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

PolarPlot[-2 t, {t, 0, 2 Pi}, AxesLabel -> {x, y}]

enter image description here

When $t=\pi/4$, I have:


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.

  • $\begingroup$ 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. $\endgroup$
    – Kuba
    Commented Nov 14, 2018 at 10:27
  • $\begingroup$ Hello @Kuba, can you explain your comment please? If the angle is unknown, how can I do the reflection? Maybe the angle is $\pi/4$? $\endgroup$ Commented Nov 14, 2018 at 10:30
  • $\begingroup$ I think the confusion here is what exactly does GetCoordinates give for PolarPlot $\endgroup$
    – Lotus
    Commented Nov 14, 2018 at 10:33
  • $\begingroup$ 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. $\endgroup$
    – Kuba
    Commented Nov 14, 2018 at 10:36
  • $\begingroup$ @Kuba Ok, your example is clear. I have the inverse problem: given $r=-1.57$, what is the angle (without knowing x and y)? $\endgroup$ Commented Nov 14, 2018 at 10:42

1 Answer 1


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}, 
 PlotLegends -> "Expressions"]

enter image description here

  • $\begingroup$ Hello @AlexTrounev, thank you very much for your excellent explanation. $\endgroup$ Commented Nov 14, 2018 at 16:05
  • 1
    $\begingroup$ You're welcome! $\endgroup$ Commented Nov 14, 2018 at 16:22

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