Deriving curvature formula How do you derive the formula for unsigned curvature of a curve $\gamma (t) = (x(t), y(t)$ which is not necessarily parameterised by arc-length. 
The formula given is $$|\kappa (t)| = \left|\frac{x'' y' - y'' x'}{((x')^2 + (y')^2)^{3/2}}\right|$$
All the definitions of curvature in my notes give it in terms for an arc-length parameterisation $T'(s) = \kappa(s) N(s)$. This is apparently meant to be 'easily derivable', am I missing something? 
 A: Assume that your curve $\gamma(t) = (x(t),y(t))$ is parametrised by a parameter of your choice. We need to relate $t$ to the arc-length parameter $s$ and, more importantly, relate their derivatives. 
The arc-length parameter is defined as follows:
$$s(t) = \int ||\dot{\gamma}|| \ dt = \int \sqrt{\dot{x}^2+\dot{y}^2} \ dt \, , $$
where $\dot{\gamma}$ (etc) represents $d\gamma/dt$. From this, we can see that:
$$\frac{ds}{dt} = \sqrt{\dot{x}^2+\dot{y}^2} \, . $$
Using the chain rule, we see that:
$$\begin{array}{cccCC}
\frac{d\gamma}{dt} &=& \frac{ds}{dt} \, \frac{d\gamma}{ds} &=& \frac{ds}{dt}{\bf T} \\
\frac{d^2\gamma}{dt^2} &=& \frac{d^2s}{dt^2} \, \frac{d\gamma}{ds} + \left(\frac{ds}{dt}\right)^{\! \! 2} \, \frac{d^2\gamma}{ds^2} &=& \frac{d^2s}{dt^2} {\bf T} + \left(\frac{ds}{dt}\right)^{\! \! 2} \kappa{\bf N}
\end{array}$$
We can take the scaler (dot) product of both sides of this last expression with ${\bf N}$. Since ${\bf T}$ and ${\bf N}$ are both unit length and are perpendicular we have $\langle {\bf T},{\bf T}\rangle = \langle {\bf N},{\bf N}\rangle = 1$ and $\langle {\bf T},{\bf N}\rangle = 0$. Thus: 
$$\left\langle \frac{d^2\gamma}{dt^2},{\bf N} \right\rangle = \kappa \left(\frac{ds}{dt}\right)^{\! \! 2} \, .$$
We know that $ds/dt = (\dot{x}^2+\dot{y}^2)^{1/2}$ and so $(ds/dt)^2 = \dot{x}^2+\dot{y}^2$. We now have:
$$\begin{array}{ccc}
\kappa(\dot{x}^2+\dot{y}^2) &=& \left\langle \frac{d^2\gamma}{dt^2},{\bf N} \right\rangle \\
\kappa &=& \frac{1}{\dot{x}^2+\dot{y}^2} \left\langle \left(\ddot{x},\ddot{y}\right),\frac{(-\dot{y},\dot{x})}{\sqrt{\dot{x}^2+\dot{y}^2}} \right\rangle \\
 &=& \frac{1}{\dot{x}^2+\dot{y}^2} \left( \frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}} \right) \\
 &=& \frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\left(\dot{x}^2+\dot{y}^2\right)^{3/2}} 
\end{array}$$
If you want the unsigned curvature then just take the absolute value of both sides.
A: Let 
$$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)$$
be a regular $C^2$-parametrization of a curve $\gamma$. Then $s'(t)=\sqrt{x'^2(t)+y'^2(t)}>0$ for all $t$. The function
$$\theta(t):={\rm arg}\bigl(x'(t),y'(t)\bigr)$$
gives the polar angle of the tangent vector along $\gamma$. From
$$\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2},\>{x\over x^2+y^2}\right)$$ we obtain
$$\theta'(t)={-y'\over x'^2+y'^2}x''+{x'\over x'^2+y'^2}y''={x'y''-x''y'\over x'^2+y'^2}$$
and therefore
$$\kappa:={d\theta\over ds}={\theta'\over s'}={x'y''-x''y'\over (x'^2+y'^2)^{3/2}}\ .$$
