Writing the polar equations $r = 2\cos 3\theta$ in Cartesian form

Question

For an example problem, in my textbook, the author wanted to demonstrate how to graph a polar function. Deeming it most convenient, my author took the polar function$$r=2\cos 3\theta ,$$and re-wrote it as the parametric equations$$\begin{cases} x=2\cos 3\theta \cos \theta\\ y=2\cos 3\theta \sin \theta .\end{cases}$$

How did the author arrive at these equations?

Answer 1

You can simply use the standard transformation from Cartesian coordinates to polar coordinate:

Given $r>0\;\text{and}\;\theta\in [0,2\pi)$

$$x=r\cos(\theta),\tag{$x$}$$ $$y=r\sin(\theta)\tag{$y$}$$

You are given that $$r=2\cos (3\theta)\tag{$r$}$$ So, replacing $r$ in each of the equations $(x)$ and $(y)$ with its equivalent $r = [2\cos(3\theta)]$, gives you:

$$x=[2\cos (3\theta)] \cos (\theta)\;\;\text{and}\;\; y=[2\cos (3\theta)] \sin (\theta).$$

---

Here's a nice image to help make sense of the "standard transformation" from Cartesian to polar coordinates: The Cartesian point $(x, y)$ is the furthest point from the origin along the blue line (length $r$), so given $\theta$ and the radius $r$, $(x, y) = (r\cos \theta, r \sin\theta)$.

It also helps to note the right triangle: $$\cos\theta = \dfrac{x}{r} \iff x = r \cos\theta\;\text{ and}\;\sin\theta = \dfrac yr \iff y = r\sin\theta.$$ This image is a good reminder as to how to transform $x, y$ into polar coordinates.

See also Polar coordinate system.

Answer 2

Hint: Use this fact that there is a transformation from Cartesian coordinate to Polar coordinate: $$x=r\cos(\theta), y=r\sin(\theta),\;\; r>0\;\;\text{and}\;\; \theta\in[0,2\pi]$$

Answer 3

Generaly if $r(\theta)$ is the polar function, the cartesian coordinates are given by $x(\theta)=r(\theta)\cos \theta$ and $y(\theta)=r(\theta)\sin \theta$. These are your parametric equations.