Calculus: Early Transcendentals, by Anton, 7th Edition,ch11

Calculus - Early Transcendentals

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447 CHAPTER 11 Analytic Geometry in Calculus EXERCISE SET 11.1 1. (1, 6 ) (3, 3 ) (4, e ) (–1, r ) 0 p / 2 (5, 8 ) (–6, – p ) 2. ( , L ) 3 2 0 p / 2 ( 3, i ) ( 5, @ ) (2, $ ) (0, c ) (2, g ) 3. (a) (3 3 , 3) (b) ( 7 / 2 , 7 3 / 2) (c) (3 3 , 3) (d) (0 , 0) (e) ( 7 3 / 2 , 7 / 2) (f) ( 5 , 0) 4. (a) ( 4 2 , 4 2) (b) (7 2 / 2 , 7 2 / 2) (c) (4 2 , 4 2) (d) (5 , 0) (e) (0 , 2) (f) (0 , 0) 5. (a) both (5 , π ) (b) (4 , 11 π/ 6) , (4 , π/ 6) (c) (2 , 3 π/ 2) , (2 , π/ 2) (d) (8 2 , 5 π/ 4) , (8 2 , 3 π/ 4) (e) both (6 , 2 π/ 3) (f) both ( 2 , π/ 4) 6. (a) (2 , 5 π/ 6) (b) ( 2 , 11 π/ 6) (c) (2 , 7 π/ 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) r 2 = x 2 + y 2 = 4; circle (b) y = 4; horizontal line (c) r 2 = 3 r cos θ , x 2 + y 2 = 3 x , ( x 3 / 2) 2 + y 2 = 9 / 4; circle (d) 3 r cos θ + 2 r sin θ = 6, 3 x + 2 y = 6; line 10. (a) r cos θ = 5, x = 5; vertical line (b) r 2 = 2 r sin θ , x 2 + y 2 = 2 y , x 2 + ( y 1) 2 = 1; circle (c) r 2 = 4 r cos θ + 4 r sin θ, x 2 + y 2 = 4 x + 4 y, ( x 2) 2 + ( y 2) 2 = 8; circle (d) r = 1 cos θ sin θ cos θ , r cos 2 θ = sin θ , r 2 cos 2 θ = r sin θ , x 2 = y ; parabola 11. (a) r cos θ = 7 (b) r = 3 (c) r 2 6 r sin θ = 0, r = 6 sin θ (d) 4( r cos θ )( r sin θ ) = 9, 4 r 2 sin θ cos θ = 9, r 2 sin 2 θ = 9 / 2 12. (a) r sin θ = 3 (b) r = 5 (c) r 2 + 4 r cos θ = 0, r = 4 cos θ (d) r 4 cos 2 θ = r 2 sin 2 θ , r 2 = tan 2 θ , r = tan θ
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448 Chapter 11 13. 0 p / 2 -3 -3 3 3 r = 3 sin 2 θ 14. -3 3 -2.25 2.25 r = 2 cos 3 θ 15. 0 p / 2 -4 4 -1 r = 3 4 sin 3 θ 16. 0 p / 2 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 11.2.1(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) x 2 + ( y + 3 / 2) 2 = 9 / 4 , r = 3 sin θ (c) Figure 11.1.18, a = 1 , n = 3 , r = sin 3 θ 19. (a) Figure 11.1.18, a = 3 , n = 2 , r = 3 sin 2 θ (b) From (8-9), symmetry about the y -axis and Theorem 11.1.1(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, r 2 = 9 cos 2 θ 20. (a) Example 6 rotated through π/ 2 radian: a = 3 , r = 3 3 sin θ (b) Figure 11.1.18, a = 1 , r = cos 5 θ (c) x 2 + ( y 2) 2 = 4, r = 4 sin θ
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Exercise Set 11.1 449 21. Line 2 22. Line ( 23. Circle 3 24. 4 Circle 25. 6 Circle 26. 1 2 Cardioid 27. Circle 1 2 28. 4 2 Cardioid 29. Cardioid 3 6 30. 5 10 Cardioid 31. 4 8 Cardioid 32. 1 3 1 Limaçon 33. 1 2 Cardioid 34. 1 7 4 Lima ç on 35. 3 2 1 Lima ç on 36. 4 2 3 Limaçon 37. Lima ç on 3 1 7 38. 2 5 8 Lima ç on 39. 3 5 Lima ç on 7 40. 3 1 7 Lima ç on 41. Lemniscate 3 42. 1 Lemniscate 43. Lemniscate 4 44. Spiral 2 p 4 p 6 p 8 p
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450 Chapter 11 45. Spiral 2 p 4 p 6 p 8 p 46. 2 p 6 p 4 p Spiral 47. 1 Four-petal rose 48. 3 Four-petal rose 49. 9 Eight-petal rose 50. 2 Three-petal rose 51. -1 1 -1 1 52. 1 -1 -1 1 53. 3 -3 -3 3 54.
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