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Thread: Potential Function...

  1. #1
    s7b
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    Potential Function...

    How do you find a potential function of the field:

    F= (ysinz)i + (xsinz)j + (xycosz)k

    I'm lost on questions like this...
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  2. #2
    Flow Master
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    Quote Originally Posted by s7b View Post
    How do you find a potential function of the field:

    F= (ysinz)i + (xsinz)j + (xycosz)k

    I'm lost on questions like this...
    Let $\displaystyle F = \nabla \phi$. Then:

    $\displaystyle \frac{\partial \phi}{\partial x} = y \sin z$ .... (1)

    $\displaystyle \frac{\partial \phi}{\partial y} = x \sin z$ .... (2)

    $\displaystyle \frac{\partial \phi}{\partial z} = xy \cos z$ .... (3)

    These pde's have to be solved simultaneously.

    From (1): $\displaystyle \phi = x y \sin z + g(y, z)$ where $\displaystyle g(y, z)$ is an arbitrary function of y and z.

    Substitute this into (2): $\displaystyle x \sin z + \frac{\partial g}{\partial y} = x \sin z \Rightarrow \frac{\partial g}{\partial y} = 0 \Rightarrow g(y, z) = C$ where C is an arbitrary constant.

    So you can say that $\displaystyle \phi = x y \sin z$. Check that this function does this job.
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  3. #3
    MHF Contributor chisigma's Avatar
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    If we have a vector field described as…

    $\displaystyle \overrightarrow{f} =f_{x} (x,y,z)\cdot $$\displaystyle \overrightarrow{i} + f_{y} (x,y,z)\cdot$$\displaystyle \overrightarrow{j} + f_{z} (x,y,z)\cdot$$\displaystyle \overrightarrow{k}$

    … and it is conservative the following relation holds…

    $\displaystyle \overrightarrow{f}= \nabla {U(x,y,z)} $$\displaystyle = \frac{\partial U}{\partial x}\cdot$$\displaystyle \overrightarrow{i} $$\displaystyle +\frac{\partial U}{\partial y}\cdot $$\displaystyle \overrightarrow{j} + $$\displaystyle \frac{\partial U}{\partial z}\cdot $$\displaystyle \overrightarrow{k}$

    … where $\displaystyle U(*,*,*)$ is a scalar function which is called potential. In the case you have proposed is…

    $\displaystyle f_{x}= y\cdot \sin z$

    $\displaystyle f_{y}= x\cdot \sin z$

    $\displaystyle f_{z}= x\cdot y\cdot \cos z$

    … so that you can verify that the scalar function…

    $\displaystyle U(x,y,z)= x\cdot y\cdot \sin z$

    … satisfies the conditions to be the potential of the field given by you…

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$

    P.S. Again i didn't note that fantastic had posted before me... i apologize for that...
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