Finding the Arc Length of a Cycloid

Find the length \(s\) of one arch of the cycloid: \[ \begin{equation*} x(t) =a(t-\sin t)\qquad y(t) =a(1-\cos t)\quad a>0 \end{equation*} \]

Figure 17 shows the graph of the cycloid.

Solution One arch of the cycloid is obtained when \(t\) varies from \(0\) to \(2\pi .\) Since \[ \begin{equation*} \dfrac{dx}{dt}=a-a\cos t=a(1-\cos t)\qquad\hbox{and}\qquad \dfrac{dy}{dt}=a\sin t \end{equation*} \]

are both continuous and are never simultaneously \(0\) on \(( 0,2\pi) \), the cycloid is smooth on \([ 0,2\pi ]\). The arc length \(s\) is \[ \begin{eqnarray*} s&=&\int_{a}^{b}\sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt} \right) ^{2}}~dt=\int_{0}^{2\pi } \sqrt{a^{2}(1-\cos t)^{2}+a^{2}\sin ^{2}t} \\ &=&a \int^{2\pi}_0 \sqrt{( 1-2\cos t+\cos ^{2}t) +\sin ^{2}t} ~dt = a \int^{2\pi}_0 \sqrt{1-2\cos t+1}~dt\\ &=& \sqrt{2} a \int^{2\pi}_0 \sqrt{1-\cos t}~dt \end{eqnarray*} \]

To integrate \(\sqrt{1-\cos t}\) we use a half-angle identity. Since \(\sin \dfrac{t}{2}\geq 0\) if \(0\leq t \leq 2\pi ,\) we have \(\sin \dfrac{t}{2}=\sqrt{\dfrac{1-\cos t}{2}}.\) Then \[ \begin{equation*} \sqrt{1-\cos t}=\sqrt{2}\sin \frac{t}{2} \end{equation*} \]

Now the arc length \(s\) from \(t=0\) to \(t=2\pi \) is \[ s= \sqrt{2}a \int_{0}^{2\pi } \sqrt{1 - \cos t}\, dt=\sqrt{2} a \int_{0}^{2\pi }\sqrt{2} \sin \frac{t}{2}~dt=2a\left[ -2\cos \frac{t}{2}\right] _{0}^{2\pi }=8a \]