# Interpretation of PolarPlot

I tried to plot the function $$r=-2t$$ with $$t$$ in $$[0,2\pi]$$:

PolarPlot[-2 t, {t, 0, 2 Pi}, AxesLabel -> {x, y}] When $$t=\pi/4$$, I have:

$$r=-\pi/2=-1.57$$

Through Get Coordinates, I saw the only point so that $$r=-1.57$$ is $$("-"1.57, \,\, 3.9...)$$. Why is the angle $$3.9..=\pi+(\pi/4)$$ different from $$t=\pi/4$$?

How can I get the above angle by using math formulas? I can't use the classic relationship $$tan(t)=y/x$$ because I don't know x and y (without looking the plot).

Thank you so much in advance.

• That is your problem, the point you measured has r = 1.57. If you want to get an angle for r = -1.57 you need to reflect that point through the origin. – Kuba Nov 14 '18 at 10:27
• Hello @Kuba, can you explain your comment please? If the angle is unknown, how can I do the reflection? Maybe the angle is $\pi/4$? – Gennaro Arguzzi Nov 14 '18 at 10:30
• I think the confusion here is what exactly does GetCoordinates give for PolarPlot – Lotus Nov 14 '18 at 10:33
• If you have a point (1,1) cartesian, is it (sqrt(2), Pi/4) or (-sqrt(2), 3Pi/4)? My point is that information that you use negative radius is gone so you have to think about it yourself. – Kuba Nov 14 '18 at 10:36
• @Kuba Ok, your example is clear. I have the inverse problem: given $r=-1.57$, what is the angle (without knowing x and y)? – Gennaro Arguzzi Nov 14 '18 at 10:42

Here we have an example of incorrect use of the polar radius function r[t], which by definition must be positive or zero, $$r\ge 0$$. Mathematica offers a kind of solution for $$r<0$$ using the Euler formula $$-1=e^{i\pi}$$, and interpreting multiplication by $$-1$$ as a rotation of $$\pi$$. In this case,$$\pi$$ is added to the argument t. This is where the strange data comes from.
PolarPlot[{-2*t, 2*t} , {t, 0, Pi}, AxesLabel -> {x, y}, 