- #26

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^{k}(3+k/r (x

^{2}+y

^{2}+z

^{2}))

And x

^{2}+y

^{2}+z

^{2}is (r

^{k})

^{2/k}

So I can probably do something with that... Yep.

- Thread starter whatisreality
- Start date

- #26

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And x

So I can probably do something with that... Yep.

- #27

BiGyElLoWhAt

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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.

- #28

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r^{k}(3+kr)!

- #29

BiGyElLoWhAt

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

- #30

BiGyElLoWhAt

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- #31

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The derivatives are right according to wolfram. So that's one thing ruled out.

- #32

BiGyElLoWhAt

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##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

- #33

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I used the chain rule on (x

##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

- #34

BiGyElLoWhAt

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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.

- #35

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- #36

BiGyElLoWhAt

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

- #37

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

- #38

BiGyElLoWhAt

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XD

- #39

BiGyElLoWhAt

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

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