I'm working with coordinate transformations, and tried initially with some simple transform from cartesian to polar in the plane. So I imagined having a vector field $v=\left({3x+y},{x+y}\right)$. One way to transform it into polar is to use matrices, i.e.:
$$\left(\begin{matrix}v^\rho \\ v^\theta\end{matrix}\right)=\left(\begin{matrix}\cos\theta && \sin\theta \\ -\sin\theta && \cos\theta\end{matrix}\right)\left(\begin{matrix}v^x \\ v^y\end{matrix}\right)=\left(\begin{matrix}\cos\theta && \sin\theta \\ -\sin\theta && \cos\theta\end{matrix}\right)\left(\begin{matrix}3x+y \\ x+y\end{matrix}\right)$$
Performing all the calculations, I end up with:
$$\left(\begin{matrix}v^\rho \\ v^\theta\end{matrix}\right)=\left(\begin{matrix}3\rho\cos^2\theta+2\rho\sin\theta\cos\theta+\rho\sin^2\theta \\ \rho\cos^2\theta-2\rho\sin\theta\cos\theta-\rho\sin^2\theta\end{matrix}\right)$$
However, I could also use the formula for the conversion of contravariant vectors:
$$\tilde{v}^a=v^b\frac{\partial\tilde{x}^a}{\partial{x^b}}$$
Considering that:
$$\rho=\sqrt{x^2+y^2}$$ $$\theta=\arctan\left(\frac{y}{x}\right)$$
I find out:
$$v^\rho=v^x\frac{\partial\rho}{\partial{x}}+v^y\frac{\partial\rho}{\partial{y}}=3\rho\cos^2\theta+2\rho\sin\theta\cos\theta+\rho\sin^2\theta$$ $$v^\theta=v^x\frac{\partial\theta}{\partial{x}}+v^y\frac{\partial\theta}{\partial{y}}=\cos^2\theta-2\sin\theta\cos\theta-\sin^2\theta$$
So, while the value of $v^\rho$ coincides between the two method, there seems to be a factor $\rho$ between the two values of $v^\theta$. I've checked my calculations several times but can't find any mistake, can anyone help and suggest what I'm doing wrong here?
Thanks