Differential geometry of planar curves

In this blog we explore some basic concepts of the geometry of smooth curves in the plane. A moving point \(p\) is described mathematically by letting \(p\) be a function of time \(t\), so that we can write \[p(t)=(x(t),y(t)).\] The velocity is the first derivative of \(p\) with respect to \(t\), that is \[v(t)=\frac{dp}{dt}(t)=\left(\frac{dx}{dt}(t),\frac{dy}{dt}(t)\right),\] For a given time \(t\), the norm of the velocity vector is a scalar known as the speed. We can measure speed in more familiar terms by introducing the arc-length parameter \(s\). More concretely, the arc-length of the curve segment between time \(0\) and \(t\) is \[s(t)=\int_{0}^{t}\left|\frac{dp}{dt}(u)\right|du,\] where \[\left|\frac{dp}{dt}\right|=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}.\] We conclude that speed is the instantaneous rate of change of the distance traveled along the curve; this is what the speedometer of a car measures. By the chain rule \[\frac{dp}{dt}=\frac{dp}{ds}\frac{ds}{dt}\] and hence \[\left|\frac{dp}{ds}\right|=\left|\frac{dp}{dt}\right|/\left|\frac{ds}{dt}\right|=1.\] For any given \(s,\) the vector \[T(s)=\frac{dp}{ds}(s).\] is a unit tangent vector to the curve parametrized by \(p\), and we will refer to \(T\) as the unit tangent field. The field \[N=\left(-\frac{dy}{ds},\frac{dx}{ds}\right)\] is the known as the unit normal field and is defined so that the pair \(T,N\) form an orthonormal field of vectors along the curve. Taking the derivative of the formula \[T\cdot T=1,\] we get \[\frac{dT}{ds}\cdot T=0.\] This means that there must exist a function \(\kappa=\kappa(s)\) such that \[\frac{dT}{ds}=\kappa N.\] The function \(\kappa\) is know as the curvature function and it can be computed as \[\kappa=\frac{dT}{ds}\cdot N.\] We will now explore the meaning of this function.

Physical interpretation of curvature

By using the chain rule we get

\[\begin{aligned} \frac{dp}{dt} & =\frac{dp}{ds}\frac{ds}{dt},\\ \frac{d^{2}p}{dt^{2}} & =\frac{dp}{ds}\frac{d^{2}s}{dt^{2}}+\frac{d^{2}p}{ds^{2}}\left(\frac{ds}{dt}\right)^{2}.\end{aligned}\] From the above formulas we obtain a formula for the acceleration of \(p\) as follows \[\frac{d^{2}p}{dt^{2}}=a_{T}T+a_{N}N,\] where the tangential and normal accelerations \(a_{T}\) and \(a_{N}\) are \[a_{T}=\frac{d^{2}s}{dt^{2}}\text{ and }a_{N}=\kappa\left(\frac{ds}{dt}\right)^{2}.\] If we let \(t=s\) we have that \[\frac{ds}{dt}=1,\] and hence \[\kappa=a_{N}.\] In words: the curvature is the normal component of the acceleration of a particle when that particle moves at unit speed.

Geometric interpretation of curvature

We can obtain a second interpretation of the curvature function as follows. Since \(T\) has unit norm, we can find a function \(\theta=\theta(s)\) such that \[T=(\cos(\theta),\sin(\theta))\] Taking the derivative with respect to \(s\) we obtain \[\frac{dT}{ds}=\frac{d\theta}{ds}(-\sin(\theta),\cos(\theta))=\frac{d\theta}{ds}N\] and hence \[\kappa=\frac{dT}{ds}\cdot N=\frac{d\theta}{ds}.\] We conclude that the curvature is the rate of change of the angle of the tangent vector with respect to the arc-length parameter.

Curve from curvature

We can construct a curve with a given curvature function \(\kappa\) as follows. From the above formulas we have that \[\theta(s)=\theta_{0}+\int_{s_{0}}^{s}\kappa(u)du,\] and \[\begin{aligned} p(s) & =p_{0}+\int_{s_{0}}^{s}(\cos(\theta(u),\sin(\theta(u))du\\ & =p_{0}+\left(\int_{s_{0}}^{s}\cos(\theta(u))du,\int_{s_{0}}^{s}\sin(\theta(u))du\right)\end{aligned}\] The scalar \(\theta_{0}\) and the vector \(p_{0}\) determine the position and orientation of our curve.

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