# How does regularity of curves prevent "cusps"?

1. Feb 16, 2015

### center o bass

A regular curve on a manifold $M$ is a curve $\gamma:I \to M$ such that $\dot \gamma(t) \neq 0$ for any $t \in I$. In John Lee's "Introduction to Curvature" he says that this intuitively means that we prevent the curve from having "cusps" and "kinks".

How can I see that this is the case? I.e. how could $\dot \gamma(t) = 0$ for some t lead to a "kink" or "cusp"?

2. Feb 16, 2015

### wabbit

It doesn't. $\dot \gamma(t) = 0$ means a stationary point and can happen just from parametrization. But cusps or kinks do require $\dot \gamma(t) = 0$, so forbidding it ensures you get none.

$\dot \gamma(t) \neq 0$ is needed for two other reasons : it ensures $\dot \gamma(t)$ gives you the direction of the tangent vector to the curve at every point, and that the parametrization is locally one-to-one / doesn't backtrack.

Last edited: Feb 16, 2015
3. Feb 17, 2015

### center o bass

But why -- intuitively -- do cusps or kinks require $\dot gamma = 0$?

4. Feb 17, 2015

### wabbit

One way to see it is to think of the curve as traced by a moving point at $\gamma(t)$; then $\dot\gamma(t)$ is the velocity of that point. If it's not zero it has a direction. What is that direction at a kink ?