0
$\begingroup$

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.

4 Answers 4

3

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

enter image description here

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:

enter image description here

  • 0
    Yes, you are absolutely correct...2012-11-16
1

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 $x=r\cos\theta,~~y=r\sin\theta$.

  • 0
    ALso needs an UV +1\2013-11-28
0

Hint: compute the radius and the angle as a function of $t$! ;-)

The radius is the square root of the sum of the squares of the two components; the angle can be determined using the inverse of the tangent function.