C u v is not a vector d the dot product of a scalar k

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Unformatted text preview: √ = 2 2π π sin θ(1 − cos θ)3/2 dθ = 0 2√ 2 2π (1 − cos θ)5/2 5 π = 32π/5 0 π 2πa sin(θ)a dθ = 4πa2 30. S = 0 31. (a) r3 cos3 θ − 3r2 cos θ sin θ + r3 sin3 θ = 0, r = π /(2n) 32. (a) A = 2 0 πa2 12 a cos2 nθ dθ = 2 4n 3 cos θ sin θ cos3 θ + sin3 θ π /(2n) (b) A = 2 0 2 (c) πa 1 × total area = 2n 4n (d) πa2 12 a cos2 nθ dθ = 2 4n 1 πa2 × total area = n 4n 33. If the upper right corner of the square is the point (a, a) then the large circle has equation r = and the small circle has equation (x − a)2 + y 2 = a2 , r = 2a cos θ, so π /4 √ 1 (2a cos θ)2 − ( 2a)2 dθ = a2 = area of square. area of crescent = 2 2 0 √ 2a Exercise Set 12.4 2π 34. A = 0 422 1 (cos 3θ + 2)2 dθ = 9π/2 2 π /2 35. A = 0 1 4 cos2 θ sin4 θ dθ = π/16 2 1 3 -3 3 0 1 –1 -3 EXERCISE SET 12.4 1. (a) (c) 4px = y 2 , point (1, 1), 4p = 1, x = y 2 a = 3, b = 2, y2 x2 + =1 9 4 (b) −4py = x2 , point (3, −3), 12p = 9, −3y = x2 (d) a = 3, b = 2, x2 y2 + =1 4 9 (e) asymptotes: y = ±x, so a = b; point (0, 1), so y 2 − x2 = 1 (f ) asymptotes: y = ±x, so b = a; point (2, 0), so x2 y2 − =1 4 4 2. (a) part (a), vertex (0, 0), p = 1/4; focus (1/4, 0), directrix: x = −1/4 part (b), vertex (0, 0), p = 3/4; focus (0, −3/4), directrix: y = 3/4 √ √ √ (b) part (c), c = a2 − b2 = 5, foci (± 5, 0) √ √ √ part (d), c = a2 − b2 = 5, foci (0, ± 5) √ √ √ (c) part (e), c = a2 + b2 = 2, foci at (0, ± 2); asymptotes: y 2 − x2 = 0, y = ±x √ √ √ √ y2 x2 − = 0, y = ±x part (f), c = a2 + b2 = 8 = 2 2, foci at (±2 2, 0); asymptotes: 4 4 y 3. (a) y (b) 5 3 y= F ( 3 , 0) 2 -3 x x -5 3 x=– 3 -3 2 5 -5 y 4. (a) 9 4 ( 9 F 0, – 4 ) y (b) F(0, 1) ( 5 2 F – ,0 ) x x y = –1 x= 5 2 423 Chapter 12 y 5. (a) x= 6 y (b) 1 2 x -4 V(–2, –2) V(2, 3) () 7 F 2, 3 y=– 7 4 ( F –2, – 9 x 4 ) -4 6 y 6. (a) x= y (b) 23 4 F x y= F ( 9 , –1) 4 (1 , 3 ) 22 1 2 x V(4, –1) V y 7. (a) 4 (b) ( 2) x=– ( 1 , 1) 2 y 9 4 V 2, 5 4 V(–2, 2) ( ) 7 F –4, 2 y=3 F(2, 2) x x 2 4 8. (a) y x = –9 y (b) 2 V(–4, 3) ( 7 2 F – ,3 ( F –1, ) V (–1, 1) 17 16 x y= 15 16 x 9. (a) c2 = 16 − 9 = 7, c = √ 7 (b) c2 = 9 − 1 = 8, c = y (0, 3) (–4, 0) y2 x2 + =1 1 9 (√7, 0) √ 8 y (0, 3) (0, √ 8) x (4, 0) x (– √7, 0) (0, –3) (–1, 0) (1, 0) (0, – √ 8) (0, –3) ) Exercise Set 12.4 424 10. (a) c2 = 25 − 4 = 21, c = (0, 5) √ 21 (b) y y2 x2 + =1 9 