Divergence of vector field: Del operator/nabla

[whatisreality]

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]

Aha! It is in the derivative (rather the substitution). See if you can find it.

Spoiler: No peeking until you've tried

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

 

[whatisreality]

Aha! It is in the derivative (rather the substitution). See if you can find it.

Spoiler: No peeking until you've tried

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]

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.

 

[whatisreality]

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]

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

 

[fourier jr]

who's serving things up on a silver platter now?

 

[BiGyElLoWhAt]

XD

 

[BiGyElLoWhAt]

Eh, we got to "the answer", it was just wrong. Substitution errors and whatnot haha.