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Let $y=f(x)$ in Cartesian coordinates. Swapping the $x$ coordinate with the $y$ coordinate has the effect of reflection about the axis $y=x$.

But if $r=f(\theta )$ in polar coordinates (where $r$ is the modulus and $\theta$ is the argument), what happens when we swap $r$ with $\theta$?

For example, what's the difference between the polar graphs of $r=\theta ^2$ and $\theta =r^2$? Do they look the same? Or is there some kind of reflection as well?

BIRA
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  • We rectangular coordinates measure the same things in different dimensions so swapping will flip the image. But polar coordinates measure different things angles versus distance. Swapping them would be twisting distance into angles and stretching angles into distance. Simple things like a line segment $\theta = c$ turns into a circle $r= c$. A spiral $r=c\theta$ becomes and scaled spiral $\theta=cr$. $r=\theta^2$ is spiral be that spins a lot at first then bursts out whereas $\theta=r^2$ is a spiral that grows fast at first and then spins fast. – fleablood Aug 28 '20 at 23:45

2 Answers2

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Polar coordinates we call well known mapping $\mathbb{R}^2 \to \mathbb{R}^2$, from $(x,y)$ to $(\theta, r)$ using formulas $x = r\cos \theta$, $y = r\sin \theta$, $r \geqslant 0,\theta \in [0, 2\pi) $.

As to plane $(\theta, r)$, then it is usual cartesian coordinates, usual $\mathbb{R}^2$, and you can think about it exactly as you think about $(x,y)$. $r=\theta^2$ is exactly parabola. $\theta=r^2$ is both branches of square root.

We use polar coordinate, when some function/ curve looks "difficult" for $(x,y)$ and by mapping it to $(\theta, r)$ plane we obtain more "easy" case. Most known example is circle $x^2+y^2=r^2$, which by polar coordinate moves to interval $[0, 2\pi) \times \{1\}$. Disk $x^2+y^2\leqslant r^2$ is mapped to rectangle $[0, 2\pi) \times [0,1]$.

Addition. Now about swapping variables. By definition axial symmetry is not identical Orthogonal Transformation which have line of fixed points. This line is called symmetry axis. To obtain for point $M$ symmetrical point $M'$ with respect to symmetry axis one need to draw perpendicular line to symmetry axis from $M$ and take point $M'$ on this perpendicular on other side of symmetry axis on same distanse as $M$.

For example, if we consider $y=x$ as symmetry axis, then for point $(a,b)$ symmetrical point is $(b,a)$.

So, on $\mathbb{R}^2$ swapping coordinates i.e. having graph $y=f(x)$ and considering $x=f(y)$ is exactly creating symmetry with respect to line $y=x$. Same is, of course if we speak about $r=f(\theta)$ and considering $\theta=f(r)$ - they are symmetric with respect to line $r=\theta$.

Another question is what gives swapping variables for $(x,y)$ in $(\theta, r)$ and reverse. Let's consider firstly "polar plane". As is stated above, swapping variables there means symmetry with respect to line $r=\theta$. Last is well known Archimedean spiral on "cartesian plane". So swapping coordinates $\theta$ and $r$ gives on plane $(x,y)$ graphs "symmetric" with respect to spiral $r=\theta$ which is same as $\sqrt{x^2+y^2}=\arctan \frac{y}{x}$. For example parabola $r=\theta^2$, which is some type of spiral on $(x,y)$, after swapping gives $\theta=r^2$, or taking its one branch, $r=\sqrt{\theta}$ is again some spiral on $(x,y)$.

Summing up:

  1. parabola $y=x^2$ is axial symmetric with respect to square root $x=y^2$ using symmetry axis line $y=x$.

  2. In "polar" language spiral $r=\theta^2$ is "spirally" symmetric with respect to spiral $\theta=r^2$ using symmetry "axis" spiral $r=\theta$

Second example. Let's take in polar plane $r=\tan\theta$ i.e. points $(\theta,\tan\theta)$. Swapping variables give $\theta=\tan r$ i.e. points $(\tan r,r)$. Obviously $(\theta,\tan\theta)$ is axially symmetrical to $(\tan r,r)$ with respect to symmetry axis $\theta=r$. Now if we consider corresponding points on $(x,y)$ plane, then symmetry axis $\theta=r$ creates spiral, while $r=\tan\theta$ and $\theta=\tan r$ create some corresponding curves on $(x,y)$: $\sqrt{x^2+y^2}=\frac{y}{x}$ and $\arctan \frac{y}{x}=\tan \sqrt{x^2+y^2}$. Obviously $(x,y)$ curves are not axially symmetrical.

If it sounds acceptable, we can call "spirally" symmetrical on plane $(x,y)$ such points, which preimages are axially symmetrical on plane $(\theta, r)$ with respect to symmetry axis $\theta=r$.

