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Math Help - Finding the Cartesian equation for a set of parametric equations

  1. #1
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    Finding the Cartesian equation for a set of parametric equations

    I'm trying to solve this problem, but I am having trouble. Can someone give me some pointers?

    A curve is defined by the parametric equations:

    x = 2t / (1 + t^2)
    y = (1 - t^2) / (1 + t^2)

    By considering the expression x^2 + y^2, eliminate the parameter t to obtain the Cartesian (x-y) equation of the curve.

    Thanks
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  2. #2
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    Re: Finding the Cartesian equation for a set of parametric equations

    so ... what did you determine for the expression x^2+y^2 in terms of t ?
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    Re: Finding the Cartesian equation for a set of parametric equations

    This is the equation I formed: x^2 + y^2 = (4t^2 - t^4 + 1) / (1 + t^4). I don't know how to eliminate t from this point.
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    Re: Finding the Cartesian equation for a set of parametric equations

    first off, note that ...

    (1+t^2)^2 \ne 1 + t^4



    x^2 = \frac{4t^2}{(1 + t^2)^2}

    y^2 = \frac{1 - 2t^2 + t^4}{(1 + t^2)^2}

    try adding x^2+y^2 again ...
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  5. #5
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    Re: Finding the Cartesian equation for a set of parametric equations

    Oops, stupid error! Ok, got it now. The ratio cancels out and I get x2 + y2 = 1. Many thanks.
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    Re: Finding the Cartesian equation for a set of parametric equations

    Hello, everyone!

    This problem (and solution) involves a paradox.


    Paramtric equations: . \begin{Bmatrix}x &=& \dfrac{2t}{1+t^2} & [1] \\ \\[-4mm] y &=& \dfrac{1-t^2}{1+t^2} & [2] \end{Bmatrix}

    Cartesian equation: . x^2 + y^2 \:=\:1\;\;[3]


    From [3], we have a circle with center at the Origin and radius 1.
    . . It has x-intercepts (\pm1,\,0) and y-intercepts (0,\,\pm1)


    Let's determine the intercepts using the parametric equations.


    For x-intercepts, let y = 0\!: \;\;\frac{1-t^1}{1+t^2} \:=\:0 \quad\Rightarrow\quad 1-t^2 \:=\:0 \quad\Rightarrow\quad t^2 \:=\:1 \quad\Rightarrow\quad t \:=\:\pm1

    Substitute into [1]: . x \:=\:\frac{2(\pm1)}{1+(\pm1)^2} \;=\;\frac{\pm2}{2} \;=\;\pm1

    The x-intercepts are: . (1,\,0)\text{ and }(-1,\,0)


    For y-intercepts, let x = 0\!:\;\;\frac{2t}{1+t^2} \:=\:0\quad\Rightarrow\quad 2t \:=\:0 \quad\Rightarrow \quad x \:=\:0

    Substitute into [2]: . y \:=\:\frac{1-0^2}{1+0^2} \quad\Rightarrow\quad y \:=\:1

    The y-intercept is (0,1) . . . There is only one!


    Where is the other y-intercept?
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