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In the precalculus curriculum I am teaching (using Stewart's book Precalculus: Mathematics for Calculus, 7th ed.), we do a bit of polar graphing, which includes discussion of symmetry on polar graphs. We teach the students to test for symmetry in standard ways:

  1. If the equation is invariant under the transformation $(r,\theta)\mapsto(r,-\theta)$, then we have symmetry across the polar axis. This is equivalent to the rectangular transformation $(x,y)\mapsto(x,-y)$.
  2. If the equation is invariant under the transformation $(r,\theta)\mapsto(r,\pi-\theta)$, then we have symmetry across the line $\theta=\frac{\pi}{2}$. This is equivalent to the rectangular transformation $(x,y)\mapsto(-x,y)$.
  3. If the equation is invariant under either of the the transformations $(r,\theta)\mapsto(r,\pi+\theta)$ or $(r,\theta)\mapsto(-r,\theta)$, then we have symmetry across the pole. This is equivalent to the rectangular transformation $(x,y)\mapsto(-x,-y)$.

These are all great, but then there was a homework question about the graph given by $r^2=\sin\theta$.

It's clear that we can replace $r$ with $-r$, so we have symmetry number 3. Additionally, $\sin\theta=\sin(\pi-\theta)$, we have symmetry number 2.

The function appears to fail the test for symmetry number 1, because $\sin(-\theta)=-\sin\theta$. However, any function with two of these symmetries automatically has the third. (Together with the identity transformation, we're just looking at the Klein 4-group here.)

enter image description here

Symmetry number 1 can be detected algebraically, but to do so, you have to use the equivalent transformation $(r,\theta)\mapsto(-r,\pi-\theta)$, which is clearly just the composition of symmetries 2 and 3. Similarly, symmetry number 2 could also be detected by the composition of 1 and 3: $(r,\theta)\mapsto(-r,-\theta)$

Anyway, the online WebAssign homework was only counting symmetries 2 and 3 right. I emailed them, and they say they've replaced the question, and I'll see the new version next time I make this assignment. Great, but that leaves math/teaching questions unanswered.

  • Has anyone else had this issue come up, and how did you handle it?
  • What is the easiest way to account for the failure of this equation, or one like it, when testing for symmetry number 1?
  • If we applied two tests for each symmetry (one with $r$ and one with $-r$), are there any equations that would still "get away" undetected?
  • Is it sufficient to state (A) that any function with two of these symmetries automatically has the third, and (B) any such function will pass two of the traditional tests, thus letting us know?
  • Is there anything else students should be on the lookout for, when trying to obtain symmetries of polar graphs from analyzing equations in $r$ and $\theta$?
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  • $\begingroup$ I believe that the answer to your sufficiency question is yes because the Klein 4 group is generated by any two of its nonzero elements. I've never tried to teach this topic before, however, so I'm not about to try to answer the rest of it. $\endgroup$
    – Opal E
    Oct 15, 2019 at 3:16
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    $\begingroup$ This is a good question which strikes at some of the trouble with introducing $r<0$ for polar graphing. My initial thought is that $r^2 = \sin \theta$ only allows solutions with $0 \leq \theta \leq \pi$ (restricting $\theta \in [0, 2\pi)$ for brevity). So, geometrically such solutions correspond to $-r$ with $\theta$ outside $[0, \pi]$, but those are not solutions to the given equation. Such points merely correspond geometrically to the direct algebraic solutions. $\endgroup$ Oct 15, 2019 at 4:39
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    $\begingroup$ For example, $\theta = \pi/6$ allows $r = \pm 1/\sqrt{2}$. Then, geometrically, $r=-1/ \sqrt{2}$ and $\theta = \pi/6$ we can trade for $r = 1/\sqrt{2}$ and $\theta = 7\pi/6$ (which does not directly solve $r^2 = \sin \theta$). So, what constitutes a solution ? $\endgroup$ Oct 15, 2019 at 4:42
  • $\begingroup$ @Maesumi, yes, thank you. I've edited that now. $\endgroup$ Mar 15 at 20:45

4 Answers 4

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The problem underlying the discussion in the question can be summarized as that it is necessary to choose a branch cut to define a complex logarithm (or arctangent).

It is a mistake and pedagogically a bad practice to allow negative values of $r$.

