Why Does the Derivative Transformation Use Cosine in Spherical Coordinates?

[dingo_d]

Homework Statement

There is one thing I'm not getting in this problem. It's about uniformly magnetized sphere. In solving the problem with formula for magnetic potential:

\phi_M=-\nabla\cdot\int\frac{\vec{M}}{|\vec{r}-\vec{r}'|}d^3r'

At one point a change of variable for differentiation is made

\frac{\partial}{\partial z}=\frac{\partial r}{\partial z}\frac{\partial}{\partial r}

And he says that \frac{\partial r}{\partial z}=\cos\theta. But I can't see that. If I'm using the formula for spherical coordinate system transformation: z=r\cos\theta I don't get just cosine, I get 1/cosine :\

So what formula is he using?

Thanks

 

[rafaelpol]

He is using r = \sqrt {x^2 + y^2+z^2}, and then
<br /> \frac{\partial r}{\partial z} = \frac{z}{r} = cos{(\theta)}<br />

 

[fzero]

Since r = \sqrt{ x^2 + y^2 + z^2},

\frac{\partial r }{\partial z} = \frac{z}{r},

which leads to the formula from the text. I think you're making a mistake because the partial derivatives in spherical and Cartesian coordinates are related by a 3x3 matrix, so it's not enough to merely compare reciprocals.

 

[dingo_d]

oh! I see, thanks :D