## Calculus: Early Transcendentals 8th Edition

The slope of the tangent line is $~1$
$r = cos~2\theta$ $x = r~cos~\theta = cos~2\theta~cos~\theta$ $y = r~sin~\theta = cos~2\theta~sin~\theta$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-2~sin~2\theta~sin~\theta+cos~2\theta~cos~\theta}{-2~sin~2\theta~cos~\theta-cos~2\theta~sin~\theta}$ When $\theta = \frac{\pi}{4}$: $\frac{dy}{dx} = \frac{-2~sin~\frac{\pi}{2}~sin~\frac{\pi}{4}+cos~\frac{\pi}{2}~cos~\frac{\pi}{4}}{-2~sin~\frac{\pi}{2}~cos~\frac{\pi}{4}-cos~\frac{\pi}{2}~sin~\frac{\pi}{4}}$ $\frac{dy}{dx} = \frac{-2(1)~(\frac{\sqrt{2}}{2})+(0)~(\frac{\sqrt{2}}{2})}{-2~(1)~(\frac{\sqrt{2}}{2})-(0)~(\frac{\sqrt{2}}{2})}$ $\frac{dy}{dx} = \frac{-\sqrt{2}}{-\sqrt{2}}$ $\frac{dy}{dx} = 1$ The slope of the tangent line is $~1$