It is a mistake because pairs $(r, \theta)$ with $r$ possibly negative have no right to be called coordinates. Coordinates is a synonym for diffeomorphism (of whatever regularity). A system of coordinates on a domain gives a way to label uniquely each point of the domain. The uniqueness is essential for any labeling system to be properly called coordinates.

It is bad practice pedagogically because it generates unnecessary confusion.

The map $(x, y) \to (\sqrt{x^{2} + y^{2}}, \arctan(y/x))$ is a diffeomorphism from $(0, \infty)\times (—\pi, \pi]$ to $\mathbb{R}^{2}\setminus\{(x, y): x \geq 0\}$. Its inverse is the restriction to its image of the complex exponential, written in real coordinates. The branch cut along some axis is needed to obtain a diffeomorphism. The choice of the negative real axis as the branch is customary.

Allowing negative values of $r$ introduces redundancy. It destroys the unique labeling property.

The equation $r^{2} = \sin \theta$ has no solutions when $\theta \in (\pi, 2\pi)$.

The picture drawn in the question is wrong because negative values of $r$ should not be allowed. Those points on the curve with $r < 0$ do not correspond to points on the curve $$(x^{2} + y^{2})^{3/2} = y.$$ The curve should only be the top part of the curve indicated. Of the symmetries considered only the map - $(r, \theta) \to (r, \pi - \theta)$ (reflect across the vertical axis) is valid.

The map $(r, \theta) \to (r, \pi + \theta)$ is not a symmetry of the equation. If $\sin \theta$ is positive, then $\sin(\pi + \theta)$ is negative, and there is no solution. When negative values of $r$ are allowed, then $(-r, \theta) \sim (r, \pi + \theta)$ so, for a point $(r, \theta)$ of the curve with $r > 0$, the "solution" $(-r, \theta) \sim (r, \theta + \pi)$ is not a solution! Put another way, the equation $r^{2} = \sin \theta$ is not well defined on the quotient of $\mathbb{R}\setminus\{0\} \times (0, 2\pi)$ by the identification $(r, \theta) \sim (-r, \pi + \theta)$. The solutions with $r$ negative do not correspond to any points in the $(x, y)$-plane.

The two-lobed curve drawn in the question corresponds to the Cartesian equation $$(x^{2} + y^{2})^{3} = y^{2},$$ which corresponds to the two polar equations $r^{2} = \pm \sin \theta$. This curve does admit all the symmetries discussed in the question.

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    $\begingroup$ This is a good answer so I upvoted. But still, I think a framework where $r<0$ is needed. Typically, polar coordinates appear with $r$ given as a function of $\theta$, so $r=f(\theta)$. I really want the polar graph of $r=\cos(\theta)$ to be a complete circle, traced continuously as theta varies from $0$ to $\pi$! If we have to abide by the convention $r>0$, then we are stuck with a semi-circle. $\endgroup$
    – user52817
    Oct 16, 2019 at 19:22
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    $\begingroup$ I very much like this counter-cultural answer. It has long annoyed me that we allow $r<0$ in the Calculus II curriculum I taught. I usually grow tired of it by Calculus III where I insist that $r = \sqrt{x^2+y^2}$. But, I must admit there exist calculus texts which do not require $r \geq 0$. $\endgroup$ Oct 18, 2019 at 1:20
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    $\begingroup$ In my view, "polar coordinates" is the map $P(r,\theta) = (r\cos(\theta), r\sin(\theta))$. Calling this map "coordinates" is a bit of a misnomer, but it is not terrible because this map is a local diffeomorphism away from $r=0$: when you restrict to any open rectangle $(r_1, r_2) \times (\theta_1, \theta_2)$ with $0 \not\in (r_1, r_2)$ and $\theta_2 - \theta_1 < 2\pi$ you get a diffeomorphism onto the image. When we write $r^2 = \sin(\theta)$ we mean the image of the graph of $r^2 = \sin(\theta)$ under $P$. $\endgroup$ Mar 1 at 12:26
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    $\begingroup$ @JamesS.Cook, I don't see this answer as a counter-cultural at all; I've never seen $r<0$ allowed and I suspect it doesn't exist beyond a particular set of textbooks. $\endgroup$
    – Kostya_I
    Mar 7 at 15:43
  • $\begingroup$ @Kostya_I I think it is allowed in popular graphing websites. For instance, I just graphed $r=-1$ on www.desmos.com and it gives me a circle. Pretty much all the mainline calculus texts I've looked at do this for polar graphing as it is commonly taught in American university Calculus II. I wish it were not so, but it is what it is. $\endgroup$ Mar 7 at 23:00
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A given point with polar coordinates $(r, \theta)$ has a doubly infinite set of representations $(r,\theta+2n\pi)$ and $(-r,\theta+(2n+1)\pi)$, for integer $n$. When checking for symmetry they should all be considered.

