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May 5th, 2012, 06:09 PM   #1
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Partials in Different Coordinates

Let and

Show that

and

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May 6th, 2012, 03:26 AM   #2
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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
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May 6th, 2012, 09:55 AM   #3
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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.
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May 6th, 2012, 12:47 PM   #4
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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
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May 6th, 2012, 12:59 PM   #5
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Re: Partials in Different Coordinates

So then is dG/dr = dF/dr? Why?
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May 6th, 2012, 01:24 PM   #6
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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
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May 6th, 2012, 02:13 PM   #7
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Re: Partials in Different Coordinates

I guess I didn't expect it to be that simple. Thanks for your help!
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May 6th, 2012, 02:22 PM   #8
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Re: Partials in Different Coordinates

Haha, sometimes it's just too easy to see it Happy to help
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