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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 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.

2 Answers 2

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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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    I guess it's an assumption about $\alpha$ (regular) (we want a constant speed).2012-02-11
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Hint: What is the curvature of the circle of radius $\vert\alpha(t_0)\vert$ centered at the origin?