Differential Geometry - Computation Help I'm trying to learn differential geometry through one of MIT's online courses (lecture notes found here:  http://ocw.mit.edu/courses/mathematics/18-950-differential-geometry-fall-2008/lecture-notes/ch1_revised.pdf) and am stuck with what should be an easy question.  The question asks to show the following:  Suppose $c(s)$ is a regular curve ($c'(s) \ne 0$) in the plane with $|c(s)| \le 1$ and suppose there is a point t with $|c(t)| = 1$.  Then $|\kappa(t)| \ge 1$.  Here, $\kappa$ is defined as 
$\frac{det(c', c'')}{||c'||^3}.$
It's should be straightforward if we write $c$ as the graph of a function and rotate so that $c(t) = (0,1)$.  However, I feel as if there should be a more elegant solution using only what's found in the first day's notes (found in the link).  I just don't see it.    

Let me assume, without loss of generality, that $c$ is parametrized by arclength, so that $|c'|\equiv 1$, and let me also assume $t_0=0$ (where $c(t_0)$ has length one). Furthermore, upon reparametrizing $c$ backwards, we may assume that $\mathcal{B}=(c(0),c'(0))$ is a direct orthonormal basis of $\Bbb R^2$ (this frame is indeed orthonormal, as follows from the fact that $|c|^2$ attains a local maximum at $0$, and the the preceding hypothesis).

The curvature then takes on the form 
$$\kappa(0)=\det(c'(0),c''(0))=\det_{\mathcal B}(c'(0),c''(0))=-c(0)\cdot c''(0)$$
Also, the Taylor expansion of $c$ near $0$ is $c(t)=c(0)+tc'(0)+\frac12t^2c''(0)+o(t^2)$, and thus $$\begin{array}{ccl}
|c(t)|^2&=&\underbrace{c(0)\cdot c(0)}_{=1}+2t\underbrace{c(0)\cdot c'(0)}_{=0}+(\underbrace{2\times\frac12 c''(0)\cdot c(0)}_{=-\kappa(0)}+\underbrace{c'(0)\cdot c'(0)}_{=1})t^2+o(t^2)\\
&=&1+(1-\kappa(0))t^2+o(t^2)
\end{array}$$
This expression has to stay $\leq 1$ near $0$, which in turn forces $1-\kappa(0)\leq 0$
