Consider the formula for curve (arc-) length in integral calculus:
$$\int_a^b\sqrt{1+\left(\frac{\partial y}{\partial x}\right)^2}dx$$
What would the geometrical opposite thing to measure be? Some kind of leaping curve-radius? I imagine allowing a normal line for every point on the curve:
$$y=kx+m, \\ k = -\left(\frac{\partial y}{\partial x}\right)^{-1}\\ m = f(x)-kx$$
And then finding some radius which maps our function down to a shortest radially displaced function in some sense. Simplest example a circle would give constant radius and one point in the middle for any point on the arc.
How could this be made more formal?
I have been thinking about some optimization problem. Since in the normal lines above, $k$ and $m$ are decided for each $(x,y)$ for $y=f(x)$. Then we can build a system of lines equations minimizing 2-norm should give us (endpoint,radius) pairs.