Where does the 3rd power of the denominator of the curvature of the graph of a function come frome?

Question

According to Wikipedia, the curvature of the graph of a function $f$ is given by the following ratio (assuming second-differentiability).

$$\mathscr{k}_f(x) = \frac{f^{\prime\prime}(x)}{(1 + [f^{\prime}(x)]^2)^{\frac{3}{2}}}$$

The denominator of $\mathscr{k}_f(x)$ is nearly of the form $(a^2 + b^2)^{\frac{1}{2}}$ that makes me think about it as the radius of a circle, but this intuition is cut short by taking that "radius" to the third power. Where does this third power come from in the derivation or definition of this functional?

Answer 1

The curvature $K$ is defined as (the absolute value of) the rate of change of the angle of the curve's tangent with respect to the curve's arc length. That is, for $$y = f(x)$$ $$K = |\frac{d\theta}{ds}|$$

where $\theta$ is the angle of the tangent and $s$ is the arc length.

Now $$\tan\theta = f'(x) = y'$$ so $$\sec^2\theta\frac{d\theta}{dx} = y''$$ $$(1+\tan^2\theta)\frac{d\theta}{dx} = y''$$ $$(1+y'^2)\frac{d\theta}{dx} = y''$$ Therefore $$\frac{d\theta}{dx} = \frac{y''}{1+y'^2}$$

From Pythagoras' theorem, $$ds^2 = dx^2 + dy^2$$ so $$\frac{ds}{dx} = (1 + y'^2)^\frac12$$

And since $$K = |\frac{d\theta}{ds}| = |\frac{d\theta}{dx} \frac{dx}{ds}|$$ we have $$K = \frac{|y''|}{(1+y'^2)^\frac32}$$

(The denominator is guaranteed to be positive).

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You should verify that a circle has constant curvature, which is the reciprocal of its radius.

Answer 2

It is the numerator that should be looked at as it influences shape more strongly.

To me it appears that you are thinking more on rotation up from the vicinity's lowest tangent position at bottom position of a parabola, circle, catenary etc.

Curvature is defined as the speed or rate at which slope changes. When the rotation $\phi$ goes all the way up to 90 degrees then to include same curvature it is necessary to augment/ modify the parabola curvature away from the starting “shallow” situation.

The curvature is not proportional to $\sec \phi$ but should be proportional to $\sec^3 \phi$

If that change is not made, it keeps getting flatter and flatter.E.g., if we had only the ode

$$ \dfrac{y''}{\sqrt{1 +y^{‘2}}}$$

then on integration we find this to be a catenary that refuses to curl up. With a circle, curvature is increased to:

$$ \dfrac{y’’}{[1 +y^{‘2}]^\frac32}$$

Similarly leave alone the numerator and looking at the denominator as sum of two squares $ dx^2+dy^2 $ for constant

$$ \dfrac{y’’}{[1 +y^{'2}]}$$

we have a catenary of uniform strength where vertical limbs straighten up fast by this slope

In the more exaggerated case of a Cornu's spiral for example where the curvature is proportional to the arc length

$$ \dfrac{y’’}{[1 +y^{‘2}]^\frac32}= s \text{(=arc length )} $$

it curls up almost indefinitely like a creeper or ivy.

The behavior of slope / curvature would be clear on seeing graphs of some of these curves very roughly sketched: