How do I graph a rose curve?

1 Answer
Apr 11, 2016

Answer:

The polar equation of a rose curve is either #r = a cos ntheta or r = a sin ntheta#. The number of rose petals will be n or 2n according as n is an odd or an even integer. See explanation.

Explanation:

Having seen that there were more than 1 K viewers in a day, I now add more.

The 2-D polar coordinates #P ( r, theta)#, r = #sqrt (x^2 + y^2 ) >= 0#. It represents length of the position vector #< r, theta >. theta# determines the direction. It increases for anticlockwise motion of P about the pole O. For clockwise rotation, it decreases. Unlike r, #theta# admit negative values.

r-negative tabular values can be used by artists only.

The polar equation of a rose curve is either #r = a cos ntheta or r = a sin ntheta#.

n is at your choice. Integer values 2,, 3, 4.. are preferred for easy counting of the number of petals, in a period. n = 1 gives 1-petal circle.

To be called a rose, n has to be sufficiently large and integer + a fraction, for images looking like a rose. For integer values, the petals might be redrawn, when the drawing is repeated over successive periods.
.
The period of both #sin ntheta and cos ntheta# is #2pi/n#.

The number of petals for the period #[0, 2pi/n]# will be n or 2n ( including r-negative n petals ) according as n is odd or even, for #0 <= theta <= 2pi#. Of course, I maintain that r is length #>=0#, and so non-negative. For Quantum Physicists, r > 0.

Foe example, consider #r = 2 sin 3theta#. The period is #2pi/3# and the number of petals will be 3. In continuous drawing. r-positive and r-negative petals are drawn alternately. When n is odd, r-negative petals are same as r-positive ones. So, the total count here is 3.

Prepare a table for #(r, theta)#, in one period #[0, 2pi/3]#, for #theta = 0, pi/12, 2pi/12, 3pi/12, ...8pi/12#. Join the points by smooth curves, befittingly. You get one petal. You ought to get the three petals for #0 <= theta <= 2pi.#.

For #r = cos 3theta#, the petals rotate through half-petal angle = #pi/6#, in the clockwise sense.

A sample graph is made for #r = 4 cos 6theta#, using the Cartesian
equivalent. It is r-positive 6-petal rose, for #0 <=theta <=2pi#.

graph{(x^2+y^2)^3.5-4(x^6-15x^2y^2(x^2-y^2)-y^6)=0}