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Math Help - Find the potential function f for the field F

  1. #1
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    Find the potential function f for the field F

    F=\frac{1}{z}i-6j-\frac{x}{z^{2}}k

    We're assuming that it has already been shown that the field is conservative.

    I don't think I'm quite understanding how to do this.

    So far, I have:

    \frac{\partial f}{\partial x}=\frac{1}{z}

    \frac{\partial f}{\partial y}=-6

    \frac{\partial f}{\partial z}=-\frac{x}{x^{2}}

    So, then I take the first one and integrate it with respect to x and get:

    f(x,y,z)=\frac{x}{z}+g(y,z)

    Then I find:

    \frac{\partial f}{\partial y}=\frac{x}{z}+g(y,z)

    =>0+\frac{\partial g}{\partial y}

    and set this equal to \frac{\partial f}{\partial y} from above:

    \frac{\partial g}{\partial y}=-6

    So I take the integral of this and get:

    -6y+h(z)

    I'm not sure of what to do now.

    The final answer to the problem is \frac{x}{z}-6y+C
    Last edited by downthesun01; December 6th 2010 at 12:44 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    We have:

    f(x,y,z)=\dfrac{x}{z}-6y+h(z)

    Now:

    \dfrac{{\partial f}}{{\partial x}}=\dfrac{1}{z},\quad\forall\;h(z)=C (constant function).

    Regards.

    Fernando Revilla
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  3. #3
    MHF Contributor

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    Quote Originally Posted by downthesun01 View Post
    F=\frac{1}{z}i-6j-\frac{x}{z^{2}}k

    We're assuming that it has already been shown that the field is conservative.

    I don't think I'm quite understanding how to do this.

    So far, I have:

    \frac{\partial f}{\partial x}=\frac{1}{z}

    \frac{\partial f}{\partial y}=-6

    \frac{\partial f}{\partial z}=-\frac{x}{x^{2}}

    So, then I take the first one and integrate it with respect to x and get:

    f(x,y,z)=\frac{x}{z}+g(y,z)

    Then I find:

    \frac{\partial f}{\partial y}=\frac{x}{z}+g(y,z)

    =>0+\frac{\partial g}{\partial y}

    and set this equal to \frac{\partial f}{\partial y} from above:

    \frac{\partial g}{\partial y}=-6

    So I take the integral of this and get:

    -6y+h(z)

    I'm not sure of what to do now.
    Excellent! Now just continue what you were doing: since g(y,z)= -6y+ h(z), f(x,y,z)= \frac{x}{z}+ g(y,z)= \frac{x}{z}- 6y+ h(z) so that by differentiating with respect to z,
    \frac{\partial f}{\partial z}= -\frac{x}{z^2}+ h'(z)
    but we know that \frac{\partial f}{\partial z}= -\frac{x}{z^2}
    so we have -\frac{x}{z^2}+ h'(z)= -\frac{x}{z^2} and h'(z)= 0. Since h is a function of the single variable, z, h'(z)= 0 means that h(z) is, in fact, a constant, C.

    f(x, y, z)= \frac{x}{z}- 6y+ h(z)= \frac{x}{z}- 6y+ C

    The final answer to the problem is \frac{x}{z}-6y+C
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