# Finding a Potential Function for a Vector Field

1. May 14, 2013

### Niruli

Hi everyone! I've been having a hard time figuring this one out for a while, so any help will be appreciated!

1. The problem statement, all variables and given/known data

$\textbf{F}$= <(2zx)/(x^2+e^z*y^2), (2ze^z*y)/(x^2+e^z*y^2), log(x^2+e^z*y^2) + (ze^z*y^2)/(x^2+e^z*y^2)>
(a) Where is the following vector field defined?
(b) Is this region simply connected?
(c) Find a potential function for $\textbf{F}$.

2. Relevant equations

$\textbf{F}$= ∇V, where V is the potential function.

3. The attempt at a solution

I answered (a) and (b). For (c), I computed the x- and y-components of V by integrating the x- and y-components of $\textbf{F}$, but I have no idea how to integrate the z-component of $\textbf{F}$ to find the z-component of V.

dV/dx= ∫(2zx)/(x^2+e^z*y^2)dx= zln(x^2+e^z*y^2) + f(y,z)

dV/dy= ∫(2ze^z*y)/(x^2+e^z*y^2)dy= zln(x^2+e^z*y^2) + g(x,z)

dV/dz= ∫(log(x^2+e^z*y^2) + (ze^z*y^2)/(x^2+e^z*y^2))dz

2. May 14, 2013

### haruspex

The equations you've written aren't really correct (LHS should be e.g. ∫(∂V/∂x).dx), but I get the idea.
From the first two, aren't you looking for an answer like $z \ln(x^2+e^z y^2) + h(z)$?

3. May 14, 2013

### Dick

I'd say from what you've got so far that V=z*ln(x^2+e^z*y^2) looks like a good candidate for a potential function. Just try finding dV/dz and see it works. In finding a potential function, you are allowed to just guess.

4. May 14, 2013

### LCKurtz

You have an unknown potential function $V(x,y,z)$. After your first integration, you know a little more about it, so it isn't completely unknown:$$V(x,y,z) = z\ln(x^2+e^zy^2) + f(y,z)$$Instead of doing the same thing for the $y$ and $z$ variable, you should use this $V$ for your next step$$V_y = \frac z {x^2+e^zy^2}\cdot 2e^zy + f_y(y,z) = \frac {2ze^zy}{x^2+e^zy^2}$$This tells you that $f_y(y,z) = 0$ so $f$ is purely a function of $z$, let's call it $g(z)$. So now you know more about $V$:$$V(x,y,z) = z\ln(x^2+e^zy^2) + g(z)$$Now differentiate this with respect to $z$ and compare it with the third component to see what you get for $g(z)$.

This is the preferred method to do these problems, because it eliminates the guesswork and eliminates two integrations by doing differentiations instead.

5. May 14, 2013

### Dick

Not to disagree with you, but 'eliminating the guesswork' is not always what you want. If you can guess the potential function early in the process you can skip the whole formalism, by just checking that it works. And it often does work for this kind of problem. Not always, but often.

6. May 14, 2013

### LCKurtz

I'm not against shortcuts, but too often students try to work the problem just like the OP did. Then they never learn how to do it properly. I used to show my students a 2-D example where when they integrated both components of the vector and took the "common part" for their answer, it was incorrect.

7. May 14, 2013

### Dick

Ok, sure. Assuming the "common part" will work without checking it is wrong. I get your point. Still checking the "common part" as are you going along will often give you a shortcut. Works for a lot of problems. Like this one. That was my only point. In this case the OP had a good candidate on the first integration.

Last edited: May 14, 2013
8. May 24, 2013

### Niruli

haruspex, yes, I was looking for exactly that, but I didn't understand what to do with the integrand for the z-component. It turns out my professor just made an error and the term was in fact supposed to be in terms of the natural logarithm, which makes my problem disappear.
Dick and LCKurtz, thank you so much for your help! :)