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Partial Derivatives

Name: Anonymous 2008-09-06 12:52

Sorry to bother again. http://i34.tinypic.com/17wq2u.jpg

Either I'm missing something about dealing with these expressions or this makes no sense whatsoever. I can upload the rest of the question if necessary; it's about transforming the expression "∂2f/∂x2 + ∂2f/∂y2" to the variables g, ρ and Φ where:

f(x,y) is the function in the Cartesian plane
g(ρ,Φ) gives the function in polar coordinates
and
x = ρ cos Φ
y = ρ sin Φ

But the thing in the pic is the only part I don't get.

Name: Anonymous 2008-09-06 13:05

The d/dΦ is an operator.

You can multiply out the brackets but remember when you "multiply" the sinΦ/p.d/dΦ term by the cos Φ . dg/dρ term the operator will kind of produce two terms because of the product rule,

so you get (sinΦ/p.d/dΦ)(cos Φ . dg/dρ)

= sin Φ cos Φ dg/dρdΦ - sin ^2/p dg / dρ


So for example the 1/ρ^2 term is coming in because the operator d/dρ is going to act on the term 1/ρ as well as the term dg/dΦ

That make it clear for you?

Name: Anonymous 2008-09-06 13:07

ps. what is this stuff for?
First year undergrad?

Name: Anonymous 2008-09-06 13:57

>>2

Oh, I see what's happening then. One thing though, here:

>>so you get (sinΦ/p.d/dΦ)(cos Φ . dg/dρ)

>>= sin Φ cos Φ dg/dρdΦ - sin ^2/p dg / dρ

Why is the d/dΦ acting only on the cos Φ and not the sin Φ as well? (ie. isn't it a triple product so shouldn't it be "triple product" ruled?

>>3

Yeah, but reading in advance of the start of term (as instructed), so I'm on my own for a while.

Name: Anonymous 2008-09-06 14:01

Oh, it doesn't act on its own term, gotcha. Thanks a lot!

Name: Anonymous 2008-09-06 17:08

>>5

Simple answer, only acts on things to the right of it.

Complex answer wouldn't help you.

Don't change these.
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