4
$\begingroup$

Suppose $\alpha$ is a curve parametrized by arc length and there is some $s_0$ such that $||\alpha(s)||\le ||\alpha(s_0)||$, $\forall s$ near $s_0$. Show that: $$\kappa(s_0) \ge \frac{1}{||\alpha(s_0)||}.$$

I know the curvature that is parametrized by arc length equals $\kappa = ||\frac{dT}{ds}||$ and since $||\alpha(s)||\le ||\alpha(s_0)||$ then the point $s_0$ is a local maximum of $f(s)=||\alpha(s)||?$

$\endgroup$
  • $\begingroup$ Do you mean the curvature at $s_0$? $\endgroup$ – lhf Sep 17 '13 at 2:03
7
$\begingroup$

You correctly observed that the key fact is that $\Vert \alpha(s) \Vert$ has a local maximum at $s_0$. Thus we have (using $\Vert\alpha(s)\Vert^2$ instead for computational convenience) in particular

$$\frac12 \frac{d^{2}}{ds^{2}}\bigg|_{s=s_{0}}\left\Vert \alpha\left(s\right)\right\Vert ^{2}=\left\Vert \alpha'\left(s_{0}\right)\right\Vert ^{2}+\left\langle \alpha\left(s_{0}\right),\alpha''\left(s_{0}\right)\right\rangle \le0.$$

The first term here is $1$ because the curve is arclength parametrized; thus we have $\left\langle \alpha\left(s_{0}\right),\alpha''\left(s_{0}\right)\right\rangle \le-1$ and so from Cauchy-Schwarz we have $\left\Vert \alpha''\left(s_{0}\right)\right\Vert \left\Vert \alpha\left(s_{0}\right)\right\Vert \ge1$. Noting that the curvature of $\alpha$ at $s_{0}$ is just $\left\Vert \alpha''\left(s_{0}\right)\right\Vert $ (once again because $\alpha$ is arclength parametrized), we are done.

$\endgroup$
  • $\begingroup$ Thank you! I would have never guess to use $$\frac{d^{2}}{ds^{2}}\bigg|_{s=s_{0}}\left\Vert \alpha\left(s\right)\right\Vert ^{2}=\left\Vert \alpha'\left(s_{0}\right)\right\Vert ^{2}+\left\langle \alpha\left(s_{0}\right),\alpha''\left(s_{0}\right)\right\rangle \le0.$$ How did you derive that? $\endgroup$ – Lays Sep 17 '13 at 8:06
  • $\begingroup$ @Lays: the fact that the second derivative is non-positive is from the local maximum. The formula I found for the second derivative comes from the product rule $\langle u, v \rangle' = \langle u',v \rangle + \langle u, v' \rangle$ applied twice. Also I left a factor of two, I'll correct my answer. $\endgroup$ – Anthony Carapetis Sep 17 '13 at 8:52
  • $\begingroup$ I see that you divided by $1/2$, where did that come from? Because when I took the product rule I got $$2\langle \alpha'(s), \alpha'(s)\rangle + \langle \alpha''(s), \alpha(s)\rangle + \langle \alpha(s), \alpha''(s)\rangle?$$ $\endgroup$ – Lays Sep 17 '13 at 19:43
  • 1
    $\begingroup$ @Lays: that's correct: the inner product is commutative, so you should be able to factor out a 2 from the whole expression. I just stuck a $1/2$ on the other side because it was the fastest way to fix it up with an edit. $\endgroup$ – Anthony Carapetis Sep 18 '13 at 2:02

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.