Is there geometric intuition for the $\frac{3}{2}$ exponent in radius of curvature formula?

Question

I can follow a derivation of radius of curvature

$$\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}$$

but I can't see intuitively how there should be a $\frac{3}{2}$ exponent in there.

Is there a diagram or explanation that makes this intuitive?

Answer 1

Maybe I can't account for the $3/2$, but I can see where a $3$ would come from.

Consider the relation

$\frac{dx}{dy}=\frac{1}{dy/dx}$

Differentiate with respect to $y$:

$\frac{d^2x}{dy^2}=-\frac{(d^2y/dx^2)(dx/dy)}{(dy/dx)^2}=-(\frac{d^2y}{dx^2})(\frac{dx}{dy})^3$

Now consider the case where $|dy/dx|>>1$. Then the curvature becomes proportional to $\frac{d^2x}{dy^2}$ which, from the above, contains $(\frac{dx}{dy})^3$. The radius of curvature, being the reciprocal of the curvature, then becomes proportional to $(\frac{dy}{dx})^3$ in this limiting case.

Answer 2

A square root appears because the differential elements are computed using the arc length, $$ds=\sqrt{dx^2+dy^2}.$$

Hence further derivatives imply half-integer exponents.

Investigate the derivative of $\dfrac{dx}{ds}=\dfrac1{\sqrt{1+y'^2}}$.

Remnants of Pythagoras.