Bellaard.com

# Curvature

By: Gijs Bellaard

## Osculating Circle

Consider a curve $$\gamma$$ within the plane. The curvature $$\kappa$$ of the curve $$\gamma$$ at a point $$p$$ is, historically, defined as the reciprocal of the radius of the circle that best fits the curve at $$p$$. This circle is known as the osculating circle. More precisely, we can create a circle through $$p$$ and any other point $$q$$ on $$\gamma$$ with its tangent at $$p$$ equal to the tangent of the curve at $$p$$. The osculating circle is then defined as the limit of these circles when we let $$q$$ go to $$p$$. Geometrically this definition is very satisfying, but algebraically it's quite cumbersome.

## Parametrized Curves

Again, consider a curve $$\gamma(t):[0,1] \to \bbR^n$$ within $$n$$-dimensional Euclidean space. We will denote a time derivative using Newton notation: $\dot \gamma := \frac{d\gamma}{dt}, \quad \ddot \gamma := \frac{d^2\gamma}{dt^2}.$ If the curve is arc-length parameterized, which means the magnitude of the velocity is equal to 1, i.e. $$\|\dot \gamma\|=1$$, then the curvature can be shown to be equal to: \begin{equation} \label{kappa1} \kappa = \|\ddot \gamma\|, \end{equation} where $$\|\cdot\|$$ is the Euclidean length. In other words, the curvature is the magnitude of the acceleration. However, typically, a curve is not arc-length parameterized. In this case the curvature becomes the following "nasty" expression: \begin{equation} \label{kappa2} \kappa = \frac{\|\dot \gamma \wedge \ddot \gamma\|}{\|\dot \gamma\|^3}. \end{equation} If one is unfamiliar with the wedge product: $$\|v \wedge w\|$$ refers the area of the parallelogram the vectors $$u$$ and $$v$$ make. An equivalent expression would be $$\|v \wedge w\| = \|v\|\|w\| \sin \alpha$$ where $$\alpha$$ is the angle between $$u$$ and $$v$$. Computationally this expression is efficient, but interpretively it's, in my opinion, awful. This is why I will introduce an equivalent definition of curvature that is interpretable. Before we continue, one can show that expression \eqref{kappa1} is indeed equal to \eqref{kappa2} in the case of an arc-length parameterized curve. To do this we first notice that by differentiation we have: $\|\dot \gamma\|^2 = 1 \Rightarrow \dot \gamma \cdot \dot \gamma = 1 \Rightarrow \dot \gamma \cdot \ddot \gamma = 0,$ where the central dot is the normal dot product on $$\bbR^n$$. This in turn means that the angle between $$\dot \gamma$$ and $$\ddot \gamma$$ is ninety degrees, and thus $$\|\dot \gamma \wedge \ddot \gamma\|$$ simplifies to just $$\|\dot \gamma\| \|\ddot \gamma\|$$. Showing that \eqref{kappa1} and \eqref{kappa2} are equal is now trivial, using again that $$\|\dot \gamma\|=1$$.

## An Interpretable Definition

To give this new definition of curvature we first need to define the direction $$d$$ of a curve. The direction of a curve is defined to be the normalized velocity vector of the curve: $d = \frac{\dot \gamma}{\|\dot \gamma\|}.$ This makes intuitive sense: by normalizing the velocity vector we have extracted its direction and omitted its magnitude. Using the direction of the curve we can simply define curvature as: \begin{equation} \label{kappa3} \kappa = \frac{\|\dot d\|}{\|\dot \gamma\|}. \end{equation} This definition is very much interpretable: the numerator is a direct measure of change of direction and the denominator is just speed. This means curvature can be said to be equal to change of direction over speed. Nice! To show \eqref{kappa3} is equivalent to \eqref{kappa2} we will first take a look at $$\dot d$$. Using the quotient rule we have: $\dot d = \frac{\|\dot \gamma\| \ddot \gamma - \dot \gamma \frac{\dot \gamma \cdot \ddot \gamma}{\|\dot \gamma\|}}{\|\dot \gamma\|^2} = \frac{\ddot \gamma}{\|\dot \gamma\|} - \left( d \cdot \frac{\ddot \gamma}{\|\dot \gamma\|} \right)d.$ If one is familiar with the Gram-Schmidt process, this last expression can be identified as the orthonormalization of $$\ddot \gamma/\|\dot \gamma\|$$ with respect to $$d$$. Because $$\|d\|=1$$ we also have that $$\|\dot d\| = \|d \wedge \dot d\|$$. Combining these two facts, and using properties of the wedge product, we see that: $\|\dot d\| = \|d \wedge \dot d\| = \left\| d \wedge \frac{\ddot \gamma}{\|\dot \gamma\|} \right\| = \frac{\|\dot \gamma \wedge \ddot \gamma\|}{\|\dot \gamma\|^2}$ From which we immediately see that \eqref{kappa3} is equal to \eqref{kappa2}.