1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taking partial derivative in polar coordinates

  1. May 11, 2014 #1
    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!!!
     
  2. jcsd
  3. May 11, 2014 #2
    I claim that theta is a function of x and y, can you find what theta is then?
     
    Last edited: May 11, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Taking partial derivative in polar coordinates
Loading...