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How would you graph this

  1. Sep 22, 2009 #1
    In polar equations, how can I graph r^2= sin [tex]\theta[/tex] on TI-89?

    The r is squared. I tried to graph r = sqrt(sin [tex]\theta[/tex] ) and r = -sqrt(sin [tex]\theta[/tex] ) separately on [0, [tex]\pi[/tex] ], however I am not sure this is the correct answer?

  2. jcsd
  3. Sep 22, 2009 #2


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    Hi Bachelier! :smile:

    (have a theta: θ and a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

    (I don't know anything about the TI-89, but … )

    If this r and θ are the usual polar coordinates, then r can't be negative, and so you don't need r = -√(sinθ). :wink:
  4. Sep 22, 2009 #3
    the formula in polar coordinates is that of a Lemniscate:

    r squared equals sin (theta)

    I am used to graphing R, not R squared.
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