Question

Let a(s) = (x(s), y(s), 0) be a unit speed curve. Prove that

K = |x'y" – x''y'l.

Answer 1

Consider the identity

$(\dot{x}^2+\dot{y}^2)(\ddot{x}^2+\ddot{y}^2)=(\dot{x}\ddot{x}+\dot{y}\ddot{y})^2+(\dot{x}\ddot{y}-\ddot{x}\dot{y})^2$

With x=x(s), y=y(s) and dots denoting differentiation over the variable s, we have

$\dot{x}^2+\dot{y}^2=1$ by condition

$\ddot{x}^2+\ddot{y}^2=|\ddot{a}(s)|^2=K^2$ by definition

$\dot{x}\ddot{x}+\dot{y}\ddot{y}=\frac{1}{2}\frac{d}{ds}(\dot{x}^2+\dot{y}^2)=\frac{1}{2}\frac{d}{ds}(1)=0$

Then we obtain $K^2=(\dot{x}\ddot{y}-\ddot{x}\dot{y})^2$ or, equivalently, $K=|\dot{x}\ddot{y}-\ddot{x}\dot{y}|$, as required.