I'm having trouble with a question. I don't know where to start, but have made an attempt.
A curve has polar equation $r^2=a^2\sin2\theta$. What is the area of the smallest possible square which encloses the whole curve?
I tried with simple values such as a = 1, and tried to find the "width" of the curve to no avail (the maximum x point in a Cartesian sense).
Any help is greatly appreciated.
After a bit of trying things out, I finally got the answer.
$$y=r\sin\theta=a\sqrt{\sin2\theta}\sin\theta \\\frac{dy}{d\theta} = a(\sqrt{\sin2\theta}\cos\theta + \frac{1}{2} \times 2\cos2\theta(\sin2\theta)^{-\frac{1}{2}}\sin\theta)\\ =a\left(\cos\theta\sqrt{\sin2\theta} + \frac{\cos2\theta\sin\theta}{\sqrt{\sin2\theta}}\right)\\ \frac{dy}{d\theta} = 0 \Rightarrow \cos\theta = -\frac{\cos2\theta\sin\theta}{\sqrt{\sin2\theta}}\\ \Rightarrow \cos\theta\sin2\theta + \cos2\theta\sin\theta = 0\\ \Rightarrow \sin3\theta = 0\\ \Rightarrow \theta = 0, \pm\frac{\pi}{3} $$ From the graph, $\theta \neq 0$, so the tangent parallel to the initial line (in the first quadrant) is where $\theta = \frac{\pi}{3}$.
The tangent is therefore $y=a\sqrt{\sin\frac{2\pi}{3}}\sin\frac{\pi}{3}$, and by symmetry the tangent perpendicular to the initial line in the first quadrant is $x=a\sqrt{\sin\frac{2\pi}{3}}\sin\frac{\pi}{3}$. Therefore the area is $$\left(2a\sqrt{\sin\frac{2\pi}{3}}\sin\frac{\pi}{3}\right)^2\\ = 4a^2\sin\frac{2\pi}{3}\sin^2\frac{\pi}{3}\\ =4a^2 \times \frac{\sqrt{3}}{2} \times \frac{3}{4}\\ = \frac{3\sqrt{3}}{2}a^2$$