Q: A particle is following the path C: f(t)=(2cos(t), 2sin(t), t), t>=0, and flies off on the tangent line at time t=3pi/2. Find the position of the particle at time t=5pi/2. Solution: f'(t)=(-2sint,2cost,1) f'(3pi/2)=(2,0,1) f(3pi/2)=(0,-2,3pi/2) Equation of the tangent line: l(s)=(0,-2,3pi/2) + s(2,0,1) s=0, (0,-2,3pi/2) So s=5pi/2 - 3pi/2 = pi gives position at t=5pi/2 l(pi)=(0,-2,3pi/2) + pi (2,0,1) = (2pi, -2, 5pi/2) [answer] I don't understand the red part. How come s=pi gives position at t=5pi/2 ? What is the relation between s and t? Are they realted linearly? It would be nice if someone can explain this part. Thanks!