## Calculus: Early Transcendentals 8th Edition

The tangent line is horizontal at the points $~~(\frac{3}{2},\frac{\pi}{3}),~~$ $(\frac{3}{2},\frac{5\pi}{3}),~~$ and $~~(0,\pi)$ The tangent line is vertical at the points $~~(2,0),~~$ $(\frac{1}{2}, \frac{2\pi}{3}),~~$ and $~~(\frac{1}{2}, \frac{4\pi}{3})$
$r = 1+~cos~\theta$ Note that $r \geq 0$ for all values of $\theta$. Thus we need to consider values of $\theta$ from $0 \leq \theta \lt 2\pi$. For other values of $\theta$, the points on the curve are repeated. $x = r~cos~\theta = cos~\theta+~cos^2~\theta$ $y = r~sin~\theta = sin~\theta+sin~\theta~ cos~\theta$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{cos~\theta+cos^2~\theta-sin^2~\theta}{-sin~\theta-2~cos~\theta~sin~\theta}$ When the tangent line is horizontal, $\frac{dy}{dx} = 0$ We can find the values of $\theta$ when $\frac{dy}{dx} = 0$: $\frac{dy}{dx} = \frac{cos~\theta+cos^2~\theta-sin^2~\theta}{-sin~\theta-2~cos~\theta~sin~\theta} = 0$ $cos~\theta+cos^2~\theta-sin^2~\theta = 0$ $cos~\theta+cos^2~\theta-(1-cos^2~\theta) = 0$ $2~cos^2~\theta+cos~\theta-1 = 0$ $(2~cos~\theta-1)(cos~\theta+1) = 0$ $cos~\theta = \frac{1}{2}~~~$ or $~~~cos~\theta = -1$ $\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \pi$ When $\theta = \frac{\pi}{3}$: $r = 1+cos~\frac{\pi}{3} = \frac{3}{2}$ When $\theta = \frac{5\pi}{3}$: $r = 1+cos~\frac{5\pi}{3} = \frac{3}{2}$ When $\theta = \pi$: $r = 1+cos~\pi = 0$ The tangent line is horizontal at the points $~~(\frac{3}{2},\frac{\pi}{3}),~~$ $(\frac{3}{2},\frac{5\pi}{3}),~~$ and $~~(0,\pi)$ 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 $0$: $-sin~\theta-2~cos~\theta~sin~\theta = 0$ If $sin~\theta \neq 0$: $2~cos~\theta~sin~\theta = -sin~\theta$ $cos~\theta = -\frac{1}{2}$ $\theta = \frac{2\pi}{3}, \frac{4\pi}{3}$ If $sin~\theta = 0,~~$ then $\theta = 0, \pi$ However, above we stated that the tangent line is horizontal when $\theta = \pi$ We can use L'Hospital's Rule to verify the following limit: $\lim\limits_{\theta \to \pi}\frac{dy}{dx}$ $= \lim\limits_{\theta \to \pi}\frac{cos~\theta+cos^2~\theta-sin^2~\theta}{-sin~\theta-2~cos~\theta~sin~\theta}$ $= \lim\limits_{\theta \to \pi}\frac{-sin~\theta-2~cos~\theta~sin~\theta-2~sin~\theta~cos~\theta}{-cos~\theta+2~sin^2~\theta-2~cos^2~\theta}$ $= \frac{0}{-1}$ $= 0$ Both the numerator and denominator approach $0$ as $\theta \to \pi$, but since $\lim\limits_{\theta \to \pi}\frac{dy}{dx} = 0$, the tangent line is horizontal at $\theta = \pi$ When $\theta = 0$: $r = 1+cos~0 = 2$ When $\theta = \frac{2\pi}{3}$: $r = 1+cos~\frac{2\pi}{3} = \frac{1}{2}$ When $\theta = \frac{4\pi}{3}$: $r = 1+cos~\frac{4\pi}{3} = \frac{1}{2}$ The tangent line is vertical at the points $~~(2,0),~~$ $(\frac{1}{2}, \frac{2\pi}{3}),~~$ and $~~(\frac{1}{2}, \frac{4\pi}{3})$