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Math Help - Eliminate the Parameter Problem

  1. #1
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    Eliminate the Parameter Problem

    I have to eliminate the parameter of:

    x=e^(t) + e^(-t)
    y=6-2t

    It has to be Function Y in terms of X.

    Everything I do has everything cancel. I can't figure out how to do this. I even tried figuring out what T was in the Y equation, substituting that into the X equation and then getting it in terms of Y. I can't figure it out.
    Last edited by thegame189; March 18th 2008 at 09:14 AM.
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  2. #2
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    Hyperbolic

    thegame189,

    Are you familiar with hyperbolic functions?

    --Kevin C.
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  3. #3
    Eater of Worlds
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    One thing you could do is note that e^{t}+e^{-t}=2cosh(t)

    Therefore, x=2cosh(t), \;\ t=cosh^{-1}(\frac{x}{2})

    Sub into y and get y=6-2cosh^{-1}(\frac{x}{2})
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  4. #4
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    Can someone check the answers I got for these?

    Find the EXACT length of C. I got:

    L = e^(3) - e^(-3)

    Find the EXACT area of the surface obtained by rotating C about the x-axis.
    I got:

    A = 2*Pi*[4e^3 - 1]


    I used the formula L = sqrt of [(dx^2) + (dy^2)], which then factored factored into [e^(t) + e^(-t)]... I then put the 3 and 0 in to get the Length. Any mistakes?

    For Area, I did the same thing but then multiplied it by 2Pi*Y, which is 2Pi*(6-2t)... I then FOILed and had to integrate: 6e^(t) -te^(t) + 6e^(-t) -te^(-t). I got: 6e^(t) - te^(t) + e^(t) - 6e^(-t) + te^(-t) + te^(-t) and evaluated it from 0 to 3. I got: [6e^3 - 3e^3 + e^3 - 6e^(-3) + 6e^(-3). My final was: [2Pi(4e^3 - 1)]

    See any mistakes?
    Last edited by thegame189; March 18th 2008 at 11:38 AM.
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