1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Mar 29, 2012 #1
    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. jcsd
  3. Mar 29, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    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'.
  4. Mar 29, 2012 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    In other words,
    [itex]\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)\,.[/itex]
  5. Mar 29, 2012 #4
    Facepalm! Yep, that will do it, thanks for the tip.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook