Answer
There are horizontal tangents at : $(\frac 1 2, -1)$ and at $(-\frac 1 2, 1)$
Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{dy}{dt} = \frac{d(cos(3\theta))}{dt} = -3sin(3\theta)$
$\frac{dx}{dt} = \frac{d(cos(\theta))}{dt} = -sin(\theta)$
Therefore: $\frac{dx}{dy} = \frac{-3sin(3\theta)}{-sin(\theta)} = \frac{3sin(3\theta)}{sin(\theta)}$
2. There will be a horizontal tangent when $3sin(3\theta) = 0$ and $sin(\theta) \neq 0$:
$3sin(3\theta) = 0$
$sin(3\theta) = 0$
** Notice: $sin(t) = 0;$ So: $t = 0, t = \pi \space , \space t = 2\pi, \space t = 3\pi$ . (For values between 0 and $3\pi$)
If $3 \theta$ is equal to $t$:
$3\theta = 0 \longrightarrow \theta = 0$
$3\theta = \pi \longrightarrow \theta = \frac {\pi} 3$
$3\theta = 2\pi \longrightarrow \theta = \frac {2\pi} 3$
$3\theta = 3\pi \longrightarrow \theta = \pi$
- Check if $sin(\theta) \neq 0 $
$sin(0) = 0$ - Invalid.
$sin(\frac {\pi} 3 ) = \frac {\sqrt 3} 2$
$sin(\frac {2\pi} 3 ) = \frac {\sqrt 3} 2$
$sin(\pi) = 0$ - Invalid
Therefore, the curve has horizontal tangent at $\theta = \frac {\pi} 3$ and at $\theta = \frac {2\pi} 3$
3. There will be a vertical tangent when $sin(\theta) = 0$ and $3sin(3\theta) \neq 0$:
$sin(\theta) = 0;$ So: $\theta = 0, \theta = \pi \space , \space \theta = 2\pi, \space \theta = 3\pi$ . (For values between 0 and $3\pi$)
Check if $3sin(3\theta) \neq 0$
$3sin(3(0)) = 0$
$3sin(3(\pi)) = 0$
$3sin(3(2\pi)) = 0$
$3sin(3(3\pi)) = 0$
Therefore, the curve doesn't have vertical tangents.
4. Find the points:
$\theta = \frac {\pi} 3$
$x = cos(\frac {\pi} 3) = \frac{1}{2}$
$y = cos(3(\frac {\pi} 3)) = cos(\pi) = -1$
$\theta = \frac {2\pi} 3$
$x = cos(\frac {2\pi} 3) = -\frac{1}{2}$
$y = cos(3(\frac {2\pi} 3)) = cos(2\pi) = 1$
