Question

\[w = f(x,y)\\x(r, \theta) = r \cos \theta\\y(r, \theta) = r \sin \theta\] prove that \[{\partial{w} \over \partial{x}} = {\partial{w} \over \partial{r}} \cos \theta - {\partial{w} \over \partial\theta}{\sin \theta \over r}\]

Answer 1

this is just an application of the chain rule...where are you stuck?

Answer 2

use

\[ {\partial{w} \over \partial{r}}= {\partial{w} \over \partial{x}} {\partial{x} \over \partial{r}}+ {\partial{w} \over \partial{y}} {\partial{y} \over \partial{r}}\]

\[ {\partial{w} \over \partial{\theta}}= {\partial{w} \over \partial{x}} {\partial{x} \over \partial{\theta}}+ {\partial{w} \over \partial{y}} {\partial{y} \over \partial{\theta}}\]

the follows quickly after that

Answer 3

\[{\partial{w} \over \partial{x}}\] is where I am confused. I can't figure out how to take the partial of \[w = f(x, y)\]

Answer 4

\[{\partial{w} \over \partial{r}} \cos \theta - {\partial{w} \over \partial\theta}{\sin \theta \over r}\]
\[=\left[{\partial{w} \over \partial{x}} {\partial{x} \over \partial{r}}+ {\partial{w} \over \partial{y}} {\partial{y} \over \partial{r}}\right] \cos \theta - \left[{\partial{w} \over \partial{x}} {\partial{x} \over \partial{\theta}}+ {\partial{w} \over \partial{y}} {\partial{y} \over \partial{\theta}}\right]{\sin \theta \over r}\]

\[=\cdots\]

Answer 5

I assume you can find \[\frac{\partial x}{\partial r},\frac{\partial y}{\partial r},\frac{\partial x}{\partial \theta},\frac{\partial y}{\partial \theta}\]

Answer 6

yes, the ones are easy for me

Answer 7

those ones*

Answer 8

ok...replace those in the expression I have above and simplify.

Answer 9

ok, I'm working it out

Answer 10

Okay, it all simplified down to:
\[{\partial{w} \over \partial{x}} = {\partial{w} \over \partial{x}}\]

It makes sense, just plug in the partials that I know how to find and it worked out!

Answer 11

this is what I did...

\[=\left[{\partial{w} \over \partial{x}} {\partial{x} \over \partial{r}}+ {\partial{w} \over \partial{y}} {\partial{y} \over \partial{r}}\right] \cos (\theta) - \left[{\partial{w} \over \partial{x}} {\partial{x} \over \partial{\theta}}+ {\partial{w} \over \partial{y}} {\partial{y} \over \partial{\theta}}\right]{\sin(\theta) \over r}\]

\[=\left[{\partial{w} \over \partial{x}} \cos(\theta)+ {\partial{w} \over \partial{y}} \sin(\theta)\right] \cos (\theta) - \left[{\partial{w} \over \partial{x}} (-r\sin(\theta))+ {\partial{w} \over \partial{y}} r\cos(\theta)\right]{\sin (\theta) \over r}\]

\[={\partial{w} \over \partial{x}} \cos^2(\theta)+ {\partial{w} \over \partial{y}} \sin(\theta) \cos \theta+ {\partial{w} \over \partial{x}} \sin^2(\theta))- {\partial{w} \over \partial{y}} \cos(\theta)\sin(\theta)\]

\[={\partial{w} \over \partial{x}} \cos^2(\theta)+ {\partial{w} \over \partial{x}} \sin^2(\theta)={\partial{w} \over \partial{x}}\]

Answer 12

Thanks for the help. That was only part a of the problem, I think that I can do the rest now

Answer 13

yw