relationship between cartesian velocity and polar velocity I am trying to convert an equation from Cartesian to polar coordinates. I know for a given $x$ and $y$ Cartesian, we can get polar $r=\sqrt{x^2+y^2}$ and $\theta = \arctan(y/x)$. 
However, for a given $v_x$ and $v_y$, I want to get $v_r$ and $v_\theta$. Can we write $v_r = \sqrt{v_x^2+v_y^2}$ and $v_\theta = \arctan{(v_y/v_x)}$?
I know its' probably not that simple, but a simple derivation and explanation will be really helpful.
 A: If $r(t) = \sqrt{x(t)^2+y(t)^2}$ , then
$$\dot r = \frac{x\dot x+y\dot y}{\sqrt{x^2+y^2}}$$
If $\theta(t) = \arctan\left[y(t)/x(t)\right]$ , then
$$\dot\theta = \frac{x\dot y - \dot x y}{x^2+y^2}$$
Alternatively:
If $x(t) = r(t)\cos\left[\theta(t)\right]$ , then
$$\dot x = \dot r \cos \theta - r\dot\theta\sin\theta$$
If $y(t) = r(t)\sin\left[\theta(t)\right]$ , then
$$\dot y = \dot r\sin\theta + r\dot\theta \cos\theta$$
A: In polar coordinates the position and the velocity of a point are expressed using the orthogonal unit vectors $\mathbf e_r$ and $\mathbf e_\theta$, that, are linked to the orthogonal unit cartesian vectors $\mathbf i$ and $\mathbf j$ by the relations: 
$$
\mathbf e_r=\mathbf{i}\cos \theta +\mathbf{j}\sin \theta 
$$
$$
\mathbf e_\theta=-\mathbf{i}\sin \theta +\mathbf{j}\cos \theta 
$$
The position of a point is given by $\mathbf r=r\mathbf e_r$, where $r=\sqrt{x^2+y^2}$. from this we can find the velocitiy as:
$$
\mathbf v= \frac{d}{dt}(r\mathbf e_r)=\dot r\mathbf e_r+r\dot{\mathbf e}_r
$$
with abit of calculus we see that this gives:
$$
\mathbf v=\dot r\mathbf e_r+r\dot \theta \mathbf e_\theta
$$
so the two components of this vector,  in polar coordinates, are:
$$
v_r=\dot r \qquad v_\theta=r\dot \theta
$$
and you can see that 
$$
v_r=\dot r=\frac{d}{dt}\sqrt{x^2+y^2}=\frac{x\dot x+y\dot y}{\sqrt{x^2+y^2}}\ne \sqrt{\dot x^2+\dot y^2}=\sqrt{v_x^2+v_y^2}
$$
You can find  by yourself the expression of $v_\theta$ in terms of the cartesian components of the velocity (that is not simple as in your question).
