all.
So with a parametric curve $\vec{r}=\langle x(t),y(t)\rangle$, curvature is given by $$\kappa=\frac{|x'y''-x''y'|}{(x'^2+y'^2)^{3/2}}.$$ When we have constant arc-length, an alternate expression is $$\kappa=|\vec{r}''(t)|=\sqrt{x''^2+y''^2}.$$ So I see why these are both valid expressions (I can derive them both), but when we have constant arc-length, I don't see why, when we have constant arc-length $$|x'y''-x''y'|=x''^2+y''^2$$ is true. Care to enlighten me?
For a unit speed curve, $1 = (x')^2 + (y')^2$. Differentiating and dividing by $2$ gives $0 = x'' x' + y'' y'$. Geometrically, the velocity $(x', y')$ and acceleration $(x'', y'')$ are orthogonal. Since the velocity is non-vanishing by hypothesis, there exists a real $\kappa$ such that $(x'', y'') = \kappa(-y', x')$; taking magnitudes shows $|\kappa|$ is in fact the curvature. Substituting $x'' = -\kappa y'$ and $y'' = \kappa x'$ gives $$ |x'y'' - y'x''| = |\kappa(x')^2 + \kappa(y')^2| = |\kappa| = \sqrt{(x'')^2 + (y'')^2}. $$