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Grad operation

  1. May 13, 2008 #1
    [SOLVED] grad operation

    1. The problem statement, all variables and given/known data

    Calculate grad([tex]\Phi[/tex]) in terms of r and r, where [tex]\Phi[/tex]=[tex]\frac{1}{r^{3}}[/tex]

    r= xi + yj + zk
    r = magnitude (r)

    The r cubed term is scalar by the way, it comes out looking bold for some reason.

    3. The attempt at a solution

    [tex]\Phi[/tex] = ( [tex]\sqrt{x^{2} + y^{2} + z^{2}}[/tex] [tex])^{-3}[/tex] = [tex](x^{2} + y^{2} + z^{2})[/tex][tex]^{\frac{-3}{2}}[/tex]

    grad[tex]\Phi[/tex] = [tex]\frac{\partial \Phi}{\partial x}[/tex]i + [tex]\frac{\partial \Phi}{\partial y}[/tex]j + [tex]\frac{\partial \Phi}{\partial z}[/tex]k

    [tex]\frac{\partial \Phi}{\partial x}[/tex] [tex]\Rightarrow[/tex] let [tex](x^{2} + y^{2} + z^{2})[/tex] = u

    [tex]\frac{\partial \Phi}{\partial u}[/tex] = [tex]\frac{-3}{2}u^{\frac{-5}{2}}[/tex]

    [tex]\frac{\partial u}{\partial x}[/tex] = 2x

    [tex]\frac{\partial \Phi}{\partial x}[/tex] = 2x([tex]\frac{-3}{2}u^{\frac{-5}{2}}[/tex])

    = -3x([tex](x^{2} + y^{2} + z^{2})[/tex][tex])^{\frac{-5}{2}}[/tex] = -3x( [tex]\frac{1}{r^{5}}[/tex] )

    differentials with respect to y and z will be identical except swapping -3x for -3y and -3z.

    These will be the terms for i j and k

    Giving a final answer of

    -[tex]\frac{3}{r^{5}}[/tex] (r).


    EDIT (again the r^5 term is the magnitude not the vector)

    Is this the correct solution? can anybody help?
     
  2. jcsd
  3. May 13, 2008 #2

    cristo

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    Looks fine to me.

    Another way to go about this problem is to use the formula for grad in spherical polar coordinates (see this .pdf for example). In this case, it will reduce to [tex]\nabla\left(\frac{1}{r^3}\right)=\hat{\bold{r}}\frac{\partial}{\partial r}(r^{-3})=\hat{\bold{r}}\cdot\frac{-3}{r^4}[/tex] which is the same result.
     
  4. May 13, 2008 #3
    [tex]r= \sqrt{x^{2} + y^{2} + z^{2}}[/tex]

    you will need to rethink your unit vector r.
     
  5. May 13, 2008 #4
    Thank you very much

    EDIT: what do you mean phrak? i used the equation you give for the magnitude of r, and it is not a unit vector in my question.
     
  6. May 13, 2008 #5

    cristo

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    His r isn't a unit vector, it is [itex]\bold{r}=(x,y,z)[/itex]. The unit vector, which I denoted with a hat, is then [tex]\hat{\bold{r}}=\frac{\bold{r}}{r}[/tex]
     
  7. May 13, 2008 #6
    Thanks, Cristo. my bad. I didn't read far enough.
     
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