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Math Help - Div of f(r), what is the secret?

  1. #1
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    nabla f(r), what is the secret?

    Hey all!

    I'm currently reviewing my multivariable calculus and vector analysis for re-exams in August. This is a problem that has been showing up more than once and back before I thought I got it but now I realize I can not calculate this example simply because I don't know what to do.

    The Question:

    Solve
    (nablaf(r)) * r, where r = |r| and r = xi+yi+zj.
    Clarification:
    Bold is vector
    * = dot

    Attempt at solution
    I know that nabla = (d/dx,d/dy,d/dz) and appying this to f(r) won't work (is my assumption). Thus I assume that nabla = d/dr in this case.

    so nablaf(r) = ( d/dr f(r), d/dr f(r) , d/dr f(r) ) = f'(r)(d/dr,d/dr,d/dr)

    That is pretty much as far as I get because I don't know what to do. Applying div,grad or curl to any fuction that is simply f(x,y,z) is something i know very well, but this has me baffeld.

    I would appreciate any help that I can get on this.
    Last edited by Hanga; July 1st 2011 at 06:01 AM.
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  2. #2
    A Plied Mathematician
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    Re: Div of f(r), what is the secret?

    Are you sure that it's div? Because divergence is a scalar quantity, and therefore cannot be dotted into the vector r. Are you sure it isn't the gradient of the scalar function f?
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  3. #3
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    Re: Div of f(r), what is the secret?

    Yes, you are correct. It's supposed to be (nabla f(r)) * r
    This question in first part of vector differenctial identitiy nabla * (f(r)r = (nablaf(r)) * r + f(r)(div r)
    I'll edit my main question so that it is correct. I still need help though
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  4. #4
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    Re: Div of f(r), what is the secret?

    Right. So I'm assuming it's not the dot product in the expression

    \nabla f(r)\cdot \mathbf{r}

    that's the problem, but the gradient. Right? Here's the trick: view r=|\mathbf{r}| as a function of the three coordinates thus:

    r=\sqrt{x^{2}+y^{2}+z^{2}}.

    That is, you'll need to use the chain rule on, for example, the x component of the gradient thus:

    \frac{\partial}{\partial x}\,f\!\left(\sqrt{x^{2}+y^{2}+z^{2}}\right).

    Can you continue?
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  5. #5
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    Re: Div of f(r), what is the secret?

    Indeed I can!

    df/dr dr/dx = f'(r) * x/r in the x component and same for the rest. In the end I got r * r which is r^2 and the answer becomer f'(r)r.
    Thanks that really helped me out!
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