PolarPlot the function $r=-2t$ with $t$ in $[0,2\pi]$:

Questions
1

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

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

48500.png
48500.png737×522 6.89 KB

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.

2

This is a duplicate from the site https://mathematica.stackexchange.com/questions/185957/interpretation-of-polarplot

I will postpone my answer.
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"]

polarplot.png
polarplot.png668×352 24 KB

3

PolarPlot[r[t], {t, 0, 2 Pi}] is basically the same as ParametricPlot[r[t] {Cos[t], Sin[t]}, {t, 0, 2 Pi}]. When r[t]<0 the point is placed in the opposite direction as the one you would expect from the value of the angle t. Try this:

Manipulate[
 PolarPlot[-2 t, {t, 0, 2 Pi}, 
  Epilog -> {PointSize[Large], Point[-2 s*AngleVector[s]], 
    Arrow[{{0, 0}, Pi/2 AngleVector[s]}]}],
 {{s, Pi/4}, 0, 2 Pi}]
4

Thank you for your code @GianlucaGorni, it’s very useful.