So I have computed $$\frac{dx}{d\theta} = \frac{\cos{3\theta}}{r}$$ and $$\frac{dy}{d\theta} = \frac{\sin{3\theta}}{r}$$
I set these equal to zero and found the horizontal tangents to occur in the following points:
$$(0,0), (\left(\frac{3}{4}\right)^{\frac{1}{4}},\frac{\pi}{3})\text{ and }(\left(\frac{3}{4}\right)^{\frac{1}{4}},\frac{4\pi}{3})$$
However if $r=0$, then $\frac{dy}{d\theta}$ is undefined, however the limit as $\theta$ approaches $0$ of $\frac{dy}{dx}$ appears to be $0$, so that would indicate a horizontal tangent.
So is $(0,0)$ a point of a horizontal tangent? Likewise the same is true for $\frac{dx}{d\theta}$ (that is, $r=0$ makes $\frac{dx}{d\theta}$ undefined). So does that mean then that we have $(0,0)$ has both a horizontal and vertical tangent? It also appears from the graph, that this is possible, but I am just not sure. Any help would be greatly appreciated.
Best Answer
$$\dfrac{dy}{d\theta}\cdot \frac{d\theta}{dx} = \dfrac{dy}{dx}$$ and we want to know when $\dfrac{dy}{dx} = 0$.
$$\frac{dx}{d\theta} = \frac{\cos3\theta}{r},\quad \frac{dy}{d\theta} = \frac{\sin 3\theta}{r} \implies \dfrac{dy}{dx} = \frac{\sin{3\theta}}{r}\cdot \dfrac{r}{\cos 3\theta}$$
Now solve for when $$\require{cancel} \dfrac{dy}{dx} = \frac{\sin{3\theta}}{\cancel{r}}\cdot \dfrac{\cancel{r}}{\cos 3\theta} = = \tan(3\theta)= 0$$
That's going to happen when $\sin 3\theta = 0$ (and note that when $\sin{3\theta} = 0 \iff 3\theta = k \pi \iff \theta = k\pi/3\;$ ($k$ any integer), we have that $\cos 3\theta = \pm 1\iff 3\theta = k\pi \iff \theta = k\pi/3$.)
The points at which the tangents are horizontal and/or vertical are given in polar coordinates, $(r, \theta)$.