Polar to cartesian transformation

Question

I came across an interesting animation the other day.

What's going on here? Why is this possible?

From the author:

A function in the Cartesian plane can be transformed into polar coordinates by wrapping one axis around itself and collapsing it to a point.

1. Start with Cartesian graph.

2. Clip the graph to satisfy $y>0$ (not necessary in the example $y=\sin(6x)+2$).

3. Reflect in the line $y=x$.

4. Bend it to backwards on itself, as shown in the animation, to obtain the polar graph.

What is meant by bend it backwards on itself and wrapping one axis around itself and collapsing it to a point? Why are steps #2 and #3 necessary?

Answer 1

It was a bit tricky to derive the transformation demonstrated in the animation.

Let us start from a point $(x, y)$ of $y = f(x)$. There are two major transformations:

1. $T_1: \mathbb{R} \times \mathbb{R}_+ \to \mathbb{R}^2 $, $ (a,b) = T_1(x,y)$. Reflection with respect to $y = x$. Therefore, $T_1(x,y) = (y,x)$.
2. $T_2: \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}^2 $, $ (p,q) = T_2 (a,b)$. Transformation into polar coordinates by wrapping one axis around itself and collapsing it to a point as Wikipedia described. We want to have $(y\cos x, y\sin x) = (T_2 \circ T_1)(x,y)$.

Let us derive the second transformation in the animation. Let $R > 0$ and consider the circle centered at $(-R, 0)$ with the radius $R+a$. Then, the circle intersects the $x$-axis at $(a, 0)$. Draw a concentric circle with the radius $R+1$. Draw a line segment between $(-R, 0)$ and $(a, 0)$ and rotate it taking $(-R, 0)$ as a pivot, CCW if $b\ge 0$ and CW if $b < 0$, such that the arclength $AB$ equals $|b|$. Therefore, the angle $\theta$ in the figure is $b/(R+1)$. Let $r = (R+a)$. Then, the new coordinate is obtained by $$ \begin{aligned} (p,q) &= (r\cos \theta - R,~r\cos \theta)\\ &= \left((R+a)\cos\dfrac{b}{R+1} -R,~ (R+a)\sin\dfrac{b}{R+1}\right)\\ &= :U_R(a,b) \end{aligned} $$ $T_2$ can be thought of as the limit of $U$: $$ T_2 = \lim_{R \searrow 0} U_R $$ which follows from the definition of $U$. This is just one of the possible instances of $U$.

Note that $\lim_{R \to \infty}U_R$ is an identity map.

For all $(a,b) \in \mathbb{R}_+ \times \mathbb{R}$, $$ \lim_{R \searrow 0} U_R(a,b) = (a \cos b, a \sin b) $$

Comparing the result with the desired one $(y\cos x, y \sin x)$, we know that $T_1$ is indeed needed.

You can think of the circle with the radius $R$ as a bent axis, which collapses to a point (circle degenerates to a point at the origin) as $R \searrow 0$.

The animation shows the result of changing the values of $R$.

Answer 2

The graphs 1,3 in the animation have been labelled red OK.

Bending back on itself etc explanation ... introduces imho unnecessary confusion in graph 2 at this stage of learning definition while introducing polar coordinates.

The author is trying to imho explain invariance of "neutral line" of constant 2 radius in equation $ y=2+ \sin 6 x;\;x- C = \sin ^{-1}(y) \; r = 2+ \sin 6 \theta ; $ in the respective animation graphs.

Interchanging x-, y- axes is equivalent to defining an inverse function $ y=f(x)\to x= f^{-1}(y) $

Sketched in graph 2, $ x-C= \sin^{-1}y$ in which is a multi-valued function... the value $ C=2$ was chosen for illustration.

Bending back on itself etc... implies that y-axis lines that meet at $ y=-\infty$ are brought together from a single remote point ( parallel lines) gradually reducing the enclosed radius of curvature at reference point from "$\infty$" to 2 units ( in Mechanics beam theory called uniform bending where normals to any curve remain normal) for all straight lines between $ (p,q) $ and center $(R,0)$ in Kirchhoff hypothesis where length between two circles and two radial lines of a ring sector remains the same..

A 2-D complex variable mapping can be found if desired.