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=2cosθi−3sinθj+2k at θ=π2.
drdθ=2cos90i−3sin90j+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=1√13.(−3j+2k)
=−3j+2k√13
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