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Verify partial differentiation.

  1. Oct 29, 2013 #1
    I just want to verify

    For Polar coordinates, ##r^2=x^2+y^2## and ##x=r\cos \theta##, ##y=r\sin\theta##

    ##x(r,\theta)## and## y(r,\theta)## are not independent to each other like in rectangular.

    In rectangular coordinates, ##\frac{\partial y}{\partial x}=\frac{dy}{dx}=0##

    But in Polar coordinates,
    [tex]\frac{\partial r}{\partial x}=\cos\theta,\;\frac{\partial \theta}{\partial x}=-\frac{\sin\theta}{r}[/tex]
    [tex]\frac{\partial y(r,\theta)}{\partial x(r,\theta)}=\frac{\partial y(r,\theta)}{\partial r}\frac{\partial r}{\partial x(r,\theta)}+\frac{\partial y(r,\theta)}{\partial \theta}\frac{\partial \theta}{\partial x(r,\theta)}=

    (\cos\theta) \frac{\partial y(r,\theta)}{\partial r}-\left(\frac{\sin\theta}{r}\right)\frac{\partial y(r,\theta)}{\partial \theta}[/tex]

    Thanks
     
  2. jcsd
  3. Oct 29, 2013 #2

    mfb

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    Staff: Mentor

    If you calculate the partial derivatives at the end of your last equation, you'll see that the result is zero.
     
  4. Oct 29, 2013 #3
    So ##\frac{\partial y}{\partial x}=0## for Polar coordinates.

    Then I am even more confused!!

    [tex]r^2=x^2+y^2\Rightarrow\; r\frac{\partial r}{\partial x}=x+y\frac{\partial y}{\partial x}[/tex]
    [tex]\Rightarrow\; \frac{\partial r}{\partial x}=\frac{x}{r}+\frac{y}{r}\frac{\partial y}{\partial x}[/tex]

    If ##\frac{\partial y}{\partial x}=0##, then ##\frac{\partial r}{\partial x}=\frac{x}{r}##

    Let's just use an example where ##\theta =60^o##, so to every unit change of ##x##, ##r## will change for 2 unit. So ##\frac{\partial r}{\partial x}=2##

    The same reasoning, ##r=2x##. So using this example, ##\frac{\partial r}{\partial x}=\frac{x}{r}=0.5## which does not agree with the example I gave.

    Please help, I've been stuck for over a day.

    Thanks
     
  5. Oct 29, 2013 #4

    mfb

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    Staff: Mentor

    If you fix θ, x and y get dependent, but then you are no longer in the general polar coordinates.
     
  6. Oct 29, 2013 #5
    Thanks for your patience, I really don't get this. Even if I don't fix the ##\theta##,
    [tex]\frac{\partial r}{\partial x}=\cos\theta\;\hbox { as }\;\frac{\partial y}{\partial x}=0 [/tex]

    You can now put in ##90^o\;>\;\theta\;>45^o##, you'll get ##r>x## and ##\frac{\partial r}{\partial x}>1##

    Here I am not fixing ##\theta##, I just use ##\frac{\partial y}{\partial x}=0## only.

    I think I am serious missing something.
     
    Last edited: Oct 29, 2013
  7. Oct 31, 2013 #6
    Now I am officially lost!!! I since posted this question in two different math forums, people there both asked and clarified, then it's been 12 hours with no response just like here!!!

    I have one book and at least one article derived the equations like in my first post, but obviously the example I gave does not agree with the first post and I triple checked my example. I don't think I did anything wrong.......apparently I have not get any suggestion otherwise from three forums!!! I am pretty sure I am missing something as the book I used is a text book used in San Jose State and I studied through 7 chapters and yet to find a single mistake until the question here.

    Anyone has anything to say?

    From my example,
    [tex]\frac{\partial{r}}{\partial{x}}=\frac{r}{x}=\frac{1}{\cos\theta}[/tex]
    And this answer makes a lot more sense.

    Thanks
     
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