In polar coordinates, can r be negative?

I'm getting different answers for this. Many websites say that when you get a negative value of r, you flip the coordinate 180 degrees across the pole. However my teacher says that you cannot have a negative value of r because the function should pick up this point when a value of theta that is 180 degrees greater is used. For example, when you have $r=asin(2\theta)$, my teacher suggests that you should only you draw two petals- one in the first quadrant and one in the third quadrant, because r is only positive for these values of theta. However I have seen some websites draw 4 petals for this graph- one in each quadrant. Which is the correct graph?

I think that the graph of $r = a \sin( 2 \theta)$ should have two petals only. It's not wrong to draw four petals if you define negative $r$, as $(r,\theta) = (-r,\theta + \pi)$ if $r<0$. The problem with this definition is that polar coordinates are now no longer a bijection. The same point in cartesian coordinates will have two different polar coordinates. This is not a problem if you only want to draw graphs, but it is a serious problem in more advanced applications of Calculus. For instance you cannot use this coordinate change in a double integral if the transformation is not bijective (the point $r=0$ is not a problem because it is a set with measure $0$).

I think everyone will agree that $r = 2 \cos(\theta)$ is one circle. If you allow negative $r$, you will draw each point of the circle twice. This is not a problem if you're just graphing, but if you want the arc length you can get twice the correct answer.

Now let's return to $r = a \sin(2\theta)$ and see the corresponding cartesian equation.

$r = a \sin(2\theta) = 2 a \cos\theta \sin\theta$.

Multiplying each side by $r^2$ we get

$r^3 = 2 a ( r \cos \theta ) ( r \sin \theta)$,

which in cartesian coordinates is

$( x^2 + y^2 )^{\frac32} = 2 a x y$.

Since the left-hand side is positive, you should only draw points if $2a x y \geq 0$, so only in first and third quadrant if $a>0$. If we square both sides than we get

$( x^2 + y^2 )^{3} = 4 a^2 x^2 y^2$

and this graph has indeed four petals, but we introduced extraneous (or spurious) solutions, like squaring $x=1$ to obtain $x^2=1$ which now allows $x=-1$ as a solution. If we start with this last equation and try to derive the corresponding polar coordinates we get

$r^6 = 4 a^2 (r \cos\theta)^2 (r \sin \theta) ^2$.

$r^2 = a^2 ( 2 \cos \theta \sin \theta)^2 = a^2 ( \sin(2\theta) )^2$

So

$r = a \sqrt{\sin^2(2 \theta)} = a |\sin(2\theta)|$.

And $r = a |\sin(2\theta)|$ has indeed four petals, but $|\sin(2\theta)| \neq \sin(2\theta)$ if $\sin(2\theta)<0$, which happens in the second and fourth quadrant. So I would draw $r = a \sin(2\theta)$ with two petals only and would use $r = a |\sin(2\theta)|$ if I wanted the curve with four petals. I wouldn't say it's wrong to draw points with negative $r$ if you know what you're doing and you're not using these points to evaluate lengths and areas.

I agree with you. Saying $r$ when given as a function of $\theta$ must be positive is, in my opinion, as bad as saying that if $y=f(x)$ then $y$ must be positive. It certainly takes away from the beauty and symmetry of polar graphs. I'm actually wondering what he thinks the graph of $r=\cos\theta$ looks like.

I hope you've misunderstood your teacher, because his statement is not true. Though I wouldn't be surprised if this is some crazy new 'common core' approach.

To be honest, you should just use the definitions your teacher gives. It is less hassle and if you later want to argue a point which depends on those definitions then you have a leg to stand on. That doesn't mean you shouldn't press the issue and see if you can't get a reason from em or try to convert em, but in the meantime E is still the one grading your paper.

The question of which one is "right" is, unfortunately, probably meaningless. Both have good justifications and unless you can find some internal inconsistency with using one or the other, they both seem reasonable.

I would agree that the function should have two petals only.

Polar coordinates are usually defined such that every point $(x,y)$ in $\mathbb{R}^2$ can be reassigned some value $r\geq 0$ and $0 \leq \theta < 2\pi$. In this formulation, $r$ is a distance. It cannot be negative. It's the variation of $\theta$ that moves you into different quadrants, not $r$.

That said, it's possible to bend the rules. Wolfram alpha gives 4 petals

https://www.wolframalpha.com/input/?i=polar+plot+r%3Dsin(2%5Ctheta)

although I would argue that (strictly speaking) this isn't correct.