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.
Best Answer
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.