# a question about differential geometry, how to prove that $|k(t_{0})|>{1\over |\alpha(t_{0})|}$

Let $\alpha:(a,b)\to\Bbb R^2$ be a regular parametrized plane curve. Assume that there exits $t_{0}$, $a<t_{0}<b$, 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 $|k(t_{0})|>{1\over |\alpha(t_{0})|}$.

My thoughts: Without loss of generalization, I let t be viewed as the arc-length of $\alpha$, and then we can just prove $|k(s_{0})|>{1\over |\alpha(s_{0})|}$. Then suppose $\alpha(s)=(x(s),y(s))$, next I let $f(s)=x(s)^2+y(s)^2$, if so ${df(s)\over ds }={2\alpha(s)\dot\alpha(s)}$ . Then, I am stuck here, since I have no idea how to construct the relation between $k(s_{0})$ and $\alpha(s_{0})$. Can someone tell me how to prove it?

## 1 Answer

Let $d(s)=\langle\alpha(s),\alpha(s)\rangle$, $d$ has a maximum at $s_0$ thus the first order derivative $\langle\dot\alpha, \alpha\rangle=0$ vanishes and the second order $\langle\ddot\alpha,\alpha\rangle+\langle\dot\alpha,\dot\alpha\rangle\le0$. Since $s$ is the arc-length parameter we have $\langle\dot\alpha,\dot\alpha\rangle=1$, which leads to the desired result:
$1\le|\langle\ddot\alpha,\alpha\rangle|\le|\ddot\alpha||\alpha|=|k||\alpha|$

• why $\langle\ddot\alpha,\alpha\rangle+\langle\dot\alpha,\dot\alpha\rangle\le0$, I am a little confused – python3 Feb 4 '15 at 1:23
• the function d achieves a maximum at a point s0 so it's first order derivative vanishes and second order derivative is non-positive. – Xipan Xiao Feb 4 '15 at 1:39
• Oh-------,got it – python3 Feb 4 '15 at 1:48