# B zx 2x y 2 zy x 2y 4 solve the system 2x y

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Unformatted text preview: since t = τ 3 , = 0 when τ = 0. (b) dτ dt dτ dτ 36. (a) 37. (a) g (τ ) = πτ (b) g (τ ) = π (1 − τ ) 38. t=1−τ 39. Represent the helix by x = a cos t, y = a sin t, z = ct with a = 6.25 and c = 10/π , so that the radius of the helix is the distance from the axis of the cylinder to the center of the copper cable, and the helix makes one turn in a distance of 20 in. (t = 2π ). From Exercise 29 the length of the helix is 2π 6.252 + (10/π )2 ≈ 44 in. 3 40. r(t) = cos ti + sin tj + t3/2 k, r (t) = − sin ti + cos tj + t1/2 k 2 1√ 4 + 9t (a) r (t) = sin2 t + cos2 t + 9t/4 = 2 (b) 1√ ds = 4 + 9t dt 2 2 (c) 0 √ 2 1√ (11 22 − 4) 4 + 9t dt = 2 27 41. r (t) = (1/t)i + 2j + 2tk (a) (b) r (t) = 1/t2 + 4 + 4t2 = (2t + 1/t)2 = 2t + 1/t 3 ds = 2t + 1/t dt (c) (2t + 1/t)dt = 8 + ln 3 1 42. If r(t) = x(t)i + y (t)j + z (t)k is smooth, then r (t) is continuous and nonzero. Thus the angle between r (t) and i, given by cos−1 (x (t)/ r (t) ), is a continuous function of t. Similarly, the angles between r (t) and the vectors j and k are continuous functions of t. 43. Let r(t) = x(t)i + y (t)j and use the chain rule. EXERCISE SET 14.4 1. (a) y (b) y x x 503 2. Chapter 14 y x 3. r (t) = 2ti + j, r (t) = √ 4t2 + 1, T(t) = (4t2 + 1)−1/2 (2ti + j), T (t) = (4t2 + 1)−1/2 (2i) − 4t(4t2 + 1)−3/2 (2ti + j); 2 2 1 1 2 T(1) = √ i + √ j, T (1) = √ (i − 2j), N(1) = √ i − √ j. 5 5 55 5 5 4. r (t) = ti + t2 j, T(t) = (t2 + t4 )−1/2 (ti + t2 j), T (t) = (t2 + t4 )−1/2 (i + 2tj) − (t + 2t3 )(t2 + t4 )−3/2 (ti + t2 j); 1 1 1 1 1 T(1) = √ i + √ j, T (1) = √ (−i + j), N(1) = − √ i + √ j 2 2 22 2 2 5. r (t) = −5 sin ti + 5 cos tj, r (t) = 5, T(t) = − sin ti + cos tj, T (t) = − cos ti − sin tj; √ √ √ 1 1 1 3 3 3 i + j, T (π/3) = − i − j, N(π/3) = − i − j T(π/3) = − 2 2 2 2 2 2 √ 1 + t2 , T(t) = (1 + t2 )−1/2 (i + tj), t 1 e i+ √ j, T (t) = (1 + t2 )−1/2 (j) − t(1 + t2 )−3/2 (i + tj); T(e) = √ 1 + e2 1 + e2 1 e 1 (−ei + j), N(e) = − √ i+ √ j T (e) = (1 + e2 )3/2 1 + e2 1 + e2 1 6. r (t) = i + j, r (t) = t 1 7. r (t) = −4 sin ti + 4 cos tj + k, T(t) = √ (−4 sin ti + 4 cos tj + k), 17 1 4 1 T (t) = √ (−4 cos ti − 4 sin tj), T(π/2) = − √ i + √ k 17 17 17 4 T (π/2) = − √ j, N(π/2) = −j 17 8. r (t) = i + tj + t2 k, T(t) = (1 + t2 + t4 )−1/2 (i + tj + t2 k), T (t) = (1 + t2 + t4 )−1/2 (j + 2tk) − (t + 2t3 )(1 + t2 + t4 )−3/2 (i + tj + t2 k), T(0) = i, T (0) = j = N(0) 1 9. r (t) = et [(cos t − sin t)i + (cos t + sin t)j + k], T(t) = √ [(cos t − sin t)i + (cos t + sin t)j + k], 3 1 T (t) = √ [(− sin t − cos t)i + (− sin t + cos t)j], 3 1 1 1 1 1 1 T(0) = √ i + √ j + √ k, T (0) = √ (−i + j), N(0) = − √ i + √ j 3 3 3 3 2 2 Exercise Set 14.4 504 √ 10. r (t) = sinh ti + cosh tj + k, r (t) = sinh2 t + cosh2 t + 1 = 2 cosh t, 1 1 T(t) = √ (tanh ti + j + sech tk), T (t) = √ (sech2 ti − sech t tanh tk), at t = ln 2, 2 2 4 3 3 1 4 tanh(ln 2) = and sech(ln 2) = so T(ln 2) = √ i + √ j + √ k, 5 5 52 2 52 3 4 4 T (ln 2) = √ (4i − 3k), N(ln 2) = i − k 5 5 25 2 11. From the remark, the line is parametrized by normalizing v, but T(t0 ) = v/ v , so r = r(t0 ) + tv becomes r = r(t0 ) + sT(t0 ). 12. r (t) = 1, 2t t=1 t=1 r = 1, 1 + s 1 2 = 1, 2 , and T(1) = √ , √ , so the tangent line can be parametrized as 5 5 2 s 2s 1 √ , √ , so x = 1 + √ , y = 1 + √ . 5 5 5 5 13. r (t) = cos ti − sin tj + tk, r (0) = i, r(0) = j, T(0) = i, so the tangent line has the parametrization x = s, y = 1. √ t 1 17 k, r (1) = i + j − √ k, r (1) = √ , so the tangent 8 8 9 − t2 √ √ √ 1 s8 1 i+j− √ k . line has parametrizations r = i + j + 8k + t i + j − √ k = i + j + 8k + √ 8 17 8 14. r(1) = i + j + 15. T = √ 8k, r (t) = i + j − √ 3 4 4 4 3 3 cos ti − sin tj + k, N = − sin ti − cos tj, B = T × N = cos ti − sin tj − k 5 5 5 5 5 5 1 1 16. T (t) = √ [(cos t + sin t)i + (− sin t + cos t)j ], N = √ [(− sin t + cos t)i − (cos t + sin t)j], 2 2 B = T × N = −k 17. r (t) = t sin ti + t cos tj, r = t, T = sin ti + cos tj, N = cos ti − sin tj, B = T × N = −k √ 18. T = (−a sin ti + a cos tj + bk)/ a2 + b2 , N = − cos ti − sin tj, √ B = T × N = (b sin ti − b cos tj + ak)/ a2 + b2 √ √ √ 2 2 2 2 i+ j + k, T = − sin ti + cos tj = (−i + j), N = −(cos ti + sin tj) = − (i + j), 19. r(π/4) = 2 2 2 2 √ B = k; the rectifying, osculating, and normal planes are given (respectively) by x + y = 2, z = 1, −x + y = 0. √ 1 1 1 20. r(0) = i + j, T = √ (i + j + k), N = √ (−j + k), B = √ (2i − j − k); the rectifying, osculating, 3 2 6 and normal planes are given (respectively) by −y + z = −1, 2x − y − z = 1, x + y + z = 2. 21. (a) By formulae (1) and (11), N(t) = B(t) × T(t) = r (t) r (t) × r (t) × . r (t) × r (t) r (t) (b) Since r is perpendicular to r × r it follows from Lagrange’s Identity (Exercise 32 of S...
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