I'm currently taking a Calculus course, and studying curves. When learning about regular parametrizations, where a condition asked is that the derivative of the parametrization isn't null, I started to think about what conceptually meant that. And by doing so, I thought about the following example:
- $\sigma_1 : [-1,1] \rightarrow C $ , where $\sigma_1(t)= (t,t^2)$
- $\sigma_2 : [-1,1] \rightarrow C$ , where $\sigma_2(t)= (t^3,t^6)$
They both descrbe the same curve, however when you take the derivatives you get:
- $\sigma_1'(0)= (1,0)$
- $\sigma_2'(0)= (0,0)$
And I can understand why derivatives of different parametrizations will give different results, but I cant figure out the intuition behind the fact that one can be equal to $(0,0)$ and the other one not. Specially as the curve described is a parabola which, when thinking about it as a function, it has a minimum thus it makes sense that the derivative of the parametrization is null, and not otherwise.
The intuition comes from physics. Think of the curve as a trajectory, and of a parametrization as describing a point moving along it. Then $\sigma(t)$ gives the position at time $t$ and $\sigma'(t)$ is the velocity vector at time $t$. Points can move along the same curve with different velocities, and that $\sigma'(t)=0$ means that the point stops moving at time $t$. A parametrization is regular if your point never stops, if it moves so that its velocity is non-zero at all times.
This implies, in particular, that it will always be moving in the same direction, because to reverse direction it would have to stop at some point and then go back. Which, in turn, implies that the map $t\mapsto\sigma(t)$ is injective unless the curve crosses itself, at different times the point must be in different places along the curve.