Potential function for conservative vector field

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SUMMARY

The discussion focuses on finding a potential function for the conservative vector field F = . The solution process involves integrating the components of the vector field, leading to the expression f = (1/2)x² + xy + C(y,z) after integrating with respect to x. The participants emphasize the importance of correctly identifying the variables for differentiation and integration, ultimately suggesting that integrating with respect to y first may yield a more straightforward path to the solution.

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  • Understanding of conservative vector fields
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  • Experience with potential functions in vector calculus
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kasse
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[SOLVED] Potential function for conservative vector field

Homework Statement



Find a potential function for the conservative vector field F = <x + y, x - z, z - y>
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2. The attempt at a solution

OK, we know that

(1) fx = x + y
(2) fy = x - z og
(3) fz = z - y

We can then integrate (1) with respect to x, and we get

(4) f = (1/2)x2 + xy + C(y,z)

We then differentiate (4) wrt y and get:

(5) fy = x + C'(y,z)

Comparing (2) and (5):

(6) C'(y,z) = -z

Integrate this wrt z:

(7) C(y,z) = -(1/2)z2 + C(y)

We then have the following expression for the potential function so far:

(8) f = (1/2)x2 + xy + -(1/2)z2 + C(y)

Then I'm stuck. Is my method correct so far?
 
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In step (6) when you write C'(y,z)=(-z) the prime means derivative wrt y. So I wouldn't integrate wrt z. You're on the right track. Just pay more attention to what variable the derivatives are taken wrt.
 
kasse said:
OK, we know that

(1) fx = x + y
(2) fy = x - z og
(3) fz = z - y

We can then integrate (1) with respect to x, and we get

(4) f = (1/2)x2 + xy + C(y,z)

We then differentiate (4) wrt y and get:

(5) fy = x + C'(y,z)

Comparing (2) and (5):

(6) C'(y,z) = -z

I was with you up to here. Why did you compare (2) and (5) and then integrate with respect to z? It seems to me that if you'd integrated with respect to y first, you'd have gotten

C(y,z) = -yz + C(z) ,

from which point you should be all right...

I guess I prefer integrating each partial differential equation separately and then intercomparing all three. As long as the equations are too complicated, things usually fall into place pretty well.
 

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