Answer
1. Tangent line is vertical when $t = \left( {2n + 1} \right)\pi $ for $n = 0, \pm 1, \pm 2, \pm 3,...$
For $0 \le t \le 4\pi $, the points are $\left( {\pi ,\pi } \right)$, $\left( {3\pi ,3\pi } \right)$.
2. Tangent line is horizontal when $t = \pm \frac{\pi }{3} + 2\pi n$ for $n = 0, \pm 1, \pm 2, \pm 3,...$
For $0 \le t \le 4\pi $, the points are
$\left(1.913,-0.685\right)$, $\left(4.370,6.968\right)$, $\left(8.196,5.598\right)$ and $\left(10.653,13.251\right)$.
Work Step by Step
We have
$x\left( t \right) = t + \sin t$, ${\ \ }$ $x'\left( t \right) = 1 + \cos t$
$y\left( t \right) = t - 2\sin t$, ${\ \ }$ $y'\left( t \right) = 1 - 2\cos t$
Using Eq. (8) of Section 12.1, the slope of the tangent line is
$\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{1 - 2\cos t}}{{1 + \cos t}}$
1. The tangent line is vertical if $\frac{{dy}}{{dx}}$ is infinite. This occurs when $1 + \cos t = 0$. So, the solution is $t = \left( {2n + 1} \right)\pi $ for $n = 0, \pm 1, \pm 2, \pm 3,...$
For $0 \le t \le 4\pi $, substituting $t$ in $\left( {t + \sin t,t - 2\sin t} \right)$ we obtain the points $\left( {\pi ,\pi } \right)$, $\left( {3\pi ,3\pi } \right)$.
2. The tangent line is horizontal if $\frac{{dy}}{{dx}} = 0$. So, we solve the equation
$\frac{{dy}}{{dx}} = \frac{{1 - 2\cos t}}{{1 + \cos t}} = 0$
$1 - 2\cos t = 0$, ${\ \ \ }$ $\cos t = \frac{1}{2}$.
The solutions are $t = \pm \frac{\pi }{3} + 2\pi n$ for $n = 0, \pm 1, \pm 2, \pm 3,...$
For $0 \le t \le 4\pi $, substituting $t$ in $\left( {t + \sin t,t - 2\sin t} \right)$ we obtain the points $\left(1.913,-0.685\right)$, $\left(4.370,6.968\right)$, $\left(8.196,5.598\right)$ and $\left(10.653,13.251\right)$.
