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May 24th, 2013, 01:37 PM   #1
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Find the potential function

Hi, here is the question and the solution, but there is something i did not understand, i'd like someone to explain to me.

We are given the conservative vector field Find the potential function of the vector field.

Solution:


Where did C'(y) go???
Why??
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May 25th, 2013, 02:17 PM   #2
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Re: Find the potential function

Quote:
Originally Posted by OriaG
Hi, here is the question and the solution, but there is something i did not understand, i'd like someone to explain to me.

We are given the conservative vector field Find the potential function of the vector field.
A potential function for a vector is a function H(x,y) such that and
Here, we must have and

Quote:
Solution:

I would have phrased it slightly differently. Since and the partial derivative with respect to x is done treating y like a constant we integrate, treating y as a constant: (with a= y and b= 1+ y, constants)
The first is just while the second can be done integrating "by parts": let u= x, . The du= dx, and so the integral is equal to . Combining those, the integral, with respect to x, is . Plus a constant of integration, of course, but, here, since we were treating y as a constant, that "constant of integration" may be an arbitrary function of y: , just as your solution has.

Quote:
Where did C'(y) go???
C'(y) didn't "go" anywhere- it was never there. The first two terms above, are the result of differentiating the formula above (that I have called H(x,y)- I don't want to call it f because you had already written "" for the gradient.) while the last term, is from the orignal ", the being the second component: . The fact that those are equal tells us that which means that C really is a constant, not a function of y after all.

Quote:
Why??
HallsofIvy is offline  
May 25th, 2013, 03:59 PM   #3
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Re: Find the potential function

Got it, thanks again HallsofIvy
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