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