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2-2024 Quran and mathematics
7 months ago
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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)));
ReplyDeletehaving first turned on plots using
with(plots):