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 May 24th, 2013, 01:37 PM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 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 $\vec{F}= (e^{x-y}(1+x+y),e^{x-y}(1-x-y))$ Find the potential function of the vector field. Solution: $f(x,y)=\int e^{x-y}(1+x+y)dx+C(y) = e^{x-y}+e^{x-y}(x-1)+ye^{x-y}+C(y) = e^{x-y}(x+y)+C(y).$ $\frac{df(x,y)}{dy}= -e^{x-y}(x+y)+e^{x-y}+C'(y) = e^{x-y}(1-x-y)+C#39;(y) = e^{x-y}(1-x-y)$ Where did C'(y) go??? $C'(y) = 0$ Why?? $y= c,f(x,y) = e^{x-y}(x+y)+c$
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 $\vec{F}= (e^{x-y}(1+x+y),e^{x-y}(1-x-y))$ Find the potential function of the vector field.
A potential function for a vector $f(x,y)\vec{i}+ g(x,y)\vec{j}$ is a function H(x,y) such that $\frac{\partial H}{\partial x}= f(x,y)$ and $\frac{\partial H}{\partial y}= g(x,y)$
Here, we must have $\frac{\partial H}{\partial x}= e^{x- y}(1+ x+ y)$ and $\frac{\partial H}{\partial y}= e^{x- y}(1- x- y)$

Quote:
 Solution: $f(x,y)=\int e^{x-y}(1+x+y)dx+C(y) = e^{x-y}+e^{x-y}(x-1)+ye^{x-y}+C(y) = e^{x-y}(x+y)+C(y).$
I would have phrased it slightly differently. Since $\frac{\partial H}{\partial x}= e^{x- y}(1+ x+ y)$ and the partial derivative with respect to x is done treating y like a constant we integrate, treating y as a constant: $\int e^{x- a}(b+ x)dx= \int be^{x- a} dx+ \int xe^{x- a}dx$ (with a= y and b= 1+ y, constants)
The first is just $be^{x- a}= (1+ y)e^{x- y}$ while the second can be done integrating "by parts": let u= x, $dv= e^{x- a}$. The du= dx, and $v= e^{x- a}$ so the integral is equal to $\int udv= uv- \int vdu= xe^{x- a}- \int e^{x- a}dx= xe^{x- a}- e^{x- a}= e^{x- a}(x- 1)= e^{x- y}(x- 1)$. Combining those, the integral, with respect to x, is $(1+ y)e^{x- y}+ e^{x- y}(x- 1)= (x+ y)e^{x- y}$. 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: $H(x,y)= (x+ y)e^{x- y}+ C(y)$, just as your solution has.

Quote:
 $\frac{df(x,y)}{dy}= -e^{x-y}(x+y)+e^{x-y}+C'(y) = e^{x-y}(1-x-y)+C#39;(y) = e^{x-y}(1-x-y)$ Where did C'(y) go???
C'(y) didn't "go" anywhere- it was never there. The first two terms above, $-e^{x-y}(x+y)+e^{x-y}+C'(y)= e^{x-y}(1-x-y)+C#39;(y)$ 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 "$\vec{F}$" for the gradient.) while the last term, $e^{x- y}(1- x- y)$ is from the orignal "$\vec{F}= (e^{x-y}(1+x+y),e^{x-y}(1-x-y))$, the $\frac{\partial H}{\partial y}$ being the second component: $e^{x-y}(1- x- y)$. The fact that those are equal tells us that $C'(y)= 0$ which means that C really is a constant, not a function of y after all.

Quote:
 $C'(y) = 0$ Why?? $y= c,f(x,y) = e^{x-y}(x+y)+c$

 May 25th, 2013, 03:59 PM #3 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Find the potential function Got it, thanks again HallsofIvy

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