Business Calc Homework w answers_Part_81

Business Calc - Section 10.1 Quick Review 10.1 1(cos(0 sin(0 2 cos 3 x 2 3 3 sin 2 2 dy dx d 2y dx 2 dy dx dy/dt dx/dt dy/dt dx/dt 0 sin t 3 sin t

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Unformatted text preview: Section 10.1 Quick Review 10.1 1. (cos (0), sin (0)) 2. cos 3. x 2 3 3 , sin 2 2 dy dx d 2y dx 2 dy dx dy/dt dx/dt dy /dt dx/dt 0 sin t 3 sin t sin t 401 2. (a) (b) 1) 3. (a) y 3 (1, 0) (0, 1) sin2 t 0 y2 1 (since cos2 t 4. The portion in the first three quadrants, moving counterclockwise as t increases. 5. x t, y t2 1, 1 t 3 6. The graph is a circle with radius 2 centered at (2, 3). Modify the x cos t, y sin t parameterization correspondingly: x 2 cos t 2, y 2 sin t 3, 0 t 2 . 7. dy dx dy/dt dx/dt 3 cos t , 2 sin t 3 3( equals 4 2( 3 ,x 4 2/2) 2/2) 3 . 2 y dy/dt dx/dt 3/(2 3t) 1/(2 t 1) 3 t 1 3t 3 d 2y dx 2 dy /dt dx/dt 3 t 3/2 3 t 3/(2t 2 3 1/(2 t 3/t) 1) 3 t t2 3 1 3/t (b) which at t 3 8. y x C. For t 2 3 2 3 ( 2) 2 2 3 Thus, y x 3 2 2 and y 3 2. 3 2 2 , so 4. (a) (b) C and C 2. 3 ,x 4 dy dx d 2y dx 2 dy dx d 2y dx 2 y dy/dt dx/dt dy /dt dx/dt 1/t 1/t 2 1 1/t 2 t t2 9. y 3 2 2 2 x 3 C. For t 2 ( 3 2 x 3 2 and y 5 6 2 3 2 , so 2 5. (a) (b) y dy/dt dx/dt 3t 2 2t 3 2) C and C . Thus, y 3 2 5 2 . 6 dy /dt dx/dt [(2t 3)(6t) 12t 2 18t (2t 3)3 6t 2 18t (2t 3)3 (3t 2)(2)]/(2t 2t 3 6t 2 3)2 10. y x, so 3 Length 0 3 1 1 0 3 2 2 x dx 6. (a) 3 0 9 x dx 4 9 3/2 x 4 313/2 8 . 27 dy dx d 2y dx 2 y dy/dt dx/dt 2t 2t 1 1 8 1 27 (b) dy /dt dx/dt [(2t 1)(2) 4 (2t 1)3 cos t sin t Section 10.1 Exercises 1. (a) (b) dy dx d 2y dx 2 (2t 1)(2)]/(2t 2t 1 1)2 y dy/dt dx/dt dy /dt dx/dt 2 sin t 4 cos t 1 sec2 t 2 1 tan t 2 1 sec3 t 8 7. 4 cos t dy dx dy/dt dx/dt cot t k (k any integer). Then k , 1 sin 2 (a) cot (x, y) (2, 0 when t 2 1 cos 2 2 k 2). 1). The points are (2, 0) and (2, (b) cot is undefined when t (x, y) (2 cos (k ), 1 k (k any integer). Then sin (k )) 1). (2 1, 1). The points are (1, 1) and (3, 402 dy dx Section 10.1 dy/dt dx/dt sec2 t sec t tan t 8. csc t 13. x t 2, y 1 t, so (t 2)2 t 2 dt 1 dt 1)3/2 1) 1 1 0 (a) Nowhere, since csc t never equals zero. (b) csc t is undefined when t k (k any integer). Then (x, y) (sec (k ), tan (k )) ( 1, 0). The points are (1, 0) and ( 1, 0). 9. dy dx dy/dt dx/dt 3t 2 Length 0 1 0 t t2 4 1 4 3t 2 4 3 2 2 3 8 3 3 3 1 2 (t 3 1 3/2 (2 3 (a) 4 3t 2 0 when t 2 2 2 3 2 3 . 