Archimedean spiral in cartesian coordinates I know that the Archimedean spiral can be represented using the polar coordinate system very easily. But I was wondering if it can be represented using the Cartesian coordinate system, and if so what is the function?
 A: Here is a solution for a double Archimedean spiral (see figure below).
Let us consider the simplest Archimedean spiral with polar equation:
$$\tag{1}r=\theta.$$
Using the following formulas:
$$\tag{2}\begin{cases}r^2=x^2+y^2\\\tan{\theta}=\tfrac{y}{x}\\\end{cases},$$
(1) can be transformed into the following implicit cartesian equation:
$$\tag{3}\arctan(\tfrac{y}{x})=\sqrt{x^2+y^2} \ \ \ \ \ \ \ \ \ \ (x \neq 0).$$
Taking $\tan$ on both sides gives the solution:

$$\tag{4}y= x \tan\sqrt{x^2+y^2}.$$

Remarks:

*

*Note that, by multiplying by $x$, restriction $x \neq 0$ is no longer needed. This shouldn't appear as magics : on the contrary, condition $x \neq 0$ in (3) was depriving the curve from points $(0,(4k+1)\tfrac{\pi}{2}), \ k \in \mathbb{Z},$ that have been returned to their owner under the form (4).


*In fact, equation (4) defines a double Archimedean spiral (changing $(x,y)$ into $(-x,-y)$ doesn't change this equation). See picture below where the red curve is the Archimedean spiral, strictly speaking, and the magenta curve is its copy through a central symmetry.


*From (4), it doesn't look possible to extract explicit cartesian equations $y=f_n(x)$ (there would be of course an infinite number of such equations).


*(4) cannot be transformed into a polynomial implicit equation $P(x,y)=0$. A simple reason : any straight line intersects an Archimedean spiral in an infinite number of points, which is impossible for a polynomial equation.


*It is easy from there to obtain a cartesian equation analogous to (4) for the general Archimedean spirals $r=a\theta.$

