# How to derive velocity in polar coordinates

In one of my aerodynamics classes i need to use the following derivation to convert the velocity components u and v to a polar coordinate system:

$$v_r = u \cos(\theta) + v \sin(\theta) \\ v_t = v \cos(\theta) - u \sin(\theta)$$

I'm trying to get to this derivation to understand it but I cant figure it out. Can someone show me how they converted the velocity in Cartesian coordinate's to polar coordinates?

• The position vector $\mathbf{r}$ is defined as $[~x~~y~]$ in the standard basis or in polar coordinates as $(~r\cos\theta~~r\sin\theta~)$. Take time derivatives of each component and use the chain rule. Dec 18, 2020 at 15:44
• physics.stackexchange.com/questions/258629/… Dec 18, 2020 at 15:45

Let us introduce $$[x, y]^T = [r\cos\theta, r\sin\theta]^T$$. The velocity is given by time derivative of this position vector.

$$\begin{bmatrix} \dot{x}\\ \dot{y}\\ \end{bmatrix} = \begin{bmatrix} \dot{r}\cos\theta -r\sin \theta~\dot{\theta} \\ \dot{r}\sin\theta + r\cos\theta~\dot{\theta}\\ \end{bmatrix}$$

$$\implies \begin{bmatrix} \dot{x}\\ \dot{y}\\ \end{bmatrix} =\begin{bmatrix} \cos\theta & -r\sin \theta\\ \sin\theta &+ r\cos \theta \\ \end{bmatrix}\begin{bmatrix}\dot{r}\\ \dot{\theta}\end{bmatrix}$$

Now, invert the matrix on the right hand side and you will arrive at the expression in your question.

$$\implies \begin{bmatrix} \dot{r}\\ \dot{\theta}\\ \end{bmatrix} =\dfrac{1}{r}\begin{bmatrix} r\cos\theta & r\sin \theta\\ -\sin\theta &+ \cos \theta \\ \end{bmatrix}\begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix}$$

$$\implies \begin{bmatrix} \dot{r}\\ \dot{\theta}\\ \end{bmatrix} =\begin{bmatrix} \cos(\theta)~\dot{x} + \sin(\theta)~\dot{y}\\ -\dfrac{1}{r}\sin(\theta)~\dot{x}+ \dfrac{1}{r}\cos(\theta)~\dot{y} \\ \end{bmatrix}$$

$$\implies \begin{bmatrix} \dot{r}\\ r\dot{\theta}\\ \end{bmatrix} =\begin{bmatrix} \cos\theta~\dot{x} + \sin\theta~\dot{y}\\ -\sin\theta~\dot{x}+ \cos\theta~\dot{y} \\ \end{bmatrix}$$

Note that $$v_t=r\dot{\theta}$$.