3 2 2 3 2 3 0.609 Then (x, y) , , 4 8 3 14. x 8t cos t, y /2 0 8t sin t, so (8t cos t)2 (8t sin t)2 dt , Length which evaluates to (3.155, 3.079). (b) Nowhere, since 4 10. dy dx dy/dt dx/dt 3 cos t 3 sin t (0.845, 2 3.079) and /2 8t dt 0 3t is never undefined. cot t k (k any integer). 15. x y Length 4t 2 0 /2 2 sec t tan t sec2 t sec t tan t cos t sec t cos t, (a) cot t 0 when t sin t, so /3 2 Then (x, y) 2 ( 2, 1 (b) 3 cos 2 (sec t 0 /3 /3 cos t)2 1 dt ( sin t)2 dt k ,1 3 sin 2 k 2). sec2 t tan t dt /3 0 0 3). The points are ( 2, 4) and ( 2, ln sec t cot t is undefined when t (x, y) ( 2 ( 2 11. x length 0 ln 2 0 k (k any integer). Then 3 sin (k )) 16. x et 2t, y 2 3 cos (k ), 1 1 (e t e t, so 2t)2 (1 e t 2 3, 1). The points are (1, 1) and ( 5, 1). 1 cos t, so (1 cos t)2 dt Length 1 ) dt , which using NINT evaluates to sin t, y ( sin t)2 2(1 0 4.497. 17. x Area sin t, y 2 cos t, so sin t) ( sin t)2 sin t) dt 2 cos t) dt t dt 2 2 (2 0 2 cos2 t dt 4 cos2 0 2 0 (2 2t 1 t 2 cos t 0 8 2 t 2 cos dt 2 0 t 2 2 sin 4 2 0 18. x Area t, y 2 , so ( 1 dt 1)3/2 1) 2 0 2 0 12. x Length 2t 3 3, y ( 0 3 1 2t 2) dt t, so 3)2 1 2 t 2 2 3/2 t 3 2 0 t)2 1 2 t dt (1 2t t)2 dt 3 0 4 3 t t2 (t 0 21 2 4 1 2 (t 3 3 4 (5 5 9 14.214 19. x Area 1, y 3 0 2t, so 2 (t 1) 1 (2t)2 dt, 178.561. which using NINT evaluates to Section 10.1 20. x sec t 0 0 403 cos t (see Ex. 15), y /3 sin t, so cos t)2 ( sin t)2 dx 25. In the first integral, replace t with x. Then becomes dx dx dx dt Length /3 2 (cos t) (sec t 2 cos t /3 1. g(y), y y, c y d. The tan t dt 26. Parameterize the curve as x . /3 2 2 0 sin t dt cos t 0 2 21. (a) x(t) (b) x Area 0 parameter is y itself, so replace t with y in the general t 1 formula. Then dy dy becomes dt dy 2t, y(t) 2, y 1 t 1, 0 1. 1, so 2 (t 1 27. x 1) 22 1) dt 1 t, y 2t 4 1, so t2 ( 2t 1)2 dt 12 dt Total length 0 4 2 2 3 (c) Slant height Area (1 5 (t 0 (t 0 1) dt 4 1 5 t2 2 t 0 5 22 2) 5 12 3 5 r x, h 1 2 t 2 t 0 m 12. 12 for m: 2 5, so Now solve 1 2 m 2 1 2 t 2 t 0 m 2 6, or m2 4 2 48 2m 1 12 0, and 22. (a) Because these values for x(t) and y(t) satisfy y m 13. Take the positive 13 1)3/2 1, which gives (3.394, 5.160). which is the equation of the line through the origin and (h, r), and this range of t-values gives the correct initial and terminal points. 