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Math Help - more parametric equations part 2

  1. #1
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    more parametric equations part 2

    Eliminate the parameter, t to obtain the cartesian equation in x and y. Sketch the curve, taking note of any restrictions


    d) x = t + \frac{1}{t}, ~  y = t^2 - \frac{1}{t^2}


    e) x = t + \frac{1}{t}, ~  y= t - \frac{1}{t}


    f) x  = \frac{1 - t^2}{1 + t^2}, ~ y = \frac{2t}{1+t^2}


    I know it's a lot and i really dont expect this all to be answered, i just put all the questions which i had serious trouble with and i would be SO grateful to anyone who could give me a few pointers on how to solve these because i just dont know how to get rid of the t OR find the restrictions. again , i know there are alot of questions and im not just expecting answers or a cheap way to get answers for hw, i realli just dont understand it and would seriously appreciate any help .
    thankyou guyz !
    Last edited by mr fantastic; June 7th 2009 at 03:31 AM. Reason: Moved question form original thread. re-formatted equations using latex
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  2. #2
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    Quote Originally Posted by THSKluv View Post
    Eliminate the parameter, t to obtain the cartesian equation in x and y. Sketch the curve, taking note of any restrictions


    d) x = t + \frac{1}{t}, ~ y = t^2 - \frac{1}{t^2}


    e) x = t + \frac{1}{t}, ~ y= t - \frac{1}{t}


    f) x = \frac{1 - t^2}{1 + t^2}, ~ y = \frac{2t}{1+t^2}


    I know it's a lot and i really dont expect this all to be answered, i just put all the questions which i had serious trouble with and i would be SO grateful to anyone who could give me a few pointers on how to solve these because i just dont know how to get rid of the t OR find the restrictions. again , i know there are alot of questions and im not just expecting answers or a cheap way to get answers for hw, i realli just dont understand it and would seriously appreciate any help .
    thankyou guyz !
    d) Same technique as c) in this thread: http://www.mathhelpforum.com/math-he...-part-1-a.html


    e) x = t + \frac{1}{t} \Rightarrow x^2 = t^2 + 2 + \frac{1}{t^2} \Rightarrow x^2 - 2 = t^2 + \frac{1}{t^2} .... (1)

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

    Equate equations (1) and (2).


    f) Square each equation. Add. Simplify. This question is similar to e) in this thread: http://www.mathhelpforum.com/math-he...ions-help.html
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  3. #3
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    Hello, THSKluv!

    I understand your diffculty.
    These are tricky, requiring special techniques.


    Eliminate the parameter t to obtain the cartesian equation in x and y.
    Sketch the curve, taking note of any restrictions.


    d)\;\;\begin{array}{cccc}x &=& t + \dfrac{1}{t} & [1] \\ \\[-3mm] y &=& t^2 - \dfrac{1}{t^2} & [2] \end{array}
    \begin{array}{cccccc}\text{Square [1]:} & x^2 &=& t^2 + 2 + \dfrac{1}{t^2} \\ \\[-3mm] \text{Add [2]:} & y &=& t^2 \;\;-\;\; \dfrac{1}{t^2} \end{array}

    And we have: . x^2 + y \:=\:2t^2+2 \quad\Rightarrow\quad t^2 \:=\:\frac{x^2+y-2}{2}

    Substitute into [2]\!:\quad y \;=\;\frac{x^2+y-2}{2} - \frac{2}{x^2+y-2}

    Therefore: . \boxed{y \;=\;\frac{(x^2+y-2)^2-4}{2(x^2+y-2)}}


    I have no idea what the graph looks like.

    I note that: . x^2+y-2 \:\neq\:0
    That is, the graph does not exist anywhere on the parabola: y \:=\:2-x^2




    e)\;\;\begin{array}{cccc}x &=& t + \dfrac{1}{t} & [1] \\ \\[-3mm] y&=& t - \dfrac{1}{t} & [2] \end{array}
    Add [1] and [2]: . x + y \:=\:2t \quad\Rightarrow\quad t \:=\:\frac{x+y}{2}

    Substitute into [1]: . x \;=\;\frac{x+y}{2} - \frac{2}{x+y}

    Multiply by 2(x+y)\!:\quad 2x(x+y) \:=\:(x+y)^2-4

    . . which simplifies to: . \boxed{y^2-x^2 \:=\:4}


    We have a "vertical" hyperbola . \begin{array}{c}_{\cup} \\ ^{\cap} \end{array}
    . . I'm not sure of the restrictions.




    f)\;\;\begin{array}{cccc}x  &=& \dfrac{1 - t^2}{1 + t^2} & [1]\\ \\[-3mm] y &=& \dfrac{2t}{1+t^2} & [2]\end{array}
    I just happen to be familiar with this one . . .


    \begin{array}{ccccc}\text{Square [1]:} & x^2 &=& \dfrac{(1-t^2)^2}{(1+t^2)^2} \\ \\[-3mm]\text{Square [2]:} & y^2 &=& \dfrac{4t^2}{(1+t^2)^2} \end{array}

    Add: . x^2+y^2 \;=\;\frac{(1-t^2)^2}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2} \;=\;\frac{1-2t^2+t^4 + 4t^2}{(1+t^2)^2} \;=\;\frac{1+2t^2+t^4}{(1+t^2)^2}

    . . x^2+y^2 \;=\;\frac{(1+t^2)^2}{(1+t^2)^2} \quad\Rightarrow\quad \boxed{x^2 + y^2 \:=\:1}


    We have a unit circle centered at the Origin.
    . . It has a "hole" at (-1,0). .(Why?)

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