Ch 5 solutions

# F 52 so exercise exercise set 52 231 12 1 3 4 2 2

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Unformatted text preview: = x + C1 x + C2 . 3 15 4 + x2 y 59. (a) 130 (b) 6 Chapter 5 4 53. dy/dx = 2x + 1, y = (2x + 1)dx = x2 + x + C ; y = 0 when x = −3, so (−3)2 + (−3) + C = 0, C = −6 thus 2 ber 10,10, 2008 16:01 2 ”ISM ETET chapter 5” Sheet number 7 Page number 231 mber 2008 16:01 ”ISM chapter 5” Sheet number 7 Page number 231 black black y = x + x − 6. x 4 − 6 − 4 −2 2 6 55. Set x) = m = − sin x− 2 f (x) = (− sin x)dx = cos x + C ; f (0) = 2 = 1 + C , so C = 1, f (x) 231cos x + 1. f ( 5.2 , so = Exercise Exercise Set 5.2 231 12 1 3 − 4 2 2 3x 2 x 58.57. dy/dxf)(x) =xdx2= x x2 + C1 . The+ + 2of cos 3tangent line is −3 so dy/dx = −3 when x = 1. Thus 3(1)2 + C1 = −3, (a) f 58. (a) (x== 36x sinsin 3x 27 sinsin 3slopecos the− 0.251607 x 3− − x 9 x 3x x − 0.251607 3 27 9 −6 ￿￿ 42 − 6, y = 4 −6 C = −= 6 =x (3x2 − 6)dx = x3 − 6x + C2 . If x = 1, then y = 5 − 3(1) = 2 so (b) 1 f (x)(x) so 4dy/dxx2 √3√ (b) f = + x2 + + 4+ −6 (c) f (x) = x2 /2 − 1 4 + x2 x2 (1)2 − 6(1) + C = 2, C = 4 + 7 thus y = x3 − 6x + 7. 2 2 y 59. (a) 59. (a) 60. (a) 66 6 44 4 22 2 (b) (b) (b) y y x x x − 6 − 6− 4 − 4 2 − 2 − 2 24 46 6 4 − 6 − 4 −2 2 6 −2 −2 −2 −4−4 −4 − 6 −chapter 5” ember 10, 2008 (a) ”ISM ET−6 Sheet number 8 Page number 232 6 (b) 59. 16:01 22 (c)(c) (x)(x) x x/2 /21 1 f f = =x − − (c) ( ) ( + 1)/ 1. (c) f yx= =ex2 /2 − 2 232 black Chapter 5 yy 60.61. This slope ﬁeld is zero along the y -axis, and so corresponds to (b). (a) (b) 61. This 60. (a) slope ﬁeld is zero along the y -axis, and so corresponds to (b). (b) y y 66 10 10 44 5 5 22 –4 –3 –2 –1 12 x x 34 x x –5 4− − 6 − 6−–5− 4 2 − 2 2 24 46 6 −2 −2 –10 –10 −4−4 63. This slope ﬁeld has a negative value along the y -axis, and thus corresponds to (c). −6 a 63. This slope ﬁeld has − 6negative value along the y -axis, and thus corresponds to (c). 62. This slope ﬁeld is independent of y , is near zero for large negative values of x, and is very large y fory = (e positive x. (c)(c) large x(+ 1)/1)/2 It must correspond to (d). y = 9ex + 2 3 61. This slope ﬁeld is zero along thethe y -axis, and so corresponds (b). 