Sin 215 cos tan 1 45 34 154 1 15 csc 2 sec

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Unformatted text preview: , 4 cos φ ; rφ ×rθ = 2 6 sin2 φ cos θ, 4 sin2 φ sin θ, 3 cos φ sin φ , rφ × rθ = 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ, 2π π 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ dφ dθ ≈ 111.5457699 S= 0 0 CHAPTER 17 Topics in Vector Calculus EXERCISE SET 17.1 1. (a) III because the vector field is independent of y and the direction is that of the negative x-axis for negative x, and positive for positive (b) IV, because the y -component is always positive, and the x-component is positive for positive x, negative for negative x 2. (a) I, since the vector field is constant (b) II, since the vector field points away from the origin 3. (a) true (b) true (c) true 4. (a) false, the lengths are equal to 1 (c) false, the x-component is then zero y 5. 6. (b) false, the y -component is then zero y 7. y x x x y 8. 9. y (b) 12. (a) (b) φ = φx i + φy j = y x x 11. (a) 10. x y x i+ j = F, so F is conservative for all x, y 1 + x2 y 2 1 + x2 y 2 φ = φx i + φy j = 2xi − 6y j + 8z k = F so F is conservative for all x, y φ = φx i + φy j = (6xy − y 3 )i + (4y + 3x2 − 3xy 2 )j = F, so F is conservative for all x, y φ = φx i + φy j + φz k = (sin z + y cos x)i + (sin x + z cos y )j + (x cos z + sin y )k = F, so F is conservative for all x, y 608 609 Chapter 17 13. div F = 2x + y , curl F = z i 14. div F = z 3 + 8y 3 x2 + 10zy , curl F = 5z 2 i + 3xz 2 j + 4xy 4 k 15. div F = 0, curl F = (40x2 z 4 − 12xy 3 )i + (14y 3 z + 3y 4 )j − (16xz 5 + 21y 2 z 2 )k 16. div F = yexy + sin y + 2 sin z cos z , curl F = −xexy k 2 17. div F = 18. div F = x2 + y2 + z2 , curl F = 0 x z 1 + xzexyz + 2 , curl F = −xyexyz i + 2 j + yzexyz k x x + z2 x + z2 19. · (F × G) = · (−(z + 4y 2 )i + (4xy + 2xz )j + (2xy − x)k) = 4x 20. · (F × G) = · ((x2 yz 2 − x2 y 2 )i − xy 2 z 2 j + xy 2 z k) = −xy 2 21. ·( × F) = · (− sin(x − y )k) = 0 22. ·( × F) = · (−zeyz i + xexz j + 3ey k) = 0 23. ×( × F) = × (xz i − yz j + y k) = (1 + y )i + xj 24. ×( × F) = × ((x + 3y )i − y j − 2xy k) = −2xi + 2y j − 3k 25. div (k F) = k ∂q ∂h ∂f +k +k = k div F ∂x ∂y ∂z ∂ h ∂g − ∂y ∂z 26. curl (k F) = k ∂h ∂f − ∂z ∂x i+k j+k ∂f ∂g − ∂x ∂y k = k curl F 27. Let F = f (x, y, z )i + g (x, y, z )j + h(x, y, z )k and G = P (x, y, z )i + Q(x, y, z )j + R(x, y, z )k, then div (F + G) = = ∂P ∂f + ∂x ∂x ∂Q ∂g + ∂y ∂y + ∂g ∂h ∂f + + ∂x ∂y ∂z + + ∂ h ∂R + ∂z ∂z ∂Q ∂R ∂P + + ∂x ∂y ∂z = div F + div G 28. Let F = f (x, y, z )i + g (x, y, z )j + h(x, y, z )k and G = P (x, y, z )i + Q(x, y, z )j + R(x, y, z )k, then curl (F + G) = ∂ ∂ ∂ ∂ (h + R) − (g + Q) i + (f + P ) − (h + R) j ∂y ∂z ∂z ∂x ∂ ∂ (g + Q) − (f + P ) k; ∂x ∂y expand and rearrange terms to get curl F + curl G. + 29. div (φF) = =φ φ ∂f ∂φ + f ∂x ∂x +φ ∂g ∂h ∂f + + ∂x ∂y ∂z = φ div F + φ·F ∂g ∂φ ∂h ∂φ + g+φ + h ∂y ∂y ∂z ∂z + ∂φ ∂φ ∂φ f+ g+ h ∂x ∂y ∂z Exercise Set 17.1 610 ∂ ∂ ∂ ∂ ∂ ∂ (φh) − (φg ) i + (φf ) − (φh) j + (φg ) − (φf ) k; use the product ∂y ∂z ∂z ∂x ∂x ∂y rule to expand each of the partial derivatives, rearrange to get φ curl F + φ × F 30. curl (φF) = 31. div(curl F) = ∂ ∂x ∂ h ∂g − ∂y ∂z + ∂ ∂y ∂f ∂h − ∂z ∂x ∂ ∂z + ∂g ∂f − ∂x ∂y ∂2g ∂2f ∂2h ∂2g ∂2f ∂2h − + − + − = 0, ∂x∂y ∂x∂z ∂y∂z ∂y∂x ∂z∂x ∂z∂y assuming equality of mixed second partial derivatives = ∂2φ ∂2φ ∂2φ ∂2φ − i+ − ∂y∂z ∂z∂y ∂z∂x ∂x∂z of mixed second partial derivatives 32. curl ( φ) = 33. · (k F) = k · F, · (F + G) = 34. × (k F) = k × F, · F+ × (F + G) = j+ · G, ×F+ ∂2φ ∂2φ − ∂x∂y ∂y∂x · (φF) = φ × G, · F+ k = 0, assuming equality φ · F, ·( × (φF) = φ × F + φ × F, × F) = 0 × ( φ) = 0 37. (a) curl r = 0i + 0j + 0k = 0 (b) x2 + y 2 + z 2 = r= x x2 + y2 + z2 i+ y x2 + y2 + z2 j+ z x2 + y2 + z2 k= r r 38. (a) div r = 1 + 1 + 1 = 3 1 = r (b) 39. (a) (x2 + y 2 + z 2 )−1/2 = − f (r) = f (r) ∂r ∂r ∂r f (r) i + f (r) j + f (r) k = f (r) r = r ∂x ∂y ∂z r (b) div[f (r)r] = f (r)div r + f (r) · r = 3f (r) + 40. (a) curl[f (r)r] = f (r)curl r + (b) 2 f (r) = div[ f (r)] = div =3 xi + y j + z k r =− r3 (x2 + y 2 + z 2 )3/2 f (r) r · r = 3f (r) + rf (r) r f (r) × r = f (r)0 + f (r) r×r=0+0=0 r f (r) f (r) r= div r + r r f (r) ·r r f (r) rf (r) − f (r) f (r) + + f (r) r·r=2 r r3 r 41. f (r) = 1/r3 , f (r) = −3/r4 , div(r/r3 ) = 3(1/r3 ) + r(−3/r4 ) = 0 42. Multiply 3f (r) + rf (r) = 0 through by r2 to obtain 3r2 f (r) + r3 f (r) = 0, d[r3 f (r)]/dr = 0, r3 f (r) = C, f (r) = C/r3 , so F = C r/r3 (an inverse-square field). 43. (a) At the point (x, y ) the slope of the line along which the vector −y i + xj lies is −x/y ; the slope of the tangent line to C at (x, y ) is dy/dx, so dy/dx = −x/y . (b) ydy = −xdx, y 2 /2 = −x2 /2 + K1 , x2 + y 2 = K 611 Chapter 17 44. dy/dx = x,...
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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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