4 √ c2 = 9 − 4 = 5, c = 5 (0, √ 21) y (0, 2) x (–2, 0) (√ 5, 0) (–3, 0) (2, 0) x (0, – √ 21) (0, –5) 11. (a) (3, 0) (– √ 5, 0) (y − 3)2 (x − 1)2 + =1 16 9 √ c2 = 16 − 9 = 7, c = 7 (b) (0, –2) (y + 1)2 (x + 2)2 + =1 4 3 c2 = 4 − 3 = 1, c = 1 y y (1 – √7, 3) (1, 6) (–2, –1 + √ 3) (1 + √7, 3) x (–3, –1) (–3, 3) (0, –1) (–4, –1) (5, 3) (–1, –1) x (–2, –1 – √ 3) (1, 0) 12. (a) (y − 5)2 (x + 3)2 + =1 16 4 √ c2 = 16 − 4 = 12, c = 2 3 (b) (y + 2)2 x2 + =1 4 9 √ c2 = 9 − 4 = 5, c = 5 y (–3 –2 √ 3, 5) y (0, 1) (–3, 7) (0, –2 + √ 5) x (–7, 5) (1, 5) (–3, 3) (–3 +2 √ 3, 5) (–2, –2) (2, –2) x (0, –5) (0, –2 – √ 5) 425 13. (a) Chapter 12 (y − 1)2 (x + 1)2 + =1 9 1 √ c2 = 9 − 1 = 8, c = 8 (b) (y − 5)2 (x + 1)2 + =1 4 16 √ c2 = 16 − 4 = 12, c = 2 3 y y (–1, 9) (–1 – √8, 1) (–1, 2) (–1, 5 + 2 √3) (2, 1) (–3, 5) x (–4, 1) (–1, 0) (1, 5) (–1, 5 – 2 √3) (–1 + √8, 1) x (–1, 1) 14. (a) (y − 3)2 (x + 1)2 + =1 4 9 √ c2 = 9 − 4 = 5, c = 5 (b) (y + 3)2 (x − 2)2 + =1 9 5 c2 = 9 − 5 = 4, c = 2 y (–1, 6) (2, –3 + √5) y (–3, 3) (1, 3) (4, –3) (–1, –3) (0, –3) x (2, –3 – √5) (–1, 3 – √5) (–1, 0) √ 15. (a) c2 = a2 + b2 = 16 + 4 = 20, c = 2 5 y (0, √ 13) y = –2 x 3 (b) y 2 /4 − x2 /9 = 1 √ c2 = 4 + 9 = 13, c = 13 y (0, 2) y= 2x 3 x y = –1 x 2 (–4, 0) y= 1x (4, 0) 2 x (0, – √ 13) x (–1, 3 + √5) (0, –2) (–2 √ 5, 0) (2 √ 5, 0) (5, –3) Exercise Set 12.4 426 16. (a) c2 = a2 + b2 = 9 + 25 = 34, c = √ 34 y (0, √ 34) (0, 3) 3 5 y=– x y= (b) x2 /25 − y 2 /16 = 1 √ c2 = 25 + 16 = 41, c = 41 y 4 5 y=– x 3 x 5x y= 4 x 5 (–5, 0) (5, 0) x (0, – √ 34) (0, –3) (– √41, 0) 17. (a) c2 = 9 + 4 = 13, c = y √ 13 (√41, 0) (b) (y + 3)2 /36 − (x + 2)2 /4 = 1 √ c2 = 36 + 4 = 40, c = 2 10 y – 4 = 2 (x – 2) 3 (2 – √13, 4) (–2, –3 + 2 √10) (2 + √13, 4) (5, 4) y (–2, 3) (–1, 4) x y + 3 = – 3(x + 2) x y + 3 = 3(x + 2) (–2, –9) y – 4 = – 2 (x – 2) 3 (–2, –3 – 2 √10) √ 18. (a) c2 = 3 + 5 = 8, c = 2 2 y+4= y √3 (x – 2) 5 (2, – 4 + 2 √ 2) (b) (x + 1)2 /1 − (y − 3)2 /2 = 1 √ c2 = 1 + 2 = 3, c = 3 y − 3 = −√2(x + 1) y x (–2, 3) (2, – 4 + √ 3) (-1 − √3, 3) (2, – 4 – √ 3) √3 (x – 2) 5 (-1 + √3, 3) x (2, – 4 – 2 √ 2) y+4=– (0, 3) y − 3 = √2(x + 1) 427 Chapter 12 19. (a) (x + 1)2 /4 − (y − 1)2 /1 = 1 √ c2 = 4 + 1 = 5, c = 5 (b) (x − 1)2 /4 − (y + 3)2 /6...
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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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