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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?

Thanks.

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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 $$f'(t)=2\left\langle \alpha'(t),\alpha(t)\right\rangle$$ and $$f''(t)=2\left\langle\alpha''(t),\alpha(t)\right\rangle+2|\alpha'(t)|^2$$ so $$1=|\alpha'(t_0)|^2<- \left\langle\alpha''(t_0),\alpha(t_0)\right\rangle\leqslant \left|\alpha''(t_0)\right|\cdot \left|\alpha \left(t_0\right)\right|,$$ and we get the result.

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  • $\begingroup$ Thank you so much. But I was wondering how do you think of using the second derivative of $|\alpha(t)|^2?$ $\endgroup$ – Vladimir Feb 11 '12 at 19:47
  • $\begingroup$ 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. $\endgroup$ – Davide Giraudo Feb 11 '12 at 19:50
  • $\begingroup$ Yes... And why $|\alpha'(t_0)|^2=1$? $\endgroup$ – Vladimir Feb 11 '12 at 19:54
  • $\begingroup$ I guess it's an assumption about $\alpha$ (regular) (we want a constant speed). $\endgroup$ – Davide Giraudo Feb 11 '12 at 19:56
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Hint: What is the curvature of the circle of radius $\vert\alpha(t_0)\vert$ centered at the origin?

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