61. This slope ﬁeld is zero along y -axis, and so corresponds to to (b). x –2 y y 10 –3 10 5 1 3 5 –9 –3 –3 –1 –1 x 1 13 x 3 –5 –5 65. This slope ﬁeld appears to 0 · 0 dx = 0 0 dx + C 64. Theorem 5.2.3(a) says that becf (x) dx =(approximately 2), means thatcorresponds to diﬀerential, so the ”proof” constant cF (x) + C , which and thus –10 –10 equation (a). is not valid. y 10 5 62. This slope ﬁeld is independent of y , is near zero for large negative values of x, and is very large Exercise Set 5.3 131 67. (a) F (x) = 1 , G (x) = + 1 + x2 1 x2 1 1 = = F (x). 2 1 + 1/x 1 + x2 (b) F (1) = π/4; G(1) = − tan−1 (1) = −π/4, tan−1 x + tan−1 (1/x) = π/2. √ (c) Draw a triangle with sides 1 and x and hypotenuse 1 + x2 . If α denotes the angle opposite the side of length x and if β denotes its complement, then tan α = x and tan β = 1/x, and sin(α + β ) = sin α cos β + sin β cos α = 1 x·1 1·x x2 + = 1, and cos(α + β ) = cos α cos β − sin α sin β = − = 0, so the cosine of α + β is 2 2 2 1+x 1+x 1+x 1 + x2 −1 −1 zero and the sine of α + β is 1; consequently α + β = π/2, i.e. tan x + tan (1/x) = π/2. 69. (sec2 x − 1)dx = tan x − x + C . 71. (a) 1 2 (1 − cos x)dx = 1087 73. v = √ 2 273 1 (x − sin x) + C . 2 1 2 (b) (1 + cos x) dx = 1 (x + sin x) + C . 2 1087 1/2 1087 1/2 T −1/2 dT = √ T + C , v (273) = 1087 = 1087 + C so C = 0, v = √ T ft/s. 273 273 Exercise Set 5.3 u23 du = u24 /24 + C = (x2 + 1)24 /24 + C . 1. (a) u3 du = −u4 /4 + C = −(cos4 x)/4 + C . (b) − 3. (a) 1 4 sec2 u du = (b) 1 4 u1/2 du = 7. (a) u9 du = 1 10 1 u +C = (1 + sin t)10 + C . 10 10 (u − 1)2 u1/2 du = 2 (1 + x)3/2 + C . 3 (b) 9. (a) (b) − 1 3/2 1 u + C = (1 + 2y 2 )3/2 + C . 6 6 1 1 u du = − u2 + C = − cot2 x + C . 2 2 5. (a) − (b) 1 1 tan u + C = tan(4x + 1) + C . 4 4 (u5/2 − 2u3/2 + u1/2 )du = csc2 u du = − cot u + C = − cot(sin x) + C . 1 du = ln |u| + C = ln | ln x| + C . u 1 5 1 1 eu du = − eu + C = − e−5x + C . 5 5 11. (a) u = x3 , 1 3 du 1 = tan−1 (x3 ) + C . 2 1+u 3 2 4 2 7/2 4 5/2 2 3/2 u − u + u + C = (1 + x)7/2 − (1 + x)5/2 + 7 5 3 7 5 132 Chapter 5 √ (b) u = ln x, 15. u = 4x − 3, 1 7 17. u = 7x, 1 4 1 du = sin−1 (ln x) + C . 1 − u2 1 1 sin u du = − cos u + C = − cos 7x + C . 7 7 1 4 19. u = 4x, du = 4dx; 21. u = 2x, du = 2dx; 23. u = 2x, 1 2 1 10 1 u +C = (4x − 3)10 + C . 40 40 u9 du = √ 1 2 sec u tan u du = eu du = 1 1 sec u + C = sec 4x + C . 4 4 1u 1 e + C = e2x + C . 2 2 1 1 du = sin−1 (2x) + C . 2 1 − u2 25. u = 7t2 + 12, du = 14t dt; 1 14 u1/2 du = 1 3 /2 1 u +C = (7t2 + 12)3/2 + C . 21 21 1 1 du = (−3) − u3 2 27. u = 1 − 2x, du = −2dx, −3 1 3 1 +C = + C. u2 2 (1 − 2x)2 du 11 1 du = − +C...
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