28. solution. The midpoint is at t (x, y) ( 13 1)2 1 , (2 3 2 13 (b) x Area h, y 1 0 r, so 2 (rt) h2 r2 t 2 h2. r2 h2, so Area 2 cos 2t, so 2 sin 2t)2 . [0, 9] by [ 3, 3] r 2 dt 1 0 [ 4, 4] by [0, 10] r h2 r (c) Slant height 23. (a) x Length 0 0 /2 Use the right half of the curve, 0 x r r2 h2 Area 0 t . r2 3 cos t, y 6 cos 2t, so (3 cos t)2 (6 cos 2t)2 dt, 159.485. 2 (3 sin t) which using NINT evaluates to 2 sin 2t, y /2 ( 2 dt (2 cos 2t)2 dt 29. (b) x Length cos t, y 1/2 1/2 1/2 sin t, so ( cos t)2 dt . ( sin t)2 dt Use the top half of the curve, and make use of the shape's symmetry. x 3 cos t, y /2 1/2 6 cos 2t, so (6 cos 2t)2 dt 24. x Length 3 sin t, y 2 0 4 cos t, so ( 3 sin t)2 (4 cos t)2 dt 22.103. Area 2 0 2 (3 sin 2t) (3 cos t)2 which using NINT, evaluates to 144.513. which using NINT evaluates to 404 30. y so Section 10.1 0 for t 0 and t 2 .x a a cos t, y a sin t, (b) x t cos t, y 2 0 2 t sin t, so (t cos t)2 (t sin t)2 dt 2 Length 2 [a(1 2 0 2 t dt 0 Area 2 a cos t)] (a 2 a cos t)2 (a sin t)2 dt (a) x 2 2 34. All distances are a times as big as before. (1 cos t) 2 cos t dt a(cos t 2a t sin t), y 2 2 a2 8 a 2 0 2 0 2 0 2 0 a(sin t t cos t) t 2 sin2 2 t sin3 dt 2 t 4 sin2 dt 2 (b) Length For exercises 3538, x t 2 t dt 2 2 2 t cos3 3 2 0 v0 cos v0 sin and y v0 sin 32t, and y 0 for t 0 or t 8 a2 8 a 2 1 cos2 sin in mid-flight at t v sin /16 0 0 16 v0 sin 32 . The maximum height is attained . To find the path length, evaluate 32t)2 dt using NINT. To t 2 cos 2 (v0 cos )2 (v0 sin 64 a 3 2 find the maximum height, calculate cos t) ymax (v0 sin ) 31. dx dt a(1 v0 sin 32 16 v0 sin 32 2 . 0. (Note: integrate with respect to x from 0 to 2a ; integrate with respect to t from 0 to 2 .) 2a 35. (a) The projectile hits the ground when y y t t(150 sin 20 0 or t 16t) 0 3.206 150 sin 20 Area 0 2a 0 y dx a(1 2 cos t)a(1 2 cos t t 2 cos t) dt x cos t) dt 1 sin 2t 4 2 0 75 sin 20 8 150 cos 20 , y (75 sin 20 )/8 0 32t (150 sin 20 461.749 ft 32t)2 a 2 (1 0 2 Length 3 a2 (150 cos 20 )2 a2 t dx 32. dt 2 sin t cos t), so 2 dt which, using NINT, evaluates to (b) The maximum height of the projectile occurs when a(1 y cos t)] a(1 3 cos t 2 0, 75 75 sin 20 , y sin 20 16 16 Volume 0 [a(1 a 3 3 2 cos t) dt 2 so t 3 41.125 ft (1 0 3 cos t 3 sin 2t 4 2 0 cos t) dt 36. (a) (b) 37. (a) (b) 641.236 ft 5625 64 a t 3 sin t sin t 3 t 2 1 sin3 3 87.891 ft t 840.421 ft 16,875 64 5 2 3 a 263.672 ft 33. (a) QP has length t, so P can be obtained by starting at Q and moving t sin t units right and t cos