Higher order differential equations and the chain rule (2 variables)

simba_

1. The problem statement, all variables and given/known data

The function F is defined by F (r, θ) = f (x(r, θ), y(r, θ)), where f is twice continuously
differentiable and
x(r, θ) = r cos θ, y(r, θ) = r sin θ.
Use the chain rule to find

d²F/dθ²


2. Relevant equations



3. The attempt at a solution

I know that dF/dθ = (df/dx)(dx/dθ) + (df/dy)(dy/dθ)
and i can solve this

I know the next step is (d/dθ)(dF/dθ) but this is were i get lost. I must of missed the class this was explained in :(

HallsofIvy

[tex]\frac{\partial F}{\partial \theta}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}[/tex]

Since [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex], [itex]x_\theta= -r sin(\theta)[/itex] and [itex]y_\theta= r cos(\theta)[/itex]

So
[tex]\frac{\partial F}{\partial \theta}= - r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta) \frac{\partial f}{\partial y}[/tex]
which, I presume, is what you got.

Now,
[tex]\frac{\partial^2 F}{\partial \theta^2}= \frac{\partial}{\partial \theta}\left(\frac{\partial F}{\partial \theta}\right)[/tex]

But the formula
[tex]\frac{\partial \phi}{\partial \theta}= \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial \theta}[/tex]
is true for any function, [itex]\phi[/itex] and, in particular, for
[tex]\phi= \frac{\partial F}{\partial \theta}= - r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta) \frac{\partial f}{\partial y}[/tex]


That is,
[tex]\frac{\partial^2 F}{\partial \theta}= \frac{\partial}{\partial \theta}\left(-r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta)\frac{\partial f}{\partial y}\right)[/tex]
[tex]+ \frac{\partial}{\partial\theta}\left((-r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta)\frac{\partial f}{\partial y}\right)[/tex]
[tex]= - r cos(\theta)\frac{\partial f}{\partial x}- r sin(\theta)\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)[/tex]
[tex]- r sin(\theta)\frac{\partial f}{\partial y}+ r cos(\theta)\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)[/tex]

This is getting awkward to write as a single formula so just note that you do
[tex]\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)[/tex]
and
[tex]\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)[/tex]
using that same "chain rule formula":
[tex]\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)[/tex]
[tex]= -r sin(\theta)\frac{\partial^2 f}{\partial x^2}+ r cos(\theta)\frac{\partial^2 f}{\partial x\partial y}[/tex]
and
[tex]\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)[/tex]
[tex]= r cos(\theta)\frac{\partial f}{\partial x\partial y}- r sin(\theta)\frac{\partial^2 f}{\partial y^2}[/tex]

simba_

Thanks for that, I know it must of taken a while to write out.
I had a good look at it there now but some parts still dont make sense to me. I'll have another look at it tomorrow with fresh eyes :)