Using this term we can call $\sqrt{x^2+y^2}=\frac{y}{x}$ and $\arctan \frac{y}{x}=\tan \sqrt{x^2+y^2}$ "spirally" symmetrical on plane $(x,y)$.

zkutch
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  • Name "Polar coordinates" is used for mapping, which is in first paragraph. It is function $\mathbb{R}^2 \to \mathbb{R}^2$. First $\mathbb{R}^2$ we imagine as $(x,y)$ plane, second as $(r,\theta)$. But they are only 2 exemplars of one and same object $\mathbb{R}^2$. Practically $\mathbb{R}^2$ is mapped to itself. – zkutch Aug 28 '20 at 23:04
  • Maybe I mised something, but I don't know how that answers the question. The first paragraph is a well-known prerequisite, I would say; the second paragraph is about Cartesian coordinates. And the third paragraph is about how a disk is mapped to a rectangle... – BIRA Aug 28 '20 at 23:07
  • Sorry if there is some misunderstanding, but in you question is sentence "what's the difference between the polar graphs of $r=\theta^2$ and $\theta=r^2$? Do they look the same?" - in second paragraph is answer. May be you mean something other? – zkutch Aug 28 '20 at 23:12
  • Oh, so do the graphs really look the same? I wasn't sure... – BIRA Aug 28 '20 at 23:14
  • They are exactly same as $y=x^2$ and $x=y^2$. – zkutch Aug 28 '20 at 23:19
  • Well, it seems that there was a misapprehension. $y=x^2$ and $x=y^2$ are the same, in a sense, but $x=y^2$ looks like a reflected version of $y=x^2$, and does not coincide with the other graph everywhere. In polar coordinates, $r=\theta ^2$ and $\theta =r^2$ (which is $r=\sqrt{\theta}$) look very different... It's not as simple as a reflection, it seems like squishing or something else... So I wouldn't consider these graphs to be exactly the same per se. – BIRA Aug 28 '20 at 23:25
  • $r=\tan\theta$ and $r=\arctan\theta$ is a very good example. – BIRA Aug 28 '20 at 23:43
  • Letters, is it $(x,y)$ or is it $(r,\theta)$, does not create any difference when you look at graphs on $\mathbb{R}^2$ - this is what I want to say. $\mathbb{R}^2_{(x,y)} $ is same as $\mathbb{R}^2_{(r,\theta)} $. $x=y^2$ is not even function with respect to $x$. As graph it gives both branches of square root, as I wrote in answer, positive and negative. Your last example can be written as $y=\tan x$ and $y = \arctan x$. May be you mean interdependency of $y=f(x)$ and $x=f(y)$ for same $f$? Then they are symmetrical with respect to $y=x$. – zkutch Aug 29 '20 at 00:30
  • No, I meant the graphs of $r=\tan\theta$ and $r=\arctan \theta$ (where $r$ is modulus and $\theta$ is argument) in polar coordinates: they look very different... unlike the graphs of $y=\tan x$ and $y=\arctan x$ in Cartesian coordinates. – BIRA Aug 29 '20 at 00:48
  • Definitely you mean something other, but expressed it not clear. Of course $r=\tan \theta$ on plane $(r,\theta)$ is same as $y=\tan x$ on plane $(x,y)$. May be you mean image of $y=\tan x$ on plane $(r,\theta)$? Then it is $r\sin \theta = \tan \frac{\sin \theta}{\cos \theta}$. Is my guess correct? – zkutch Aug 29 '20 at 00:57
  • Polar graphs, polar graphs... – BIRA Aug 29 '20 at 01:40
  • @BIRA. Poetry? I thought little about swapping variables and added it to answer. Is this what you mean? – zkutch Aug 29 '20 at 09:47
  • What does "spirally symmetric" mean? – BIRA Aug 29 '20 at 12:13
  • It is not known term, I especially used it in quotes. I mean it play same role in polar terms as $y=x$ in cartesian. Spirals $r=\theta^2$ and $\theta=r^2$ are symmetrically on both sides of spiral $\theta=r$. – zkutch Aug 29 '20 at 12:22
  • Are you sure about the spiral symmetry? If I compare $r=\tan\theta$ and $\theta =\tan r$, I wouldn't say they are "spirally symmetric" at all. – BIRA Aug 29 '20 at 13:21
  • I am sure, but let me little time to formally prepare everything. – zkutch Aug 29 '20 at 13:49
  • @BIRA. Wrote promised. – zkutch Aug 30 '20 at 11:50
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enter image description here

The polar plots of $r= \theta^2, \sqrt \theta, \theta$ are shown in blue,red and grey respectively.

The gray line does not reflect red/blue profiles in a way presently imaginable by me.

This is perhaps because $(r,\theta)$ do not have the same physical dimension and belong to a common type or genre.

$Z=r e^{i \theta}$ is a complex number. I do not know what $U = \theta e^{ir}$ really is.

A very good question, but my answer may be is no good. That is indirectly suggested geometric reciprocal requirement is not answered.

Narasimham
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  • What do you mean by physical dimension? Well, of course, $r=\theta ^2$ is just an example. But I wonder if something can be inferred about all the polar graphs, similar to what can be inferred about all the inverted Cartesian graphs of $y=f(x)$ (i.e. the reflection property). Nevertheless, thanks for your answer with the illustration. – BIRA Aug 28 '20 at 23:02
  • In Cartesian case,$x,y$ are linear dimension. In Polar case $r,\theta$ have a linear dimension and no dimension (angle is a ratio of lengths) respectively. – Narasimham Aug 28 '20 at 23:06
  • Oh, I know what you mean now. But is the graph of $r=\sqrt{\theta}$ the same as the graph of $\theta =r^2$? If we take these as ordered pairs... – BIRA Aug 28 '20 at 23:10
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    Yes, indeed they are same. – Narasimham Aug 28 '20 at 23:13