Relations between curvature and area of simple closed plane curves.
Let $\gamma$ be a simple closed plane curve. We know that a curve with constant curvature $\kappa$ will trace a circle in the plane. The radius of this circle is the inverse of its curvature. Now, let's say we had a lower bound $L$ on the curvature $\kappa(\gamma)$ of $\gamma$. Then intuitively it seems that the entirety of $\gamma$ should be contained in some circle of curvature $L$. However, I can't seem to prove this. This leads me to ask the following:
If $\gamma$ is a simple closed plane curve with $\kappa(\gamma)$ bounded below by some constant $L \in \mathbb{R}$, how is the minimum radius of a circle containing $\gamma$ related to $\kappa(\gamma)$? In other words, what bound (if any) does $L$ place on the area $A(\gamma)$ of $\mathrm{int}(\gamma)$?
Solution 1:
We need two ingredients, the first one is the total curvature formula
$$
\int_C k(s)ds= 2\pi
$$
for every smooth simple closed planar curve $C$. Here $k(s)$ is the curvature function. If $k(s)\ge K>0$ for all $s$ (I do not like using $L$ for a curvature bound), we obtain
$$
K L(C)\le \int_C k(s) ds=2\pi,
$$
where $L=L(C)$ is the length of $C$. Thus,
$$
L\le \frac{2\pi}{K}.
$$
The second ingredient is the isoperimetric inequality:
$$
4\pi A(C)\le L^2(C)
$$
for every simple closed planar curve $C$, where $A(C)$ is the area bounded by $C$. (The equality is attained if and only if $C$ is a round circle.)
Combining these, we obtain:
$$
A(C)\le \frac{L^2}{4\pi} \le \frac{4\pi^2}{4\pi K^2}= \frac{\pi}{K^2}.
$$
This inequality is sharp, consider for instance $C$ which is a round circle. The converse is also true: if the equality is attained then the curve is a round circle.