# Reparametrize a helix

• Apr 2nd 2009, 02:48 PM
arbolis
Reparametrize a helix
Hi,
I'm learning about reparametrizing curves and I don't understand what my teacher did in the following example :
Reparametrize the helix $(\cos t, \sin t, t)$ from $(1,0,0)$ in the same direction than $t$ increases.
Solution: $s(t)=\int_0^t |r'(u)| du$.
Then she continues from this and I understand all she did.
What I don't understand is why the lower bound of the integral is 0.
It's sometimes different from 0, so how do I know what lower bound to chose? I'm sure it's because of the (1,0,0) point, but why 0 as lower bound? Is it because the z-coordinate is worth 0 in (1,0,0)?
• Apr 4th 2009, 01:24 AM
Calculus26
think time
You are reparameterizing in terms of arc length so s(t) is the distance traveled in time t.

In most instances you are at the initial point when t= 0 which is why the bottom limit is 0. you are at the point (1,0,0) at t = 0. It has nothing to do with the spatial coordinates but rather the time coordinate.

If for instance you were at the initial point (1,0,0) at t= t0 then t0 would be the lower limit of integration