θ = a, a family of vertical lines (b) r = b/ sin θ, y = r sin θ = b, a family of horizontal lines 60. The image of (r0 , θ0 ) under a rotation through an angle α is (r0 , θ0 + α). Hence (f (θ), θ) lies on the original curve if and only if (f (θ), θ + α) lies on the rotated curve, i.e. (r, θ) lies on the rotated curve if and only if r = f (θ − α). √ 2 (cos θ + sin θ) 2 (b) r = 1 + cos(θ − π/2) = 1 + sin θ 61. (a) r = 1 + cos(θ − π/4) = 1 + (c) r = 1 + cos(θ − π ) = 1 − cos θ √ (d) r = 1 + cos(θ − 5π/4) = 1 − 2 (cos θ + sin θ) 2 62. r2 = 4 cos 2(θ − π/2) = −4 cos 2θ 63. Either r − 1 = 0 or θ − 1 = 0, so the graph consists of the circle r = 1 and the line θ = 1. p/2 u=1 r=1 0 64. (a) r2 = Ar sin θ + Br cos θ, x2 + y 2 = Ay + Bx, (x − B/2)2 + (y − A/2)2 = (A2 + B 2 )/4, which 1 A2 + B 2 . is a circle of radius 4 (b) Formula (4) follows by setting A = 0, B = 2a, (x − a)2 + y 2 = a2 , the circle of radius a about (a, 0). Formula (5) is derived in a similar fashion. 65. y = r sin θ = (1 + cos θ) sin θ = sin θ + sin θ cos θ, dy/dθ = cos θ − sin2 θ + cos2 θ = 2 cos2 θ + cos θ − 1 = (2 cos θ − 1)(cos θ + 1); dy/dθ = 0 if cos θ = 1/2 or if cos θ = −1; θ = π/3 or π . √ √ If θ = π/3, π , then y = 3 3/4, 0 so the maximum value of y is 3 3/4 and the polar coordinates of the highest point are (3/2, π/3). 66. x = r cos θ = (1 + cos θ) cos θ = cos θ + cos2 θ, dx/dθ = − sin θ − 2 sin θ cos θ = − sin θ(1 + 2 cos θ), dx/dθ = 0 if sin θ = 0 or if cos θ = −1/2; θ = 0, 2π/3, or π . If θ = 0, 2π/3, π , then x = 2, −1/4, 0 so the minimum value of x is −1/4. The leftmost point has polar coordi...
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