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Thread: Parameter Elimination Problem - # 3

  1. #1
    MHF Contributor Jason76's Avatar
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    Parameter Elimination Problem - # 3

    $\displaystyle x = t^{2} - 2t$

    $\displaystyle y = t $

    Solving for t

    $\displaystyle x = t^{2} - 2t $

    $\displaystyle 2t = t^{2} $

    $\displaystyle 2 = \dfrac{t^{2}}{2t} $

    $\displaystyle 2 = t $

    $\displaystyle y = \sqrt{2} $ - This must be converted into an "x =" equation

    Ultimately this last equation needs to be converted to an "x =" thing, because it's part of a bigger area problem in terms of the y axis not x axis. However, something went wrong on this thing.
    Last edited by Jason76; Jul 26th 2014 at 08:20 PM.
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    Re: Parameter Elimination Problem - # 3

    Quote Originally Posted by Jason76 View Post
    $\displaystyle x = t^{2} - 2t$

    $\displaystyle y = t $
    $x=t^2-2t=(t-1)^2-1=(y-1)^2-1$

    that do it?
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    MHF Contributor Jason76's Avatar
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    Re: Parameter Elimination Problem - # 3

    Quote Originally Posted by romsek View Post
    $x=t^2-2t=(t-1)^2-1=(y-1)^2-1$

    that do it?
    How did you get that? Was that completing the square?
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    Re: Parameter Elimination Problem - # 3

    Quote Originally Posted by Jason76 View Post
    How did you get that? Was that completing the square?
    yes
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    Re: Parameter Elimination Problem - # 3

    Quote Originally Posted by Jason76 View Post
    $\displaystyle x = t^{2} - 2t$

    $\displaystyle y = t $

    Solving for t

    $\displaystyle x = t^{2} - 2t $

    $\displaystyle 2t = t^{2} $

    $\displaystyle 2 = \dfrac{t^{2}}{2t} $

    $\displaystyle 2 = t $

    $\displaystyle y = \sqrt{2} $ - This must be converted into an "x =" equation

    Ultimately this last equation needs to be converted to an "x =" thing, because it's part of a bigger area problem in terms of the y axis not x axis. However, something went wrong on this thing.
    Surely you can see that you can just replace the "t" with "y"...
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  6. #6
    MHF Contributor Matt Westwood's Avatar
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    Re: Parameter Elimination Problem - # 3

    Quote Originally Posted by Jason76 View Post
    $\displaystyle x = t^{2} - 2t$

    $\displaystyle y = t $

    Solving for t

    $\displaystyle x = t^{2} - 2t $

    $\displaystyle 2t = t^{2} $

    $\displaystyle 2 = \dfrac{t^{2}}{2t} $

    $\displaystyle 2 = t $

    $\displaystyle y = \sqrt{2} $ - This must be converted into an "x =" equation

    Ultimately this last equation needs to be converted to an "x =" thing, because it's part of a bigger area problem in terms of the y axis not x axis. However, something went wrong on this thing.
    What you've done here is: at the lines:

    $\displaystyle x = t^{2} - 2t $

    $\displaystyle 2t = t^{2} $

    you are effectively trying to solve for t where x = 0. So at the end of this piece, you have established that $\displaystyle y = \sqrt 2$ when $\displaystyle x = 0$, or that the curve moves through the point $\displaystyle (0, \sqrt 2)$.
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