Parametric Arclength

Parametric Arclength is the length of a curve given by parametric equations. For instance, the curve in the image to the right is the graph of the parametric equations $x(t) = t^2 + t$ and $y(t) = 2t - 1$ with the parameter $t$. One could wish to find the arclength of curve between the points $t =-\frac{1}{2}$ and $t=1$, as noted by the thicker red curve in the image to the right.

Generalized, a parametric arclength starts with a parametric curve in $\mathbb{R}^2$. This is given by some parametric equations $x(t)$, $y(t)$, where the parameter $t$ ranges over some given interval. The following formula computes the length of the arc between two points $a,b$.

Consider a parametric curve $(x(t), y(t))$, where $t\in [a,b]$. The length of the arc traced by the curve as $t$ ranges over $[a,b]$ is $$\int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} \, dt .$$

Consider the curve given by parametric equations $x(t)$, $y(t)$, where $t$ ranges over $[0,1]$. To compute the arclength of this curve, one can approximate the curve with polygonal paths and take the limit of the lengths of these paths as they become more refined.

A curve in the plane, approximated a polygonal path consisting of straight line segments.

If $0 \le t_1 < t_2 \le 1$, the straight line segment between $(x(t_1), y(t_1)$ and $(x(t_2), y(t_2))$ has length $$\sqrt{(x(t_1) - x(t_2))^2 + (y(t_1) - y(t_2))^2}.$$ Using the mean value theorem, there are $c, d \in (t_1, t_2)$ such that $$x(t_2) - x(t_1) = (t_2 - t_1) x'(c),$$ $$y(t_2) - y(t_1) = (t_2 - t_1) y'(d).$$ Thus, the length of this line segment equals $$(t_2 - t_1) \sqrt{x'(c)^2 + y'(d)^2}.$$ As the $t_1$ and $t_2$ are made closer (in order to make the line segments approximating the curve shorter), this approaches the element $$\sqrt{x'(t)^2 + y'(t)^2 } \, dt.$$ Thus, the arclength of the curve over $[0,1]$ is $$\int_{0}^{1} \sqrt{x'(t)^2 + y'(t)^2} \, dt.$$

In particular, if $f: \mathbb{R} \to \mathbb{R}$ is differentiable, the graph of $f$ can be parametrized as $t \mapsto (t, f(t))$. Then, the arclength of the segment of this graph bounded by $(a, f(a))$ and $(b,f(b))$ equals $$\int_{a}^{b} \sqrt{1+(f'(t))^2} \, dt.$$

Suppose the graph of $y = f(x)$ is rotated about the $x$-axis to obtain a solid. By the above formula for arclength, the infinitesimal arclength of this graph at some $x\in \mathbb{R}$ is $\sqrt{1+(f'(x))^2} \, dx$. When this arc of the graph is rotated about the $x$-axis, the solid formed is an infinitesimal cylinder of surface area $2\pi f(x) \sqrt{1+(f'(x))^2} \, dx$. Integrating this over the desired interval, one obtains the formula below.

Let $\gamma$ denote the arc of the graph $y = f(x)$ bounded by the points $(a,f(a))$ and $(b,f(b))$. The surface area of the solid formed by rotating $\gamma$ about the $x$-axis is $$2\pi \int_{a}^{b} f(x) \sqrt{1+(f'(x))^2} \, dx.$$