Thread: Div of f(r), what is the secret?

Hey all!

I'm currently reviewing my multivariable calculus and vector analysis for re-exams in August. This is a problem that has been showing up more than once and back before I thought I got it but now I realize I can not calculate this example simply because I don't know what to do.

The Question:

Solve
(nablaf(r)) * r, where r = |r| and r = xi+yi+zj.
Clarification:
Bold is vector
* = dot

Attempt at solution
I know that nabla = (d/dx,d/dy,d/dz) and appying this to f(r) won't work (is my assumption). Thus I assume that nabla = d/dr in this case.

so nablaf(r) = ( d/dr f(r), d/dr f(r) , d/dr f(r) ) = f'(r)(d/dr,d/dr,d/dr)

That is pretty much as far as I get because I don't know what to do. Applying div,grad or curl to any fuction that is simply f(x,y,z) is something i know very well, but this has me baffeld.

I would appreciate any help that I can get on this.

Are you sure that it's div? Because divergence is a scalar quantity, and therefore cannot be dotted into the vector r. Are you sure it isn't the gradient of the scalar function f?

Yes, you are correct. It's supposed to be (nabla f(r)) * r
This question in first part of vector differenctial identitiy nabla * (f(r)r = (nablaf(r)) * r + f(r)(div r)
I'll edit my main question so that it is correct. I still need help though

Right. So I'm assuming it's not the dot product in the expression



that's the problem, but the gradient. Right? Here's the trick: view as a function of the three coordinates thus:



That is, you'll need to use the chain rule on, for example, the x component of the gradient thus:



Can you continue?

Indeed I can!

df/dr dr/dx = f'(r) * x/r in the x component and same for the rest. In the end I got r * r which is r^2 and the answer becomer f'(r)r.
Thanks that really helped me out!