Calculus: Early Transcendentals, by Anton, 7th Edition,ch16

# Calculus - Early Transcendentals

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657 CHAPTER 16 Topics in Vector Calculus EXERCISE SET 16.1 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 constant, and the x -component varies priodically with 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 (b) false, the y -component is then zero (c) false, the x -component is then zero 5. x y 6. x y 7. x y 8. x y 9. x y 10. x y 11. (a) φ = φ x i + φ y j = y 1 + x 2 y 2 i + x 1 + x 2 y 2 j = F , so F is conservative for all x, y (b) φ = φ x i + φ y j = 2 x i 6 y j + 8 z k = F so F is conservative for all x, y 12. (a) φ = φ x i + φ y j = (6 xy y 3 ) i + (4 y + 3 x 2 3 xy 2 ) j = F , so F is conservative for all x, y (b) φ = φ x i + φ y j + φ z 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 13. div F = 2 x + y , curl F = z i 14. div F = z 3 + 8 y 3 x 2 + 10 zy , curl F = 5 z 2 i + 3 xz 2 j + 4 xy 4 k

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658 Chapter 16 15. div F = 0, curl F = (40 x 2 z 4 12 xy 3 ) i + (14 y 3 z + 3 y 4 ) j (16 xz 5 + 21 y 2 z 2 ) k 16. div F = ye xy + sin y + 2 sin z cos z , curl F = xe xy k 17. div F = 2 x 2 + y 2 + z 2 , curl F = 0 18. div F = 1 x + xze xyz + x x 2 + z 2 , curl F = xye xyz i + z x 2 + z 2 j + yze xyz k 19. · ( F × G ) = · ( ( z + 4 y 2 ) i + (4 xy + 2 xz ) j + (2 xy x ) k ) = 4 x 20. · ( F × G ) = · (( x 2 yz 2 x 2 y 2 ) i xy 2 z 2 j + xy 2 z k ) = xy 2 21. · ( ∇ × F ) = · ( sin( x y ) k ) = 0 22. · ( ∇ × F ) = · ( ze yz i + xe xz j + 3 e y k ) = 0 23. ∇ × ( ∇ × F ) = ∇ × ( xz i yz j + y k ) = (1 + y ) i + x j 24. ∇ × ( ∇ × F ) = ∇ × (( x + 3 y ) i y j 2 xy k ) = 2 x i + 2 y j 3 k 27. Let F = f i + g j + h k ; div ( k F ) = k ∂f ∂x + k ∂g ∂y + k ∂h ∂z = k div F 28. Let F = f i + g j + h k ; curl ( k F ) = k ∂h ∂y ∂g ∂z i + k ∂f ∂z ∂h ∂x j + k ∂g ∂x ∂f ∂y k = k curl F 29. 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 ) = ∂f ∂x + ∂P ∂x + ∂g ∂y + ∂Q ∂y + ∂h ∂z + ∂R ∂z = ∂f ∂x + ∂g ∂y + ∂h ∂z + ∂P ∂x + ∂Q ∂y + ∂R ∂z = div F + div G 30. 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 ) = ∂y ( h + R ) ∂z ( g + Q ) i + ∂z ( f + P ) ∂x ( h + R ) j + ∂x ( g + Q ) ∂y ( f + P ) k ; expand and rearrange terms to get curl F + curl G . 31. Let F = f i + g j + h k ; div ( φ F ) = φ ∂f ∂x + ∂φ ∂x f + φ ∂g ∂y + ∂φ ∂y g + φ ∂h ∂z + ∂φ ∂z h = φ ∂f ∂x + ∂g ∂y + ∂h ∂z + ∂φ ∂x f + ∂φ ∂y g + ∂φ ∂z h = φ div F + φ · F
Exercise Set 16.1 659 32. Let F = f i + g j + h k ; curl ( φ F ) = ∂y ( φh ) ∂z ( φg ) i + ∂z ( φf ) ∂x ( φh ) j + ∂x ( φg ) ∂y ( φf ) k ; use the product rule to expand each of the partial derivatives, rearrange to get φ curl F + φ × F 33. Let F = f i + g j + h k ; div(curl F ) = ∂x ∂h ∂y ∂g ∂z + ∂y ∂f ∂z ∂h ∂x + ∂z ∂g ∂x ∂f ∂y = 2 h ∂x∂y 2 g ∂x∂z + 2 f ∂y∂z 2 h ∂y∂x + 2 g ∂z∂x 2 f ∂z∂y = 0 ,

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