Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\alpha:(a,b)\rightarrow\mathbb R^2$ be a regular parametrized plane curve. Assume that there exists $t_0$, $a<t_0<b,$ such that the distance $|\alpha (t)|$ from the origin to the trace of $\alpha$ will be a maximum at $t_0$. Prove that the curvature $k$ of $\alpha$ at $t_0$ satisfies $k(t_0)\geq1/|\alpha(t_0)|$.

I am confused about how to use the condition "the distance $|\alpha (t)|$ from the origin to the trace of $\alpha$ will be a maximum at $t_0$". Any suggestions?


share|cite|improve this question
Take the second derivative of $|\alpha(t)|^2$. – Davide Giraudo Feb 11 '12 at 17:01
up vote 3 down vote accepted

By regular I assume that $\alpha$ is two times differentiable with respect to the arc length $t$. Since $f(t):=|\alpha(t)|^2$ reaches a maximum at $t_0$ in the interior of $[a,b]$, then the second derivative at $t_0$ is negative. We have $$f'(t)=2\langle \alpha'(t),\alpha(t)\rangle$$ and $$f''(t)=2\langle\alpha''(t),\alpha(t)\rangle+2|\alpha'(t)|^2$$ so $$1=|\alpha'(t_0)|^2<-2\langle\alpha''(t_0),\alpha(t_0)\rangle\leq |\alpha''(t_0)|\cdot |\alpha(t_0)|,$$ and we get the result.

share|cite|improve this answer
Thank you so much. But I was wondering how do you think of using the second derivative of $|\alpha(t)|^2?$ – Vladimir Feb 11 '12 at 19:47
We know that the curvature is expressed thanks to the second derivative of $\alpha$ and the square is here to make the expression easier to differentiate. – Davide Giraudo Feb 11 '12 at 19:50
Yes... And why $|\alpha'(t_0)|^2=1$? – Vladimir Feb 11 '12 at 19:54
I guess it's an assumption about $\alpha$ (regular) (we want a constant speed). – Davide Giraudo Feb 11 '12 at 19:56

Hint: What is the curvature of the circle of radius $\vert\alpha(t_0)\vert$ centered at the origin?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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