Scalar potential of F = e^yz(1, xz, xy) when curl(F) ≠ 0

[Adyssa]

Homework Statement

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.

Homework Equations

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

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)

|    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?

 

[Dick]

Homework Statement

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.

Homework Equations

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

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)

|    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?

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'.

 

[SammyS]

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

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.

 

[Adyssa]

Facepalm! Yep, that will do it, thanks for the tip.