1
$\begingroup$

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!

$\endgroup$
4
  • 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

Reset to default
2
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.