# 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.