Find the unit tangent vector at the point $(2,0,\pi)$ for the curve with parametric equations $x=2\sin\theta; y=3\cos\theta; z=2\theta$, we see that the point $(2,0,\pi)$ corresponding to $\theta=\frac{\pi}{2}$.

Solution:

In vectors $r=xi+yj+zk$. Therefore, 
$r=2\sin\theta{i}+3\cos\theta{j}+2\theta{k}$
Where $\theta=\frac{\pi}{2}$
Take the derivative of $r$ in respect to $\theta$. 
$\frac{dr}{d\theta}=2\cos\theta{i}-3\sin\theta{j}+2k$ at $\theta=\frac{\pi}{2}$. 

$\frac{dr}{d\theta}=2\cos{90i}-3\sin{90j}+2k= - 3j+2k$
Now we take the modulus of the derivative. 
$|\frac{dr}{d\theta}|=\sqrt{(-3)^2+2^2)}=\sqrt{13}$. 
The unit tangent vector is thus:
$T=\frac{1}{|\frac{dr}{d\theta}|}\times\frac{dr}{d\theta}$. 
$T=\frac{1}{\sqrt{13}}.(-3j+2k)$
$=\frac{-3j+2k}{\sqrt{13}}$