F 1 2 3 level curve 4x 2y 3 3 2x y 0 f x y

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Unformatted text preview: ure of II. 34. (a) II takes the value zero at x = 0, yet the curvature of I is large there; hence I is the curvature of II. (b) I has constant zero curvature; II has constant, positive curvature; hence I is the curvature of II. 35. (a) 0 0 1 (b) 1 5 0 5 0 Exercise Set 14.5 508 4 36. (a) 4 (b) -1 1 -1 1 -4 -4 |12x2 − 4| 37. (a) κ = (1 + (4x3 − y (b) 3/2 4x)2 ) 8 k f(x) x -2 2 (c) f (x) = 4x3 − 4x = 0 at x = 0, ±1, f (x) = 12x2 − 4, so extrema at x = 0, ±1, and ρ = 1/4 for x = 0 and ρ = 1/8 when x = ±1. y 38. (a) (c) κ(t) = 30 t2 + 2 (t2 + 1)3/2 (d) lim κ(t) = 0 t→+∞ x -30 30 -30 −r sin θ + cos θ 39. r (θ) = r (θ) = dr dθ −r cos θ − 2 sin θ r2 + 2 dr dθ 2 κ= r2 + dr dθ −r i + r cos θ + sin θ d2 r dr + cos θ 2 dθ dθ dr dθ j; i + −r sin θ + 2 cos θ j; d2 r dθ2 2 3/2 . 40. Let r = a be the circle, so that dr/dθ = 0, and κ(θ) = 3 3 , κ(π/2) = √ 41. κ(θ) = √ 1/2 2 2(1 + cos θ) 22 43. κ(θ) = d2 r dr + sin θ 2 dθ dθ 10 + 8 cos2 3θ 2 , κ(0) = 2 θ )3/2 3 (1 + 8 cos 1 1 = r a 42. κ(θ) = √ 44. κ(θ) = 1 1 , κ(1) = √ 2θ 5e 5e2 θ2 + 2 3 , κ(1) = √ 2 + 1)3/2 (θ 22 509 Chapter 14 45. The radius of curvature is zero when θ = π , so there is a cusp there. 46. d2 r 3 dr = − sin θ, 2 = − cos θ, κ(θ) = 3/2 √ dθ dθ 2 1 + cos θ 47. Let y = t, then x = 1/|2p| t2 and κ(t) = 2 ; 4p [t /(4p2 ) + 1]3/2 t = 0 when (x, y ) = (0, 0) so κ(0) = 1/|2p|, ρ = 2|p|. ex (1 − 2e2x ) ex , κ (x) = ; κ (x) = 0 when e2x = 1/2, x = −(ln 2)/2. By the first 2x )3/2 (1 + e (1 + e2x )5/2 √ 1 1 derivative test, κ(− ln 2) is maximum so the point is (− ln 2, 1/ 2). 2 2 48. κ(x) = 49. Let x = 3 cos t, y = 2 sin t for 0 ≤ t < 2π , κ(t) = 6 so (9 sin t + 4 cos2 t)3/2 2 1 1 (9 sin2 t + 4 cos2 t)3/2 = (5 sin2 t + 4)3/2 which, by inspection, is minimum when 6 6 t = 0 or π . The radius of curvature is minimum at (3, 0) and (−3, 0). ρ(t) = 6(1 − 45x4 ) 6x for x > 0, κ (x) = ; κ (x) = 0 when x = 45−1/4 which, by the 4 )3/2 (1 + 9x (1 + 9x4 )5/2 first derivative test, yields the maximum. 50. κ(x) = 51. r (t) = − sin ti + cos tj − sin tk, r (t) = − cos ti − sin tj − cos tk, √ √ r (t) × r (t) = − i + k = 2, r (t) = (1 + sin2 t)1/2 ; κ(t) = 2/(1 + sin2 t)3/2 , √ √ ρ(t) = (1 + sin2 t)3/2 / 2. The minimum value of ρ is 1/ 2; the maximum value is 2. √ 52. r (t) = et i − e−t j + 2k, r (t) = et i + e−t j; √ √ √ 1 2 , ρ(t) = √ (et + e−t )2 = 2 2 cosh2 t. The minimum value of ρ is 2 2. κ(t) = 2t −2t + 2 e +e 2 53. From Exercise 39: dr/dθ = aeaθ = ar, d2 r/dθ2 = a2 eaθ = a2 r; κ = 1/[ 1 + a2 r]. 