# Parametric Arclength

Suppose a parametric curve in \(\mathbb{R}^2\) is given by \(x(t)\), \(y(t)\), where the parameter \(t\) ranges over some given time interval. The curve traces out some arc in the plane, and one may wish to compute the length of this arc. This can be done using the following formula:

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 \[ \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} \, dt .\]

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## Derivation Of Formula

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.

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 \[\int_{a}^{b} \sqrt{1+(f'(t))^2} \, dt.\]

## Surface Area

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 \[2\pi \int_{a}^{b} f(x) \sqrt{1+(f'(x))^2} \, dx.\]

**Cite as:**Parametric Arclength.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/parametric-equations-arc-length-basic/