# Polar coordinates - parallel to initial line question

I have the polar curve $r=a\sqrt{\sin(2\theta)}$, where $a \neq 0$ (graph shown above). I want to find the points where the tangent to the curve is parallel to the initial line, $\theta = 0$.

I have written $y=r\sin\theta$ and worked out the solutions to $\frac{dy}{d\theta} = 0$. I get to the stage where one of my solutions is $\sin \theta = 0$ and another solution which leads to $\theta = -\frac{\pi}{3}$.

My question is why do we get the solution $\sin\theta = 0$. I don't see how we have a tangent line parallel to the initial line at $\theta = \frac{\pi}{2}$, for example... So if someone can share a light on the origin of this solution, that would be helpful.

Thanks

$(a\sin\theta\sqrt{\sin2\theta})'=a(\cos\theta\sin2\theta+\sin\theta\cos2\theta)/\sqrt{\sin2\theta}=a\sin3\theta/\sqrt{\sin2\theta}=0$

So solutions as you said are $\theta_1=0, \theta_2=\pi/3, \theta_3=\pi, \theta_4=4\pi/3$ , no solutions at $\pi/2$ (maybe you misspelled). Solutions in $2\pi/3$ and $5\pi/3$ are excluded cos curve isn't defined there.

You see that curve reaches 4 times the point $(x,y)=(0,0)$, once when $\theta=0$, second when $\theta=\pi/2$, then when $\theta=\pi$ and when $\theta=3\pi/2$

In the first and 3rd case curve is parallel to x axis, in the 2nd and 4th case it is parallel to y axis.

Hope this clears it for you...

First, your $r(\theta)$ is only valid for $\theta \in [0,\pi/2]\cup [\pi, 3\pi/2]$, as you need $\sin(2\theta)\geq 0$, so I'm not sure where you're getting $-\pi/3 \equiv 5\pi/3$ as a solution.

Next, we have $y = \sqrt{2} (\sin \theta)^{3/2} (\cos \theta)^{1/2},$ so $dy/d\theta$ vanishes when $$3 (\sin \theta)^{1/2}(\cos \theta)^{3/2} - (\sin \theta)^{5/2}(\cos\theta)^{-1/2} = 0.$$

Two solutions are $\theta\in\{0,\pi\}$ and two non-solutions (since the derivative diverges) are $\theta\in\{\pi/2, 3\pi/2\}$. Away from these special values we can multiply by $(\cos \theta)^{1/2}(\sin\theta)^{-1/2}$ to yield $$3\cos^2\theta - \sin^2\theta = 0$$ or $$\sin(\theta) = \pm \frac{\sqrt{3}}{2},$$ which in $[0,\pi/2]\cup [\pi, 3\pi/2]$ has solutions $\theta = \pi/3$ and $\theta = 4\pi/3$. Therefore $$\theta \in \{0,\pi, \pi/3, 4\pi/3\}$$ which makes sense given the plot.