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I have $r^2=2\cos2\theta$ and I'm being asked to convert this equation to rectangular coordinates. So I'm using double angle trigonometric identities to get:

$$ r^2=2\cos2\theta \\ r^2=2(\cos^2\theta-\sin^2\theta) \\ r^2=2\cos^2\theta-2\sin^2\theta $$

Then I do the following substitutions based on the facts that $y=r\sin\theta$ and $x=r\cos\theta$:

$$ r^2=\frac{2x^2}{r^2}-\frac{2y^2}{r^2} \\ (r^2)^2=2x^2-2y^2 \\ (x^2+y^2)^2=2x^2-2y^2 $$

However this doesn't feel right, since I would be getting a fourth-degree equation. Can you find any error in my development? Or how should I go on solving this problem? Thanks in advance.

Rushabh Mehta
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Fernando Gómez
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  • Everything you did is correct. You can't say something doesn't "feel" right when you don't have any experience with it yet to compare it against. – Ninad Munshi Nov 18 '19 at 22:51
  • @Ninad Munshi I was asked to graph the equation and the program couldn't plot any graph. Hence why I don't felt it right. – Fernando Gómez Nov 18 '19 at 22:54
  • You were probably asked to plot this in polar. That doesn't require converting it to rectangular. – Ninad Munshi Nov 18 '19 at 22:55
  • No, I wasn't asked to plot it in polar. Please stop guessing and assuming things, and do not pollute this thread. – Fernando Gómez Nov 18 '19 at 22:56
  • "Polluting" is a harsh word, you were the one who did not add the detail that you needed to graph the curve, and we are not mind readers. Regardless, I still have an answer for you. – Ninad Munshi Nov 18 '19 at 23:09
  • @FernandoGómez: Please stay polite if people are trying to help you. If you judge just from graphing calculators, you should put that in your question. Many, many polar equations turn into fourth-degree equations, and experience will in fact show that. – Ted Shifrin Nov 18 '19 at 23:21
  • @Ninad Munshi if you're not mind readers then please ask politely to clarify my question, instead of trying to patronize me. If what I did is correct then "Everything you did is correct" is all I needed, thanks. The second part in your comment and everything that followed was uncalled for and is off topic. – Fernando Gómez Nov 19 '19 at 00:23
  • @Ted Shifrin I can stay polite but please stay on topic. I didn't mention the graphing because it wasn't needed. The question is clear: am I doing it right or wrong, or how should I do it, so the answer should revolve around that. Everything else was uncalled for and off topic. – Fernando Gómez Nov 19 '19 at 00:23
  • I'm sorry you felt patronized, it was not my intention. It was only a statement of fact, experience with equations can lead to feeling right or wrong. Going back on topic as you suggest, instructors do not expect their students to know the equation for a lemniscate off the top of their heads. I usually asked students to derive the lemniscate equation by plotting in polar, then converting to rectangle, not by asking them to recognize a fourth degree curve. – Ninad Munshi Nov 19 '19 at 00:29
  • @Ninad Munshi "instructors do not expect" I don't actually have an instructor, it's just me going through books and finding exercises online and then trying to solve them. Perhaps I came across this exercise I wasn't supposed to do at this point of my studies (quadric surfaces), which is a downside of not having an instructor. I assumed the answer was going to be the equation of quadric surface, which now I see was an error. That's why when I saw the fourth degree curve and then my plotter didn't work, I thought I had something wrong. Anyway thanks for the answer and explanation. – Fernando Gómez Nov 19 '19 at 00:34

1 Answers1

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In order to plot this curve, first consider the equation

$$r^2 = 2\cos 2\theta$$

Notice that the left hand side is always positive, so that induces a domain restriction on $\theta$.

$$|2\theta| \leq \frac{\pi}{2} \implies -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$$

Then notice that there is a symmetry $\theta \mapsto -\theta$, which means that the curve is symmetric about the $x$ axis (which is the ray $\theta = 0$), so we only need to know what the curve looks like on the top half and we can reflect it down to get the rest.

From here, it is best to pick example points and tabulate them:

$$\begin{array}{c | c} \theta & r \\ \hline 0 & \pm\sqrt{2} \\ \frac{\pi}{12} & \pm \sqrt[4]{3} \\ \frac{\pi}{8} & \pm \sqrt[4]{2} \\ \frac{\pi}{6} & \pm 1 \\ \frac{\pi}{4} & 0 \\ \end{array} $$

Plot the points and connect them as smoothly as you can, then reflect it across the $x$ axis. Drawing this out, we get a familiar shape, a lemniscate (or infinity symbol) figure.

One can confirm this is indeed a lemniscate by looking at the equation you derived. It is in the correct mathematical form for a lemniscate, with $c=1$.

Ninad Munshi
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  • Thank you, this I understood. So what I gather is, my error was assuming the problem revolved around quadric surfaces (which is the topic I'm currently studying), when clearly I should have not. Thanks again. – Fernando Gómez Nov 19 '19 at 00:29
  • @FernandoGómez I see what you mean. Would you like a reference? Are you trying to just recognize basic quadric surfaces, or more challening stuff like rotated quadric surfaces i.e. $0 = z^2 - z + x + 2xz + x^2 + y^2$ ? – Ninad Munshi Nov 19 '19 at 00:40
  • Both I think, but for the moment I'm just trying to recongnize them and doing a few exercises. And yes please if you have references those would come in handy no doubt. – Fernando Gómez Nov 19 '19 at 00:55