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


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


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.

No comments

(optional field, I won't disclose or spam but it's necessary to notify you if I respond to your comment)
All html tags except <b> and <i> will be removed from your comment. You can make links by just typing the url or mail-address.
Anti-spam question: