How do I prove that two equations in Cartesian and Polar coordinates are equivalent?

Question

I'm asked to verify that the set of points described by the cartesian equation $$(x-1)^2 + y^2 = 1$$ is equal to the set of points described by the polar equation $$r = 2 \cos{\theta}, \cos{\theta} > 0$$ Clearly the point described by $(0,0)$ under the Cartesian coordinate system is an element of the first set, since $(0-1)^2 + 0^2 = 1$. However, this point corresponds to polar coordinates of the form $(0, \theta)$, for arbitrary $\theta$, but $0 = 2\cos{\theta} \iff \cos{\theta} = 0 \iff \theta=\frac{\pi}{2} + n\pi$; so there are plenty of pairs $(r, \theta)$ that do satisfy the second condition, but there are plenty of pairs that do not. Do I just need to prove that there exists at least one such pair in the second set? How do I think about problems like this formally, and what would a formal proof of this look like?

Let $(x, y)$ such that $(x-1)^2 + y^2 = 1$. The pair $(r, \theta)$ which corresponds to $(x, y)$ satisfies $x = r\cos{\theta}$, $y = r\sin{\theta}$. We must show that such a pair also satisfies $r = 2\cos{\theta}$.

If $r \neq 0$, we have that $1 = (x-1)^2 + y^2 \iff 1 = (r\cos{\theta} - 1)^2 + (r\sin{\theta})^2 \iff 0 = r^2 - 2r\cos{\theta} \iff r = 2\cos{\theta}$.

My problem is with the case of $r=0$. Is the argument below formal enough?

If $r = 0$, then we must have had that $x = r\cos{\theta} = 0$, $y = r\sin{\theta} = 0$. The pair $(0, \frac{\pi}{2})$ in polar coordinates satisfies $r = 2\cos{\theta}$, and so the origin is also contained in the second set.

Answer 1

In $\mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way: $$\begin{cases} x = \rho \cos \theta \\ y = \rho \sin \theta. \end{cases} $$ Of course I need to specify more about $\rho$ and $\theta$, for now $\rho \geq 0$ and $\theta \in [0,2\pi)$. Note that $$\begin{cases} \mathrm{d}x = \cos \theta \,\mathrm{d}\rho -\rho\sin \theta \,\mathrm{d}\theta \\ \mathrm{d}y = \sin \theta \,\mathrm{d}\rho + \rho \cos \theta \,\mathrm{d}\theta, \end{cases}$$ hence \begin{align} \mathrm{d}x \wedge \mathrm{d}y & = (\cos \theta \,\mathrm{d}\rho -\rho\sin \theta \,\mathrm{d}\theta) \wedge (\sin \theta \,\mathrm{d}\rho + \rho \cos \theta \,\mathrm{d}\theta) \\ & = \rho \,\mathrm{d}\rho \wedge \mathrm{d}\theta. \end{align} Therefore the transformation is singular when $\rho = 0$. Also, reversing the equations above you get $$ \begin{cases} \rho^2 = x^2+y^2 \\ \tan \theta = y/x, \end{cases} $$ so when $\theta = \pi/2+k\pi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $\rho > 0$ and $\theta \neq \pi/2 +k\pi$. This way the coordinate transformation is invertible.

Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $ becomes \begin{align} (\rho \cos \theta-1)^2+\rho^2\sin^2\theta -1 & = 0 \\ \rho^2-2\rho\cos \theta& = 0 \\ \rho(\rho-2\cos \theta)&=0, \end{align} which implies $\rho = 2\cos \theta$, because $\rho \neq 0$. Actually $\rho > 0$, which implies $\cos \theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $\rho = 0$ is also a solution of $\rho(\rho-2\cos \theta)=0$, thus just study this case on its own. It is trivial, because $\rho = 0$ corresponds to the origin $(x,y) = (0,0)$.

Answer 2

Note that your original circle (in Cartesian coordinates) is a circle centered at $(1,0)$ of radius $1$. So the entire circle is in quadrants 1 and 4. So $\theta$ must be in the interval $[-\pi/2,\pi/2]$. Otherwise you just go over the circle again and again. You could have also noticed that you were given $\cos\theta\gt 0$, so it will also restrain your solutions.

The way I tend to approach this problem is to explicitly use $x=r\cos\theta$ and $y=r\sin\theta$ in the first equation on the left hand side. If you get $1$, then the proof is done.

Answer 3

Those sets are not equal.

The point $(0, 0) \in \mathbb{R}^2$ is in your first set of points, but not on the second. Proof: if $x = y = 0$, then $r = 0$, and since the second equation says $r = 2 \cos \theta$, then we must have $\cos \theta = 0$, violating the restriction that $\cos \theta > 0$.

That said, you probably meant $\cos \theta \geq 0$, so let's assume that.

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As you said yourself, you have to prove that the set of points described by one equation is equal to the set of points described by another equation. So you have to prove equality of two sets. There is one super standard way to do this rigorously: to prove that two sets $A$ and $B$ are equal, you prove that $A \subseteq B$ and $B \subseteq A$.

