# Cardioid: converting parametric form into polar coordinates

I am interested in converting parametric equations:

$$\varphi=\left(\varphi_{1},\varphi_{2}\right)=\left(2\cos{t}-\cos{2t},2\sin{t}-\sin{2t}\right)$$

which describe a cardioid, into polar coordinates. What would be a good beginning for this type of an operation? I would be thankful for some hints.

-

The parametric equations describe $(x,y)(t) = (2 \cos(t) - \cos(2t), 2 \sin(t) - \sin(2t))$:

In order to convert this into polar coordinates, express the radius, and the angle in terms of $x$ and $y$ first: $$r(t)^2 = x(t)^2 + y(t)^2$$ this would be a simple expression in terms of $\cos(t)$. You could them express the polar angle using atan2 as follows $\theta(t) = \arctan(x(t), y(t))$. The radial representation would be obtained if it were possible to solve for $t = t(\theta)$, and substituted into $r(t)$ to obtain $r(\theta)$. But the equation is not linear and does not admit a simple closed-form.

However, seeing that $r(t)$ is much simpler, we can invert it to find $t=t(r)$ and then $\theta(r) = \theta(t(r))$. Doing so will only allow us to parametrize half of the cardoid, due explicit symmetry $r(t) = r(2\pi - t)$.

If you permit the use of software, here is a way to get the radial representation:

-
Well, first of all, thank you for a thorough answer. By letting $\theta(r)=\theta(t(r))$ and solving for $r$ in terms of $\theta$ I arrive at $r=\sqrt{5-4\cos{\theta}}$. This, as you mentioned, is the parametrization of only half of the cardioid... indeed, the full polar form of cardioid generated by a circle with radius $1$ seems to be more simple ($r=2+2\cos{\theta}$). Is there any way we can obtain a full representation without using software? –  Johnny Westerling Nov 15 '12 at 23:05
The simple equation is for the shifted cardoid, that is the one parameterized as $(2\cos(t) - 1 - \cos(2t), 2 \sin(t) - \sin(2t))$. If you start with this parametrization, I am confident you can recover the simple answer you are looking for, i.e. $r = 2 - 2 \cos(\theta)$. –  Sasha Nov 16 '12 at 1:00
Yes, you are absolutely correct... –  Johnny Westerling Nov 16 '12 at 5:10
I was working to this one by very tedious manipulating relations and some substitutions in Maple that I saw this old question. I can see that we get $$(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$$ The final step indeed is to use $x=r\cos\theta,~~y=r\sin\theta$.
Hint: compute the radius and the angle as a function of $t$! ;-)