Rose Petal Polar Plot With 6 Petals

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The functional form for a rose petal with 6 petals can be represented as r = sin(3θ/2) or r = √(sin(6θ). A rose curve's petal count depends on the integer value of n; for even n, the formula yields 2n petals, while for odd n, it yields n petals. The discussion highlights that using n = 3/2 can also create a six-petaled rose, although the width of the petals may not be uniform. The equation r^2 = 3sin(2θ) was suggested but identified as resulting in a two-petaled lemniscate. Overall, multiple equations can achieve the desired six petals, albeit with variations in petal width.
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What is the functional form of rose petal with 6 petals? I am asked to graph this function with matlab, but it seems impossible according to my calculus textbook. According to my textbook, a rose curve can have the form r = a \cos n \theta or r = a \sin n \theta. When n is even, then there are 2n petals; when n is odd, then there are n petals.

Is there any way of accomplishing this, graphing a rose petal with 6 petals?
 
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Bashyboy said:
What is the functional form of rose petal with 6 petals? I am asked to graph this function with matlab, but it seems impossible according to my calculus textbook. According to my textbook, a rose curve can have the form r = a \cos n \theta or r = a \sin n \theta. When n is even, then there are 2n petals; when n is odd, then there are n petals.

Is there any way of accomplishing this, graphing a rose petal with 6 petals?

Experiment with nonintegers. Try n=3/2.
 
I shall assume you are edicted to pick n as an integer.
Let r = sin(nθ), and write α = π/n. The first petal is from θ=0 to θ=α, the second from α to 2α. But if you look at where these appear, the second will look like the first rotated about the origin by an angle - what angle (as a multiple of α)?
The third petal will look like the second, but rotated by that same angle. What will the total of these angles be when you stop getting new petals?
 
You can make a six-petaled rose with the equation ##r^2 = 3sin(2θ)##. I am not sure if the 3 changes the number of petals. But that will do it for you.
 
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simpsonsruler said:
You can make a six-petaled rose with the equation ##r^2 = 3sin(2θ)##. I am not sure if the 3 changes the number of petals. But that will do it for you.
No, that is a two petaled lemniscate.

Depending on your taste, you might try ##r=\sqrt{sin(6\theta)}## or you might try ##r = sin(3\theta/2)##. Those each have 6 petals, but the width of each petal is not ##(2\pi)/6## the way you might like.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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