Ok, I'm trying to understand the newton's law in terms of polar coordinate.
As far as I understand, in polar coordinate the vectors are represented as r direction and $\theta$ direction.
My question is, when we differentiate the position vector $\hat{r}$ in order to find velocity vector, $\dot{\hat{r}} = \frac{d}{dx}(r\hat{r}) = \dot{r}\hat{r}+r\dot{\hat{r}} = \dot{r}\hat{r} +r\dot{\theta}\hat{\theta} $, right?
I can understand the process here.
My confusion is, maybe this is really stupid question, if position is represented as r direction and $\theta$ direction, then isn't it should be like, $\hat{r} = (r,\theta) = r\hat{r} + \theta\hat{\theta}$ ?
Then why velocity vector is not $\dot{\hat{r}} = \frac{d}{dx}(r\hat{r}+\theta\hat{\theta}) = \dot{r}\hat{r}+r\dot{\hat{r}}+\dot{\theta}\hat{\theta}+\theta\dot{\hat{\theta}} $ ???
Why we considering only r direction when we calculate velocity?
the position is always $\vec{r} = r \hat{r}$ where $\hat{r}$ is a unit vector pointing toward the object $\hat{r} = (cos(\theta), sin(\theta))$, $\hat{\theta}$ is defined to be always perpendicular to $\hat{r}$ (i.e. $\hat{\theta} = \hat{k} \times \hat{r}$)
your confusion is this statement "if position is represented as r direction and θ direction" which is wrong, position is just $r$ magnitude in the direction of $\hat{r}$ which is defined to always point directly towards the object.
another thing is the derivative is always taken with respect to time $t$ not to $x$ (which could be confused with the x coordinate)