What does $d \theta$ mean? i'm doing elementary differential geometry homework and i came across a notation that i cannot understand. I need to show that if 
$\gamma(t) = (x(t), y(t))$ then if we put $\gamma$ into polar coordinates $x= r \cos \theta$;  $y=r \sin \theta$ we get:
If
$$x^2 + y^2  = x \frac{dy}{d \alpha} - y \frac{dx}{d \alpha}$$ then $$r^2(1- \frac{ d \theta}{d \alpha})=0.$$ What puzzles me is the meaning of $d \theta$. What does that quantity represent? Is it the deriivative $$\frac{\partial \arctan (\frac{y(t)}{x(t)})}{\partial t}?$$
 A: A literal answer to your question is that 
$$d \theta = \frac{x dy - y dx}{x^2+y^2}.$$
However, this doesn't answer why this is called $d \theta$. There is no global continuous function $\theta: \mathbb{R}^2 \setminus \{ (0,0) \} \to \mathbb{R}$ which sends each point to its angle. For example, $(x,y) \mapsto \tan^{-1}(y/x)$ is undefined when $x=0$ and, if we use the standard convention that $-\pi/2 < \tan^{-1}( \ )< \pi/2$, only gives the right answer when $x>0$. However, on open sets of $\mathbb{R}^2 \setminus \{ (0,0) \}$, such functions can be defined. For example, $\tan^{-1}(y/x)$ works on $\{ x>0 \}$, or $2 \tan^{-1} \left( \frac{y}{x+\sqrt{x^2+y^2}} \right)$ works everywhere except on the negative real axis. For any one of these functions $\theta$, we have $d \theta$ given by the above formula.
To see why this should happen, let $U$ be a connected open set and let $\theta_1
$ and $\theta_2$ be two angle functions on $U$. Then $\theta_1 - \theta_2$ must be an integer multiple of $2 \pi$ and, since $U$ is connected, it must always be the same integer multiple of $2 \pi$. So $\theta_1 - \theta_2$ is a constant and $d \theta_1 - d \theta_2=0$.
A: $d\theta$ is a differential element, just like $dr, dx, dy$ and $dt$.
$d\theta$ cannot equal the partial derivative you wrote down. But a first-order full or partial derivative can be viewed (loosely) as the ratio of two differential elements.
This is basically an application of product rule followed by a bit of algebra to get things to cancel out. It's really quite simple. I can post again to edit this and guide you through it step-by-step if you are unable to do the manipulation.
