Find a cartesian equation of the curve $r=1 - \cos \theta$ with $0 \le \theta \le 2 \pi$

Question

I'm trying to do this problem but I cannot get it. Here is what I have done so far: Start with $$r=1 - \cos \theta \;\;\;\text{with} \;\;\; 0 \le \theta \le 2 \pi$$ So I multiply both sides by r: $${r^2} = r-r\cos \theta$$ Now I have: $${x^2+y^2}=r-x$$ Correct?

Next: $${x^2+y^2}=\sqrt{x^2+y^2}-x$$ And that is where I'm stuck. What should I do? Does anyone know the final answer so I know what I am heading for?

Thanks.

Answer 1

As indicated in the comments, it doesn’t seem this can be reduced any further. It would help to notice that this curve is a limaçon, which is partially why I know you can’t do much more with it.

The curve can be parametrized, however, as $$\left\{ \begin{array}{} x = (1-\cos t)\cos t \\ y = (1-\sin t)\sin t \\ \end{array} \right.$$

To read more about equations for limaçons, turn here.

Summary of the comments: As a student, you have demonstrated that you understand the relationship between polar and rectangular coordinates, and trying to write this expression as a piecewise function $x\mapsto y$ is absurd.

Answer 2

Looks fine to me. If you want to simplify it into the form $y = f(x)$ you're going to have a rough time, because examining the graph, it's clear that will take at least 4 different pieces.