Curvature k of a 1D function graph y = f(x)

Result

With $f:\mathbb R \rightarrow \mathbb R$ $$\kappa(t) = \frac{|f''(t)|}{ \left ( \sqrt{1+f'(t)^2} \right)^3}$$

Derivation

We treat $y = f(x)$ as a special case of a parametrized 2D curve $\vec s(t) = [x(t), y(t)]$:

$$\begin{aligned} x(t) &= t \\ y(t) &= f(t) \end{aligned}$$

Recall the formula of the curvature of a curve:

$\kappa(t) = \frac{ \| s'(t) \times s''(t) \| }{ \| s'(t) \|^3 }$

Unpack the terms for the 2D case:

$\kappa = \frac{x'y'' - y'x''}{ \left( \sqrt{x'(t)^2 + y'(t)^2} \right)^3 }$

Now we just plug in the parametric version of our function graph:

$\begin{aligned} x'(t) &= 1 \\ x''(t) &= 0 \end{aligned}$

$\kappa(t) = \frac{|y''(t)|}{ \left ( \sqrt{1+y'(t)^2} \right)^3}$

And voila.

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Dwarves quietly sabotage giant’s labour.
(Fr: Les nains sapent sans bruit l’ouvrage des géants.)
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