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17.1 Ex 19-32

# 17.1 Ex 19-32 - S E C T I O N 17.1 Vector Fields(ET Section...

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S E C T I O N 17.1 Vector Fields (ET Section 16.1) 1061 19. F = x , 0 , z SOLUTION This vector field is shown in (A) (by process of elimination). 20. F = x x 2 + y 2 + z 2 , y x 2 + y 2 + z 2 , z x 2 + y 2 + z 2 SOLUTION The unit radial vector field is shown in plot (D), as these vectors are radial and of uniform length. 21. F = 1 , 1 , 1 SOLUTION The constant vector field 1 , 1 , 1 is shown in plot (C). In Exercises 22–25, find a potential function for the vector field F by inspection. 22. F = x , y SOLUTION We must find a function ϕ ( x , y ) such that ϕ x = x and ϕ y = y . We choose the following function: ϕ ( x , y ) = 1 2 x 2 + 1 2 y 2 . 23. F = ye xy , xe xy SOLUTION The function ϕ ( x , y ) = e xy satisfies ϕ x = ye xy and ϕ y = xe xy , hence ϕ is a potential function for the given vector field. 24. F = yz 2 , xz 2 , 2 xyz SOLUTION We choose a function ϕ ( x , y , z ) such that ϕ x = yz 2 , ϕ y = xz 2 , ϕ z = xyz The function ϕ ( x , y , z ) = x yz 2 is a potential function for the given field. 25. F = 2 xze x 2 , 0 , e x 2 SOLUTION The function ϕ ( x , y , z ) = ze x 2 satisfies ϕ y = 0, ϕ x = 2 xze x 2 and ϕ z = e x 2 , hence ϕ is a potential function for the given vector field. 26. Find potential functions for the vector fields F = e r r 3 and G = e r r 4 in R 3 . SOLUTION We use the gradient of r , r = e r , and the Chain Rule for Gradients to write 1 2 r 2 = r 3 r = r 3 e r = e r r 3 = F 1 3 r 3 = r 4 r = r 4 e r = e r r 4 = G Therefore ϕ 1 ( r ) = − 1 2 r 2 and ϕ 2 ( r ) = − 1 3 r 3 are potential functions for F and G , respectively.

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17.1 Ex 19-32 - S E C T I O N 17.1 Vector Fields(ET Section...

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