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# Polar Plots - extra loops appear

Posted 10 years ago
 When plotting under Graphics,  PolarPlot[{1-2cos[3t]},{t,0,2pi}] I get the expected cloverleaf of 3 loops.However, surprisingly,  in each loop appears another smaller loop.How can the presence of these smaller loops be explained ?
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Posted 10 years ago
 Try this to show how 3 parts of the plot fit togetherShow[PolarPlot[{1 - 2 Cos[3 t]}, {t, 0, 2 Pi/3}, PlotStyle -> Black], PolarPlot[{1 - 2 Cos[3 t]}, {t, 2 Pi/3, 4 Pi/3}, PlotStyle -> Red], PolarPlot[{1 - 2 Cos[3 t]}, {t, 4 Pi/3, 2 Pi}, PlotStyle -> Blue]]
Posted 10 years ago
 When limits are set to [0,2Pi/3] ,indeed only one big loop is shown and no small loop inside of it.And now arises a new question: why small loops with limits [0,2Pi] and none with limits [0,2Pi/3] ?The presence of  small loops seems odd: for every given t in [0,2Pi] , there is only 1 value for r = 1-2Cos[3t]
Posted 10 years ago
 You can also get a sense sense of how this function unfolds by exploring it with a Manipulate: Manipulate[  Column[{    PolarPlot[{1 - 2 Cos[3 t]}, {t, 0, angle},      PlotRange -> {{-3, 3}, {-3, 3}}, Frame -> True,      PlotLabel -> N@angle],    Plot[{1 - 2 Cos[3 t]}, {t, 0, angle},      PlotRange -> {{0, 2 Pi}, {-1.1, 3}}, Frame -> True,      PlotLabel -> N@angle]    }  ], {{angle, 0.01}, 0.01, 2 Pi} ]Just drag the slider to see Professor Blinder's separate expressions unnfold.
Posted 10 years ago
 After a night's sleep, I think I found myself a rather simple explanation:Normally, negative values of r ( = 1-2Cos3t  in this case) are not considered in polar coordinates.However, they indeed seem to not have been discarded here, but presented in the opposite direction of the pole.So for instance ( when plotting with limits [0 , 2pi] ) , for t = 4pi/3 , an angle in the third quadrant, r = -1 , and this value is given as the top of the small loop in the first quadrant.No wonder then, that limiting t to [0, 2Pi/3] , no small loop appears in the big one, because inside this latter, r is only negative for values of t outside this region of t.Another interesting way to see and understand this is asking for  r = 1-2Cos3t  with limits for t : [-pi/9 , 7pi/9]
Posted 10 years ago
 For 3 simple loops tryPolarPlot[Sin[3 t], {t, 0, Pi}]orPolarPlot[Cos[3 t], {t, 0, Pi}]