# 008040 755505013725 r dz dr d 4a3 3 0 2 2 a 2 sin

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Unformatted text preview: x sec y tan y j, −→ √ √ f (0, π/4) = 2(−i + j), P O = −(π/4)j, u = −j, Du f = − 2 xey yey √ xyey + √ f (x, y ) = √ i + 2 xy 2 xy j, f (1, 1) = (e/2)(i + 3j), u = −j, Du f = −3e/2 40. then u and f (x, y ) = −y (x + y )−2 i + x(x + y )−2 j, f (2, 3) = (−3i + 2j)/25, if Du f = 0√ orthogonal, by inspection 2i + 3j is orthogonal to f (2, 3) so u = ±(2i + 3j)/ 13. 41. f (x, y ) = 8xy i + 4x2 j, √ u = ±(−4i + j)/ 17. 42. f (x, y ) = (6xy − y )i + 3x2 − x j, √ P so u = ±(−33i + 10j)/ 1189. f are f (1, −2) = −16i + 4j is normal to the level curve through P so f (2, −3) = −33i + 10j is normal to the level curve through 43. Solve the system (3/5)fx (1, 2) − (4/5)fy (1, 2) = −5, (4/5)fx (1, 2) + (3/5)fy (1, 2) = 10 for fx (1, 2) and fy (1, 2) to get fx (1, 2) = 5, fy (1, 2) = 10. For (c), f (1, 2) = 5i + 10j, √ √ u = (−i − 2j)/ 5, Du f = −5 5. 44. −→ √ √ f (−5, 1) = −3i + 2j, P Q = i + 2j, u = (i + 2j)/ 5, Du f = 1/ 5 45. √ √ f (4, −5) = 2i − j, u = (5i + 2j)/ 29, Du f = 8/ 29 46. Let u = u1 i + u2 j where u2 + u2 = 1, but Du f = f · u = u1 − 2u2 = −2 so u1 = 2u2 − 2, 1 2 2 (2u2 − 2) + u2 = 1, 5u2 − 8u2 + 3 = 0, u2 = 1 or u2 = 3/5 thus u1 = 0 or u1 = −4/5; u = j or 2 2 3 4 u = − i + j. 5 5 551 Chapter 15 47. (a) At (1, 2) the steepest ascent seems to be in the direction i + j and the slope in that direction √ √ 1 1 seems to be 0.5/( 2/2) = 1/ 2, so f ≈ i + j, which has the required direction and 2 2 magnitude. y (b) 5 −∇f (4, 4) x 5 48. (a) 200 300 400 500 100 0 ft P Depart from each contour line in a direction orthogonal to that contour line, as an approximation to the optimal path. (b) 500 200 300 400 100 0 ft P At the top there is no contour line, so head for the nearest contour line. From then on depart from each contour line in a direction orthogonal to that countour line, as in Part (a). 49. z = 6xi − 2y j, 50. z = 3i + 2y j, z= z= 36x2 + 4y 2 = 6 if 36x2 + 4y 2 = 36; all points on the ellipse 9x2 + y 2 = 9. 9 + 4y 2 , so z= 4y 9+ 4y 2 =√ 8 8 = 5 9 + 16 √ 51. r = ti − t2 j, dr/dt = i − 2tj = i − 4j at the point (2, −4), u = (i − 4j)/ 17; √ z = 2xi + 2y j = 4i − 8j at (2, −4), hence dz/ds = Du z = z · u = 36/ 17. 