t units downward. (If either quantity is negative, the corresponding direction is reversed.) Since Q (cos t, sin t), the coordinates of P are x cos t t sin t and y sin t t cos t. y Q 38. (a) It is not necessary to use NINT. 75/8 Length 5625 8 0 (150 32t) dt 150t 16t 2 75/8 0 703.125 ft 351.5625 b (b) t sin t t t t t cos t P(x, y) x 5625 16 39. In the integral a 2 y dx 2 dt dy 2 dt, replace t with x dt and y with f (x). Then dx dx becomes dt dx (1, 0) 1. Section 10.2 dy dx 3 0 405 40. e x, so Area 2 ex 1 (e x)2 dx which using NINT evaluates to dy 41. dx 1 , so Area x2 1273.371. 4 5. (a) 2u 3v 2u (b) 12 2 2(3), 2( 2) 6, 4 3( 2), 3(5) 6, 15 3v 6 ( 6), 4 15 ( 19) 2 12, 19 505 2 1 1 x 1 1 2 dx, which, x2 6. (a) using NINT, evaluates to dy 42. dx 9.417. (b) x 2u 2(3), 2( 2) 6, 4 5v 5( 2), 5(5) 10, 25 2u 5v 6 ( 10), 4 25 ( 16) 3 u 5 4 v 5 3 u 5 2 16, 29 29 2 1097 9 , 5 6 5 8 ,4 5 8 , 5 197 5 5 ( 2) 13 24 60 , 13 13 24 10 , 13 13 6421 13 60 13 15 10 , 13 13 6 5 (ln 2)2 2 2 x (ln 2)2 (ln 2)(2 1 x 2 ), so 2 x)2 dx which 7. (a) x Area 2 (2 x 2 x) (ln 2)2(2 x 3 3 (3), ( 2) 5 5 4 4 ( 2), (5) 5 5 4 v 5 1 2 5 9 5 14 2 5 5 (3), 13 using NINT evaluates to 116.687. 4 s Section 10.2 Vectors in the Plane (pp. 520529) Quick Review 10.2 1. 2. 3 5 1 14 , 5 5 (b) (5 2 1 3 5 1)2 1 4 b 3 (3 2)2 17 8. (a) 5 u 13 12 v 13 12 12 ( 2), (5) 13 13 12 v 13 70 2 13 15 13 3. Solve 4: b 11. 0 0 1 1 3 5 2 6 0 and a 4. a b and b 6. 8 5 u 13 3, 70 13 3 1 5 5. Slope of AB = Slope of CD, so 3 4. Slope of AB = Slope of CD, so (b) 9. 2 10. 2 ( 3)2 1, ( 4) 2 1 3 0, 3 1, 0 1 2 1, 3 4 0 3 1, 1 3 1, ( 1) 1 0, 0 1, 1 6. (a) (b) 7. (a) (b) 8. (a) (b) 9. c2 c 10. 242 cos 32 120 2 3 11. 0 30 6 2, 0 2 2 CD E 2, ( 1) 12. E AB CD E AB E 13. cos 14. cos 2(3)(5) cos (30 ) 34 15 3, so ( 1), 2 1 ( 1), 1 45 4 2 2 , sin 3 3 3 , sin 4 3 4 1 3 , 2 2 1 2 , 1 2 52 15 34 272 242 3 2.832 2(27)(19) cos , so 15. This is the unit vector which makes an angle of 120 90 210 with the positive x-axis; 3 2 1 2 192 272 192 257 and 2(27)(19) 513 257 cos 1 1.046 radians or 59.935 . 513 cos 210 , sin 210 16. cos 135 , sin 135 , 1 1 2 , 2 Section 10.2 Exercises 1. (a) 3(3), 3( 2) = 9, (b) 2. (a) (b) 3. (a) 3 (b) 4. (a) 3 (b) 52 12 92 42 ( 6)2 2(5) 2 2 5 10 5 74 5, 7 ( 10) 2 ( 2), 32 ( 2), ( 7)2 2( 2), 6 117 4, 116 3 10 2 29 1, 3 13 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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