How to plot $r^2 = 36\cos(2\phi)$ in Cartesian coordinates?

Question

How would I plot $r^2 = 36\cos(2\phi)$ in $x$ and $y$ coordinate system?

I know that $r^2 = x^2 + y^2$, and I know that $x = r\cos(\phi)$ and $y = r\sin(\phi)$. However I'm not sure how to proceed.

I get $x^2 + y^2 = 36\cos(2\phi)$. But what next?

I know circle equation $x^2 + y^2 = constant$, but my equation has a non constant on the right hand side.

Any suggestions and ideas appreciated. :)

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Edit: I'm adding 100 reputation bounty to anyone who could answer my post and show how to plot this function on paper step by step.

In particular I know how to plot $y = f(x)$, but this function is completely different beast, it has both x and y quadratic.

Any help would be appreciated. I'm learning about polar coordinates.

Answer 1

The implicit equation is $$ \begin{align} r^2 &=36\cos(2\phi)\\ &=36[\cos^2(\phi)-\sin^2(\phi)]\\ r^4 &=36[r^2\cos^2(\phi)-r^2\sin^2(\phi)]\\ x^4+2x^2y^2+y^4 &=36x^2-36y^2 \end{align} $$ But parametrically, to plot the curve, $$ \begin{align} x&=6\cos(\phi)\sqrt{\cos(2\phi)}\\ y&=6\sin(\phi)\sqrt{\cos(2\phi)} \end{align} $$ Since we must have $\cos(2\phi)\ge0$, $\phi\in[-\pi/4,\pi/4]$ (right lobe) or $\phi\in[3\pi/4,5\pi/4]$ (left lobe):

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Answer 2

Using $r^2=x^2+y^2,x=r\cos\phi,y=r\sin\phi$, and the trig identity $\cos(2\phi)=\cos^2\phi-\sin^2\phi$, we have

$$r^2=36\cos(2\phi)$$ $$\implies r^4=36r^2(\cos^2\phi-\sin^2\phi)=(6r\cos\phi)^2-(6r\sin\phi)^2$$

$$\implies(x^2+y^2)^2=36x^2-36y^2$$ $$\implies {x}^{4}+2\,{x}^{2}{y}^{2}+{y}^{4}-36\,{x}^{2}+36\,{y}^{2}=0$$

Now let $x^2=a,y^2=b$. We now have

$$a^2+2ab+b^2-36a+36b=0$$ which is just the equation of a parabola. Now plot this. I will allow you to figure out how graphing parabolas works. Once you have this, erase all parts of the curve that are not in the first quadrant. You now have a plot of a set of points $(a,b)=(x^2,y^2)$. So now to convert this to $(x,y)$, just take the positive square roots of those points. For example, the point $(a,b)=(x^2,y^2)=(36,0)$ on the parabola turns into $(6,0)$ on the new curve. Do this until you have a plot of the curve in the first quadrant. Now notice that $(x^2,y^2)=((-x)^2,y^2)$, which means that if a point $(x,y)$ lies on the curve, so does $(-x,y)$. Similarly for $(x,-y)$ and $(-x,-y)$. Plotting the rest of these points in, you should get a curve that looks like an infinity sign. This is called a lemniscate.

Answer 3

This is Matlap's script. Note: this is plotting the real values not the imaginary ones( you might get warning in Matlab, to solve this problem, use real())

phi = 0:0.01:pi;
r = sqrt( 36*cos(2*phi) );
x = r.*cos( phi );
y = r.*sin( phi );
plot(x,y)

This is the result

Is this what you're looking for?

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Edit: For drawing them both

clear all
clc
phi = 0:0.01:pi;
r = sqrt( 36*cos(2*phi) );
x = r.*cos( phi );
y = r.*sin( phi );

subplot(2,1,1)
plot(x,y)
title('Cartesian Coordinates (x, y)');
xlabel('x-axis');
ylabel('y-axis');
grid on


subplot(2,1,2)
polar(phi, r)
title('Polar  Coordinates (r, \phi)');

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Updated:

The following results for $\phi$ from $\pi$ to $2\pi$. For the code, you need to change only this line phi = 0:0.01:pi; to phi = pi:0.01:2*pi;

And for this period phi = 0:0.01:2*pi; the result is as follows