54. Use implicit differentiation on r2 = a2 cos 2θ to get 2r again to get r dr dθ r2 + 2 r2 + dr dθ d2 r + dθ2 2 −r dr dθ 2 = −2a2 cos 2θ so r d2 r = 3 r2 + dθ2 2 = r2 + dr dθ d2 r =− dθ2 2 ,κ= [r2 dr dr = −2a2 sin 2θ, r = −a2 sin 2θ, and dθ dθ dr dθ 2 − 2a2 cos 2θ = − dr dθ 2 − 2r2 , thus a2 sin 2θ dr 3 =− so ; 2 ]1/2 dθ r + (dr/dθ) a4 sin2 2θ r4 + a4 sin2 2θ a4 cos2 2θ + a4 sin2 2θ a4 3r = = = 2 , hence κ = 2 . r2 r2 r2 r a Exercise Set 14.5 510 55. (a) d2 y/dx2 = 2, κ(φ) = |2 cos3 φ| √ √ (b) dy/dx = tan φ = 1, φ = π/4, κ(π/4) = |2 cos3 (π/4)| = 1/ 2, ρ = 2 y (c) 3 x -2 56. (a) 1 5 5 , 0 , 0, − 3 2 (b) clockwise (c) it is a point, namely the center of the circle 57. κ = 0 along y = 0; along y = x2 , κ(x) = 2/(1 + 4x2 )3/2 , κ(0) = 2. Along y = x3 , κ(x) = 6|x|/(1 + 9x4 )3/2 , κ(0) = 0. y 58. (a) (b) For y = x2 , κ(x) = 2 (1 + 4x2 )3/2 4 so κ(0) = 2; for y = x4 , κ(x) = x -2 12x2 so κ(0) = 0. (1 + 16x6 )3/2 κ is not continuous at x = 0. 2 59. κ = 1/r along the circle; along y = ax2 , κ(x) = 2a/(1 + 4a2 x2 )3/2 , κ(0) = 2a so 2a = 1/r, a = 1/(2r). |y | so the transition will be smooth if the values of y are equal, the values of y (1 + y 2 )3/2 are equal, and the values of y are equal at x = 0. If y = ex , then y = y = ex ; if y = ax2 + bx + c, then y = 2ax + b and y = 2a. Equate y , y , and y at x = 0 to get c = 1, b = 1, and a = 1/2. 60. κ(x) = 61. The result follows from the definitions N = T (s) and κ = T (s) . T (s) dB dB = 0 because B(s) = 1 so is perpendicular to B(s). ds ds dB dT dB dT + · T(s) = 0, but = κN(s) so κB(s) · N(s) + · (b) B(s) · T(s) = 0, B(s) · ds ds ds ds dB dB · T(s) = 0 because B(s) · N(s) = 0; thus is perpendicular to T(s). T(s) = 0, ds ds dB dB is perpendicular to both B(s) and T(s) but so is N(s), thus is parallel to N(s) and (c) ds ds hence a scalar multiple of N(s). 62. (a) B · (d) If C lies in a plane, then T(s) and N(s) also lie in the plane; B(s) = T(s) × N(s) so B(s) is always perpendicular to the plane and hence dB/ds = 0, thus τ = 0. 511 63. Chapter 14 dT dB dN = B× + × T = B × (κN) + (−τ N) × T = κB × N − τ N × T, but B × N = −T and ds ds ds dN = −κT + τ B N × T = −B so ds 64. r (s) = dT/ds = κN so r (s) = κdN/ds + (dκ/ds)N but dN/ds = −κT + τ B so r (s) = −κ2 T + (dκ/ds)N + κτ B, r (s) × r (s) = T × (κN) = κT × N = κB, [r (s) × r (s)] · r (s) = −κ3 B · T + κ(dκ/ds)B · N + κ2 τ B · B = κ2 τ , τ = [r (s) × r (s)] · r (s)/κ2 = [r (s) × r (s)] · r (s)/ r (s) 2 and B = T × N = [r (s) × r (s)]/ r (s) 65. r = a cos(s/w)i + a sin(s/w)j + (cs/w)k, r = −(a/w) sin(s/w)i + (a/w) cos(s/w)j + (c/w)k, r = −(a/w2 ) cos(s/w)i − (a/w2 ) sin(s/w)j, r = (a/w3 ) sin(s/w)i − (a/w3 ) cos(s/w)j,...
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