Parametric curves applications

[kingwinner]

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!

 

[hotvette]

$$s = t - \frac{3\pi}{2}$$

It just represents the starting time when the particle leaves the curve.

 

[kingwinner]

But I don't get WHY...
Why not, say, s = t^2 -3pi/2 or s= t - 2* 3pi/2?

 

[hotvette]

It's simple variable substitution. Here's an analogy. Say you have the equation $y = m(x - x_0) + y_0$. For convenience, I can let $s = x - x_0$, which means $y = ms + y_0$. Same thing is happening here.

Though I haven't seen parametric equations written in this form before , I think the equation of the tangent line could have been written (w/o variable substitution) as:

l(t)=(0,-2,3pi/2) + (t-3pi/2)(2,0,1)

letting s = t-3pi/2 results in:

l(s)=(0,-2,3pi/2) + s(2,0,1)

When t = 5pi/2, s = 5pi/2-3pi/2 = pi

 

[HallsofIvy]

If it is 3:00 now, how long will be until it is 5:00?

Would you even consider saying s = (5)² -3 or s= 5 - 2(3)?

 

[kingwinner]

Why would the position on the tangent line be moved by (2,0,1) per second?[/color]
Say, l(s)=(0,-2,3pi/2) + s(8,0,4), I changed the direction vector, but this would still represent the same line, how can I figure out how s and t are related now?

Thanks!