Finding the Slope of the Tangent Line to a Cycloid

Consider the cycloid defined by \[ \begin{equation*} x(t) =a(t-\sin t)\qquad y(t) =a(1-\cos t)\qquad 0<t<2\pi \qquad a>0 \end{equation*} \]

(a) Show that the slope of the tangent line to the cycloid is given by \(\dfrac{\sin t}{1-\cos t}\).

(b) Find any points where the tangent line to the cycloid is horizontal.

Solution (a) \[ \begin{align*} x(t)&=at-a\sin t& y(t)=a-a\cos t \\ \dfrac{dx}{dt}&=a-a\cos t&\dfrac{dy}{dt}=a\sin t \end{align*} \]

For \(0<t<2\pi, {\dfrac{dx}{dt}} = a(1-\cos t) \neq 0\). Then the slope of the tangent line is \[ \begin{equation*} \frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{a\sin t}{a(1-cost)}=\frac{\sin t}{1-\cos t} \end{equation*} \]

(b) The cycloid has a horizontal tangent line when \(\dfrac{dy}{dt} =a\sin t=0,\) but \(\dfrac{dx}{dt}\neq 0.\) For \(0<t<2\pi\), we have \(a\sin t=0\) when \(t=\pi\). Since \(\dfrac{dx}{dt}\neq 0\) for \(0< t< 2\pi ,\) the cycloid has a horizontal tangent line when \(t=\pi ,\) at the point \((\pi a,2a)\). The equation of the horizontal tangent line is \(y=2a\), as shown in Figure 14.