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Little help with spherical coordinates

  1. Nov 28, 2009 #1

    Oddbio

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    Gold Member

    Here is a small screenshot of something I'm reading:
    http://img262.imageshack.us/img262/3585/sphericalcoords.png [Broken]

    The first six equations are ok. (I don't think anyone actually needs the figure right? It's just general spherical coordinates). φ is the angle in the x-y plane.

    I get how the seventh equation is found, because:
    [tex]r=\sqrt{x^{2}+y^{2}+z^{2}}[/tex]
    so they simply take the derivative of that and then they have the derivative in terms of x y and z and then they simply change it to spherical coordinates using one of the first three equations in the image.

    But I cannot figure out how they get the eighth and ninth equations. (last 2).
    Because:
    [tex]\theta=cos^{-1}\left(\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)[/tex]

    and the derivative of that with respect to x (partial derivative) is:
    [tex]\frac{\partial\theta}{\partial x}=\left(\frac{xz}{\left(x^{2}+y^{2}+z^{2}\right)^{3/2}\sqrt{1-\frac{z^{2}}{x^{2}+y^{2}+z^{2}}}}\right)[/tex]
    and similarly for y and z.
    OR:
    [tex]\frac{\partial\theta}{\partial x}=\left(\frac{xz}{\sqrt{\left(x^{2}+y^{2}+z^{2}\right)^{3}-z^{2}\left(x^{2}+y^{2}+z^{2}\right)^{2}}}\right)[/tex]

    I don't think I'm doing this right, then I'd still have to solve for [itex]\frac{\partial\phi}{\partial x}[/itex] and with y and z too.


    The manner in which that document I'm reading goes over this topic leads me to believe that the method of doing this should be much simpler.
    Can anyone offer some advice please?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 28, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Oddbio! Welcome to PF! :smile:
    No, they don't get equation 7 from a derivative,

    they get it simply by solving equations 4 to 6 as ordinary linear equations

    (which you can verify by substituting equations 4 to 6 into the RHS of equation 7).

    Same for 8 and 9. :wink:

    (and in the √ in your denominator, you seem to have missed that it's √(x2 + y2)/r)
     
  4. Nov 28, 2009 #3

    Oddbio

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    Gold Member

    Oh I don't know why I didn't follow my first instinct.

    Thank you, and thanks for welcoming me to the forum.
     
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