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.


2 Answers 2


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 $$\dot\theta = \frac{x\dot y - \dot x y}{x^2+y^2}$$


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 $$\dot y = \dot r\sin\theta + r\dot\theta \cos\theta$$

  • $\begingroup$ So, Can we write $v_r = \sqrt{v_x^2+v_y^2}$ and $v_\theta = \arctan{(v_y/v_x)}$? $\endgroup$
    – Nosrati
    Commented Sep 25, 2017 at 19:25
  • $\begingroup$ @Nosrati I know I am very late to the party but I just wanted to point out that expression for $v_\theta$ does not even have units of velocity.. $\endgroup$
    – Luismi98
    Commented Mar 5, 2021 at 22:55
  • $\begingroup$ Also, I think this may be useful for latecomers: note the conversion from $(v_x,v_y)$ to $(v_r,v_\theta)$ is nothing but a rotation of the axis clockwise by $\theta$. The modulus of the velocity is $\sqrt{v_x^2+v_y^2}$ in $v_x v_y$ coordinates, but the $v_r v_\theta$ coordinates are nothing but a rotation of the latter, and hence the modulus of the velocity must also be $\sqrt{v_r^2+v_\theta^2}$. Therefore, $|\mathbf{v}|=\sqrt{v_x^2+v_y^2}=\sqrt{v_r^2+v_\theta^2} \neq v_r$. $\endgroup$
    – Luismi98
    Commented Mar 5, 2021 at 23:02

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).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .