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Thursday, July 26, 2007

How to construct this image in Mathematica ?

( What do you think ? )

1 comment:

  1. It's a three-petal cardiod, so in polar coordinates that's just r=cos(3t). But that'll need rotating to get the precise version you have; which is the same as offsetting the t term. So r=cos(3(t+Pi/2)) does the job. Never used mathematica, but in Maple it'd be polarplot(cos(3*(t+Pi/2)));

    having first turned on plots using


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