Question

How do I graph a rose curve?

Answer 1

Hint: In this question, we are supposed to show how a rose curve can be done on a graph, and we will solve this with the help of polar equation. We know that the 2-D polar coordinates are $$P\left(r,\theta \right),r=\sqrt{x^2+y^2}⩾0$$

Complete step by step answer:
The 2-D polar coordinates $$P\left(r,\theta \right),r=\sqrt{x^2+y^2}⩾0$$ It represents length of the position vector $$<r,\theta >$$.θ 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 $$r=a\cos n\theta$$ or$$r=a\sin n\theta$$
$$n$$ 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.
The number of petals for the period $[0,2{\displaystyle \frac{\pi }{n}}]$ will be $n$ or $2n$ (including r-negative $n$ petals) according as $n$ is odd or even, for $$0⩽\theta ⩽2\pi$$. Of course, I maintain that $r$ is length $$⩾0,$$ and so non-negative. For Quantum Physicists, $$r>0$$.
For example, consider$$r=2\sin 3\theta$$. The period is $2{\displaystyle \frac{\pi }{3}}$ 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 $[0,2{\displaystyle \frac{\pi }{3}}]$, for $$\theta =0,{\displaystyle \frac{\pi }{12}},2{\displaystyle \frac{\pi }{12}},3{\displaystyle \frac{\pi }{12}},...8{\displaystyle \frac{\pi }{12}}$$. Join the points by smooth curves, befittingly. You get one petal. You ought to get the three petals for $$0⩽\theta ⩽2\pi ..$$
For$$r=\cos 3\theta$$, the petals rotate through half-petal angle $={\displaystyle \frac{\pi }{6}}$, 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 $$0⩽\theta ⩽2\pi$$.
Graph $(x^2+y^2{)}^3\times 5-4(x^6-15x^2{y}^2(x^2-y^2)-y^6)=0$

Note: In mathematics, a rose or rhodonea curve is a sinusoid plotted in polar coordinates.
For integer values, the petals might be redrawn, when the drawing is repeated over successive periods.
The period of both $$\sin n\theta$$ and $$\cos n\theta$$ is $2{\displaystyle \frac{\pi }{n}}$.