Little help with spherical coordinates

1. Nov 28, 2009

Oddbio

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:
$$r=\sqrt{x^{2}+y^{2}+z^{2}}$$
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:
$$\theta=cos^{-1}\left(\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}\right)$$

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

I don't think I'm doing this right, then I'd still have to solve for $\frac{\partial\phi}{\partial x}$ 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.

Last edited by a moderator: May 4, 2017
2. Nov 28, 2009

tiny-tim

Welcome to PF!

Hi Oddbio! Welcome to PF!
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.

(and in the √ in your denominator, you seem to have missed that it's √(x2 + y2)/r)

3. Nov 28, 2009

Oddbio

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

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