I am slightly confused by the following curve $\gamma(t) = (e^t,0,0)$ in $\mathbb{R}^3$. Its curvature, defined as $$ \kappa(t) = \frac{\|\dot \gamma(t) \times \ddot \gamma(t)\|}{\|\dot \gamma(t)\|^3} $$ vanishes everywhere, yet if I think about the curve geometrically it does have curvature. What am I missing ? Do I need a unit-speed parametrization to obtain the correct result?

Many thanks for your help!

  • 3
    $\begingroup$ Your curve is contained in a straight line, the $x$-axis. $\endgroup$
    – Sigur
    Jan 26, 2013 at 17:45
  • 1
    $\begingroup$ @harlekin You're probably visualizing the curve $(t,e^t,0)$. $\endgroup$
    – Git Gud
    Jan 26, 2013 at 17:46
  • 2
    $\begingroup$ @Sigur, the positive portion of the $x$-axis $\endgroup$
    – Rustyn
    Jan 26, 2013 at 17:50
  • $\begingroup$ @GitGud yes, I was - I realized this the moment you pointed it out, many thanks! $\endgroup$
    – harlekin
    Jan 26, 2013 at 23:41

1 Answer 1


Your curve is exactly the interval: $(0,\infty)$ which is essentially a ray and has no curvature. Straight rays/lines have zero curvature.

Here is a cool link on curvature.

(I know you know what curvature is, I just thought the link was amusing/helpful.)


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