Transformation of derivatives from cartesian to cylindrical coordinates

Question

It is well known that for some function $\phi$ its derivatives have the following relations $$\left[\begin{array}{l} \frac{\partial \phi}{\partial r} \\ \frac{\partial \phi}{\partial \theta} \\ \frac{\partial \phi}{\partial z} \end{array}\right]=\left[\begin{array}{ccc} \cos \theta & \sin \theta & 0 \\ -r \sin \theta & r \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \frac{\partial \phi}{\partial x} \\ \frac{\partial \phi}{\partial y} \\ \frac{\partial \phi}{\partial z} \end{array}\right],\quad \left[\begin{array}{l} \frac{\partial \phi}{\partial x} \\ \frac{\partial \phi}{\partial y} \\ \frac{\partial \phi}{\partial z} \end{array}\right]=\left[\begin{array}{ccc} \cos \theta & -\frac{\sin \theta}{r} & 0 \\ \sin \theta & \frac{\cos \theta}{r} & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} \frac{\partial \phi}{\partial r} \\ \frac{\partial \phi}{\partial \theta} \\ \frac{\partial \phi}{\partial z} \end{array}\right].$$ What happens if $\phi$ is not depending on the angle $\theta$? How can I then derive $\frac{\partial \phi}{\partial r}$?

Answer 1

What happens if $\phi$ is not depending on the angle $\theta$?

Then $\frac{\partial \phi}{\partial \theta} = 0$. And you can just ignore the middle row of the left transformation matrix, and the middle column of the right transformation matrix.