<s>[INST] Curvature of plane parametric curves What is the neatest way to derive the following formula for the curvature of a parametric curve? $$\kappa =\frac{\|y'x''-y''x'\|}{(x'^2+y'^2)^{\frac{3}{2}}}  $$
 [/INST] The definition of curvature $\kappa$ is
$$
\kappa = \frac{d\alpha}{ds}
$$
where $\alpha = \arctan(\frac{dy}{dx})$ and $s$ is distance along the curve. $ds = dx\sqrt{(1 + (\frac{dy}{dx})^2)}\ $. Curvature has the units angle per length.
Start with the simplest parametric curve in two dimensions: $y = y(x)$. We then have:
$$
\begin{align*}
\kappa &= \frac{d\alpha}{ds} \\
&= \frac{d\alpha/dx}{ds/dx} \\
&= \frac{\frac{1}{1+(\frac{dy}{dx})^2} \frac{d^2y}{dx^2} }{\sqrt{(1 + (\frac{dy}{dx})^2)}} \\
&= \frac{\frac{d^2y}{dx^2}}{\left(1 + (\frac{dy}{dx})^2\right)^{3/2}}
\end{align*}
$$
Now consider a two-dimensional parameterized curve, given by $x = x(t)$ and $y = y(t)$. Denote $\frac{dx}{dt} = x'$ and $\frac{dy}{dt} = y'$ Then
\begin{align*}
\frac{dy}{dx} &= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'}{x'}\\
\end{align*}
\begin{align*}
\frac{d^2y}{dx^2} &= \frac{\frac{d}{dt}\left( \frac{dy}{dt} / \frac{dx}{dt} \right) }{\frac{dx}{dt}} \\
&= \frac{1}{\left( \frac{dx}{dt} \right)^3} \left[ \frac{dx}{dt} \frac{d^2y}{dt^2} - \frac{dy}{dt} \frac{d^2x}{dt^2} \right] \\
&= \frac{x'y'' - x''y'}{x'^3}
\end{align*}
Substituting $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the definition of $\kappa$, we get
$$
\kappa = \frac{x'y'' - x''y'}{\left( x'^2 + y'^2 \right)^{3/2}}
$$
</s>