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r = 3\cos \theta

Pictured above is the graph of $r = \cos 3 \theta$ for $0 \le \theta \le L$. Find the smallest value of $L$ that still produces the entire graph of $r = \cos 3 \theta$.

I am having trouble starting this problem. How would I start it?

Mathy Person
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    Start and $\theta = 0$ and trace it out: for what value of $\theta$ to we get to the origin for the first time, the second, the third, how to we join back up with the point for $\theta = 0$, etc. – Simon S Feb 01 '15 at 00:21
  • @SimonS: 0 radians or 0 degrees? I'm assuming radians. – Mathy Person Feb 01 '15 at 00:31
  • Equations in polar coordinates use radians. – Simon S Feb 01 '15 at 01:07
  • @SimonS: So if $\theta = 0 radians$, $r = \cos 3\theta = \cos 0 = 1$? – Mathy Person Feb 01 '15 at 01:13
  • Yes. Here are some examples, starting page 6: https://cims.nyu.edu/~kiryl/Calculus/Section_9.3--Polar_Coordinates/Polar_Coordinates.pdf – Simon S Feb 01 '15 at 01:19
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    Coincidentally, I just did an animation of the plot of $r=\sin 3 \theta$ for this answer. Perhaps it'll give you some insight. – Blue Feb 01 '15 at 03:43
  • @Blue: Do you mind clarifying: "actually plotting "backwards" through the pole for values of $\theta$ (for instance, $90 ^\circ$) when $r$ is negative."? What do you mean by "plotting backwards"? – Mathy Person Feb 01 '15 at 17:39
  • @MathyPerson: The animation shows a segment sweeping once around a circle; the segment "looks" in the direction of positive $r$ for each $\theta$. As it starts turning, it traces a petal of the curve, looking directly at the points it traces. As the segment nears $90^\circ$, it traces the lower petal, even though it's looking upward; that's because $r$ is negative for a while. Points corresponding to negative $r$ values are plotted in the direction opposite where the segment looks. This happens again just past $180^\circ$, and yet again as the segment approaches $360^\circ$. – Blue Feb 01 '15 at 20:53
  • @Blue: How would you suggest I apply that animation to this problem? – Mathy Person Feb 01 '15 at 20:56
  • @MathyPerson: Look at the animation and notice that, after a certain $\theta$, it's just re-tracing the curve. That's for instructional purposes, of course, but suppose someone considered it a waste of time and blue pixels. If I wanted to draw the $r=\sin 3\theta$ curve only once, then I should've made the animation shorter, stopping the sweeping segment early. When? For your problem, think about how a sweeping segment would trace $r=\cos 3\theta$ (starting at $\theta = 0$), and ask the same question: When should you stop the sweeping segment so that you trace the curve only once? – Blue Feb 01 '15 at 21:09
  • @Blue: At the point where the rotating line exactly hits the x-axis. – Mathy Person Feb 01 '15 at 21:19
  • @Blue: Would it be at point $\theta$, then? – Mathy Person Feb 01 '15 at 21:20
  • @MathyPerson: "$\theta$" is like "$x$" here; it's a variable, not a number. What's the angle where you'd stop the sweeping segment? Feel free to answer in degrees or radians; it doesn't really matter. (BTW: The system is suggesting that we move this discussion to chat. I don't like the chat here, so I'll just keep making comments. :) – Blue Feb 01 '15 at 21:31
  • @Blue: Alright then. :). 180 degrees – Mathy Person Feb 01 '15 at 21:46
  • Bingo! Write-up your thoughts as an answer (and "accept" it), and I'll up-vote it. :) – Blue Feb 01 '15 at 21:48
  • @Blue: Thanks for the help, and for the animation! I'm not sure if I should write up my thoughts as an answer, as I did just look at the animation and notice the point from there rather than mathematically, so I might just leave it like it is in the comments. :) – Mathy Person Feb 01 '15 at 21:55

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