1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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=2e^(ln2-lnx)
    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Elimination of parameter
  1. Eliminating parameter (Replies: 1)

Loading...