Why is the value of r in $r=\cos θ$ never negative?

Question

The graph of $r=\cos θ$ forms a circle with radius $.5$ and around $(1,0)$. However, in terms of Cartesian plane coordinates, why doesn't the $r$ value go into the second and third quadrants? Wouldn't $r$ be negative when $θ$ is in the second and third quadrants?

Answer 1

You are correct $r$ is negative when $\frac{\pi}{2} \leq \theta < \frac{3\pi}{2}$, but since $r$ is negative, it points in the opposite direction of the $\theta$. Thus, the point $(r,\theta)=(-1,\pi)$ does not map to $(x,y)=(-1,0)$, but to $(x,y) = (1,0)$

Answer 2

$r$ can be negative.

When $\theta$ is in Quadrant II or III, $r=\cos\theta$ is negative and the point is located in the opposite quadrant.

But as $\cos(\pi+\theta)=-\cos\theta$, $(r_0,\cos\theta_0)$ and $(-r_0,\cos(\theta_0+\pi))$ coincide.

Answer 3

I’d like you to first think of the equation $y=x^2$ in the Cartesian coordinate $(x,y)$-plane. That condition is a restriction over the plane that famously identifies those points whose $y$ value (height fron $x$ axis) is exactly the square of their $x$ value.

Now move back to polar coordinate $(r, \theta)$-plane. The equation $r = \cos\theta$ identifies those points whose $r$-value (distance from the origin) is exactly $\cos$ of their $\theta$ value (angle from $x$ axis). But $r$ is always greater than zero (and $\theta$ is always between $\pm\pi$), so the condition is not meaningless unless $\theta$ makes the cosine positive, that is $\theta$ is between $\pm\pi/2$.

You’d just gotten it the other way around.

---

ADDENDUM. (Technical) The reason why I require $r>0$ and $-\pi< \theta < \pi$ is because I want polar coordinates to be consistent with the usual definitions of curvilinear coordinates in analysis, or coordinate charts in differential geometry. I want my coordinates to be a homeomorphism between an open subset of the plane (in this case the plane without the negative $x$ axis and the origin) and another open subset of the plane ($(0,+\infty)\times(-\pi,+\pi)$), and in particular a bijection. Otherwise it would be meaningless to even say that the polar coordinates of a point on the plane are such and such, because multiple answers would do. Notice that my approach provides a sensible answer to the OP’s doubts without recurring to weird ideas such as “negative distance from the origin”.

Answer 4

Short summary:

In polar coordinates $r$ is the distance of a Cartesian point $(x,y)$ from the origin $$r=\sqrt{x^2+y^2}\geq 0$$ And there are no $r<0$ for whatever "going backwards through the origin".

p.s.:

But what you can do is the following: A curve $r=r(\theta)$ corresponds to the parametric curve $$(x(\theta),y(\theta))=r(\theta)\cdot(\cos(\theta),\sin(\theta))$$ This form works also for $r(\theta)<0$. So, you can see this as an extension of a polar curve to allow for "negative radii". In this case the corresponding polar curve would be $$(x(\theta),y(\theta))=|r(\theta)|\cdot(-\cos(\theta),-\sin(\theta))= |r(\theta)|\cdot(\cos(\theta-\pi),\sin(\theta-\pi))=$$ $$=\hat r(\phi)\cdot(\cos(\phi),\sin(\phi)) \mbox{ where } \phi = \theta-\pi \mbox{ and } \hat r(\phi) = |r(\theta)|$$