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

Calculate grad([tex]\Phi[/tex]) in terms ofrand 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 forijandk

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?

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# Grad operation

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