So let's take your example and tackle it rigorously. Let's start defining the two sets.

$$A = \{ \, (x,y) \in \mathbb{R}^2 \; : \; (x - 1)^2 + y^2 = 1 \, \}$$

$$B = \{ \, (x,y) \in \mathbb{R}^2 \; : \; \exists \, r,\theta \in [0,\infty) \; x = r \cos \theta \, \wedge y = r \sin \theta \, \wedge r = 2 \cos \theta \, \wedge \cos \theta \geq 0 \, \}$$

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Part 1: $A \subseteq B$

Let $(x,y) \in A$. We need to prove that $(x,y) \in B$. By definition of $B$ this happens if and only if there exists $r,\theta \in [0,\infty)$ satisfying the following four properties simultaneously:

$$x = r \cos \theta$$ $$y = r \sin \theta$$ $$r = 2 \cos \theta$$ $$\cos \theta \geq 0$$

If we can show how to construct $r, \theta$ based on $x, y$ we're done.

Our intuition says that we should probably try $r = \sqrt{x^2 + y^2}$ and $\theta = \text{atan2}(y,x)$. Note, if you are not familiar with atan2, it is a better $\arctan$. You might have thought of $\theta = \arctan(\dfrac{y}{x})$, but that would have the hassle of dealing with $x = 0$ differently. Instead, atan2 is much better because is only undefined for $x = 0$ and $y = 0$, for which case we shall explicitly choose $\theta = \dfrac{\pi}{2}$ (the reason will be clear later). Let's see if that works:

Checking property 1:

$$x = r \cos \theta$$

If $x = y = 0$, this holds because $0 = 0 \cos \dfrac{\pi}{2}$. Otherwise, this holds because $x = \sqrt{x^2 + y^2} \cos \text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).

Checking property 2:

$$y = r \sin \theta$$

If $x = y = 0$, this holds because $0 = 0 \sin \dfrac{\pi}{2}$. Otherwise, this holds because $y = \sqrt{x^2 + y^2} \sin \text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).

Checking property 3:

This one is not so immediate. What do we know about $x, y$ ? We know that $(x,y) \in A$ therefore we know that $(x - 1)^2 + y^2 = 1$. Therefore, the following is also true:

$$(r \cos \theta - 1)^2 + (r \sin \theta)^2 = 1$$ $$\iff r^2 - 2r \cos \theta + 1 = 1$$ $$\iff r^2 = 2r \cos \theta$$ $$\iff r = 0 \vee r = 2 \cos \theta$$

We wanted to prove that $r = 2 \cos \theta$, but it looks like we didn't get there. It looks like we proved something slightly weaker: that $r = 2 \cos \theta$ or $r = 0$. We now have to prove that if it happens to be the case that $r = 0$, then $r = 2 \cos \theta$ holds as well, so we can conclude that property 3 holds. This is easy: the only way for $r$ to be zero is $x = y = 0$, for which case our construction explicitly chooses $\theta = \dfrac{\pi}{2}$ which forces $r = 2 \cos \theta$ to hold.

Checking property 4:

$$\cos \theta \geq 0$$

Since we know that $x = r \cos \theta$ and $r \geq 0$ (because we defined it with a square root), we know that $\cos \theta \geq 0$ if and only if $x > 0$. So we have to prove that $x > 0$. How to do that? We know that $(x - 1)^2 + y^2 = 1$, and since $y^2 \geq 0$, this means that $(x - 1)^2 \leq 1$, which in turn means that $-1 \leq (x - 1) \leq 1$ which means implies $x \geq 0$.

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Part 2: $B \subseteq A$

Analogously, let $(x,y) \in B$. We need to prove that $(x,y) \in A$. By definition of $A$ this happens if and only if $(x - 1)^2 + y^2 = 1$ holds. We know that $(x,y) \in B$, so we know that there exists $r, \theta \in [0, \infty)$ satisfying those four properties. In particular, by the third property alone, we have:

$$r = 2 \cos \theta$$ $$\iff r^2 = 2 r \cos \theta$$ $$\iff r^2 - 2 r \cos \theta = 0$$ $$\iff r^2 - 2 r \cos \theta + 1 = 1$$ $$\iff r^2 ((\sin \theta)^2 + (\cos \theta)^2) - 2 r \cos \theta + 1 = 1$$ $$\iff r^2 (\cos \theta)^2 - 2 r \cos \theta + 1 + r^2 (\sin \theta)^2 = 1$$ $$\iff (r \cos \theta)^2 - 2 r \cos \theta + 1 + (r \sin \theta)^2 = 1$$ $$\iff (r \cos \theta - 1)^2 + (r \sin \theta)^2 = 1$$ $$\iff x^2 + y^2 = 1$$

Which completes the proof.

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Final thoughts

The proof above is probably too rigorous. But it should be enlightening anyway. We should always remember that when we make a not-so-rigorous proof, it is because we believe the proof can be made rigorous straightforwardly and we want to save both our time and our reader's time. But often we get carried away and make non-rigorous proofs without being actually convinced that we could make it rigorous if necessary.

In fact, I only noticed the problem with the strict inequality $\cos \theta > 0$ because I was going through this super rigorous proof, and on step 3 of part 1 I got stuck because I couldn't get the inequality. Then I realized that maybe they weren't equal in the first place!

At least I like to follow super rigorous proofs every now and then...