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Math Help - Curl and Divergence

  1. #1
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    Curl and Divergence

    F(x,y,z) = \frac{x}{x^2+y^2+z^2}i + \frac{y}{x^2+y^2+z^2}j+\frac{z}{x^2+y^2+z^2}k

    (a) Find curl
    (b) Find the divergence
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by purplerain View Post
    F(x,y,z) = \frac{x}{x^2+y^2+z^2}i + \frac{y}{x^2+y^2+z^2}j+\frac{z}{x^2+y^2+z^2}k

    (a) Find curl
    (b) Find the divergence
    Apply the definitions!

    (a) \nabla\times\mathbf{F}=\begin{vmatrix}\mathbf{i} & \mathbf{j}&\mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \frac{x}{x^2+y^2+z^2} & \frac{y}{x^2+y^2+z^2} & \frac{z}{x^2+y^2+z^2}\end{vmatrix}=\dots

    (b) \nabla\cdot\mathbf{F}=\frac{\partial}{\partial x}\left[\frac{x}{x^2+y^2+z^2}\right]+\frac{\partial}{\partial y}\left[\frac{y}{x^2+y^2+z^2}\right]+\frac{\partial}{\partial z}\left[\frac{z}{x^2+y^2+z^2}\right]=\dots

    Can you finish both of these problems?
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  3. #3
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    To find the derivative we will use the quotient rule correct? So,

    <br />
\frac{\partial}{\partial x}= (x^2+y^2+z^2)-2x^2\\\\<br />
\frac{\partial}{\partial y}= (x^2+y^2+z^2)-2y^2\\\\<br />
\frac{\partial}{\partial x}= (x^2+y^2+z^2)-2z^2\\\\\<br />

    Then i take the derivative and do the cross product against the original functions.
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  4. #4
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    Ok, i figured out the curl, know i need help with finding the divergence. When i used the formula to calculate the divergence, this is what i did.

    <br />
\frac{x^2+y^2+z^2-2x^2}{(x^2+y^2+z^2)^2}  + \frac{x^2+y^2+z^2)-(2y^2)}{(x^2+y^2+z^2)^2} +\frac{(x^2+y^2+z^2-2z^2)}{(x^2+y^2+z^2)^2}<br /> <br />

    Is this correct using the quotient rule? I then simplified the answer but it was still counted as wrong.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by purplerain View Post
    Ok, i figured out the curl, know i need help with finding the divergence. When i used the formula to calculate the divergence, this is what i did.

    <br />
\frac{x^2+y^2+z^2-2x^2}{(x^2+y^2+z^2)^2} + \frac{x^2+y^2+z^2)-(2y^2)}{(x^2+y^2+z^2)^2} +\frac{(x^2+y^2+z^2-2z^2)}{(x^2+y^2+z^2)^2}<br /> <br />

    Is this correct using the quotient rule? I then simplified the answer but it was still counted as wrong.
    The work is correct. Note that

    \frac{x^2+y^2+z^2-2x^2}{(x^2+y^2+z^2)^2} + \frac{x^2+y^2+z^2-2y^2}{(x^2+y^2+z^2)^2} +\frac{x^2+y^2+z^2-2z^2}{(x^2+y^2+z^2)^2} =\frac{3(x^2+y^2+z^2)-2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2}=\frac{x^2+y^2+z^2  }{(x^2+y^2+z^2)^2}=\frac{1}{x^2+y^2+z^2}.

    Does this make sense?
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  6. #6
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    Ah, messed up on my algebra, i see what you did, thanks!
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