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: Are you looking for expressions for $r$, $\dot{r}$, $\vartheta$, and $\dot{\vartheta}$ in terms of $x$, $\dot{x}$, $y$, and $\dot{y}$? Or in terms of spherical coordinate unit vectors $\hat{r}$ and $\hat{\vartheta}$?
For the former:
$r = (x^2 + y^2)^{\frac{1}{2}}$, so for the time-derivative we get
$$\dot{r} = \frac{1}{2}(x^2 + y^2)^{-\frac{1}{2}}(2x\dot{x} + 2y\dot{y}) = \frac{x\dot{x} + y\dot{y}}{r},$$
and we may write
$$\boxed{\dot{r}r = x\dot{x} + y\dot{y}}$$
I don't think we can get a clean expression for $\dot{\vartheta}$, basically because $\hat{\vartheta}$ depends on $x$ and $y$. Perhaps there is a better way to do this, but I decided to expand $\arctan \left(\frac{y}{x}\right)$, and work from there:
$\vartheta = \arctan \left( \frac{y}{x} \right) \approx \frac{y}{x} - \frac{1}{3}\left(\frac{y}{x}\right)^3 + \frac{1}{5}\left(\frac{y}{x}\right)^5 - \frac{1}{7}\left(\frac{y}{x}\right)^7 + \ldots$. The time derivative of $\frac{y}{x}$ is
$$\frac{d}{dt}\left({\frac{y}{x}}\right) = \frac{x\dot{y} - y\dot{x}}{x^2} = \zeta, $$
so for $\dot{\vartheta}$ we may write
$$\dot{\vartheta} = \zeta - \left(\frac{y}{x}\right)^2\zeta+\left(\frac{y}{x}\right)^4\zeta - \left(\frac{y}{x}\right)^6\zeta + \ldots = \zeta \left( 1 - \left(\frac{y}{x}\right)^2 +\left(\frac{y}{x}\right)^4 - \left(\frac{y}{x}\right)^6+ \ldots \right)$$
$$\dot{\vartheta} = \zeta\sum_k{\left(\frac{y}{x}\right)^{2k}\left(-1\right)^{-k}}, \textrm{ or}$$
$$\boxed{\dot{\vartheta} = \frac{x\dot{y} - y\dot{x}}{x^2}\sum_k{\left(\frac{y}{x}\right)^{2k}\left(-1\right)^{-k}}}.$$
There might be issues with this result regarding how the sign of $\arctan\left(\frac{y}{x}\right)$ varies by quadrant.
For the latter:
Start by writing $\hat{r}$ and $\hat{\vartheta}$ in terms of $\hat{i}$ and $\hat{j}$, and then differentiate to get the rates of the change with respect to $r$ and $\vartheta$. You need this because your point's position will change from ($r$, $\vartheta$) to ($r + dr$, $\vartheta + d\vartheta$), and you need the differential increments.
$$\hat{r} = cos\vartheta \hat{i} + sin\vartheta \hat{j}$$
$$\hat{\vartheta} = -sin\vartheta \hat{i} + cos\vartheta \hat{j}.$$
The derivatives tell you how the unit vectors are changing as your point moves.
$$\frac{d\hat{r}}{dr} = 0, $$
$$\frac{d\hat{\vartheta}}{dr} = 0, $$
$$\frac{d\hat{r}}{d\vartheta} = -sin\vartheta \hat{i} + cos\vartheta \hat{j} = \hat{\vartheta},$$
$$\frac{d\hat{\vartheta}}{d\vartheta} = -cos\vartheta \hat{i} - sin\vartheta \hat{j} = -\hat{r}.$$
We know that $\vec{r} = r\hat{r}$, so differentiate with respect to time to get
$$\frac{d\vec{r}}{dt} = \frac{dr}{dt}\hat{r} + r\frac{d\hat{r}}{dt} = \frac{dr}{dt}\hat{r} + r\frac{d\hat{r}}{d\vartheta}\frac{d\vartheta}{dt}.$$
Substitution gives
$$\boxed{\frac{d\vec{r}}{dt} = \vec{v} = \dot{r}\hat{r} + r\dot{\vartheta}{\hat{\vartheta}}}.$$
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
$$\boxed{\dot\theta = \frac{x \dot{y} - y \dot{x}}{x^2} \cos^2\theta} \text{.}$$
