Explanation Given information: The vortex field is given by F = 〈 − y x 2 + y 2 , x x 2 + y 2 〉 . Formula used: Potential function of a vector field F = 〈 F 1 ( x , y ) , F 2 ( x , y ) 〉 will exist if ∂ F 1 ∂ y − ∂ F 2 ∂ x = 0 That is if F is conservative. Potential function for the vector field F = 〈 F 1 ( x , y ) , F 2 ( x , y ) 〉 is given by a function f such that F = 〈 F 1 ( x , y ) , F 2 ( x , y ) 〉 = ∇ f Where ∇ f = 〈 f x , f y 〉 Calculation: Consider the given vector field. F = 〈 − y x 2 + y 2 , x x 2 + y 2 〉 Here, F 1 ( x , y ) = − y x 2 + y 2 , and F 2 ( x , y ) = x x 2 + y 2 Now check whether F is conservative or not. ∂ F 1 ∂ y − ∂ F 2 ∂ x = ∂ ∂ y ( − y x 2 + y 2 ) − ∂ ∂ x ( x x 2 + y 2 ) = [ ( x 2 + y 2 ) ⋅ ∂ ∂ y ( − y ) − ( − y ) ∂ ∂ y ( x 2 + y 2 ) ( x 2 + y 2 ) 2 ] − [ ( x 2 + y 2 ) ⋅ ∂ ∂ x ( x ) − ( x ) ∂ ∂ x ( x 2 + y 2 ) ( x 2 + y 2 ) 2 ] = ( x 2 + y 2 ) ⋅ ( − 1 )

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 16.3, Problem 32E

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The potential function for the vortex field is g(x,y)=−tan−1xy.

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Chapter 16.3, Problem 31E

Chapter 16.3, Problem 33E

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Chapter 16 Solutions

Calculus: Early Transcendentals

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The potential function for the vortex field is g ( x , y ) = − tan − 1 x y .

Solution Summary: The author explains that the potential function for the vortex field is g(x,y)=-mathrmtan-1xy

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The potential function for the vortex field is g(x,y)=−tan−1xy.

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Chapter 16 Solutions

The potential function for the vortex field is g ( x , y ) = − tan − 1 x y .

Solution Summary: The author explains that the potential function for the vortex field is g(x,y)=-mathrmtan-1xy

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To show: The potential function for the vortex field is g(x,y)=−tan−1xy.

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Chapter 16 Solutions

To show:

The potential function for the vortex field is g(x,y)=−tan−1xy.