# How do I graph a rose curve?

Apr 11, 2016

The polar equation of a rose curve is either $r = a \cos n \theta \mathmr{and} r = a \sin n \theta$. 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 $P \left(r , \theta\right)$, r = $\sqrt{{x}^{2} + {y}^{2}} \ge 0$. 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 $r = a \cos n \theta \mathmr{and} r = a \sin n \theta$.

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 $\sin n \theta \mathmr{and} \cos n \theta$ is $2 \frac{\pi}{n}$.

The number of petals for the period $\left[0 , 2 \frac{\pi}{n}\right]$ will be n or 2n ( including r-negative n petals ) according as n is odd or even, for $0 \le \theta \le 2 \pi$. Of course, I maintain that r is length $\ge 0$, and so non-negative. For Quantum Physicists, r > 0.

Foe example, consider $r = 2 \sin 3 \theta$. The period is $2 \frac{\pi}{3}$ 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 $\left(r , \theta\right)$, in one period $\left[0 , 2 \frac{\pi}{3}\right]$, for $\theta = 0 , \frac{\pi}{12} , 2 \frac{\pi}{12} , 3 \frac{\pi}{12} , \ldots 8 \frac{\pi}{12}$. Join the points by smooth curves, befittingly. You get one petal. You ought to get the three petals for $0 \le \theta \le 2 \pi .$.

For $r = \cos 3 \theta$, the petals rotate through half-petal angle = $\frac{\pi}{6}$, in the clockwise sense.

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

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