Let $E$ be the polar equation $f(r,\theta)=0$ with a graph $G$. If for some integer $n$ the following equation is equivalent to $E$ then the indicated symmetry applies.

X1: $f(r,-\theta+2n\pi)=0$ or X2: $f(-r,-\theta+(2n+1)\pi)=0$ then $G$ is symmetric with respect to $x$-axis or polar axis.

Y1: $f(r,-\theta+(2n+1)\pi)=0$ or Y2: $f(-r,-\theta+2n\pi)=0$ then $G$ is symmetric with respect to $y$-axis or $\theta=\pi/2$.

O1: $f(r,\theta+(2n+1)\pi)=0$ or O2: $f(-r,\theta+2n\pi)=0$ then $G$ is symmetric with respect to origin or pole.

For $f(r,\theta)=r^2-\sin(\theta)=0$ all three symmetries are present.

Apply X2: then $f(-r,-\theta+\pi)= (-r)^2-\sin(-\theta+\pi)=r^2-\sin(\theta)=0$.

Apply Y1: then $f(r,-\theta+\pi)=(r)^2-\sin(-\theta+\pi)=r^2-\sin(\theta)=0$.

Apply O2: then $f(-r,\theta)=(-r)^2-\sin(\theta)=r^2-\sin(\theta)=0$.

For additional examples you may want to consider $r=\sin(\theta/k)$ for some integer $k$.

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  • If we applied two tests for each symmetry (one with 𝑟 and one with −𝑟 ), are there any equations that would still "get away" undetected?

This is a question that came up recently in my class as well. The following function seems to get away from both tests for symmetry over the pole undetected: $r=\cos\frac{\theta}{2}$

If you replace $r$ with $-r$, it yields $r=-\cos\frac{\theta}{2}$

Instead, if you were to replace $\theta$ with $\theta+\pi$ (or, alternatively $\theta-\pi$), it results in $r=\sin\frac{\theta}{2}$ (or, alternatively, $r=-\sin\frac{\theta}{2}$, for the other substitution).

Taking some specific points as examples, we see that $(1,0)$ is on the graph and satisfies the equation, but the ordered pair $(-1,0)$ does not satisfy the equation. Neither do $(1,\pi)$ or $(1,-\pi)$. The ordered pair that does satisfy the equation and is across the pole from $(1,0)$ is $(-1,2\pi)$.

I suppose we could make a more rigorous version of the tests that also took into account coterminal angles.

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If a textbook gives symmetry tests for graphs of polar equations without stating that they are sufficient but not necessary, that's a glaring omission, in my opinion. I looked through several books, and I only saw this in Stewart and one other book. The rest either didn't cover symmetry tests or stated that they weren't necessary conditions. Larson seemed the most comprehensive to me. The book gives the following tests.

The graph of a polar equation is symmetric with respect to the following when the given substitution yields an equivalent equation.

  1. The line $\theta=\pi/2$: Replace $(r, \theta)$ with $(r, \pi-\theta)$ or $(-r, -\theta)$.
  2. The polar axis:      Replace $(r, \theta)$ with $(r, -\theta)$ or $(-r, \pi-\theta)$.
  3. The pole:               Replace $(r, \theta)$ with $(r, \pi+\theta)$ or $(-r, \theta)$.

Larson also gives two more "quick tests for symmetry."

  1. The graph of $r=f(\sin\theta)$ is symmetric with respect to the line $\theta = \pi/2$.
  2. The graph of $r=g(\cos\theta)$ is symmetric with respect to the polar axis.

The graph of $r=\theta+2\pi$ for $ -4\pi \leq \theta \leq 0$ is given as an example to show that these tests are not necessary for symmetry.

A graph of the polar equation r = θ + 2π between θ = -4π and θ = 0. The graph is symmetric about the line θ = π/2.

Maesumi gave necessary conditions in their answer. The equation $r=\theta+2\pi$ yields an equivalent equation under the substitution $(r, \theta) \mapsto (-r, -\theta-4\pi)$, satisfying Y2.

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