52. (a) T (x, y ) = y 1 − x2 + y 2 2 y2 ) i+ (1 + x2 + √ Du T = 1/ 9 5 √ (b) u = −(i + j)/ 2, opposite to x 1 + x2 − y 2 (1 + x2 + T (1, 1) 2 y2 ) j, √ T (1, 1) = (i + j)/9, u = (2i − j)/ 5, Exercise Set 15.6 53. (a) 552 V (x, y ) = −2e−2x cos 2y i − 2e−2x sin 2y j, E = − V (π/4, 0) = 2e−π/2 i (b) V (x, y ) decreases most rapidly in the direction of − V (x, y ) which is E. 54. z = −4xi − 8y j, if x = −20 and y = 5 then z = 80i − 40j. (a) u = −i points due west, Du z = −80, the climber will descend because z is decreasing. √ √ √ (b) u = (i + j)/ 2 points northeast, Du z = 20 2, the climber will ascend at the rate of 20 2 ft per ft of travel in the xy −plane. (c) The climber will travel a level path in a direction perpendicular to√ z = 80i − 40j, by √ inspection ±(i + 2j)/ 5 are unit vectors in these directions; (i + 2j)/ 5 makes an angle of √ tan−1 (1/2) ≈ 27◦ with the positive y -axis so −(i+2j)/ 5 makes the same angle with the negative y -axis. The compass direction should be N 27◦ E or S 27◦ W. 55. (a) (b) 56. (a) x r= x2 f (r) = + i+ y x2 + y2 j = r/r ∂f (r) ∂r ∂r ∂f (r) i+ j = f (r) i + f (r) j = f (r) r ∂x ∂y ∂x ∂y re−3r = (b) 3r2 r = y2 (1 − 3r) −3r er r f (r) 3 3 r so f (r) = 3r3 , f (r) = r4 + C , f (2) = 12 + C = 1, C = −11; f (r) = r4 − 11 r 4 4 57. ur = cos θi + sin θj, uθ = − sin θi + cos θj, z= ∂z ∂z i+ j= ∂x ∂y = 58. (a) 1 ∂z ∂z cos θ − sin θ i + ∂r r ∂θ 1 ∂z ∂z sin θ + cos θ j ∂r r ∂θ 1 ∂z ∂z 1 ∂z ∂z (cos θi + sin θj) + (− sin θi + cos θj) = ur + uθ ∂r r ∂θ ∂r r ∂θ (f + g ) = (fx + gx ) i + (fy + gy ) j = (fx i + fy j) + (gx i + gy j) = f+ g (b) (cf ) = (cfx ) i + (cfy ) j = c (fx i + fy j) = c f (c) (f g ) = (f gx + gfx ) i + (f gy + gfy ) j = f (gx i + gy j) + g (fx i + fy j) = f g + g f (d) (f /g ) = (e) (f n ) = nf n−1 fx i + nf n−1 fy j = nf n−1 (fx i + fy j) = nf n−1 f gfy − f gy g (fx i + fy j) − f (gx i + gy j) g f −f g gfx − f gx i+ j= = 2 2 2 g g g g2 dy dx = −8kx, = −2ky . Divide and solve dt dt 4 −8t −2t to get y = 256x; one parametrization is x(t) = e , y (t) = 4e . 59. r (t) = v(t) = k (x, y ) T = −8k (x, y )xi − 2k (x, y )y j; 60. r (t) = v(t) = k T = −2k (x, y )xi − 4k (x, y )y j. Divide and solve to get y = tion is x(t) = 5e−2t , y (t) = 3e−4t . 32 x ; one parametriza25 553 Chapter 15 y 61. 62. 5 4 C = –10 C = –5 (5, 3) T = 80 T = 95 T = 90 C = –15 -6 C=0 6 T = 97 x -3 -4 3 -5 z 63. (a) y x (c) f = [2x − 2x(x2 + 3y 2 )]e−(x (d) f = 0 if x = y = 0 or x = 0, y = ±1 or x = ±1, y = 0. 2 +y 2 ) i + [6y − 2y (x2 + 3y 2 )]e−(x 2 +y 2 ) j 64. dz/dt = (z/∂x)(dx/dt) + (∂z/∂y )(dy/dt) = (∂z/∂xi + ∂z/∂y j) · (dx/dti + dy/dtj) = 65. z · r (t) f (x, y ) = fx (x, y )i + fy (x, y )j, if f (x, y ) = 0 throughout t...
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