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Conservative Vector Fields and Associated Potential

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Having issues determining what I am doing wrong, so perhaps one of you can pin point it. I have the solution, and I am extremely close to the same result, however, I am nonetheless wrong.

    Find the conservative vector fields potential.

    [tex]\vec{F}(x, y, z)=[(2xy-z^2), 2yz+x^2), y^2-2zx)][/tex]



    3. The attempt at a solution
    [tex]\vartheta=\int(2xy-z^2)dx=x^2y-z^2x+C(y, z)[/tex]
    then we have
    [tex]2yz+x^2=x^2+∂C(y, z)/∂y[/tex]
    therefore,
    [tex]C(y, z)=zy^2+C(z)[/tex]
    It's at the following step that I mess something up.
    [tex]y^2-2zx=y^2+∂C(z)/∂z[/tex]
    In the solutions however, they have,
    [tex]y^2-2zx=y^2-2zx+∂C(z)/∂z[/tex]
    However, wouldn't the "-2zx" term go with ∂C(z)/∂z?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 10, 2013 #2

    CAF123

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    Your scalar function ##\phi = x^2y - z^2x + c(y,z)## Then take the partial of this with respect to ##z## and equate with the ##z## component of ##\underline{F}##. Did you mistakenly take the partial with respect to z of something else? You can write $$\frac{\partial c(z)}{\partial z} = c'(z).$$
     
  4. Jan 10, 2013 #3

    SammyS

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    I'm betting that you're going to kick yourself for this !

    So, at this point you have [itex]\displaystyle \ \ \vartheta=x^2y-z^2x+C(y, z)=x^2y-z^2x+zy^2+C(z)\ .[/itex]

    Now look at [itex]\displaystyle \ \ ∂\vartheta/∂z\ .[/itex]
     
  5. Jan 10, 2013 #4
    Perfect! Not entirely sure how I didn't see that. Figured I should ask rather than stare at it for long. Thanks!
     
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