# 00 07071 1 y 0 07071 100 07071 0 y t5 6 t4 4 t3 t2 2

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Unformatted text preview: x − π/4)) 44. (a) 3, π/2, 0 2πr x = to get x ≈ 80, 936 km. 1 29.5 √ V = 120 2 sin(120πt) 45. (b) 36/360 = 1/10 (b) 2, 2, 0 y (c) 1, 4π, 0 y 2 3 3 1 x 2 x π/2 2 4 1 -2 -2 2π 46. (a) 4, π, 0 y π/4 3π/4 5π/4 7π/4 4 2 x π/3 -2 2π/3 π x 3π/2 -0.2 9π/2 15π/2 21π/2 -2 -4 -4 47. x y 0.4 2 6π (c) 4, 6π, −6π (b) 1/2, 2π/3, π/3 y 4π (a) A sin(ωt + θ) = A sin(ωt) cos θ + A cos(ωt) sin θ = A1 sin(ωt) + A2 cos(ωt) (b) A1 = A cos θ, A2 = A sin θ, so A = √ 1 (c) A = 5 13/2, θ = tan−1 √ ; 23 √ 1 5 13 sin 2πt + tan−1 √ x= 2 23 A2 + A2 and θ = tan−1 (A2 /A1 ). 1 2 10 ^ 6 -10 x Exercise Set 1.7 48. 32 three; x = 0, x = ±1.8955 3 -3 3 –3 EXERCISE SET 1.7 1. x + 1 = t = y − 1, y = x + 2 (a) (c) t 012345 x −1 0 1 2 3 4 y 123456 (c) t 0 0.2500 0.50 0.7500 1 x 1 0.7071 0.00 −0.7071 −1 y 0 0.7071 1.00 0.7071 0 y t=5 6 t=4 4 t=3 t=2 2 t=1 t=0 x 2 2. 4 x2 + y 2 = 1 (a) y 1 t=.75 t=.5 t=.25 t=0 x t=1 -1 3. 1 t = (x + 4)/3; y = 2x + 10 4. t = x + 3; y = 3x + 2, −3 ≤ x ≤ 0 y y 12 (0, 2) -8 5. 6 x x (-3, -7) cos t = x/2, sin t = y/5; x2 /4 + y 2 /25 = 1 6. t = x2 ; y = 2x2 + 4, x ≥ 0 y y 5 8 4 x x -5 5 -5 1 33 7. Chapter 1 cos t = (x − 3)/2, sin t = (y − 2)/4; (x − 3)2 /4 + (y − 2)2 /16 = 1 8. sec2 t − tan2 t = 1; x2 − y 2 = 1, x ≤ −1 and y ≥ 0 y y 6 x -1 x 7 -2 9. cos 2t = 1 − 2 sin2 t; x = 1 − 2y 2 , −1 ≤ y ≤ 1 10. t = (x − 3)/4; y = (x − 3)2 − 9 y y 1 x x 1 -1 -1 (3, -9) 11. x/2 + y/3 = 1, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 12. y = x − 1, x ≥ 1, y ≥ 0 y y 3 1 x 2 13. 1 x x = 5 cos t, y = −5 sin t, 0 ≤ t ≤ 2π 14. x = cos t, y = sin t, π ≤ t ≤ 3π/2 5 1 -7.5 7.5 -1 -5 15. 1 –1 x = 2, y = t 16. x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ 2π 2 3 -2 0 -1 2 3 –3 Exercise Set 1.7 17. 34 x = t2 , y = t, −1 ≤ t ≤ 1 18. x = 1 + 4 cos t, y = −3 + 4 sin t, 0 ≤ t ≤ 2π 2 1 -7 0 8 1 -8 -1 19. (a) IV, because x always increases whereas y oscillates. (b) II, because (x/2)2 + (y/3)2 = 1, an ellipse. (c) (d) III because y &gt; 0. (f ) I; since x and y are bounded, the answer must be I or II; but as t runs, say, from 0 to π , x goes directly from 2 to −2, but y goes from 0 to 1 to 0 to −1 and back to 0, which describes I but not II. (a) from left to right (b) counterclockwise (c) counterclockwise (d) As t travels from −∞ to −1, the curve goes from (near) the origin in the third quadrant and travels up and left. As t travels from −1 to +∞ the curve comes from way down in the second