5
$\begingroup$

In polar coordinates $(r, \theta)$, the equation $$r = \sin\left(a \theta\right)$$ gives a rose with $a$ petals if $a$ is odd, or $2a$ petals if $a$ is even.

Thus, the number of petals generated for some values of $a$ are

   a   | petals
=======+========
   1   |   1
   2   |   4
   3   |   3
   4   |   8
   5   |   5
   6   |   12
   7   |   7

Conspicuously missing from this table are roses with $4n+2$ petals. How can you generate a rose of the same "shape" with $2, 6, 10, \ldots$ petals? If you can't, why (intuitively) is it impossible?

$\endgroup$
4
  • $\begingroup$ Offhand it seems like $r = \lvert\sin(2n\theta)\rvert$ ought to have exactly $2n$ leaves. $\endgroup$
    – MJD
    Mar 12, 2014 at 16:39
  • $\begingroup$ @MJD Could you explain why? I don't seem to get that answer on my TI-84 or on Wolfram|Alpha. $\endgroup$
    – wchargin
    Mar 12, 2014 at 16:54
  • $\begingroup$ If you don't mind the petals overlapping, look at the graphs of $r=\sin(a\theta/2)$. $\endgroup$
    – Steve Kass
    Mar 12, 2014 at 16:59
  • $\begingroup$ Sorry, I had it backwards. Here is $r=\lvert\sin(3\theta)\rvert$, with six petals. The trick only works for odd $n$, to get $2n$ leaves, but that covers exactly the cases you were asking about. $\endgroup$
    – MJD
    Mar 12, 2014 at 17:05

2 Answers 2

9
$\begingroup$

What is happening here is that $\sin n\theta$ has $n$ positive lobes and $n$ negative lobes. When $n$ is odd, the negative lobes exactly overlap the positive lobes in the graph, so you only see $n$ petals. When $n$ is even, the negative lobes appear separately, so you see $n$ positive and $n$ negative lobes, for a total of $2n$ petals.

To get $2n$ petals when $n$ is odd, you can use the absolute value function to separate the negative and positive petals. The graph of $$r=\lvert\sin n\theta\rvert$$ has exactly $2n$ petals, even in the case when $n$ is odd. So for example $r=\lvert\sin 3\theta\rvert$ has this graph:

6-petal rose

In the 3-petal rose $r = \sin3\theta$, three of the leaves are reflected across the origin onto the other three leaves, which is why the rose appears to have only 3 leaves:

3-petal rose

But actually all six petals are there; it's just that they coincide in three pairs, so you can only see three petals in the graph.

The suggestion of Tony Jacobs elsewhere in this thread, of using $r = \sin^2 3\theta$, is essentially the same; the squaring operation forces the formerly invisible negative lobes onto the opposite side of the origin. But the squaring also attenuates the petals: the petals of $\sin^2 n\theta$ are not as wide as the petals of $\sin n\theta$, whereas the petals of $\lvert\sin n\theta\rvert$ are exactly the same size and shape as the petals of $\sin n\theta$.

$\endgroup$
3
$\begingroup$

I just got a nice, 6-petaled flower with $r=\sin^2(3\theta)$. Is that what you're looking for?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .