Answer:
Provided that the cylindrical coordinate system is given by the coordinates
[tex]\rho,\,\phi, \,z [/tex]
The unit vectors are:
[tex]\hat{\rho}=\hat{x}\cos(\phi)+\hat{y}\sin(\phi)[/tex]
[tex]\hat{\phi}=-\hat{x}\sin(\phi)+\hat{y}\cos(\phi)[/tex]
[tex]\hat{z}=\hat{z}[/tex]
With their time derivatives being:
[tex]\frac{d \hat{\rho}}{dt}=\hat{\phi}\frac{d \phi}{dt}[/tex]
[tex]\frac{d \hat{\phi}}{dt}=-\hat{\rho}\frac{d \phi}{dt}[/tex]
[tex]\frac{d \hat{z}}{dt}=0[/tex]
Explanation:
Let's start by writing the coordinate transformations:
[tex]\rho=\sqrt{x^2+y^2}[/tex], [tex]x=\rho\cos(\phi)[/tex]
[tex]\phi=\atan(y/x)[/tex], [tex]y=\rho\sin(\phi)[/tex]
[tex]z=z[/tex].
For the unit vector [tex]\hat{\rho}[/tex] we have:
[tex]\hat{\rho}=\frac{\vec{\rho}}{\rho}=\frac{x\hat{x}+y\hat{y}}{\rho}=\hat{x}\cos(\phi)+\hat{y}\sin(\phi)[/tex]
For the unit vector [tex]\hat{\phi}[/tex] we have:
[tex]\hat{\phi}=\hat{x}\cos(\phi+90)+\hat{y}\sin(\phi+90)[/tex]
By adding 90 degrees to [tex]\hat{\rho}[/tex], we have (see the 2nd attachment),
[tex]\hat{\phi}=-\hat{x}\sin(\phi)+\hat{y}\cos(\phi)[/tex]
It is not hard to see that [tex]\hat{z}=\hat{z}[/tex].
From this we can write the following useful expressions that we'll use later on to determine the time derivatives:
[tex]\begin{array}{ccc}\frac{\partial \hat{\rho}}{\partial \rho}=0&\frac{\partial \hat{\phi}}{\partial \rho}=0&\frac{\partial \hat{\rho}}{\partial z}=0\\\frac{\partial \hat{\rho}}{\partial \phi}=-\hat{x}\sin(\phi)+\hat{y}\cos(\phi)=\hat{\phi}\\&\frac{\partial \hat{\phi}}{\partial \phi}=-\hat{x}\cos(\phi)-\hat{y}\sin(\phi)=-\hat{\rho}&\frac{\partial \hat{z}}{\partial \phi}=0\\\frac{\partial \hat{\rho}}{\partial z}=0&\frac{\partial \hat{\phi}}{\partial z}=0&\frac{\partial \hat{z}}{\partial z}=0\end{array}[/tex]
Now, knowing all of the above the time derivatives come in a straightforward way:
[tex]\frac{d \hat{\rho}}{d t}=\frac{\partial \hat{\rho}}{\partial \rho}.\frac{d \rho}{dt}+\frac{\partial \hat{\rho}}{\partial \phi}.\frac{d \phi}{d t}+\frac{\partial \hat{\rho}}{\partial z}.\frac{d z}{d t}=\hat{\phi}.\frac{d \phi}{d t}[/tex]
[tex]\frac{d \hat{\phi}}{d t}=\frac{\partial \hat{\phi}}{\partial \rho}.\frac{d \rho}{dt}+\frac{\partial \hat{\phi}}{\partial \phi}.\frac{d \phi}{d t}+\frac{\partial \hat{\phi}}{\partial z}.\frac{d z}{d t}=-\hat{\rho}.\frac{d \phi}{d t}[/tex]
[tex]\frac{d \hat{\rho}}{d t}=\frac{\partial \hat{z}}{\partial \rho}.\frac{d \rho}{dt}+\frac{\partial \hat{z}}{\partial \phi}.\frac{d \phi}{d t}+\frac{\partial \hat{z}}{\partial z}.\frac{d z}{d t}=0[/tex]
