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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

    Dick

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    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

    SammyS

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    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]
    Etc.
     
  5. Mar 29, 2012 #4
    Facepalm! Yep, that will do it, thanks for the tip.
     
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