# Did I sketch this polar curve correctly?

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?

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This is indeed not quite correct. When $0 < \theta < \pi/2$, what is the sign of the RHS? What does this say about $r$? –  Alon Amit May 22 '11 at 5:56
If you plug in $\pi/4$, what does it mean for $r^2$ to be $-4$? (You treated it like $r=-4$, but even that is impossible.) –  Phira May 22 '11 at 6:39

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.

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Wolfram Alpha is happy to deal with the polar coordinates directly: wolframalpha.com/input/?i=+r^2=-4sin2%CE%B8 –  Alon Amit May 22 '11 at 5:59
I am sure that's what the OP's teacher wanted him to do: have someone do the manipulations for him and then solve with a computer algebra package :-/ –  Alex B. May 22 '11 at 6:02
@Alex: I agree with the general philosophy of your criticism, but I don't think it's fairly applied in this case. I didn't do "the manipulations" for him, since the question neither asked for nor required those manipulations; I just pointed out another way to solve the problem that probably hadn't occurred to the OP; then I went on to answer the question as intended, that is, I pointed out why the OP's own result "seems off". Neither of the other answers did that; no-one else seems to have noticed that the OP mixed up sine and cosine. So I would maintain that my answer was relatively helpful. –  joriki May 22 '11 at 6:10
@Alon: That's weird -- I tried it with "plot" in front, and that doesn't work because it gets plotted in Cartesian coordinates $(r,\theta)$. That link doesn't work, by the way; here's a link that should: tiny.cc/iq75u –  joriki May 22 '11 at 6:14
@joriki: thanks for the corrected link. It's sometimes hard to guess the heuristics WA uses to interpret the input; I was lucky in this case by just copying the equation over, hoping that it would take "$r$" and "$\theta$" as a hint. –  Alon Amit May 22 '11 at 6:17

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).

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