Homework Statement

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

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

$$\Phi$$ = ( $$\sqrt{x^{2} + y^{2} + z^{2}}$$ $$)^{-3}$$ = $$(x^{2} + y^{2} + z^{2})$$$$^{\frac{-3}{2}}$$

grad$$\Phi$$ = $$\frac{\partial \Phi}{\partial x}$$i + $$\frac{\partial \Phi}{\partial y}$$j + $$\frac{\partial \Phi}{\partial z}$$k

$$\frac{\partial \Phi}{\partial x}$$ $$\Rightarrow$$ let $$(x^{2} + y^{2} + z^{2})$$ = u

$$\frac{\partial \Phi}{\partial u}$$ = $$\frac{-3}{2}u^{\frac{-5}{2}}$$

$$\frac{\partial u}{\partial x}$$ = 2x

$$\frac{\partial \Phi}{\partial x}$$ = 2x($$\frac{-3}{2}u^{\frac{-5}{2}}$$)

= -3x($$(x^{2} + y^{2} + z^{2})$$$$)^{\frac{-5}{2}}$$ = -3x( $$\frac{1}{r^{5}}$$ )

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

-$$\frac{3}{r^{5}}$$ (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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cristo
Staff Emeritus
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 $$\nabla\left(\frac{1}{r^3}\right)=\hat{\bold{r}}\frac{\partial}{\partial r}(r^{-3})=\hat{\bold{r}}\cdot\frac{-3}{r^4}$$ which is the same result.

Last edited by a moderator:
$$r= \sqrt{x^{2} + y^{2} + z^{2}}$$

you will need to rethink your unit vector r.

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.

cristo
Staff Emeritus
$$r= \sqrt{x^{2} + y^{2} + z^{2}}$$
His r isn't a unit vector, it is $\bold{r}=(x,y,z)$. The unit vector, which I denoted with a hat, is then $$\hat{\bold{r}}=\frac{\bold{r}}{r}$$