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Elimination of parameter

  1. Jan 19, 2007 #1
    1. The problem statement, all variables and given/known data

    Eliminate the parameter given
    x=2e^t, y=2e^-t

    3. The attempt at a solution

    lnx = ln2 + t, so t = lnx - ln 2

    This gives:

    y=2(e^ln2 * e^-lnx)
    y= -4x

    This does not, however, match the graph of the parametric function on my calculator. Have I made a mistake?
  2. jcsd
  3. Jan 19, 2007 #2
    [tex]e^{-\ln{x}} \neq -x[/tex]
    Last edited: Jan 19, 2007
  4. Jan 19, 2007 #3
    No, it's 1/x, so y=4/x must be the correct solution (for x bigger than 0)

    The graph doesn't seem to start at x=0 though...
    Last edited: Jan 19, 2007
  5. Jan 19, 2007 #4


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    You didn't need to go through the "ln" business. If x= 2et then xe-t= 2 so e-t= 2/x and y= 2e-t= 2(2/x)= 4/x.

    Obviously, the can't "start at x= 0"- why does that bother you? x= 2et and et is never 0.
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