How to differentiate this position Vector $\vec r=\rho\vec e_\rho+z\vec e_z$ Given the unit vectors:
$\vec e_\rho=\left(\begin{smallmatrix}
cos(\theta )\\ 
sin(\theta )\\0 
\end{smallmatrix}\right); \vec e_\phi=\left(\begin{smallmatrix}
-sin(\theta )\\ 
cos(\theta )\\0 
\end{smallmatrix}\right); \vec e_z=\left(\begin{smallmatrix}
0\\ 
0\\1 
\end{smallmatrix}\right);$
and my position vector:
$\vec r=\rho\vec e_\rho+z\vec e_z$
If $,\rho, \phi,z$ are explicit functions of time ($\rho=\rho(t),\phi=\phi(t),z=z(t) $what is $\frac{d}{dt}\vec r $ and $\frac{d^2}{d^2t}\vec r$?
I am not entirely sure but don't I just differentiate "normally"?
$\vec r=\rho(t)\left(\begin{smallmatrix}
cos(\theta )\\ 
sin(\theta )\\0 
\end{smallmatrix}\right )+z(t)\left(\begin{smallmatrix}
0\\ 
0\\1 
\end{smallmatrix}\right)$
$\frac{d}{dt}\vec r=\rho'(t)\frac{d\theta}{dt}\left(\begin{smallmatrix}
-sin(\theta )\\ 
cos(\theta )\\0 
\end{smallmatrix}\right)+z'(t)\left(\begin{smallmatrix}
0\\ 
0\\1 
\end{smallmatrix}\right)$
Or am I just writing down nonsense?
I also wasn't sure how to differentiate the z-vector. 
Maybe someone with a little more knowledge/experience can help me out here :)
Thanks in advance
Edit:
$\frac{d}{dt}\vec r=\rho'(t)\left(\begin{smallmatrix}
cos(\theta )\\ 
sin(\theta )\\0 
\end{smallmatrix}\right)+\rho(t)\frac{d\theta}{dt}\left(\begin{smallmatrix}
-sin(\theta )\\ 
cos(\theta )\\0 
\end{smallmatrix}\right)+z'(t)\left (\begin{smallmatrix}
0\\ 
0\\1 
\end{smallmatrix}\right)$

Suppose that $\theta$ depends on $t$ then $\frac{d}{dt}(\rho(t)\cos\theta)=\rho'(t)\cos\theta-\rho(t)\frac{\partial\theta}{\partial t}\sin\theta$.
It seems you missed out a term, if $\theta$ does not depend on $t$ then the extra term will be zero.
Also when you did the derivative it seems you tried to differentiate everything at once, but notice that the second term ($\rho(t)...$) is expressed as something multiplying $\vec e_\phi$.
Your deriavative of $z\vec e_z$ is fine.
