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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


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    Staff Emeritus
    Science Advisor

    Looks fine to me.

    Another way to go about this problem is to use the formula for grad in spherical polar coordinates (see http://www.mas.ncl.ac.uk/~nas13/mas2104/handout5.pdf [Broken] .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.
    Last edited by a moderator: May 3, 2017
  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


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    Staff Emeritus
    Science Advisor

    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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