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?


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