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Parametric graph

  1. May 28, 2010 #1
    x=e^(2t)

    y=t+1

    t= ( lnx ) / 2

    y= ( lnx ) / 2 + 1

    or

    blah
     
    Last edited: May 29, 2010
  2. jcsd
  3. May 28, 2010 #2

    Mark44

    Staff: Mentor

    Why did you add +- ?
     
  4. May 28, 2010 #3
    what i meant was for t<0 how to represent the graph in terms of y and x
     
  5. May 29, 2010 #4
    x=e^(-2t)
    y=t+1

    -2t=lnx
    y=1-lnx/2
    i suppose
     
  6. May 29, 2010 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Is this the same question? How did "[itex]x= e^{2t}[/itex]" become [itex]x= e^{-2t}[/itex]?

    Assuming you really did mean [itex]x= e^{-2t}[/itex], yes, that is correct.
     
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