How velocities transform from Cartesian to Polar coordinates

Consider a transformation from Cartesian to polar coordinates $(x,y)rightarrow (r,theta)$,
begin{equation}
begin{gathered}
x=rcostheta,
y=rsintheta.
end{gathered}
end{equation}
Here, we denote $x^{,mu}=(x,y)$ and $bar{x}^{,mu}=(r,theta)$.
Now, The question is the following,

In the $x^{,mu}$ coordinate system, the components of the velocity vector are $(dot{x},dot{y})$. Find out the components in the polar coordinates using vector/tensor transformation rules.

My answer:

From the coordinate transformation we have,
begin{equation}
begin{gathered}
dx=costheta dr-rsintheta dtheta,
dy=sintheta dr+rcostheta dtheta.
end{gathered}
end{equation}
Thus,
begin{equation}
begin{gathered}
frac{partial x}{partial r}=costheta=frac{x}{r};quad frac{partial x}{partial theta}=-rsintheta=-y,
frac{partial y}{partial r}=sintheta=frac{y}{r};quad frac{partial y}{partial theta}=rcostheta=x.
end{gathered}
end{equation}
The transformed components $bar{V}^{,mu}=bar{V}^{,mu}(x^{,alpha})$ reads,
begin{align}
bar{V}^{,mu}=frac{partial, bar{x}^{,mu}}{partial, x^{,beta}}V^{,beta}
end{align}
Now, for $mu=1$,
begin{align}
bar{V}^{,1}&=frac{partial, bar{x}^{,1}}{partial, x^{,beta}}V^{,beta}nonumber
&=frac{partial, bar{x}^{,1}}{partial, x^{,1}}V^{,1}+frac{partial, bar{x}^{,1}}{partial, x^{,2}}V^{,2}nonumber
&=frac{partial r}{partial x}V^{,1}+frac{partial, r}{partial y}V^{,2}nonumber
&=sectheta V^{,1}+csctheta V^{,2}nonumber
&=frac{r}{x} V^{,1}+frac{r}{y} V^{,2}
tag{1}label{eq:comptransone}
end{align}
Now, for $mu=2$,
begin{align}
bar{V}^{,2}&=frac{partial, bar{x}^{,2}}{partial, x^{,beta}}V^{,beta}nonumber
&=frac{partial, bar{x}^{,2}}{partial, x^{,1}}V^{,1}+frac{partial, bar{x}^{,2}}{partial, x^{,2}}V^{,2}nonumber
&=frac{partial theta}{partial x}V^{,1}+frac{partialtheta}{partial y}V^{,2}nonumber
&=-frac{1}{r}csctheta V^{,1}+frac{1}{r}sectheta V^{,2}nonumber
&=-frac{1}{y} V^{,1}+frac{1}{x} V^{,2}
tag{2}label{eq:comptranstwo}
end{align}

begin{equation}
begin{gathered}
dot{x}=costheta dot{r}-rsintheta dot{theta},
dot{y}=sintheta dot{r}+rcostheta dot{theta}.
end{gathered}
end{equation}
Now, we calculate the velocity components in the polar coordinates using equations ($ref{eq:comptransone}$) and ($ref{eq:comptranstwo}$),
begin{align}
v^{,r}&=sectheta dot{x}+cscthetadot{y}nonumber
&=secthetaleft(costheta dot{r}-rsintheta dot{theta}right)+cscthetaleft(sintheta dot{r}+rcostheta dot{theta}right)nonumber
&= dot{r}-rtantheta dot{theta}+dot{r}+rcottheta dot{theta}nonumber
&= 2dot{r}-r(tantheta -cottheta) dot{theta}
end{align}
begin{align}
v^{,theta}&=-frac{1}{r}csctheta dot{x}+frac{1}{r}sectheta dot{y}nonumber
&=-frac{1}{r}csctheta left(costheta dot{r}-rsintheta dot{theta}right)+frac{1}{r}sectheta left(sintheta dot{r}+rcostheta dot{theta}right)nonumber
&=-frac{1}{r}cotthetadot{r}+dot{theta}+frac{1}{r}tanthetadot{r}+dot{theta}nonumber
&=2dot{theta}+frac{dot{r}}{r}(tantheta-cottheta)
end{align}

Present question: Are the above equations which I derived correct? Shouldn’t this be something like $v^r=dot{r}$ and $v^theta=rdot{theta}$? Where am I going wrong? Help please.