# Sin 2 cos cos 2 sin sin sin 2 cos 2 sec sin cos

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Unformatted text preview: y = x2 /2 + K 45. dy/dx = 1/x, y = ln x + K y y x x 46. dy/dx = −y/x, (1/y )dy = (−1/x)dx, ln y = − ln x + K1 , y y = eK1 e− ln x = K/x x EXERCISE SET 17.2 1 1. (a) dy = 1 because s = y is arclength measured from (0, 0) 0 (b) 0, because sin xy = 0 along C ds = length of line segment = 2 2. (a) (b) 0, because x is constant and dx = 0 C 3. (a) ds = dx dt 2 + dy dt 2 1 (2t − 3t2 ) 4 + 36t2 dt = − dt, so 0 1 1 (2t − 3t2 )2 dt = 0 (b) 0 1 t(3t2 )(6t3 )2 4. (a) 1 + 36t2 + 324t4 dt = 0 1 t(3t2 )(6t3 )2 6t dt = 0 (2t − 3t2 )6t dt = − (c) 0 (c) √ 11 √ 1 4 ln( 10 − 3) − 10 − 108 36 27 864 5 1 t(3t2 )(6t3 )2 dt = (b) 0 1 2 54 5 1 648 11 t(3t2 )(6t3 )2 18t2 dt = 162 (d) 0 1 5. (a) C : x = t, y = t, 0 ≤ t ≤ 1; 6t dt = 3 0 1 (b) C : x = t, y = t2 , 0 ≤ t ≤ 1; (3t + 6t2 − 2t3 )dt = 3 0 Exercise Set 17.2 612 (c) C : x = t, y = sin(πt/2), 0 ≤ t ≤ 1; 1 [3t + 2 sin(πt/2) + πt cos(πt/2) − (π/2) sin(πt/2) cos(πt/2)]dt = 3 0 1 (d) C : x = t3 , y = t, 0 ≤ t ≤ 1; (9t5 + 8t3 − t)dt = 3 0 1 6. (a) C : x = t, y = t, z = t, 0 ≤ t ≤ 1; (t + t − t) dt = 0 1 2 1 (b) C : x = t, y = t2 , z = t3 , 0 ≤ t ≤ 1; (t2 + t3 (2t) − t(3t2 )) dt = − 0 1 60 1 (c) C : x = cos πt, y = sin πt, z = t, 0 ≤ t ≤ 1; (−π sin2 πt + πt cos πt − cos πt) dt = − 0 √ 3 3 1+t dt = 1+t 7. 0 (1 + t)−1/2 dt = 2 √ 1 5 0 0 1 √ 1 + 2t dt = 5(π/4 + ln 2) 1 + t2 1 3(t2 )(t2 )(2t3 /3)(1 + 2t2 ) dt = 2 9. t7 (1 + 2t2 ) dt = 13/20 0 0 √ 10. 8. 5 4 2π e−t dt = √ π /4 5(1 − e−2π )/4 1 −1 (8 cos2 t−16 sin2 t−20 sin t cos t)dt = 1−π 11. 0 12. 0 2 2 t − t5/3 + t2/3 dt = 6/5 3 3 3 13. C : x = (3 − t)2 /3, y = 3 − t, 0 ≤ t ≤ 3; 0 1 14. C : x = t2/3 , y = t, −1 ≤ t ≤ 1; −1 1 (3 − t)2 dt = 3 3 2 2/3 2 1/3 t − t + t7/3 dt = 4/5 3 3 π /2 15. C : x = cos t, y = sin t, 0 ≤ t ≤ π/2; (− sin t − cos2 t)dt = −1 − π/4 0 1 (−37 + 41t − 9t2 )dt = −39/2 16. C : x = 3 − t, y = 4 − 3t, 0 ≤ t ≤ 1; 0 1 (−3)e3t dt = 1 − e3 17. 2 π − 2 π 0 π /2 (sin2 t cos t − sin2 t cos t + t4 (2t)) dt = 18. 0 π6 192 π /2 cos21 t sin9 t 20. (a) (−3 cos2 t sin t)2 + (3 sin2 t cos t)2 dt 0 π /2 cos22 t sin10 t dt = =3 0 e t5 ln t + 7t2 (2t) + t4 (ln t) (b) 1 1 t 61,047 π 4,294,967,296 dt = 5 6 59 4 491 e+ e− 36 16 144 613 Chapter 17 21. (a) C1 : (0, 0) to (1, 0); x = t, y = 0, 0 ≤ t ≤ 1 C2 : (1, 0) to (0, 1); x = 1 − t, y = t, 0 ≤ t ≤ 1 C3 : (0, 1) to (0, 0); x = 0, y = 1 − t, 0 ≤ t ≤ 1 1 1 0 1 (−1)dt + (0)dt + (0)dt = −1 0 (b) C1 : (0, 0) C2 : (1, 0) C3 : (1, 1) C4 : (0, 1) to to to to 0 (1, 0); x = t, y = 0, 0 ≤ t ≤ 1 (1, 1); x = 1, y = t, 0 ≤ t ≤ 1 (0, 1); x = 1 − t, y = 1, 0 ≤ t ≤ 1 (0, 0); x = 0, y = 1 − t, 0 ≤ t ≤ 1 1 1 0 1 (−1)dt + (0)dt + 1 (−1)dt + 0 (0)dt = −2 0 0 22. (a) C1 : (0, 0) to (1, 1); x = t, y = t, 0 ≤ t ≤ 1 C2 : (1, 1) to (2, 0); x = 1 + t, y = 1 − t, 0 ≤ t ≤ 1 C3 : (2, 0) to (0, 0); x = 2 − 2t, y = 0, 0 ≤ t ≤ 1 1 1 (0)dt + 0 1 2dt + (0)dt = 2 0 0 (b) C1 : (−5, 0) to (5, 0); x = t, y = 0, −5 ≤ t ≤ 5 C2 : x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ π π 5 (−25)dt = −25π (0)dt + −5 0 1 23. C1 : x = t, y = z = 0, 0 ≤ t ≤ 1, 1 0 dt = 0; C2 : x = 1, y = t, z = 0, 0 ≤ t ≤ 1, 0 (−t) dt = − 0 1 C3 : x = 1, y = 1, z = t, 0 ≤ t ≤ 1, x2 z dx − yx2 dy + 3 dz = 0 − 3 dt = 3; C 0 5 1 +3= 2 2 24. C1 : (0, 0, 0) to (1, 1, 0); x = t, y = t, z = 0, 0 ≤ t ≤ 1 C2 : (1, 1, 0) to (1, 1, 1); x = 1, y = 1, z = t, 0 ≤ t ≤ 1 C3 : (1, 1, 1) to (0, 0, 0); x = 1 − t, y = 1 − t, z = 1 − t, 0 ≤ t ≤ 1 1 1 (−t3 )dt + 0 1 −3dt = −1/4 3 dt + 0 0 π 25. 1 (0)dt = 0 26. 0 (e2t − 4e−t )dt = e2 /2 + 4e−1 − 9/2 0 1 27. π /2 e−t dt = 1 − e−1 (7 sin2 t cos t + 3 sin t cos t)dt = 23/6 28. 0 0 29. Represent the circular arc by x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ π/2. π /2 √ √ √ √ x yds = 9 3 sin t cos t dt = 6 3 C 0 x2 + y 2 where k is the constant of proportionality, 1 1 √ √ √ x2 + y 2 ds = k et ( 2et ) dt = 2k e2t dt = (e2 − 1)k/ 2 30. δ (x, y ) = k k C 0 0 1 2 Exercise Set 17.2 31. C 614 kx ds = 15k 1 + y2 π /2 cos t dt = 5k tan−1 3 1 + 9 sin2 t 0 32. δ (x, y, z ) = kz where k is the constant of proportionality, 4 √ kzds = k (4 t)(2 + 1/t) dt = 136k/3 C 1 1 33. C : x = t2 , y = t, 0 ≤ t ≤ 1; W = 3t4 dt = 3/5 0 3 1 (t2 + 1 − 1/t3 + 1/t)dt = 92/9 + ln 3 34. W = 35. (t3 + 5t6 )dt = 27/28 W= 1 0 36. C1 : (0, 0, 0) to (1, 3, 1); x = t, y = 3t, z = t, 0 ≤ t ≤ 1 C2 : (1, 3, 1) to (2, −1, 4); x = 1 + t, y = 3 − 4t, z = 1 + 3t, 0 ≤ t ≤ 1 1 1 (−11 − 17t − 11t2 )dt = −37/2 (4t + 8t2 )dt + W= 0 0 37. Since F and r are parallel, F · r = F r , and since F is constant, 4√ √ 2dt = 16 F · dr = d(F · r) = d( F r ) = 2 C F·r= √ C √ 2(4 2) −4 C F · r = 0, since F is perpendicular to the curve 38. C 39. C : x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ π/2 π /2 0 1 − sin t + cos t dt = 3/4 4 40. C1 : (0, 3) to (6, 3); x = 6t, y = 3, 0 ≤ t ≤ 1 C2 : (6, 3) to (6, 0); x = 6, y = 3 − 3t, 0 ≤ t ≤ 1 1 0 1 6 36t2 +9 dt + 0 1 2 −12 dt = tan−1 2 − tan−1 (1/2) 36 + 9(1 − t)2 3 3 41. Represent the parabola by x = t, y = t2 , 0 ≤ t ≤ 2. 2 √ 3xds = 3t 1 + 4t2 dt = (17 17...
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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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