# 7 ft d 02180 900 radians then dh 45 a z

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Unformatted text preview: ection 13.4) that (r (t) × r (t)) × r (t) = r (t) × r (t) r (t) , and the result follows. (c) From Exercise 39 of Section 13.4, (r (t) × r (t)) × r (t) = r (t) 2 r (t) − (r (t) · r (t))r (t) = u(t), so N(t) = u(t)/ u(t) 505 Chapter 14 1 2 22. (a) r (t) = 2ti + j, r (1) = 2i + j, r (t) = 2i, u = 2i − 4j, N = √ i − √ j 5 5 π (b) r (t) = −4 sin ti + 4 cos tj + k, r ( ) = −4i + k, r (t) = −4 cos ti − 4 sin tj, 2 π r ( ) = −4j, u = 17(−4j), N = −j 2 23. r (t) = cos ti − sin tj + k, r (t) = − sin ti − cos tj, u = −2(sin ti +cos tj), u = 2, N = − sin ti − cos tj 24. r (t) = i + 2tj + 3t2 k, r (t) = 2j + 6tk, u(t) = −(4t + 18t3 )i + (2 − 18t4 )j + (6t + 12t3 )k, 1 N= √ −(4t + 18t3 )i + (2 − 18t4 )j + (6t + 12t3 )k 2 81t8 + 117t6 + 54t4 + 13t2 + 1 EXERCISE SET 14.5 1. κ ≈ 1 =2 0.5 2. κ ≈ 1 3 = 4/3 4 3. r (t) = 2ti + 3t2 j, r (t) = 2i + 6tj, κ = r (t) × r (t) / r (t) 3 = 6 t(4 + 9t2 )3/2 4. r (t) = −4 sin ti+cos tj, r (t) = −4 cos ti−sin tj, κ = r (t)×r (t) / r (t) 5. r (t) = 3e3t i − e−t j, r (t) = 9e3t i + e−t j, κ = r (t) × r (t) / r (t) 3 = 6. r (t) = −3t2 i + (1 − 2t)j, r (t) = −6ti − 2j, κ = r (t) × r (t) / r (t) 3 3 = 4 (16 sin t + cos2 t)3/2 2 12e2t 3/2 (9e6t + e−2t ) = 6 t2 (t − 1)2 (9t4 + 4t2 − 4t + 1)3/2 7. r (t) = −4 sin ti + 4 cos tj + k, r (t) = −4 cos ti − 4 sin tj, κ = r (t) × r (t) / r (t) 3 = 4/17 √ 8. r (t) = i + tj + t k, r (t) = j + 2tk, κ = r (t) × r (t) / r (t) 2 3 = t4 + 4t2 + 1 (t4 + t2 + 1)3/2 9. r (t) = sinh ti + cosh tj + k, r (t) = cosh ti + sinh tj, κ = r (t) × r (t) / r (t) 10. r (t) = j + 2tk, r (t) = 2k, κ = r (t) × r (t) / r (t) 3 = 3 = 2 (4t2 + 1)3/2 11. r (t) = −3 sin ti + 4 cos tj + k, r (t) = −3 cos ti − 4 sin tj, r (π/2) = −3i + k, r (π/2) = −4j; κ = 4i + 12k / − 3i + k 12. r (t) = et i − e−t j + k, r (t) = et i + e−t j, r (0) = i − j + k, r (0) = i + j; κ = − i + j + 2k / i − j + k 3 = 2/5, ρ = 5/2 3 = √ √ 2/3, ρ = 3/ 2 13. r (t) = et (cos t − sin t)i + et (cos t + sin t)j + et k, r (t) = −2et sin ti + 2et cos tj + et k, r (0) = i + j + k, √ √ r (0) = 2j + k; κ = − i − j + 2k / i + j + k 3 = 2/3, ρ = 3 2/2 1 2 cosh2 t Exercise Set 14.5 506 14. r (t) = cos ti − sin tj + tk, r (t) = − sin ti − cos tj + k, √ √ r (0) = i, r (0) = −j + k; κ = − j − k / i 3 = 2, ρ = 2/2 √ s 1 s 1 3 i − sin 1 + j+ k, r (s) = 1, so 15. r (s) = cos 1 + 2 2 2 2 2 1 s 1 s dT 1 dT = − sin 