The result r y x r r r r can be obtained by

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Unformatted text preview: t)i + (2 − 1/t)j + (2t − 1)k; not smooth, r (1/2) = 0 7. (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = (−3 cos2 t sin t)2 + (3 sin2 t cos t)2 + 02 = 9 sin2 t cos2 t, π /2 3 sin t cos t dt = 3/2 L= 0 π 8. (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = (−3 sin t)2 + (3 cos t)2 + 16 = 25, L = 5dt = 5π 0 √ 9. r (t) = et , −e−t , 2 , r (t) = et + e−t , L = 1 (et + e−t )dt = e − e−1 0 1 10. (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = 1/4 + (1 − t)/4 + (1 + t)/4 = 3/4, L = 11. r (t) = 3t2 i + j + √ 6 tk, r (t) = 3t2 + 1, L = 3 (3t2 + 1)dt = 28 1 12. r (t) = 3i − 2j + k, r (t) = √ 14, L = 4 √ √ 14 14 dt = 3 13. r (t) = −3 sin ti + 3 cos tj + k, r (t) = √ 10, L = √ 2π √ 10 dt = 2π 10 0 14. r (t) = 2ti + t cos tj + t sin tk, r (t) = √ 5t, L = π √ √ 5t dt = π 2 5/2 0 15. (dr/dt)(dt/dτ ) = (i + 2tj)(4) = 4i + 8tj = 4i + 8(4τ + 1)j; r(τ ) = (4τ + 1)i + (4τ + 1)2 j, r (τ ) = 4i + 2(4)(4τ + 1) 16. (dr/dt)(dt/dτ ) = −3 sin t, 3 cos t (π ) = −3π sin πτ , 3π cos πτ ; r(τ ) = 3 cos πτ, 3 sin πτ , r (τ ) = −3π sin πτ, 3π cos πτ 17. (dr/dt)(dt/dτ ) = (et i − 4e−t j)(2τ ) = 2τ eτ i − 8τ e−τ j; 2 2 r(τ ) = eτ i + 4e−τ j, r (τ ) = 2τ eτ i − 4(2)τ e−τ j 2 2 2 2 −1 √ √ ( 3/2)dt = 3 Exercise Set 14.3 500 18. (dr/dt)(dt/dτ ) = 9 1 9 1/2 t j + k (−1/τ 2 ) = − 5/2 j − 2 k; 2 τ 2τ r(τ ) = i + 3τ −3/2 j + 1 9 1 k, r (τ ) = − τ −5/2 j − 2 k τ 2 τ √ √ s s s s 2t; r = √ i + √ j, x = √ , y = √ 2 2 2 2 0 s (b) Similar to Part (a), x = y = z = √ 3 19. (a) r (t) = √ 2, s = t 2 dt = s s 20. (a) x = − √ , y = − √ 2 2 s s s (b) x = − √ , y = − √ , z = − √ 3 3 3 21. (a) r(t) = 1, 3, 4 when t = 0, t√ so s = 1 + 4 + 4 du = 3t, x = 1 + s/3, y = 3 − 2s/3, z = 4 + 2s/3 0 = 28/3, −41/3, 62/3 (b) r s=25 t√ √ 22. (a) r(t) = −5, 0, 1 when t = 0, so s = 9 + 4 + 1 du = 14t, 0 √ √ √ x = −5 + 3s/ 14, y = 2s/ 14, z = 5 + s/ 14 √ √ √ = −5 + 30/ 14, 20/ 14, 5 + 10/ 14 (b) r(s) s=10 23. x = 3 + cos t, y = 2 + sin t, (dx/dt)2 + (dy/dt)2 = 1, t du = t so t = s, x = 3 + cos s, y = 2 + sin s for 0 ≤ s ≤ 2π . s= 0 24. x = cos3 t, y = sin3 t, (dx/dt)2 + (dy/dt)2 = 9 sin2 t cos2 t, t s= 3 sin u cos u du = 0 3 sin2 t so sin t = (2s/3)1/2 , cos t = (1 − 2s/3)1/2 , 2 x = (1 − 2s/3)3/2 , y = (2s/3)3/2 for 0 ≤ s ≤ 3/2 25. x = t3 /3, y = t2 /2, (dx/dt)2 + (dy/dt)2 = t2 (t2 + 1), t u(u2 + 1)1/2 du = s= 0 x= 12 [(t + 1)3/2 − 1] so t = [(3s + 1)2/3 − 1]1/2 , 3 1 1 [(3s + 1)2/3 − 1]3/2 , y = [(3s + 1)2/3 − 1] for s ≥ 0 3 