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?

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2 Answers 2


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.

  • 2
    $\begingroup$ Not quite. If $\gamma'(t)\neq0$, that does imply there is no kink there. If $\gamma'(0)=0$, then that doesn't necessarily mean there is a kink there. Maybe there is, maybe there isn't. And no, a regular curve does not have a cusp. A cusp necessitates that the point tracing it slows down to a stop for a moment, which would make $\gamma'(t)=0$, which would make it not regular. $\endgroup$
    – 2'5 9'2
    Commented Sep 13, 2015 at 8:06
  • $\begingroup$ Ok, I think I see now. Thank you :) $\endgroup$
    – user153582
    Commented Sep 13, 2015 at 8:07
  • $\begingroup$ Also, a curve such as $\gamma(t) = (0,0,t+\sin(2t))$ has an image which is a straight line (the $z$ axis), but each time the particle stops, it changes direction (from upwards to downwards, or vice versa). Another example, where each point on the line is visited infinitely many times, is $\gamma(t) = (0,0,t\sin(2t))$. $\endgroup$ Commented Sep 13, 2015 at 13:38
  • $\begingroup$ Thanks for your nice answer.Can you explain why it is avoiding this non-regular curves,rather 0 velocity at a point case? For example as an example has been given of a straight line which is not regular.In Differential Geometry we need to study geometry of curves like curvature and torsion,which are independent of parametrization. So what it seems that a curve has a kink or cusp iff for all parametrization it is non regular! Is it true? $\endgroup$
    – Srijit
    Commented Mar 21, 2018 at 18:14

$\gamma'(t) \neq 0$ can be seen as "The particle never stops.", since the position always changes.

  • $\begingroup$ You can have a kink and have $\gamma'(t)$ exist there. $\gamma'(t)$ can be $0$ at cusps. This is about a parametrization $\gamma$ of the curve, and its derivative. You are talking about tangent lines, which is a related subject but not what this is directly about. $\endgroup$
    – 2'5 9'2
    Commented Sep 13, 2015 at 8:09
  • $\begingroup$ Arg, good point, mistake by me. $\endgroup$
    – Antitheos
    Commented Sep 13, 2015 at 8:10
  • $\begingroup$ No problem. A famous example is $(t^2,t^3)$, which traces a curve with a cusp. And the derivative at $t=0$ is $(0,0)$. If you animate this, you can see the point tracing it out slow to a stop for a brief moment. $\endgroup$
    – 2'5 9'2
    Commented Sep 13, 2015 at 8:11

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