Calculus (3rd Edition)

Solve using CAS, the points where the tangent line has slope $\frac{1}{2}$ are at $t \simeq 2.2143 + 2\pi n$ ${\ \ \ }$ for $n = 0, \pm 1, \pm 2, \pm 3,...$ For $0 \le t \le 4\pi$, we obtain the points $\left( {1.4143,1.6} \right)$ and $\left( {7.6975,1.6} \right)$.
We have $x\left( t \right) = t - \sin t$, ${\ \ }$ $x'\left( t \right) = 1 - \cos t$ $y\left( t \right) = 1 - \cos t$, ${\ \ }$ $y'\left( t \right) = \sin t$ Using Eq. (8) of Section 12.1, the slope of the tangent line to the cycloid is $\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{\sin t}}{{1 - \cos t}}$ Since the slope is $\frac{1}{2}$, so $\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{\sin t}}{{1 - \cos t}} = \frac{1}{2}$ We solve this equation using computer algebra system and the points where the tangent line has slope $\frac{1}{2}$ are at $t = \pi - {\tan ^{ - 1}}\frac{4}{3} + 2\pi n$, $t \simeq 2.2143 + 2\pi n$ ${\ \ \ }$ for $n = 0, \pm 1, \pm 2, \pm 3,...$ For $0 \le t \le 2\pi$, we obtain the point $\left( {1.4143,1.6} \right)$. For $0 \le t \le 4\pi$, we obtain the points $\left( {1.4143,1.6} \right)$ and $\left( {7.6975,1.6} \right)$.