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Grad operation

  • #1
[SOLVED] grad operation

Homework Statement



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.

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?
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,107
72
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.
 
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  • #3
4,239
1
[tex]r= \sqrt{x^{2} + y^{2} + z^{2}}[/tex]

you will need to rethink your unit vector r.
 
  • #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.
 
  • #5
cristo
Staff Emeritus
Science Advisor
8,107
72
[tex]r= \sqrt{x^{2} + y^{2} + z^{2}}[/tex]

you will need to rethink your unit vector r.
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]
 
  • #6
4,239
1
Thanks, Cristo. my bad. I didn't read far enough.
 

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