Difficulty with differentiation in polar coordinates I am trying to understand a simple calculation in polar coordinates - and I am getting totally discombobulated (the original source can be found: here).Please have a look at the following picture:

My questions(1) I don't understand how the length labeled $dr \over d \theta$ can possibly be the change of $r$ with respect to a change in $\theta$. (2) It is then said that $tan \alpha = {r \over {dr \over d \theta}}$. But how can you calculate the tan-function if you don't even have a right angled triangle - and anyway, the formular would suggest that you calculate the tan of the angle that is not labeled in the pic (because $tan={opposite \over adjecent}$)
Sorry, if this is too elementary - but it really bothers me right now... Thank you!
 A: Let me re-draw the picture

Instead of $\theta$ I wrote $s$. We are looking at a sweep from $r\to r′$ over an angle of $ds$. The arc formed by sweeping $r$ with no radial change has length $r~ds$. The change of radius is $r' - r = dr$. As you can see, when $ds$ is really small, the region formed by $dr$, the arc $r~ds$, the the un-labeled segment connecting the tip of $r$ to that of $r′$ approximates a right triangle
For the angle $b$ (which is your $\alpha$), the opposite side is the arc, of length $r~ds$, and the adjacent side is the segment $dr$. So you have $\frac1r \frac{dr}{ds}=\frac{1}{\tan b}$. 
A: (Willie beat me by a few seconds, but this is what I had written.)
No wonder you're confused. That's a horrible picture! You should draw an arc of a circle ($\approx$ a line) at right angles to the side labeled $r_0$; this will have length $r d\theta$ and split the line labeled $r$  into two pieces of lenghts $r_0$ and $dr$ (approximately, for small $d\theta$). Then you have a small right triangle such that $\tan\alpha \approx (r d\theta)/dr$.
