Changing from polar to Cartesian coordinate

Question

I'm trying to convert r = |cos θ| from polar coordinate to the Cartesian coordinate.

I have done the question for r = cos θ before, but this time I don't know how the absolute sign will affect the answer.

Answer 1

If $\theta\in[0,\pi]$, then\begin{align*}(r\cos\theta,r\sin\theta)&=(\cos^2\theta,\cos(\theta)\sin(\theta))\\&=\frac12(\cos^2\theta+\sin^2\theta+\cos^2\theta-\sin^2\theta,\sin(2\theta))\\&=\frac12(1+\cos(2\theta),\sin(2\theta)).\end{align*}So, this part of your curve is the circle centered at $(\frac12,0)$ with radius $\frac12$.

And, if $\theta\in[\pi,2\pi]$,\begin{align*}(r\cos\theta,r\sin\theta)&=-(\cos^2\theta,\cos(\theta)\sin(\theta))\\&=-\frac12(1+\cos(2\theta),\sin(2\theta)).\end{align*}So, this part of your curve is the circle centered at $(-\frac12,0)$ with radius $\frac12$.

Answer 2

I understand it as you want to write in $x$,$y$ coordinates your equation. If so, using $r=\sqrt{x^2+y^2}$ and $\theta = \arctan \frac{x}{y}$, your equation will be $$\left| \cos (\arctan \frac{x}{y})\right| = \sqrt{x^2+y^2}$$

Answer 3

Mutiply both sides by $r$

$$ r^2= r \cos \theta $$ $$ x^2 + y^2= x $$ $$ (x-\dfrac12)^2 + y^2 = (\dfrac{1}{2})^2$$

Given positive $\cos \theta $ lies in first, fourth quadrants; so we consider only in these quadrants to the right of y-axis. It is a circle of unit diameter touching y-axis at the origin to the right.

Answer 4

1) For $\theta \in [-π/2,π/2]$ : $|\cos \theta|=\cos \theta$.

$r=\cos \theta$, or $r^2=r \cos \theta$.

$x^2+y^2= x$, where $x \ge 0$.

$(x-1/2)^2+y^2=(1/2)^2$ (Completing the square).

2) For $\theta \in [π/2,(3/2)π]$: $|\cos \theta| =-\cos \theta$.

$r^2=-r \cos \theta$;

$x^2+y^2=-x$; where $x \le 0$;

$(x+1/2)^2+y^2=(1/2)^2$ (Completing the square).