Did I sketch this polar curve correctly?

Question

The equation is:

$r^2=-4 \sin(2\theta)$

I first made a reference graph in cartesian coordinates using values $\displaystyle \frac{\pi}{4}$, $\displaystyle \frac{\pi}{2}$, $\displaystyle \frac{3 \pi}{4}$, $\displaystyle \pi$. Then from that I formed this:

Something seems off about that though. Should it be across the other axis instead?

Answer 1

You can let WolframAlpha plot this by rewriting it in Cartesian coordinates:

$$r^2=-4\sin2\theta=-2\sin\theta\cos\theta\;,$$ $$r^4=-2\sin\theta r\cos\theta r=-2xy\;.$$

Concerning your own plot: It seems it's not the axes you got mixed up, but the sine and cosine.

Answer 2

Give the comment from Alon some consideration, though I wasn't able to immediately eliminate your potential graph based on that idea alone. Let's think about the interval $\frac{\pi}{2}< \theta< \pi$. On that interval, $\sin2\theta$ goes from $0$ to $-1$ to $0$, so $r^2=-4\sin2\theta$ goes from $0$ to $4$ to $0$, so $r$ goes from $0$ to $\pm 2$ to $0$ again. No matter how I interpret your graph, it says that $r=\pm4$ at $\theta=\frac{\pi}{2}$ or $\theta=\frac{3\pi}{2}$ (or something along those lines).