3a. Find the tangent vector of the parametric curve r (

3a.
Find the tangent vector of the parametric curve r (s)=( x , y , z)
Solution
r ( s)=(2 cos s , 3 sin s , 2 π −s )
This can also be represented as the unit vector as below
r͢ ͢ (s)=¿
Finding the derivative of r
r ' (s)= δ x
δ s ¿
¿ (−2 sin s ) i+¿
Now putting S=o and substituting in the derivative obtained
¿ (−2 sin( 0) ) i +¿
=¿ { ( 0 ) i , ( 3 ) j ,−k }
Therefore, the tangent vector is ¿ 0,3 ,−1> ¿
3b.
Find the equation of the tangent line to the curve at point (
0 ,−3 , π /2 ¿
Solution

Given that x=2 cos s , y=3 sin s∧¿ z=2 π−s on coordinates 0 ,−3 , π (¿ 2)
Finding the parametric values of of s that corresponds to the
coordinates (0 ,−3 , π /2 ¿
0=2 cos s , −3=3 sin s and z=2 π−s
Now, we find the parametric value for s that satisfy all the three
equations.
We find that S=270, all the three equations are satisfied.
r ( s )=¿
Now finding the derivative of r
r ' ( s )=¿)
Substituting the value of s i.e. s=270 or s= 1.5 π
r ' ( s )=¿)
r ' ( s )=¿)
Applying the point formula
¿ ¿)*(x-0, y+3, z- π /2 ¿
¿ ¿)
¿ 2 x+ 0−z + π /2
Therefore, the equation of the tangent is

F(x,y,z)¿ 2 x−z+ π /2