Calculate grad(\Phi) in r & r

[ineedmunchies]

[SOLVED] grad operation

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

Giving a final answer of

-\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?

 

[cristo]

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 .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.

 

[Phrak]

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

you will need to rethink your unit vector r.

 

[ineedmunchies]

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]

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}

 

[Phrak]

Thanks, Cristo. my bad. I didn't read far enough.