# Homework Help: Scalar potential of a function F, stuck on curl(F) != 0

1. Mar 29, 2012

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

I have a function F defined in a slightly strange way, and I'm not asked to test if curl(F) == 0, but I thought I would do this as part of my working out. Lo and behold, it looks like it doesn't == 0, and this means, as far as I know, that there is no scalar potential but I'm quite sure that this is not correct.

2. Relevant equations

F = e^yz (1, xz, xy)

3. The attempt at a solution

I multiplied the e^yz factor to each part of the function

F = (e^yz, e^yz . xz, e^yz . xy)

Then I perform the cross product of the gradient operator and the function F (dx, dy, dz are partial derivatives, please excuse my notation)

Code (Text):
|    i           j               k         |
|   dx         dy             dz        |
| e^yz  (e^yz . xz)  (e^yz . xy) |
= (z.e^yz .x - y.e^yz .x) - (e^yz .y - e^yz) + (e^yz .x - z.e^yz) != 0

am I correct in my partial differentiation of (for example) d/dy e^yz = ze^yz - I'm treating z as a constant and using the chain rule?

2. Mar 29, 2012

### Dick

I don't think you are using the product rule correctly. If you differentiating (e^yz)(xy) you need to apply the product rule (fg)'=f'g+fg'.

3. Mar 29, 2012

### SammyS

Staff Emeritus
In other words,
$\displaystyle \frac{\partial}{\partial y}\left(e^{yz} xy\right)=\left(\frac{\partial}{\partial y}e^{yz}\right)xy+e^{yz}\left(\frac{\partial}{ \partial y}xy\right)\,.$
Etc.

4. Mar 29, 2012