Let $f(t) = (x(t),y(t))$, not necessarily parametrized by arclength.
We define the unit tangent vector, $T(t) = (1/|f'(t)|)(x',y')$. Also the normal vector, $N(t) = (1/|f'(t)|)(-y',x')$, which is defined this way so that the rotation $T \rightarrow N$ is done anti-clockwise.
Since $T'$ and $N$ generate the same vector space we then define the curvature , for each $t$, as the only scalar $k(t)$ such that $T'(t) = k(t)N(t)$.
Since $|N(t)| = 1$ then we get $k(t) = T'|N$, where | is the dot product.
After computing $T'$ and that dot product i get
$k(t) = (1/(|f'(t)|^2 )) (x'y''-x''y')$
when it is supposed to be
$k(t) = (1/(|f'(t)|^3 )) (x'y''-x''y')$.
Anyone familiar with these formulae?
details:
$T(t)=(1/|f'(t)|)(x',y')$.
$T'=[\frac{\mathrm{d}}{\mathrm{d}t}(1/|f'(t)|)](x',y')+(1/|f'(t)|)(x'',y'')$
Since the first vector of the sum above is orthogonal to $N(t)$ then its dot product is $0$.
So $$T'|N = [(1/|f'(t)|)(x'',y'')] | N = [(1/|f'(t)|)(x'',y'')] | [(1/|f'(t)|)(-y',x')] = (1/(|f'(t)|^2 )) (x'y''-x''y')$$
Here how I get the formula :
We have $$T\left(t\right)=\frac{\left(x',y'\right)}{\sqrt{x'^2+y'^2}}$$ and then the velocity (the norm of $T$) is given by $$\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt{x'^2+y'^2}.$$ Hence, we can compute : $$\frac{\mathrm{d}T}{\mathrm{d}t}\left(t\right) =\frac{\left(x'',y''\right)}{\sqrt{x'^2+y'^2}}-\frac{1}{2}\left(x',y'\right)\frac{2x'x''+2y'y''}{\left(x'^2+y'^2\right)^{3/2}} =\left(x''y'-x'y''\right)\frac{\left(y',-x'\right)}{\left(x'^2+y'^2\right)^{3/2}},$$ $$k\left(t\right)=\left\Vert\frac{\mathrm{d}T}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}s}\right\Vert =\left|x''y'-x'y''\right|\frac{\sqrt{x'^2+y'^2}}{\left(x'^2+y'^2\right)^{3/2}}\times\frac{1}{\sqrt{x'^2+y'^2}} =\frac{\left|x''y'-x'y''\right|}{\left(x'^2+y'^2\right)^{3/2}}.$$