Answer :
Answer:
The parametric equation of the tangent line is:
x= -5t
y= 4
z= t + [tex]\frac{\pi}{2}[/tex]
Step-by-step explanation:
The vector r(t)=[5 cos(t) , 4 sin(t) , t]
Hence, the equation of the tangent line is the derivative of the vector r(t):
r'(t)= [-5 sin(t) , 4 cos(t) , 1]
The tangent line is given by:
x= 0 + t·[-5 sin(t)]
x= 4 + t·[4 cos(t)]
y= [tex]\frac{\pi}{2}[/tex]+t
Given the point (0, 4, π/2), t=[tex]\frac{\pi}{2}[/tex]
So we replace this value in the equation of the derivative equation:
[tex]r'(t)=[-5 sin(\frac{\pi }{2}) , 4 cos(\frac{\pi }{2} ) , 1]=[-5,0,1][/tex]
The parametric equation of the tangent line is:
x= -5t
y= 4
z= t + [tex]\frac{\pi}{2}[/tex]