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?


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:

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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