Why non-regular curves have cusp(s) I'm very confused about the author's explanation of regular curves in my calculus book. The author says that a regular curve $\gamma:[a,b]\to\mathbb{R}^3$ is a curve such that $\gamma'(t)\neq 0$ for all $t\in[a,b]$. The author then says that this is to ensure that there are no "kinks" or "cusps" in the curve, but I'm having difficulty understanding, visually, or intuitively why this is so. More precisely, does $\gamma'(t_0)=0$ if and only if the curve has a kink the point $\gamma(t_0)$? And why does it look like a kink? In my mind, I imgaine an insect flying in space, slowing down, stopping at some point $p$, and continuing it's flight from $p$. On an intuitive level, I think there could be a kink at $p$ because as the insect approaches $p$, it slows down along one line spanned by a tangent vector $v_1$ at $p$, but once it stops, it can continue its flight along another line spanned by a different tangent vector $v_2$ at $p$ (option A below). But it would also have the choice of stopping at $p$ and continuing in the same fashion it was before it slowed down and stopped (option B). So with the option B, it seems like you could have $\gamma'(t)\neq 0$ at $p$, and yet there would be no cusp at $p$. So just by looking at the graph, you wouldn't be able to tell if $\gamma$ was regular or not. Could someone please clarify?

 A: It's easy to confuse the image of the curve with the "animation" of a point tracing that curve. Think about that for a second.
You can have a perfectly smooth looking image, even a straight line, and still have $\gamma'(t)=0$ for some $t$. You just need to imagine a point tracing the line, slowing to a stop, and then picking back up again.
On the other hand, if the curve's image is smooth, you can imagine a point tracing it out at nonzero speed at all points in time. So if the curve's image is smooth, it's possible to parametrize it with a $\gamma$ that never has $\gamma'(t)=0$.
Now if there is some sort of kink, then what will happen to the point when it reaches the kink? It can either jerk suddenly to move in a different direction (making $\gamma'$ discontinuous) or it can slow down to a stop, turn, and accelrate back up again in a new direction. And this would make $\gamma'(t)=0$ at that point in time. So if there is a kink, and if $\gamma'$ is continuous, then there would have to be a $t$ with $\gamma'(t)=0$.
By outlawing $\gamma'(t)=0$ (and I assume also requiring $\gamma'$ to be continuous) the author is outlawing kinks. A side effect is that the author is also outlawing parametrizations of smooth curves where the point that is doing the tracing slows down to a stop before possibly accelerating back up again in the same direction. But that's fine. If it's a smooth curve, it admits some (other) parametrization that doesn't slow to a stop, so the author is still considering such curves.
A: $\gamma'(t) \neq 0$ can be seen as "The particle never stops.", since the position always changes.