quadrant, hits the origin at t = 0, and then makes the loop clockwise and ﬁnally approaches the origin again as t → +∞. (e) from left to right (f ) 21. VI; examine the cases t &lt; −1 and t &gt; −1 and you see the curve lies in the ﬁrst, second and fourth quadrants only. (e) 20. V, because x2 + y 2 = t2 increases in magnitude while x and y keep changing sign. Starting, say, at (1, 0), the curve goes up into the ﬁrst quadrant, loops back through the origin and into the third quadrant, and then continues the ﬁgure-eight. (a) (b) 14 -35 8 0 (c) 22. t0123 4 5 x 0 5.5 8 4.5 −8 −32.5 y 1 1.5 3 5.5 9 13.5 √ x = 0 when t = 0, 2 3. (d) √ for 0 &lt; t &lt; 2 2 (e) at t = 2 (b) 5 -2 14 0 y is always ≥ 1 since cos t ≤ 1 (c) (a) greater than 5, since cos t ≥ −1 35 23. Chapter 1 (a) o (b) 3 0 20 -1 1 O -5 24. (a) -2.3 6 (b) 1.7 2.3 -10 10 -1.7 ^ (c) x − x0 y − y0 = x1 − x0 y 1 − y0 x = 1 + t, y = −2 + 6t 26. (a) x = −3 − 2t, y = −4 + 5t 27. (a) |R − P |2 = (x − x0 )2 + (y − y0 )2 = t2 [(x1 − x0 )2 + (y1 − y0 )2 ] and |Q − P |2 = (x1 − x0 )2 + (y1 − y0 )2 , so r = |R − P | = |Q − P |t = qt. (b) t = 1/2 25. 28. 29. 30. (a) x = 2 + t, y = −1 + 2t (a) (5/2, 0) (b) (9/4, −1/2) (b) Set t = 0 to get (x0 , y0 ); t = 1 for (x1 , y1 ). (d) x = 2 − t, y = 4 − 6t (b) (c) x = at, y = b(1 − t) t = 3/4 (c) (11/4, 1/2) The two branches corresponding to −1 ≤ t ≤ 0 and 0 ≤ t ≤ 1 coincide. y − y0 y1 − y0 t − t0 to obtain = . t1 − t0 x − x0 x1 − x0 (c) x = 3 − 2(t − 1), y = −1 + 5(t − 1) (b) from (x0 , y0 ) to (x1 , y1 ) (a) Eliminate 5 0 5 -2 31. (a) y−d x−b = a c y (b) 3 2 1 1 2 3 x Exercise Set 1.7 32. 36 (a) If a = 0 the line segment is vertical; if c = 0 it is horizontal. (b) The curve degenerates to the point (b, d). y 33. 2 1.5 1 0.5 0.5 1 x 35. x = 1/2 − 4t, x = −1/2, x = −1/2 + 4(t − 1/2), y = 1/2 y = 1/2 − 4(t − 1/4) y = −1/2 x = 1/2, 34. for 0 ≤ t ≤ 1/4 for 1/4 ≤ t ≤ 1/2 for 1/2 ≤ t ≤ 3/4 y = −1/2 + 4(t − 3/4) for 3/4 ≤ t ≤ 1 (b) x = −1 + 4 cos t, y = 2 + 3 sin t (a) x = 4 cos t, y = 3 sin t 3 (c) 5 -4 4 -5 3 -3 36. -1 2 (a) t = x/(v0 cos α), so y = x tan α − gx2 /(2v0 cos2 α). y (b) 12000 10000 8000 6000 4000 2000 40000 80000 x 37. √ √ (a) From Exercise 36, x = 400 2t, y = 400 2t − 4.9t2 . 38. (a) (b) 15 -25 (b) 16,326.53 m 15 -25 25 25 –15 (c) –15 15 15 -25 25 –15 a = 3, b = 2 (c) 65,306.12 m -25 15 25 –15 a = 2, b = 3 -25 25 –15 a = 2, b = 7 37 39. Chapter 1 Assume th...
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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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