Let $V=x\frac{∂}{∂x}+y\frac{∂}{dy}$ be a vector field on the plane. Compute its coordinate representation in polar coordinates on the right half-plane $\{(x,y):x>0\}$.
What I got so far:
The question asks for a coordinate representation, which can be easily obtained by composing this vector field with the map $(x,y)\to (r\cos\theta, r\sin\theta)$. So it becomes $r\cos\theta\frac{∂}{∂x}+r\sin\theta\frac{∂}{dy}$. But this is weird. I'm pretty sure there is something wrong.
Ok since I need to go and will fell guilty if you dont find it We have by chain rule $$\frac{\partial}{\partial r}=\frac{\partial x}{\partial r}\cdot\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\cdot\frac{\partial}{\partial y} \\ \frac{\partial}{\partial\theta}=\frac{\partial x}{\partial\theta}\cdot\frac{\partial}{\partial x}+\frac{\partial y}{\partial\theta}\cdot\frac{\partial}{\partial y}$$ which reads $$\left(\begin{array}{c} \partial_r\\ \partial_\theta\end{array}\right)=\left(\begin{array}{cc} \cos\theta &\sin\theta\\ -r\sin\theta& r\cos\theta\end{array}\right) \left(\begin{array}{c} \partial_x\\ \partial_y\end{array}\right)$$ By inversing the matrix you have $$\left(\begin{array}{c} \partial_x\\ \partial_y\end{array}\right)=\left(\begin{array}{cc} \cos\theta &-\frac{1}{r}\sin\theta\\ \sin\theta& \frac{1}{r}\cos\theta\end{array}\right)\left(\begin{array}{c} \partial_r\\ \partial_\theta\end{array}\right) $$
And simple compuation give you $$V=r\partial_r$$. When one ask you about the new coordiantes of a vector field the coordinates have to be expressed in this new local basis of the tangent space (the i.e $\partial_i$ )vector)