Help with finding degrees/radians when converting rectangular to polar coordinates?

Question

I was given the problem "determine two pairs of polar coordinates for (-3,0) when theta is greater than 0 degrees and less than 360 degrees" and I know the radii are 3 and -3.

When I use arctan 0/-3, of course I get 0 degrees. What I would like to know is does the 0 degrees go with the 3 or the -3, and why.

Thank you! I hope I did this right, this is my first time on here.

Answer 1

Certainly a polar coordinate of $r=3,\theta=\pi$ obtains the point $(-3,0)$. If you want another equivalent polar coordinate, you could just get another angle equivalent to $\pi$. Can you think of one?

As for your question about arctan, be careful with that function: It will give you reliable results for the first quadrant but in other regions you'll have to use some other kind of reasoning to figure out the correct angle. That is to say, if you just tell me the value of $\arctan(y/x)$ there is no way to know exactly which point or even which quadrant $(x,y)$ is in.

To elaborate, take another example. Suppose that I have a point $(x,y)$. I know which point this is, but you don't. But I'll give you hints to figure it out. Namely, I tell you $\arctan(y/x)=1$. You, being a diligent Math student, tell me that it corresponds to an angle of $\pi/4$. Or at least that's what your calculator tells you. However, the point I really had in mind was $(-1,-1)$. This does in fact satisfy $\arctan(y/x)=\arctan(-1/-1)=\arctan(1)=\pi/4$, but it turns out that the angle of $(-1,-1)$ is $5\pi/4$.

In general the arctan does not specify the angle of the point. Unless you know which quadrant the point is in, you could always possibly be $\pi$ radians off from the correct angle. (Note that the difference of $\pi$ and $0$ is $\pi$, and also the difference of $5\pi/4$ and $\pi/4$ is $\pi$.)

Answer 2

For using polar coordinates, the anti-clockwise direction of traversing theta is taken as positive by convention. So for $r=3$, you need to go $\pi$ anti-clockwise to reach $(-3,0)$, therefore the point is $(3,\pi)$ in polar coordinates.

Also, if you take $r=-3$, you need to go $0$ radians anti-clockwise, so the point is also $(-3,0)$ in polar coordinates.

Using the $\arctan$ function would just give you an idea of how much angle you need to cover, but in which quadrant/area, you need to figure out yourself.