Find the unit tangent vector at the point (2,0,π) for the curve with parametric equations x=2sinθ;y=3cosθ;z=2θ, we see that the point (2,0,π) corresponding to θ=π2.

Solution:

In vectors r=xi+yj+zk. Therefore, 
r=2sinθi+3cosθj+2θk
Where θ=π2
Take the derivative of r in respect to θ
drdθ=2cosθi3sinθj+2k at θ=π2

drdθ=2cos90i3sin90j+2k=3j+2k
Now we take the modulus of the derivative. 
|drdθ|=(3)2+22)=13
The unit tangent vector is thus:
T=1|drdθ|×drdθ
T=113.(3j+2k)
=3j+2k13