Answer
This curve has horizontal tangent lines at these polar coordinates: $(\frac{3\sqrt 2}{2 } , \pi/4)$ and $(-\frac{3\sqrt 2}{2 } , 3\pi/4)$
And vertical tangent lines at:
$(3 , 0)$ and $(0 , \pi/2)$
Work Step by Step
1. Determine $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$.
$\frac{dr}{d\theta} = \frac{d(3cos\space \theta)}{d\theta} = -3sin\space \theta$
$\frac{dy}{d\theta} = \frac{dr}{d\theta}sin \space \theta + rcos\space \theta = (-3sin \space \theta)sin \space \theta + (3cos \space \theta)cos \space \theta$
$\frac{dy}{d\theta} = 3cos^2(\theta) - 3sin^2(\theta) = 3cos(2\theta)$
$\frac{dx}{d\theta} = \frac{dr}{d\theta}cos \space \theta - rsin\space \theta = (-3sin(\theta))cos(\theta) - (3cos(\theta))sin(\theta)$
$\frac{dx}{d\theta} -6sin(\theta)cos(\theta)= -3sin(2\theta) $
2. Identify at which points $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \neq 0$
$3cos(2\theta) = 0$
$cos(2\theta) = 0$
** cos(t) = 0 when : $t = \frac{\pi}{2}$ and $t = \frac{3\pi}{2}$ $ (0 \lt t \lt 2\pi)$
Thus:
$2\theta = \frac{\pi}{2} \longrightarrow \theta = \pi/4$
$2\theta = \frac{3\pi}{2} \longrightarrow \theta = 3\pi/4$
- Check if $\frac{dx}{d\theta} \neq 0$ for these values:
$-3sin(2(\pi/4)) = -3sin\space (\pi/2) = -3(1) = -3 \neq 0$
$-3sin(2(3\pi/4)) = -3sin (3\pi/2) = -3(-1) = 3 \neq 0$
Therefore, the tangent line is horizontal when $\theta = \pi/4$ and when $\theta = 3\pi/4$
3. Identify at which points $\frac{dx}{d\theta} = 0 \space and \space \frac{dy}{d\theta} \neq 0$
$-3sin(2\theta) = 0$
$sin(2\theta) = 0$
** sin(t) = 0 when: $t = 0$ and $t = \pi$
$2\theta = 0 \longrightarrow \theta = 0$
$2\theta = \pi \longrightarrow \theta = \pi/2$
- Check if $\frac{dy}{d\theta} \neq 0$ for these values:
$3cos(2*0) = 3(1) = 3 \neq 0 $
$3cos(2(\pi/2)) = 3cos(\pi) = 3(-1) = -3 \neq 0$
Therefore, the tangent line is vertical when $\theta = 0 \space and \space \theta = \pi/2$
4. Find the points in polar coordinates:
Horizontal:
$r = 3cos(\pi/4) = 3\frac{\sqrt 2}{2} = \frac{3\sqrt 2}{2} \longrightarrow (\frac{3\sqrt 2}{2 } , \pi/4)$
$r = 3cos(3\pi/4) = 3(-\frac{\sqrt 2}{2}) = -\frac{3\sqrt 2}{2} \longrightarrow (-\frac{3\sqrt 2}{2 } , 3\pi/4)$
Vertical:
$r = 3cos(0) = 3(1) = 3 \longrightarrow (3 , 0)$
$r = 3cos(\pi/2) = 3(0) = 0 \longrightarrow (0 , \pi/2)$
