Derivatives of polar coordiantes? I'm a undergraduate and I was reading about the polar coordinate system specifically this paper. I don't understand the term: $$\frac{de_r}{d\theta} =  e_\theta \text{,  and } \frac{de_\theta}{d\theta} = -e_r$$
I don't see how you can have a the derivative of $de_r$ over $d\theta$ since they are not the same variable, if that makes any sence.
 A: In reference to the diagram, ask yourself what happens to the unit vector $\hat{e}_r$ as you increase the angle $\varphi$ (as measured counterclockwise from the rightward pointing horizontal axis) by an infinitesimal amount $d\varphi$. Note that they use $\varphi$ in this diagram instead of $\theta$. 
In words, the tail of the vector $\hat{e}_r$ moves counterclockwise along the circumference of the circle, with its tip always pointing radially outwards from the centre. 
The difference between the transformed $\hat{e}_r$ (call it $\hat{e}_{r1}$), and the original $\hat{e}_r$ (call it $\hat{e}_{r0}$), is another vector (call it $d\hat{e}_r$), which points from the tip of $\hat{e}_{r0}$ to the tip of $\hat{e}_{r1}.$ That is, $d\hat{e}_r=\hat{e}_{r1}-\hat{e}_{r0}.$
In the limit as $d\varphi\rightarrow 0,$ that is, as $\hat{e}_{r1}$ and $\hat{e}_{r0}$ become closer and closer together, $d\hat{e}_r$ becomes closer and closer to a vector pointing perpendicular to $\hat{e}_{r0}$, i.e. it becomes close and closer to the vector $\hat{e}_\varphi$. Draw some pictures to convince yourself of this
In other words, as $d\varphi\rightarrow 0,$ $d\hat{e}_r\rightarrow \hat{e}_\varphi,$ which is to say that $$\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi.$$

Reference for the diagram
Edit:
To address @R.W.Bird's comment. I'll try to show that the magnitude of $\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi$ indeed is 1.
The magnitude of $d\hat{e}_r$ is given by
$$ \left|{d\hat{e}_r}\right|=\left|\hat{e}_{r1}-\hat{e}_{r0}\right|=\sqrt{\hat{e}_{r1}^2-2\hat{e}_{r1}\cdot\hat{e}_{r0}+\hat{e}_{r0}^2}.$$
If we use the fact that by definition $\left|\hat{e}_{r}\right|=1$, then
$$ \left|{d\hat{e}_r}\right|=\sqrt{1-2\cos(d\varphi)+1}=\sqrt{2-2\cos(d\varphi)},$$
since the angle between $\hat{e}_{r0}$ and $\hat{e}_{r1}$ is $d\varphi$. Now,
$$ \left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\left|\frac{d}{d\varphi}\sqrt{2-2\cos(d\varphi)}\right|=\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|.$$
As $d\varphi\rightarrow 0,$
$$\lim_{d\varphi\rightarrow 0}\left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\lim_{d\varphi\rightarrow 0}\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$
$$=\lim_{d\varphi\rightarrow 0}\left|\frac{d}{d\varphi}\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$
$$=\lim_{d\varphi\rightarrow 0}\left|\frac{\cos (d\varphi)}{\sqrt{2-2 \cos (d\varphi)}}-\frac{\sin ^{2}(d\varphi)}{(2-2 \cos (d\varphi))^{3 / 2}}\right|$$
$$=1$$
where from the first to second line we have used l'Hopital's rule. This argument is a little hand-wavy, but it gets us the right answer... sorry about that.
A: $\hat{e}_r$ is a radial unit vector. It obviously points in different directions for different values of $\theta$. So it has a derivative expressing that rate of change.
