Writing a Polar Equation for the Graph of an Implicit Cartesian Equation

If $(x^2+y^2)^3=4x^2y^2,$ then $r=\sin 2\theta$ for some $\theta$.

Using $r^2=x^2+y^2, x=r\cos\theta,y=r\sin\theta$, it's easy to get $r^2=\sin^22\theta$. But I don't know what to do next, since $r$ could be negative in $r=\sin2\theta.$

Actually the original problem is to show that the affine variety $V((x^2+y^2)^3-4x^2y^2)$ is contained in the four-leaved rose, whose polar equation is certainly $r=\sin 2\theta$. (Exercise 7(b), section 1.2, Ideals, Varieties and Algorithms, 3rd edition, David Cox etc.)

Solution 1:

Allowing $r$ to take both nonnegative and negative values, the graphs of $r=\sin2\theta$ and $r=-\sin2\theta$ (on $0\le\theta<2\pi$) are identical—they contain the same set of geometric points in the plane—though they are traced out differently. For $r=\sin2\theta$, as $\theta$ increases from $0$ to $2\pi$, the graph traces the first quadrant petal, then the fourth quadrant petal, then the third quadrant petal, and finally the second quadrant petal. For $r=-\sin2\theta$, as $\theta$ increases from $0$ to $2\pi$, the graph traces the third quadrant petal, then the second quadrant petal, then the first quadrant petal, and finally the fourth quadrant petal.

edit Here's what I mean:

Since the sets of points in the plane described by $r=\sin2\theta$ and $r=-\sin2\theta$ are identical, giving only one of them is sufficient to describe the same set of points as $(x^2+y^2)^3=4x^2y^2$. That's what lets us go from $r^2=\sin^22\theta$ and its equivalent of $r=\pm\sin2\theta$ to only $r=\sin2\theta$.

Solution 2:

If $x=r\cos(\theta)$ and $y=r\sin(\theta)$, then $(x^2+y^2)^3=4x^2y^2$ becomes $$ r^2=\sin^2(2\theta) $$ as you say. However, note that the curve $r=\sin(2\theta)$ is symmetric through the origin since $\sin(2(\theta+\pi))=\sin(2\theta)$, so it is the same as $r=-\sin(2\theta)$. In other words, it doesn't matter whether you take $+$ or $-$.

$\hskip{4cm}$

Solution 3:

On the variety $V\left((x^2+y^2)^3-(2xy)^2\right)$, in polar coordinates we have $$ r^6=(x^2+y^2)^3=(2xy)^2=r^4\sin^2{2\theta} $$ or $$ r^2=\sin^2{2\theta} \quad\implies\quad r=\pm\sin2\theta $$ (we can ignore the "trivial" algebraic solution $r=0$ for all $\theta$ since it is geometrically recovered for $\theta\in\pi\mathbb{Z}$). Now $\sin{2\theta}$ has period $\pi$ (furnishing the same magnitude for $r$ at antipodal points), the curve has point symmetry through the origin, allowing us to discard the sign and consider only $r\in[0,1]$. But $$ \sin^2{2\theta}=\frac{1-\cos{4\theta}}{2} $$ has period $\frac{\pi}{2}$, vanishes at the axes or "cardinal" directions, and has its maxima of $1$ when $\cos{4\pi}=-1$, i.e. at $\theta=\frac{\pi}{4}+k\frac{\pi}{2}$. This explains why the polar equation for the solution can take the forms $$ r=\left|\sin{2\theta}\right| =\left(\sin^2{2\theta}\right)^\frac{1}{2} =\left(\frac{1-\cos{4\theta}}{2}\right)^\frac{1}{2} $$ and why it gives us a four-leaved rose.

Qualitatively, the result is the same with any positive power $r=|\sin{2\theta}|^p$, but the petal thickness and area and the attenuation of the maximum radius all vary inversely with $p$.

Plotted with sage (online), here is the solution $r^2=\sin^2{2\theta}$ (with the power $p=1$) in red, with another curve $r^6=\sin^2{2\theta}$ (with $p=\frac{1}{3}$) in blue to illustrate the effect of $p$.

t=var('t')
(polar_plot(abs(sin(2*t))^(2/2), (t, 0, 2*pi), color='red')
+polar_plot(abs(sin(2*t))^(2/6), (t, 0, 2*pi), color='blue')).show()

Finally, as a refresher, a good way to remember or derive the angle-doubling step above is from the two equations $$ \begin{matrix} \cos^2t&+&\sin^2t&=&1\\ \cos^2t&-&\sin^2t&=&\cos2t. \end{matrix} $$ Adding or subtracting (and dividing by two) then yields $$ \cos^2t = \frac{1+\cos2t}{2} \qquad \text{or} \qquad \sin^2t = \frac{1-\cos2t}{2} $$ respectively.