## Calculus (3rd Edition)

The speed of the cycloid $c\left( t \right) = \left( {4t - 4\sin t,4 - 4\cos t} \right)$ at points where the tangent line is horizontal is 8.
We have $x\left( t \right) = 4t - 4\sin t$, ${\ \ \ }$ $x'\left( t \right) = 4 - 4\cos t$, $y\left( t \right) = 4 - 4\cos t$, ${\ \ \ }$ $y'\left( t \right) = 4\sin t$. By Theorem 2, the speed of the particle is $s'\left( t \right) = \sqrt {{{\left( {4 - 4\cos t} \right)}^2} + {{\left( {4\sin t} \right)}^2}} = 4\sqrt 2 \sqrt {1 - \cos t}$. Since $2{\sin ^2}\frac{t}{2} = 1 - \cos t$ ${\ \ }$ or ${\ \ }$ $\sqrt 2 \left| {\sin \frac{t}{2}} \right| = \sqrt {1 - \cos t}$, we get (1) ${\ \ \ \ \ }$ $s'\left( t \right) = 8\left| {\sin \frac{t}{2}} \right|$. Using Eq. (8) of Section 12.1, the slope of the tangent line to the cycloid is $\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{4\sin t}}{{4\left( {1 - \cos t} \right)}} = \frac{{\sin t}}{{1 - \cos t}}$ The tangent line is horizontal if $\frac{{dy}}{{dx}} = 0$. So, we solve the equation $\frac{{\sin t}}{{1 - \cos t}} = 0$ to find the $t$-values when the tangent line is horizontal. Notice that $\frac{{dy}}{{dx}}$ is infinite if $\cos t = 1$. So, the required conditions are $\sin t = 0$ and $\cos t \ne 1$. The general solution is $t = \left( {2n - 1} \right)\pi$, where $n = 0, \pm 1, \pm 2, \pm 3,...$. Let us consider the case where $t \ge 0$. So, the solution is then $t = \left( {2n - 1} \right)\pi$ for $n = 1,2,3,...$. Substituting $t = \left( {2n - 1} \right)\pi$ into equation (1) gives $s'\left( t \right) = 8\left| {\sin \frac{t}{2}} \right| = 8\left| {\sin \frac{{\left( {2n - 1} \right)\pi }}{2}} \right|$ ${\ \ \ }$ for $n = 1,2,3,...$. Since $\left| {\sin \frac{{\left( {2n - 1} \right)\pi }}{2}} \right| = 1$ for $n = 1,2,3,...$, so the speed is $s'\left( t \right) = 8$.