U x dv ex dx du dx v ex 2 u x dv e3x dx

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Unformatted text preview: dx 3 dx 9 9x2/3 √ 8 40 1 9x2/3 + 4 dx = u1/2 du, u = 9x2/3 + 4 L= 18 13 3x1/3 1 = 1 3/2 u 27 40 = 13 √ √ √ √ 1 1 (40 40 − 13 13) = (80 10 − 13 13) 27 27 or (alternate solution) x = y 3/2 , L= 1 2 3 dx = y 1/2 , 1 + dy 2 4 4 + 9y dy = 1 1 18 dx dy 2 4 + 9y 9 , =1+ y = 4 4 40 u1/2 du = 13 √ √ 1 (80 10 − 13 13) 27 273 Chapter 8 13 x − x−3 , 1 + [f (x)]2 = 1 + 4 6. f (x) = 2 3 13 x + x−3 dx = 4 L= 2 3 2 16 1 x − + x−6 16 2 16 1 x + + x−6 = 16 2 = 13 x + x−3 dx = 595/144 4 1x 1 2x 1 2x e − e−x , 1 + [f (x)]2 = 1 + e − 2 + e−2x = e + 2 + e−2x , so 2 4 4 3 13x 1 + (f (x))2 dx = e + e−x dx = (e3 − e−3 )/2 L= 20 0 7. f (x) = 1 3 1 −3 y − y , 1 + [g (y )]2 = 1 + 2 2 8. g (y ) = 4 2 1 6 1 1 −6 y− + y 4 24 1 3 1 −3 y+ y , 2 2 = 1 3 1 −3 y+ y dy = 2055/64 2 2 L= 1 1 9. (dx/dt)2 + (dy/dt)2 = (t2 )2 + (t)2 = t2 (t2 + 1), L = √ t(t2 + 1)1/2 dt = (2 2 − 1)/3 0 10. (dx/dt)2 + (dy/dt)2 = [2(1 + t)]2 + [3(1 + t)2 ]2 = (1 + t)2 [4 + 9(1 + t)2 ], 1 L= √ √ (1 + t)[4 + 9(1 + t)2 ]1/2 dt = (80 10 − 13 13)/27 0 π /2 11. (dx/dt)2 + (dy/dt)2 = (−2 sin 2t)2 + (2 cos 2t)2 = 4, L = 2 dt = π 0 12. (dx/dt)2 + (dy/dt)2 = (− sin t + sin t + t cos t)2 + (cos t − cos t + t sin t)2 = t2 , π t dt = π 2 /2 L= 0 13. (dx/dt)2 + (dy/dt)2 = [et (cos t − sin t)]2 + [et (cos t + sin t)]2 = 2e2t , π /2 L= √ 2et dt = √ 2(eπ/2 − 1) 0 4 14. (dx/dt)2 + (dy/dt)2 = (2et cos t)2 + (−2et sin t)2 = 4e2t , L = 2et dt = 2(e4 − e) 1 √ sec x tan x = tan x, 1 + (y )2 = 1 + tan2 x = sec x when 0 < x < π/4, so sec x π /4 √ sec x dx = ln(1 + 2) 15. dy/dx = L= 0 16. dy/dx = cos x = cot x, sin x π /2 L= 1 + (y )2 = √ 1 + cot2 x = csc x when π/4 < x < π/2, so √ csc x dx = − ln( 2 − 1) = − ln π/4 √ 2−1 √ √ ( 2 + 1) 2+1 = ln(1 + √ 2) 17. (a) (dx/dθ)2 + (dy/dθ)2 = (a(1 − cos θ))2 + (a sin θ)2 = a2 (2 − 2 cos θ), so 2π 2π 2(1 − cos θ) dθ (dx/dθ)2 + (dy/dθ)2 dθ = a L= 0 0 2 13 x + x−3 , 4 Exercise Set 8.4 274 18. (a) Use the interval 0 ≤ φ < 2π . (b) (dx/dφ)2 + (dy/dφ)2 = (−3a cos2 φ sin φ)2 + (3a sin2 φ cos φ)2 = 9a2 cos2 φ sin2 φ(cos2 φ + sin2 φ) = (9a2 /4) sin2 2φ, so 2π π /2 π /2 | sin 2φ| dφ = 6a L = (3a/2) 0 sin 2φ dφ = −3a cos 2φ 0 y 19. (a) = 6a 0 (b) dy/dx does not exist at x = 0. (8, 4) (-1, 1) x (c) x = g (y ) = y 3/2 , g (y ) = 3 1/2 y, 2 1 1 + 9y/4dy L= (portion for − 1 ≤ x ≤ 0) 0 4 1 + 9y/4dy + (portion for 0 ≤ x ≤ 8) 0 = 8 27 √ √ √ 8 13 √ 13 − 1 + (10 10 − 1) = (13 13 + 80 10 − 16)/27 8 27 20. For (4), express the curve y = f (x) in the parametric form x = t, y = f (t) so dx/dt = 1 and dy/dt = f (t) = f (x) = dy/dx. For (5), express x = g (y ) as x = g (t), y = t so dx/dt = g (t) = g (y ) = dx/dy and dy/dt = 1. π 2 1 + 4x2 dx ≈ 4.645975301 21. L = 1 + cos2 y dy ≈ 3.820197789 22. L = 0 0 23. Numerical integration yields: in Exercise 21, L ≈ 4.646783762; in Exercise 22, L ≈ 3.820197788. 24. 0 ≤ m ≤ f (x) ≤ M , so m2 ≤ [f (x)]2 ≤ M 2 , and 1 + m2 ≤ 1 + [f (x)]2 ≤ 1 + M 2 ; thus √ √ 1 + m2 ≤ 1 + [f (x)]2 ≤ 1 + M 2 , b b 1 + m2 dx ≤ a b 1 + [f (x)]2 dx ≤ a 1 + M 2 dx, and a (b − a) 1 + m2 ≤ L ≤ (b − a) 1 + M 2 √ 2/2 ≤ cos x ≤ 1 for 0 ≤ x ≤ π/4 so √ π π√ (π/4) 1 + 1/2 ≤ L ≤ (π/4) 1 + 1, 3/2 ≤ L ≤ 2. 4 4 25. f (x) = cos x, 26. (dx/dt)2 + (dy/dt)2 = (−a sin t)2 + (b cos t)2 = a2 sin2 t + b2 cos2 t = a2 (1 − cos2 t) + b2 cos2 t = a2 − (a2 − b2 ) cos2 t = a2 1 − a2 − b2 cos2 t = a2 [1 − k 2 cos2 t], a2 2π π /2 a 1 − k 2 cos2 t dt = 4a L= 0 1 − k 2 cos2 t dt 0 275 Chapter 8 27. (a) (dx/dt)2 + (dy/dt)2 = 4 sin2 t + cos2 t = 4 sin2 t + (1 − sin2 t) = 1 + 3 sin2 t, 2π π /2 1 + 3 sin2 t dt = 4 L= 0 1 + 3 sin2 t dt 0 (b) 9.69 4.8 1 + 3 sin2 t dt ≈ 5.16 cm (c) distance traveled = 1.5 4.6 1 + (2.09 − 0.82x)2 dx ≈ 6.65 m 28. The distance is 0 π k 0 1 2 1.84 1.83 1.832 L 1 + (k cos x)2 dx 29. L = 3.8202 5.2704 5.0135 4.9977 5.0008 Experimentation yields the values in the table, which by the Intermediate-Value Theorem show that the true solution k to L = 5 lies between k = 1.83 and k = 1.832, so k = 1.83 to two decimal places. EXERCISE SET 8.5 1 1. S = √ √ 2π (7x) 1 + 49dx = 70π 2 1 0 √ x dx = 35π 2 0 1 1 2. f (x) = √ , 1 + [f (x)]2 = 1 + 4x 2x 4 S= 1 √ 1 dx = 2π 2π x 1 + 4x √ √ x + 1/4dx = π (17 17 − 5 5)/6 4 1 √ 3. f (x) = −x/ 4 − x2 , 1 + [f (x)]2 = 1 + x2 4 = , 4 − x2 4 − x2 1 S= 1 2π −1 4 − x2 (2/ 4 − x2 )dx = 4π dx = 8π −1 4. y = f (x) = x3 for 1 ≤ x ≤ 2, f (x) = 3x2 , 2 π (1 + 9x4 )3/2 2πx3 1 + 9x4 dx = S= 27 1 2 5. S = √ √ 2π (9y + 1) 82dy = 2π 82 0 2 2 √ √ = 5π (29 145 − 2 10)/27 1 √ (9y + 1)dy = 40π 82 0 1 2πy 3 6. g (y ) = 3y 2 , S = √ 1 + 9y 4 dy = π (10 10 − 1)/27 0 7. g (y ) = −y/ 9 − y 2 , 1 + [g (y )]2 = 8. g (y ) = −(1 − y )−1/2 , 1 + [g (y )]2 = 2 2π −2 9 − y2 · 3 9 − y2 2...
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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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