# Divergence of vector field: Del operator/nabla

rk(3+k/r (x2+y2+z2))
And x2+y2+z2 is (rk)2/k
So I can probably do something with that... Yep.

BiGyElLoWhAt
Gold Member
Oh no -.-
Somewhere we lost a degree of r... Those careless mistakes. Let me finish a few things up here at work and I'll try to find the mistake.

rk(3+kr)!

BiGyElLoWhAt
Gold Member
What you have is consistent, though. So job well done on that part.

BiGyElLoWhAt
Gold Member
Yes that is consistent with what we've said so far, but according to the problem statement and wolfram the answer is (k+3)r^k, not (kr + 3) ...

It's very close to it though! I'm pretty happy, I understand that so much better.
The derivatives are right according to wolfram. So that's one thing ruled out.

BiGyElLoWhAt
Gold Member
Aha! It is in the derivative (rather the substitution). See if you can find it.
So what I actually had was
##x\frac{k}{2}u^{\frac{k}{2}-1}*2x= x^2*u^{\frac{k}{2}-1}##
Subbing back in for r (r = u^1/2) that should be ##x^2r^{k-2}## rather than k-1

Aha! It is in the derivative (rather the substitution). See if you can find it.
So what I actually had was
##x\frac{k}{2}u^{\frac{k}{2}-1}*2x= x^2*u^{\frac{k}{2}-1}##
Subbing back in for r (r = u^1/2) that should be ##x^2r^{k-2}## rather than k-1
I used the chain rule on (x2+y2+z2)0.5k though, and taking this approach I don't see where a k-2 comes from...

BiGyElLoWhAt
Gold Member
the chain rule is actually being applied twice. If you see my chain rule that I posted earlier, you ultimitely need to take the derivative with respect to u. that's where that expression comes from. So what you have there is fine, but that's u^k/2, not r. You need to substitute that back in. So you have ##\frac{k}{2}u^{0.5k - 1}## with me so far?
this is equal to ##\frac{k}{2}\frac{u^{.5k}}{u}## you need to use the fact that u = r^2 (because r has the square root in it)
##u^.5k = r^k## and ##1/u = r^{-2}## so all together it's ##\frac{k}{2}r^{k-2}## I might have lost some constants along the way, I'm kinda in a hurry atm, walking out the door at work, but that's where the 2 comes in, and that's what gets rid of the r in the lambda.

Oh, got it! And that leads to the 3+k. Phew! What a question. I really, really appreciate all your time and help, thank you :)

BiGyElLoWhAt
Gold Member
Hey not a problem, that's what we're here for.

who's serving things up on a silver platter now? BiGyElLoWhAt
Gold Member
Eh, we got to "the answer", it was just wrong. Substitution errors and whatnot haha.