2
$\begingroup$

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:

enter image description here

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

  • 0
    This is indeed not quite correct. When $0 < \theta < \pi/2$, what is the sign of the RHS? What does this say about $r$?2011-05-22
  • 0
    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.)2011-05-22

2 Answers 2

2

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.

  • 0
    Wolfram Alpha is happy to deal with the polar coordinates directly: http://www.wolframalpha.com/input/?i=+r^2=-4sin2%CE%B82011-05-22
  • 0
    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 :-/2011-05-22
  • 1
    @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.2011-05-22
  • 0
    @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: http://tiny.cc/iq75u2011-05-22
  • 0
    @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.2011-05-22
  • 0
    @joriki. I'm just confused at this point. The method I was doing it, I'm sure I was using sin when I obtained the values for my graph and yet it still came out look the opposite of what it should have been. If you have the time to want to see what my method was I took it from this video http://www.youtube.com/watch?v=mDT_DG_A0JA&feature=player_detailpage#t=61s Maybe that was the wrong way to do this? I don't know the problem seemed awfully similar to mine so I didn't see a problem with it.2011-05-22
  • 0
    @Ryan: That method is more or less OK, though the axes in the lower plot should be labelled $(x,y)$, not $(\theta,r)$, and I hadn't heard of the convention of letting negative $r$ mean opposite angle (but Wikipedia mentions it). But as user9325 pointed out, you have $r^2$, and negative values of $r^2$ don't make sense even with that convention. Concerning the sine and cosine, I don't see how you got to your graph using the sine with that method -- it looks exactly like the graph in the video, and the method clearly can't yield the same graph if you replace the cosine by the sine.2011-05-22
  • 0
    @joriki fair enough. I still think that most high school students would be better off not knowing about WA. It's never too late to learn how to bask homework questions into the computer.2011-05-22
  • 0
    @Alex: I disagree. To me, that's like saying they'd be better off not knowing about calculators. The possibility of exploring and experimenting with WA and being able to visualize and manipulate things orders of magnitude more reliably and quickly in my view far offsets the possibility of cheating and cutting corners. What they should learn is not the best way of doing mathematics in a fictitious world without technical support, but the best way of doing mathematics with exactly the means that will be at their disposal when they're going to be applying it.2011-05-22
  • 0
    @joriki: Maybe the problem exists in where I chose the intervals. In the video he set 2θ = pi/2 to get pi/4 and then formed his intervals. His equation was using cosine though, will it still be the same for sine? Is there just a better way to do this, the way it was taught in class seemed much faster but I don't understand it at all. In class they simply made the table of values in a similar way (different values for θ were chosen though). Then after getting the r values, they just plotted the curve but I can never get the proper curve. :(2011-05-22
  • 0
    @Ryan: It will still be the same that $2\theta=\pi/2$ leads to $\theta=\pi/4$, what's different is only what $r$ (or $r^2$) values you get when you plug in those angles. You can plug in any values, you don't have to choose intervals beforehand; then you can just discard the angles for which $r^2$ comes out negative. If you do this for $\theta=0,\pi/4, \pi/2, \dotsc, (2-\frac{1}{4})\pi, 2\pi$, you should get enough points to fill in the rest.2011-05-22
  • 0
    @joriki I guess continuing this discussion here would be a misuse of the comments. Let me just say that I partly agree with you, and in fact I myself teach my representation theory students how to compute things with MAGMA. Maybe we will get the opportunity some day to have a proper conversation on this topic offline.2011-05-22
0

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).