# Polar Flower

This polar equation is a quadratic. Plot the graph given by each
factor to draw the flower.

Factorise the equation $$ r^2 - r + {\textstyle{1\over 4}} \sin^2 4\theta = 0$$ and hence plot the graph given by this polar equation.

The graph of $xy=0$ is the pair of lines $x=0$ and $y=0$. Similarly you will need to draw a pair of graphs here but this time in polar coordinates.

When you plot the point $(r,\theta)$ in polar coordinates, the length of the line segment joining this point to the origin is $r$ and the angle between this line segment and the positive direction of the $x$ axis is given by $\theta$.

Elizabeth from the International School of Geneva, someone from Winchester College who gave no name and Andrei from Tudor Vianu National College, Bucharest, Romania all sent solutions to this problem.

Elizabeth factorised the expression as follows:

$$\eqalign{ r^2 - r + {\textstyle{1\over 4}} \sin^2 4\theta &= 0 \cr r^2 - r + {\textstyle{1\over 4}}(2(\sin 2\theta \cos 2\theta))^2 &= 0 \cr r^2 - r + \sin^2 2\theta \cos^2 2\theta &= 0 \cr (r - \sin^2 2\theta)(r- \cos^2 2\theta) &=0.}$$

and plotted the graphs of $r = \sin^2 {2\theta}$ and $r=\cos^2 2{\theta}$.

Andrei factorised the expression differently

$$(2r+\cos {4\theta} -1)(2r \cos {4\theta} - 1)$$

and plotted the graphs of $r = \textstyle{1\over 2}(1 + \cos 4\theta)$ and $r = \textstyle{1\over 2}(1 - \cos 4\theta)$ but you should be able to see that the results are equivalent.

Our friend from Winchester College used the difference of two squares to arrive at the same form of this result as Andrei and another method would be substitution in the formula for the solution of a quadratic equation.

Consider first the graph of $r- \sin^2 2\theta=0$ in polar coordinates where $r$ is the length of the line segment from the point from the origin and $\theta$ is the angle measured counter clockwise between this line segment and the positive $x$ axis. For points on this graph, as $\theta$ increases from 0 to ${\pi \over 4}$ we have $\sin^2 2\theta$ increases from 0 to 1. Between $\theta = {\pi\over 4}$ and ${\pi \over 2}$ the value of $r$ decreases from 1 to 0 so that the graph in the first quadrant is a 'petal' symmetrical about the line $\theta = {\pi \over 4}$.

Similarly the graph in the second quadrant is a 'petal' for $\theta $ between ${\pi \over 2}$ and $\pi$ where $r$ takes positive values corresponding to $\sin^2 \theta$. The graph in the third quadrant is a 'petal' for $\theta $ between $\pi $ and ${3\pi \over 2}$ and the graph in the fourth quadrant is a 'petal' for $\theta $ between ${3\pi \over 2}$ and $2\pi$.

Image

Next consider the graph of $r- \cos^2 2\theta=0$. This will be of the same form but rotated by ${\pi \over 4}$ corresponding to the phase shift between the graphs of $\sin 2\theta$ and $\cos 2\theta$.

For points on this graph, as $\theta$ increases from 0 to ${\pi \over 4}$ we have $\cos^2 2\theta$ decreases from 1 to 0. Between $\theta = {\pi\over 4}$ and ${\pi \over 2}$ the value of $r$ increases from 0 to 1 and between ${\pi \over 2}$ and ${3\pi \over 4}$, as $r$ decreases from 1 to 0, the 'petal' symmetrical about the y-axis is completed.

The next petal, symmetrical about the negative x-axis, is drawn for $\theta $ between ${3\pi \over 4}$ and ${5\pi \over 4}$. The next petal, symmetrical about the negative y-axis, is drawn for $\theta $ between ${5\pi \over 4}$ and ${7\pi \over 4}$ and the remaining petal is completed for $\theta $ between ${7\pi \over 4}$ and $2\pi$.

The point $P$ with polar coordinates $(r,\theta)$ is such that $OP=r$ and the angle between the x-axis and the line through $OP$ is $\theta$ measured counter clockwise. Equivalently if $P$ has Cartesian coordinates $(x,y)$ then $x=r\cos \theta$ and $y=r\sin \theta$.

The question asks for an explanation of the features of the graph. If you want to use a software package to sketch the graph you may want to download the shareware graphing software Graphmatica .