Curvature exercise, need hint. Let $\alpha: \mathbb{R} \to \mathbb{R}^3$ be a parametrization of a smooth curve in space with $\|\alpha'(s)\| = 1$ constant. Suppose that for all $s$ within a sufficiently small neighborhood of fixed real $s_0$, we have$$\|\alpha(s)\| \le \|\alpha(s_0)\| = R.$$I need to show that the curvature of this curve at $s_0$ is greater than or equal to $1/R$. I'm completely stuck. Could I have a hint in the right direction?
 A: You first have to show that the tangent vector of the curve at $s_0$ is in the tangent plane of the sphere with radius $R$. Then you can do a Taylor series kind of expansion of the curve in a neighborhood of $\alpha(s_0)$. If the curvature is less than $1/R$, then for $s$ sufficiently close to $s_0$, you will find that the curve lies outside the specified radius. It's perhaps easier to do by assuming that $\alpha(s_0)=(R,0,0)$ first. Then do an isometry of $\mathbb{R}^3$ to get the answer. Of course, what works easiest depends on what kinds of theorems you have already proved on this subject.
A: (Since you didn't react to Alan U. Kennington's answer I'm going to fill in some details.)
Let $s_0=0$. Then
$$\alpha(s)=R {\bf u} +s {\bf v}+ o(s)\qquad(s\to0)$$
with ${\bf u}$ and ${\bf v}=\dot\alpha(0)$ unit vectors. It follows that
$$|\alpha(s)|^2=R^2 + 2Rs\langle{\bf u},{\bf v}\rangle+o(s)\qquad(s\to0)\ .$$
This is $>R^2$ for certain $s$ near $0$, unless $\langle{\bf u},{\bf v}\rangle=0$, resp. ${\bf v}\perp{\bf u}$. We now look at the next Taylor approximation:
$$\alpha(s)=R {\bf u} +s {\bf v}+{s^2\over 2}\ddot\alpha(0)+o(s^2)\qquad(s\to0)\ .$$
We obtain
$$|\alpha(s)|^2=R^2 + \bigl(1+R\langle{\bf u},\ddot\alpha(0)\rangle \bigr)s^2+o(s^2)\qquad(s\to0)\ .$$ The condition $|\alpha(s)|^2\leq R^2$ for all $s$ then implies
$$1+R\langle{\bf u},\ddot\alpha(0)\rangle \leq0\ ,$$
and therefore
$$\kappa(0)=|\ddot\alpha(0)|\geq\bigl|\langle{\bf u},\ddot\alpha(0)\rangle\bigr|\geq{1\over R}\ .$$
