# Calculate grad(\Phi) in r & r

• ineedmunchies
In summary, the grad operation is used to calculate r in terms of xi, yj, and zk. r= magnitude (r) and the r^5 term is the magnitude not the vector.f

## 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?

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.

$$r= \sqrt{x^{2} + y^{2} + z^{2}}$$

you will need to rethink your unit vector r.

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 $$\hat{\bold{r}}=\frac{\bold{r}}{r}$$