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Puzzling paradox

  1. Oct 25, 2004 #1
    Hi I am having a bit of difficulty working with plane polar co-ordinates.
    We have:
    [tex] r^2 = x^2 + z^2 [/tex]
    [tex] x = r cos \theta [/tex]
    [tex] z=r sin \theta [/tex]
    I wish to find [tex] \frac {\partial r} {\partial x} [/tex]

    Using [tex] r^2 = x^2 + z^2 [/tex]
    We have:

    [tex] \frac {\partial (r^2)} {\partial x} = \frac {\partial (x^2)} {\partial x} + \frac {\partial (z^2)} {\partial x} [/tex]
    Thus [tex] 2r\frac {\partial r} {\partial x} = 2x [/tex]
    [tex] \frac {\partial r} {\partial x} = \frac {x} {r} = \frac {r cos \theta} {r} [/tex]
    Therefore [tex]\frac {\partial r} {\partial x} = cos \theta [/tex]

    But if we find [tex] \frac {\partial r} {\partial x} [/tex] using [tex] x = r cos \theta [/tex]
    We have:
    [tex] r = \frac {x} {cos \theta} [/tex]
    Therefore [tex]\frac {\partial r} {\partial x} = \frac {1} {cos \theta} [/tex]

    What is going on here? Which answer is wrong and why?
     
  2. jcsd
  3. Oct 25, 2004 #2

    Tide

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    Ooops! In your last step you [itex]\theta[/itex] is NOT a constant! You need to differentiate it wrt x as well.
     
  4. Oct 25, 2004 #3

    shmoe

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    Hi, when you differentiated with respect to x here, you treated [tex]\theta[/tex] as a constant. It's not, it's dependant on x, so you have to use the quotient rule and you'll have a [tex]\frac {\partial \theta} {\partial x}[/tex] appear via the chain rule. Use [tex]{\cos \theta}=\frac{x}{\sqrt{x^2+z^2}}[/tex] to work out [tex]\frac {\partial \theta} {\partial x}[/tex].

    eidt-Tide was quicker on the draw!
     
    Last edited: Oct 25, 2004
  5. Oct 25, 2004 #4
    I'm not sure, but I think your mistake is in the following procedure:

    [tex]r = \frac {x} {cos \theta} \rightarrow \frac {\partial r} {\partial x} = \frac {1} {cos \theta} [/tex]

    I believe that [tex] \theta [/tex] is dependent on [tex] x [/tex]. In other words [tex] \theta = \theta (x)[/tex] so you can't just treat [tex] cos \theta[/tex] as a constant when taking the derivative with respect to [tex] x [/tex].
     
  6. Oct 25, 2004 #5
    Whoops! I guess I'm pretty late on that one!

    This is the first time I used the equation formatting and it took me a while to format the post. (ha ha)

    At least, I know now that I was on the right track. :smile:
     
  7. Oct 25, 2004 #6

    Tide

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    Way to go, NS!!
     
  8. Oct 25, 2004 #7
    Thanks for your help guys!!
     
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