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Thread: 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 $\displaystyle x^2+y^2$ 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 ...

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



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

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

    try adding $\displaystyle x^2+y^2$ again ...
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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: .$\displaystyle \begin{Bmatrix}x &=& \dfrac{2t}{1+t^2} & [1] \\ \\[-4mm] y &=& \dfrac{1-t^2}{1+t^2} & [2] \end{Bmatrix}$

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


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


    Let's determine the intercepts using the parametric equations.


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

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

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


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

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

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


    Where is the other $\displaystyle y$-intercept?
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