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

Gennaro Arguzzi
Posted 7 years ago

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}]

enter image description here

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.
POSTED BY: Gennaro Arguzzi
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Thank you for your code @GianlucaGorni, it's very useful.
POSTED BY: Gennaro Arguzzi

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}]
POSTED BY: Gianluca Gorni
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