2 26. x = (1 + t)2 , y = (1 + t)3 , (dx/dt)2 + (dy/dt)2 = (1 + t)2 [4 + 9(1 + t)2 ], t √ 1 s= ([4 + 9(1 + t)2 ]3/2 − 13 13) so (1 + u)[4 + 9(1 + u)2 ]1/2 du = 27 0 √ 2/3 √ 1 1 1 + t = [(27s + 13 13) − 4]1/2 , x = [(27s + 13 13)2/3 − 4], 3 9 √ 2/3 √ √ 1 [(27s + 13 13) − 4]3/2 for 0 ≤ s ≤ (80 10 − 13 13)/27 y= 27 501 Chapter 14 t√ √ 27. x = et cos t, y = et sin t, (dx/dt)2 + (dy/dt)2 = 2e2t , s = 2 eu du = 2(et − 1) so 0 √ √ √ √ √ t = ln(s/ 2 + 1), x = (s/ 2 + 1) cos[ln(s/ 2 + 1)], y = (s/ 2 + 1) sin[ln(s/ 2 + 1)] √ for 0 ≤ s ≤ 2(eπ/2 − 1) 28. x = sin(et ), y = cos(et ), z = √t 3e , t 2eu du = 2(et − 1) so √ et = 1 + s/2; x = sin(1 + s/2), y = cos(1 + s/2), z = 3(1 + s/2) for s ≥ 0 (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = 4e2t , s = 0 29. dx/dt = −a sin t, dy/dt = a cos t, dz/dt = c, t0 t0 a2 sin2 t + a2 cos2 t + c2 dt = L= 0 a2 + c2 dt = t0 a2 + c2 0 30. x = a cos t, y = a sin t, z = ct, (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = a2 + c2 = w2 , t w du = wt so t = s/w; x = a cos(s/w), y = a sin(s/w), z = cs/w for s ≥ 0 s= 0 31. x = at − a sin t, y = a − a cos t, (dx/dt)2 + (dy/dt)2 = 4a2 sin2 (t/2), t s= 2a sin(u/2)du = 4a[1 − cos(t/2)] so cos(t/2) = 1 − s/(4a), t = 2 cos−1 [1 − s/(4a)], 0 cos t = 2 cos2 (t/2) − 1 = 2[1 − s/(4a)]2 − 1, sin t = 2 sin(t/2) cos(t/2) = 2(1 − [1 − s/(4a)]2 )1/2 (2[1 − s/(4a)]2 − 1), x = 2a cos−1 [1 − s/(4a)] − 2a(1 − [1 − s/(4a)]2 )1/2 (2[1 − s/(4a)]2 − 1), y= 32. s(8a − s) for 0 ≤ s ≤ 8a 8a dr dθ dy dr dθ dx = cos θ − r sin θ , = sin θ + r cos θ , dt dt dt dt dt dt dx dt 2 + dy dt 2 + dz dt 2 = dr dt 2 + r2 dθ dt 2 + dz dt 2 ln 2 3e2t dt = 33. (a) (dr/dt)2 + r2 (dθ/dt)2 + (dz/dt)2 = 9e4t , L = 0 3 2t e 2 ln 2 = 9/2 0 (b) (dr/dt)2 + r2 (dθ/dt)2 + (dz/dt)2 = 5t2 + t4 = t2 (5 + t2 ), 2 √ t(5 + t2 )1/2 dt = 9 − 2 6 L= 1 34. dρ dφ dθ dx = sin φ cos θ + ρ cos φ cos θ − ρ sin φ sin θ , dt dt dt dt dφ dθ dz dρ dφ dρ dy + ρ cos φ sin θ + ρ sin φ cos θ , = cos φ − ρ sin φ , = sin φ sin θ dt dt dt dt dt dt dt dx dt 2 + dy dt 2 + dz dt 2 = dρ dt 2 + ρ2 sin2 φ dθ dt 2 + ρ2 dφ dt 2 Exercise Set 14.4 502 2 35. (a) (dρ/dt)2 + ρ2 sin2 φ(dθ/dt)2 + ρ2 (dφ/dt)2 = 3e−2t , L = √ 3e−t dt = √ 3(1 − e−2 ) 0 √ 5 (b) (dρ/dt)2 + ρ2 sin2 φ(dθ/dt)2 + ρ2 (dφ/dt)2 = 5, L = √ 5dt = 4 5 1 d3 d3 d r(t) = i + 2tj is never zero, but r (τ ) = (τ i + τ 6 j) = 3τ 2 i + 6τ 5 j is zero at τ = 0. dt dτ dτ dr dt dt dr = , and...
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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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