Q:

Radius and angular derivatives expressed in Cartesian coordinates

Given a point with Cartesian coordinates $(x,y)$ and with Cartesian velocity $(\dot{x},\dot{y})$, I would like to express its radius $r$, its angle $\theta$, its radius velocity $\dot{r}$, and its angular velocity $\dot{\theta}$.
First, I know that $x = r \cos\theta$ and $y = r \sin\theta$, such that $r = \sqrt{x^2+y^2}$ and $\theta = \pm \arccos (x/r)$.
Second, I know that:
\begin{align}
\dot{x} = - r \dot\theta \sin\theta + \dot{r} \cos\theta & \quad (1)\\
\dot{y} = r \dot\theta \cos\theta + \dot{r} \sin\theta & \quad (2)
\end{align}
I should be able do derive an expression of $\dot{r}$ and $\dot\theta$ from these two equations.

Here is my first attempt, using the fact that $\cos^2\theta + \sin^2\theta = 1$:
\begin{align}
\dot{r}^2 = \dot{x}^2 + \dot{y}^2 + (r \dot\theta)^2 + 2 r \dot\theta (\dot{x} \sin\theta - \dot{y} \cos\theta) & \quad (3)\\
(r \dot\theta)^2 = \dot{x}^2 + \dot{y}^2 + \dot{r}^2 - 2 \dot{r} (\dot{x} \cos\theta + \dot{y} \sin\theta) & \quad (4)
\end{align}
Now, substituting (4) in (3), I get an equation of $\dot{r}$ where $\dot\theta$ does not appear, but which is quite intricate:
$$\dot{x}^2 + \dot{y}^2 + \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{r}^2 - 2 \dot{r} (\dot{x} \cos\theta + \dot{y} \sin\theta)} (\dot{x} \sin\theta - \dot{y} \cos\theta) - \dot{r} (\dot{x} \cos\theta + \dot{y} \sin\theta) = 0$$
I have the feeling that I am doing the right thing, but that I am missing some intuitive shortcut to properly isolate $\dot{r}$.

A:

Following the method provided by Rodney Dunning in another answer:

Taking the time-derivative of $r^2 = x^2 + y^2$, we get
$$2 r \dot{r} = 2 x \dot{x} + 2 y \dot{y} \text{,}$$
and we may write
$$\boxed{\dot{r} = \frac{x \dot{x} + y \dot{y}}{r}} \text{.}$$
Taking the time-derivative of $\tan\theta = \frac{y}{x}$, we get
$$\frac{\dot\theta}{\cos^2\theta} = \frac{x \dot{y} - y \dot{x}}{x^2} \text{,}$$
and we may write
$$\boxed{\dot\theta = \frac{x \dot{y} - y \dot{x}}{x^2} \cos^2\theta} \text{.}$$

