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Finding a potential function pt 2

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data

    find the potential function f for field F

    F = (y + z)i + (x + z)j + (x + y)k

    here is the answer

    Screenshot2012-05-30at35834AM.png

    3. The attempt at a solution

    From the videos I've seen it appears that finding the potential function is rather easy, just take the antiderivative of each equation, the first with respect to x, the next with respect to y, and the third with respect to z, then if there are duplicate terms, use only one of them once. Using that technique I get

    (y + z)x + (x+z)y + (x+y)z + C

    Clearly something else is going on that I'm not aware of.
     
  2. jcsd
  3. May 30, 2012 #2

    SammyS

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    They write thier answer in a rather curious way.

    Why not f(x, y, z) = xy + yz + zx ?
    For your solution of f(x, y, z) = (y + z)x + (x+z)y + (x+y)z + C , try taking ∂f/∂x , ∂f/∂y , ∂f/∂z and see that it doesn't work.
     
  4. May 30, 2012 #3
    Ok, I get it now, my way had several duplicate terms. And the book just uses an odd way to write xz + yx + zy
     
  5. May 30, 2012 #4

    SammyS

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    Yes, but if you're careful your way can work just fine.

    [itex]\displaystyle \frac{\partial f}{\partial x}=y+z\quad\to\quad f(x,\,y,\,z)=xy+xz+\text{Some term not containing }x\text{ but including the constant, }C[/itex]

    [itex]\displaystyle \frac{\partial f}{\partial y}=x+z\quad\to\quad f(x,\,y,\,z)=xy+yz+\text{Some term not containing }y\text{ but including the constant, }C[/itex]

    [itex]\displaystyle \frac{\partial f}{\partial z}=y+z\quad\to\quad f(x,\,y,\,z)=xz+yz+\text{Some term not containing }z\text{ but including the constant, }C[/itex]

    So by inspection we have that [itex]f(x,\,y,\,z)=xy+xz+yz+C[/itex]

    Try this on your previous problem.
     
    Last edited: May 30, 2012
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