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Elimintate the parameter problem

  1. Jul 15, 2009 #1
    elimintate the parameter
    1.
    x=lnt
    y=t^(1/2)
    e^x=t
    y=e^(x/2)

    2.
    x=2t^3
    y=1+4t-t^2
    for slope = 1

    dC/dt=(4-2t)/6t^2=(2-t)/3t^2=1
    3t^2+t-2=0
    t=-1
    t=2/3
    A(-2,-4)
    B(16/27,29/9)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 16, 2009 #2

    Office_Shredder

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    Re: parameters

    1 looks good. I'm not sure what's going on with 2... what do you mean by for slope=1? And then the whole second half?
     
  4. Jul 16, 2009 #3

    HallsofIvy

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    Re: parameters

    1) is correct. For 2) you have A, B, and C without defining them. I think what you are saying is that you want to find points where the slope of the graph given by the parametric equations is 1.
    You "dC/dt" is simply dy/dx= (dy/dt)/(dx/dt)=(2-t)/3t^2=1 and then gives 2- t= 3t^2 so 3t^2+ t- 2= 0. Yes, t= -1 and t= 2/3. Then x(-1)= -2, y(-1)= -4 and x(2/3)= 16/27, y(2/3)= 29/9 so the two points are (-2, -4) and (16/27, 29/9).

    I think your teacher would prefer that your working be not so cryptic!
     
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