## Calculus: Early Transcendentals 8th Edition

The tangent line is horizontal at the points $~~(\frac{3\sqrt{2}}{2},\frac{\pi}{4})~~$ and $~~(-\frac{3\sqrt{2}}{2},\frac{3\pi}{4})$ The tangent line is vertical at the points $~~(3,0)~~$ and $~~(0, \frac{\pi}{2})$
$r = 3~cos~\theta$ Note that this is the equation of a circle and a circle is completed as $\theta$ goes from $0 \leq \theta \lt \pi$. For other values of $\theta$, the points on the circle are repeated. $x = r~cos~\theta = 3~cos^2~\theta$ $y = r~sin~\theta = 3~sin~\theta~ cos~\theta$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3cos^2~\theta-3sin^2~\theta}{-6~cos~\theta~sin~\theta} = \frac{sin^2~\theta-cos^2~\theta}{2~cos~\theta~sin~\theta}$ When the tangent line is horizontal, $\frac{dy}{dx} = 0$ We can find the values of $\theta$ when $\frac{dy}{dx} = 0$: $\frac{dy}{dx} = \frac{sin^2~\theta-cos^2~\theta}{2~cos~\theta~sin~\theta} = 0$ $sin^2~\theta-cos^2~\theta = 0$ $sin^2~\theta = cos^2~\theta$ $tan^2~\theta = 1$ $tan~\theta = \pm 1$ $\theta = \frac{\pi}{4}, \frac{3\pi}{4}$ When $\theta = \frac{\pi}{4}$: $r = 3~cos~\frac{\pi}{4} = \frac{3\sqrt{2}}{2}$ When $\theta = \frac{3\pi}{4}$: $r = 3~cos~\frac{3\pi}{4} = -\frac{3\sqrt{2}}{2}$ The tangent line is horizontal at the points $~~(\frac{3\sqrt{2}}{2},\frac{\pi}{4})~~$ and $~~(-\frac{3\sqrt{2}}{2},\frac{3\pi}{4})$ When the tangent line is vertical, the denominator of $\frac{dy}{dx}$ is $0$ We can find the values of $\theta$ when the denominator of $\frac{dy}{dx}$ is $0$: $2~cos~\theta~sin~\theta = 0$ $sin~\theta = 0~~~$ or $~~~cos~\theta = 0$ $\theta = 0, \frac{\pi}{2}$ When $\theta = 0$: $r = 3~cos~0 = 3$ When $\theta = \frac{\pi}{2}$: $r = 3~cos~\frac{\pi}{2} = 0$ The tangent line is vertical at the points $~~(3,0)~~$ and $~~(0, \frac{\pi}{2})$