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 May 5th, 2012, 05:09 PM #1 Senior Member   Joined: Sep 2008 Posts: 105 Thanks: 0 Partials in Different Coordinates Let $x=r cos(\theta), y = r sin(\theta),$ and $G(r, \theta)= F (x,y)$ Show that $\frac{\partial G}{\partial r}= \frac{\partial F}{\partial x} cos (\theta) + \frac{\partial F}{\partial y} sin (\theta)$ and $\frac{\partial G}{\partial \theta}= -\frac{\partial F}{\partial x} r sin(\theta) + \frac{\partial F}{\partial y} r cos(\theta)$
 May 6th, 2012, 02:26 AM #2 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: Partials in Different Coordinates Chain rule in two variables Since you even ask, I suppose it is stated in your book Differentiating G with respect to r is the same as differentiating F with respect to r, where you must use the chain rule If this didn't help, just let me know and I will give you the solution
 May 6th, 2012, 08:55 AM #3 Senior Member   Joined: Sep 2008 Posts: 105 Thanks: 0 Re: Partials in Different Coordinates Thanks for answering. The section in my book is very small. I am getting confused because r depends on x, y and theta. I was trying to draw a schema to help, but I can't figure out how to do it.
 May 6th, 2012, 11:47 AM #4 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: Partials in Different Coordinates You are actually just interested in how x and y depends on r, and those relationships you have already statet above df/dr = (df/dx)(dx/dr) + (df/dy)/(dy/dr) where I don't know how to make partial d:s
 May 6th, 2012, 11:59 AM #5 Senior Member   Joined: Sep 2008 Posts: 105 Thanks: 0 Re: Partials in Different Coordinates So then is dG/dr = dF/dr? Why?
 May 6th, 2012, 12:24 PM #6 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: Partials in Different Coordinates Due to equality. Perform the same operation on two equal expressions and you must get the same result An example: 3+2=5 (3+2)/10 = 5/10 The equality still holds Differentiating both sides of an equation with respect to the same variable is no different really :P
 May 6th, 2012, 01:13 PM #7 Senior Member   Joined: Sep 2008 Posts: 105 Thanks: 0 Re: Partials in Different Coordinates I guess I didn't expect it to be that simple. Thanks for your help!
 May 6th, 2012, 01:22 PM #8 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: Partials in Different Coordinates Haha, sometimes it's just too easy to see it Happy to help

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