I am trying to self teach differential geometry and to that effect I am trying to do the Homework in the MIT open course.
The specific question I am struggling with is:
Let $c$ be a regular curve such that $|c(s)| ≤ 1$ for all $s$. Suppose
that there is a point $t$ where $|c(t)| = 1$. Prove that the curvature at that
point satisfies $|κ(t)| ≥ 1$
This is what I have so far: The curve being regular implies that it can be arc length parametrized, which means that it's curvature is just the norm of the second derivative, or in 2D $|x'y'' - x''y'|$.
The curve magnitude being upper bounded by 1 means the curve is fully contained within the unit disk.
The curve being one at $t$ means that point is ON the unit circle. Thus my intuition says this curve can only "bend inwards" at a rate equal or higher than the circle, otherwise it will be an epsilon outside of the circle at a point infinitesimally close to $t$. In other words it seems to me that a if a curve with a point on the unit circle has a smaller curvature than that of the unit circle (i.e 1) the curve will "pop out" of it for some infinitesimal amount. But I am not sure if A) this is correct B) how to formalize it.
I am looking mostly for a hint and advice, not for the full solution. Thank you lots in advance.