Plane Curves

Consider the equation of a circle $$x^2+y^2=R^2,$$ where $R$ is its radius. How do we parametrize it?

Recall that a function $f(t)$ can be parametrized an infinite number of ways. Moreover, the standard parametrization is $$r(t)=t{\partial }_x+f(t){\partial }_y.$$ However, the most common way to parametrize our circle is as follows: $$r(t)=R(\sin t{\partial }_x+\cos t{\partial }_y),$$ where $t\in [0,2\pi ).$

Note: this is not its standard parametrization.

If we want to find its perimeter, then we compute the following integral: $$\begin{array}{rl}P& ={\int }_0^{2\pi }|r^{\prime }(t)|dt\\ & ={\int }_0^{2\pi }\sqrt{R^2{\cos }^{2}t+R^2{\sin }^{2}t}dt\\ & =2\pi R,\end{array}$$ which should be familiar to you if you have some basic knowledge in geometry.

Finally, we can use the indefinite form of the above integral to express our parametric equation in terms of its arc length $s$, as follows: $$r(s)=R(\sin \frac{s}{R}{\partial }_x+\cos \frac{s}{R}{\partial }_y).$$ If we wish to compute the curvature $\varkappa$ of the circle, then we can use the above natural parametrization as follows: $$\begin{array}{rl}T& =\dot{r}\\ & =\cos \frac{s}{R}{\partial }_x-\sin \frac{s}{R}{\partial }_y,\\ \varkappa & =|\dot{T}|\\ & =\left|\frac{1}{R}(-\sin \frac{s}{R}{\partial }_x-\cos \frac{s}{R}{\partial }_y)\right|\\ & =\frac{1}{R}.\end{array}$$ This makes sense since the circle is the only plane curve with constant curvature.