chapter_11 - January 27, 2005 11:47 L24-ch11 Sheet number 1...

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January 27, 2005 11:47 L24-ch11 Sheet number 1 Page number 475 black 475 CHAPTER 11 Analytic Geometry in Calculus EXERCISE SET 11.1 1. (1, 6 ) (3, 3 ) (4, e ) (–1, r ) 0 c /2 (5, 8 ) (–6, – p ) 2. ( , L ) 3 2 0 c /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) ( 2 , 2) (b) (3 2 , 3 2) (c) (2 2 , 2 2) (d) (3 , 0) (e) (0 , 4) (0 , 0) 5. (a) (5 ) , (5 , π ) (b) (4 , 11 π/ 6) , (4 , 6) (c) (2 , 3 2) , (2 , 2) (d) (8 2 , 5 4) , (8 2 , 3 4) (e) (6 , 2 3) , (6 , 4 3) ( 2 ,π/ 4) , ( 2 , 7 4) 6. (a) (2 , 5 6) (b) ( 2 , 11 6) (c) (2 , 7 6) (d) ( 2 , 6) 7. (a) (5 , 0 . 9273) (b) (10 , 0 . 92730) (c) (1 . 27155 , 0 . 66577) 8. (a) (5 , 2 . 2143) (b) (3 . 4482 , 2 . 6260) (c) ( p 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 x ,( x 3 / 2) 2 + y 2 =9 / 4; circle (d) 3 r cos θ +2 r sin θ =6,3 x y = 6; line 10. (a) r cos θ =5, x = 5; vertical line (b) r 2 =2 r sin θ , x 2 + y 2 y , x 2 +( y 1) 2 = 1; circle (c) r 2 =4 r cos θ +4 r sin θ, x 2 + y 2 x 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 (b) r = 7 (c) r 2 +6 r sin θ =0, r = 6 sin θ (d) 9( r cos θ )( r sin θ )=4,9 r 2 sin θ cos θ =4, r 2 sin 2 θ =8 / 9 12. (a) r sin θ = 3 (b) r = 5 (c) r 2 r cos θ 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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January 27, 2005 11:47 L24-ch11 Sheet number 2 Page number 476 black 476 Chapter 11 13. 0 c /2 –3 –3 3 3 r = 3 sin 2 θ 14. 33 –2.25 2.25 r = 2 cos 3 θ 15. 0 c /2 44 –1 r =3 4 sin 3 θ 16. 0 c /2 r =2+2sin 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 11.1.1(a) eliminates the sine function, thus r = a ± b cos θ . The cartesian point ( 3 , 0) is either the polar point (3 )o r( 3 , 0), and the cartesian point ( 1 , 0) is either the polar point (1 )or( 1 , 0). A solution is a ,b = 2; we may take the equation as r 2 cos θ . (b) x 2 +( y +3 / 2) 2 / 4 = 3 sin θ (c) Figure 11.1.18, a ,n = sin 3 θ 19. (a) Figure 11.1.18, a =2 = 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 =3and5= a + b ,so b = 2 and r =3+2sin θ . (c) Example 8, r 2 = 9 cos 2 θ 20. (a) Example 6 rotated through π/ 2 radian: a 3 sin θ (b) Figure 11.1.18, a = cos 5 θ (c) x 2 y 2) 2 =4, r = 4 sin θ
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January 27, 2005 11:47 L24-ch11 Sheet number 3 Page number 477 black Exercise Set 11.1 477 21. Line 4 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. 17 4 Limaçon 35. 3 2 1 Limaçon 36. 4 2 3 Limaçon 37. Limaçon 3 38. 2 5 8 Limaçon 39. 3 5 Limaçon 7 40. 3 1 7 Limaçon 41. Lemniscate 1 42. 3 Lemniscate 43. Lemniscate 4 44. Spiral 2 c 4 c 6 c 8 c
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January 27, 2005 11:47 L24-ch11 Sheet number 4 Page number 478 black 478 Chapter 11 45.
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This note was uploaded on 06/10/2010 for the course MATH 200-177 taught by Professor Richardwhite during the Spring '10 term at Drexel.

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chapter_11 - January 27, 2005 11:47 L24-ch11 Sheet number 1...

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