- #1
nigels
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In a problem that requires converting from cartesian to polar coordinates, I need to take [itex]\frac{dr}{dx}[/itex]. I tried doing it two different ways but getting two completely different answers..
Method 1:
[itex]r=\sqrt{x^2+y^2}[/itex]
[itex]\frac{dr}{dx}=\frac{1}{2}\frac{1}{\sqrt{x^2+y^2}}2x \;\; = \frac{1}{r}r\cos\theta \;\; = \;\; \cos\theta[/itex]
Method 2:
[itex]x=r\cos\theta[/itex]
[itex]r=\frac{x}{\cos\theta}[/itex]
[itex] \frac{dr}{dx} = \frac{1}{\cos\theta}[/itex]
I know that Method 1 gives the correct answer. But, having not done derivations in a while, what's wrong with my steps in Method 2? Thanks a lot for your help!
Method 1:
[itex]r=\sqrt{x^2+y^2}[/itex]
[itex]\frac{dr}{dx}=\frac{1}{2}\frac{1}{\sqrt{x^2+y^2}}2x \;\; = \frac{1}{r}r\cos\theta \;\; = \;\; \cos\theta[/itex]
Method 2:
[itex]x=r\cos\theta[/itex]
[itex]r=\frac{x}{\cos\theta}[/itex]
[itex] \frac{dr}{dx} = \frac{1}{\cos\theta}[/itex]
I know that Method 1 gives the correct answer. But, having not done derivations in a while, what's wrong with my steps in Method 2? Thanks a lot for your help!