1 + i − cos 1 + j, κ = = ds 4 2 4 2 ds 16 16. r (s) = − 3 − 2s si + 3 2s j, r (s) = 1, so 3 1 1 dT dT i + √ j, κ = =√ = ds ds 9 − 6s 6s 1 1 + = 9 − 6s 6s 17. (a) r = x i + y j, r = x i + y j, r × r 3 2s(9 − 6s) = |x y − x y |, κ = |x y − y x | (x 2 + y 2 )3/2 (b) r = xi + y (x)j, r (x) = i + (dy/dx)j, r (x) = (d2 y/dx2 )j, so κ(x) = (d2 y/dx2 )k / i + (dy/dx)j 18. 3 = |d2 y/dx2 |/[1 + (dy/dx)2 ]3/2 . dy |y | = tan φ, (1 + tan2 φ)3/2 = (sec2 φ)3/2 = | sec φ|3 , κ(x) = = |y cos3 φ| dx | sec φ|3 19. κ(x) = | sin x| , κ(π/2) = 1 (1 + cos2 x)3/2 20. κ(x) = 2|x| , κ(0) = 0 (1 + x4 )3/2 21. κ(x) = √ 2|x|3 , κ(1) = 1/ 2 4 + 1)3/2 (x 22. κ(x) = e−x e−1 , κ(1) = (1 + e−2x )3/2 (1 + e−2 )3/2 23. κ(x) = √ 2 sec2 x| tan x| , κ(π/4) = 4/(5 5) 4 x)3/2 (1 + sec 24. By implicit diﬀerentiation, dy/dx = 4x/y , d2 y/dx2 = 36/y 3 so κ = if (x, y ) = (2, 5) then κ = 36/125 36 =√ (1 + 64/25)3/2 89 89 25. x (t) = 2t, y (t) = 3t2 , x (t) = 2, y (t) = 6t, x (1/2) = 1, y (1/2) = 3/4, x (1/2) = 2, y (1/2) = 3; κ = 96/125 26. x (t) = −4 sin t, y (t) = cos t, x (t) = −4 cos t, y (t) = − sin t, x (π/2) = −4, y (π/2) = 0, x (π/2) = 0, y (π/2) = −1; κ = 1/16 27. x (t) = 3e3t , y (t) = −e−t , x (t) = 9e3t , y (t) = e−t , √ x (0) = 3, y (0) = −1, x (0) = 9, y (0) = 1; κ = 6/(5 10) 28. x (t) = −3t2 , y (t) = 1 − 2t, x (t) = −6t, y (t) = −2, x (1) = −3, y (1) = −1, x (1) = −6, y (1) = −2; κ = 0 36/|y |3 ; (1 + 16x2 /y 2 )3/2 507 Chapter 14 29. x (t) = 1, y (t) = −1/t2 , x (t) = 0, y (t) = 2/t3 √ x (1) = 1, y (1) = −1, x (1) = 0, y (1) = 2; κ = 1/ 2 30. x (t) = 4 cos 2t, y (t) = 3 cos t, x (t) = −8 sin 2t, y (t) = −3 sin t, x (π/2) = −4, y (π/2) = 0, x (π/2) = 0, y (π/2) = −3, κ = 12/43/2 = 3/2 31. (a) κ(x) = ρ(x) = | cos x| , (1 + sin2 x)3/2 (b) κ(t) = (1 + sin2 x)3/2 | cos x| ρ(t) = 2 , (4 sin2 t + cos2 t)3/2 1 (4 sin2 t + cos2 t)3/2 , 2 ρ(0) = 1/2, ρ(π/2) = 4 ρ(0) = ρ(π ) = 1. y y c r (0) = r (c) = 1 x 1 ρ 6 =4 () x 2 1 ρ (0) = 2 32. x (t) = −e−t (cos t + sin t), κ −t y (t) = e (cos t − sin t), −t x (t) = 2e sin t, −t y (t) = −2e 6 cos t; using the formula of Exercise 17(a), 1 κ = √ et . 2 t -3 3 33. (a) At x = 0 the curvature of I has a large value, yet the value of II there is zero, so II is not the curvature of I; hence I is the curvature of II. (b) I has points of inﬂection where the curvature is zero, but II is not zero there, and hence is not the curvature of I; so I is the curvat...
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