Bachelier
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In polar equations, how can I graph r^2= sin \theta on TI-89?
The r is squared. I tried to graph r = sqrt(sin \theta ) and r = -sqrt(sin \theta ) separately on [0, \pi ], however I am not sure this is the correct answer?
thx
The r is squared. I tried to graph r = sqrt(sin \theta ) and r = -sqrt(sin \theta ) separately on [0, \pi ], however I am not sure this is the correct answer?
thx