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Math Help - Parametrization of the curve

  1. #1
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    Parametrization of the curve

    Parametrization of the curve-parametrization.jpg
    please read the question above.
    I don't know what I'm doing.
    x^2+y+1-xy=2
    x^2-1+y-xy=0
    (x-1)(x+1)-y(x-1)=0
    (x-1)(x+1-y)=0
    .
    .
    oh...
    can't find any clue to mfind the parametrimization of the curves.
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  2. #2
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    Re: Parametrization of the curve

    Hey happymatthematics.

    Does the curve meet both constraints? (If so, hint: get rid of the z term first).
    Thanks from happymatthematics
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  3. #3
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    Re: Parametrization of the curve

    is
    {z=t
    {y=s
    {x=(1-t)/s
    correct?
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  4. #4
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    Re: Parametrization of the curve

    Can you outline how you got the answer (the method is always more important than the final answer in isolation)?
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  5. #5
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    Re: Parametrization of the curve

    Hello chiro.
    I've tried in two ways.
    Please read the picture below.
    Parametrization of the curve-parametrization_with_2attempts.jpg
    Does any one of my two attempts correct?
    I don't know the reason why we need to do parametrization.
    sometimes, when I encounter some simple system of equations, I can just let x=t, y=s and then write z in terms of t and s.
    but most of the time, it is not the case.
    how do I know or how can I check if my parametrization is correct?
    Thank you.
    by the way, since I'm afraid if I wrote a long question then no one would answer me, I raise a question as short as possible.
    It's happy to heard from you that outlining my scratch work!
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  6. #6
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    Re: Parametrization of the curve

    Quote Originally Posted by happymatthematics View Post
    Click image for larger version. 

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    please read the question above.
    I don't know what I'm doing.
    x^2+y+1-xy=2
    x^2-1+y-xy=0
    (x-1)(x+1)-y(x-1)=0
    (x-1)(x+1-y)=0
    x- 1= 0 only when x= 1 so as long as x is not 1, you have x+1- y= 0 or y= x+1.
    From the original equations, you already have z= 1- xy= 1- x(x+ 1)= 1- x- x^2.
    Now take x itself as parameter: x= t, y= t+ 1, z= 1- t- t^2.

    You need to remember that there are many different "parameterizations" for the same curve so at some point you need to make a "choice" as I did in setting x= t.

    oh...
    can't find any clue to mfind the parametrimization of the curves.
    Thanks from happymatthematics
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