Applications of Definite Integrals: Arc Length

Suppose that y=f(x) is continuously differentiable on [a,b].

Figure 1

Figure 1 shows the curve y=f(x) is approximated by a PL (Piecewise Linear) curve (in red). Suppose that the line segments of the PL curve are infinitesimal and denote by dL a line element. Then it is given by \begin{align*}dL&=\sqrt{(dx)^2+(dy)^2}\\&=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\\&=\sqrt{1+[f'(x)]^2}dx\end{align*} Hence the length of the curve y=f(x), a\leq x\leq b is \begin{equation}\label{eq:arclength}L=\int_{x=a}^{x=b}dL=\int_a^b\sqrt{1+[f'(x)]^2}dx\end{equation}

Often a curve is represented by a parametric equation r(t)=(x(t),y(t)), \alpha\leq t\leq\beta such that x(t) and y(t) are continuously differentiable. Such a curve is called smooth. A typical example is a motion in the plane (or in space). For a smooth curve r(t)=(x(t),y(t)), \alpha\leq t\leq\beta, dL is \begin{align*}dL&=\sqrt{(dx)^2+(dy)^2}\\&=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\end{align*} and hence \begin{equation}\label{eq:arclength2}\int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\end{equation} If r(t) is the motion of a particle, \left|\frac{dr(t)}{dt}\right|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} is the speed of the particle and so \eqref{eq:arclength2} represents the distance traveled by the particle over the time interval \alpha\leq t\leq\beta.

Example (The circumference of a circle). Find the length of the circle of radius r.

Solution.  The equation of the upper semicircle of radius r centered at the origin is y=\sqrt{r^2-x^2}, -r\leq x\leq r. \frac{dy}{dx}=-\frac{x}{\sqrt{r^2-x^2}} and so the circumference is \begin{align*}2\int_{-r}^r\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx&=2\int_{-r}^r\frac{r}{\sqrt{r^2-x^2}}dx\\&=4\int_0^r\frac{r}{\sqrt{r^2-x^2}}dx\\&=4\int_0^r\frac{1}{\sqrt{1-\left(\frac{x}{r}\right)^2}}dx\\&=4r\int_0^1\frac{1}{\sqrt{1-u^2}}du\\&=4r[\sin^{-1}(u)]_0^1\\&=4r\left(\frac{\pi}{2}\right)=2\pi r\end{align*}

The circle x^2+y^2=r^2 of radius r centered at the origin can be described by the parametric equations x=r\cos t,\ y=r\sin t,\ 0\leq t\leq 2\pi \frac{dx}{dt}=-r\sin t and \frac{dy}{dt}=r\cos t so \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=r. Therefore the circumference is \int_0^{2\pi}rdt=2\pi r. we find that using parametric equations of the circle comes with a much simpler calculation.

Example. Find the length of the asteroid x^{\frac{2}{3}}+y^{\frac{2}{3}}=1

Solution. The name asteroid is due to its graph as seen in Figure 2.

Figure 2. Asteroid

The asteroid can be parametrized as x=\cos^3 t,\ y=\sin^3 t,\ 0\leq t\leq 2\pi \frac{dx}{dt}=-3\cos^2 t\sin t and \frac{dy}{dt}=3\sin^2 t\cos t so \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=9\cos^2t\sin^2t. For 0\leq t\leq\frac{\pi}{2}, \sin t\geq 0 and \cos t\geq 0. Hence \begin{align*}L&=4\int_0^{\frac{\pi}{2}}\sqrt{9\cos^2t\sin^2t}dt\\&=12\int_0^{\frac{\pi}{2}}\cos t\sin tdt\\&=6\int_0^{\frac{\pi}{2}}\sin 2tdt\\&=3[-\cos 2t]_0^{\frac{\pi}{2}}=6\end{align*} The integral \int_0^{\frac{\pi}{2}}\cos t\sin tdt also can be done using a substitution. Let u=\sin t. Then du=\cos tdt so \begin{align*}\int_0^{\frac{\pi}{2}}\cos t\sin tdt&=\int_0^1udu\\&=\frac{1}{2}\end{align*}

Example. Find the length of the curve y=\frac{4\sqrt{2}}{3}x^{\frac{3}{2}}-1,\ 0\leq x\leq 1

Solution. \frac{dy}{dx}=2\sqrt{2x} so \sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+8x} and \begin{align*}L&=\int_0^1\sqrt{1+8x}dx\\&=\frac{2}{3}\cdot\frac{1}{8}[(1+8x)^{\frac{3}{2}}]_0^1\\&=\frac{13}{6}\end{align*}

Example. Find the length of the curve y=\left(\frac{x}{2}\right)^{\frac{2}{3}}, 0\leq x\leq 2.

Solution. \frac{dy}{dx}=\frac{1}{3}\left(\frac{2}{x}\right)^{\frac{1}{3}}. Note that this derivative does not exist at x=0. Rewrite the function as x=2y^{\frac{3}{2}}, 0\leq y\leq 1. \frac{dx}{dy}=3\sqrt{y} and \begin{align*}L&=\int_{y=0}^{y=1}dL\\&=\int_{y=0}^{y=1}\sqrt{(dx)^2+(dy)^2}\\&=\int_0^1\sqrt{\left(\frac{dx}{dy}\right)^2+1}dy\\&=\int_0^1\sqrt{9y+1}dy\\&=\frac{1}{9}\cdot\frac{2}{3}[(9y+1)^{\frac{3}{2}}]_0^1\\&=\frac{2}{27}(10\sqrt{10}-1)\end{align*}

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