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Finding a parametric equation for y = f(x)

  1. Feb 15, 2004 #1
    I've searched the web for information on Parametric Equations, and most of them only give me information on how to find y = f(x) when given y = y(t) and x = x(t).

    Is there any sort of method for doing the reverse? I'm told that there are theoretically an infinite number of parametric equations for a given curve, but how do you go about finding any of them?

    One of the questions in our textbook says:

    Parametrize the curve y = f(x), x between [a,b].

    No further information was given. I think I need to know this in order to do the questions that come afterwards. For example, [tex]y^2=x^3[/tex], from (4,8) to (1,1). The answer ends up being y = (2-t)^3 and x = (2-t)^2. The solution manual gives me no explanations as to how they arrived at that answer.

    Help?
     
  2. jcsd
  3. Feb 15, 2004 #2
    there are an infinite number of parametrizations. one easy choice is the following:

    given a curve f(x)=y

    let y=f(t) and x=t.

    this is a sort of trivial parametrization. there are lots more. what exactly are you looking for?
     
  4. Feb 15, 2004 #3
    I'm basically looking for the thought process behind choosing a certain type of parametrization based on a restricted domain for t. The example I gave in my question:

    [tex]y^2=x^3[/tex]

    The question restricted the domain of t to be between 0 and 1. And one of the answers ended up being [tex]y(t) = (2-t)^3[/tex] and [tex]x(t) = (2-t)^2[/tex].

    How did they come up with that? Is it just sort of guessing and checking?
     
  5. Feb 16, 2004 #4

    NateTG

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    Well I would be incliend to say:
    [tex]y^2=x^3=t[/tex]
    So
    [tex]y=t^{\frac{1}{2}}[/tex]
    [tex]x=t^{\frac{1}{3}}[/tex]
    but that can be a bit ugly, so the people replaced by [tex]t'=t^6[/tex] so that the exponents are integers.

    With the new [tex]t[/tex] you get
    [tex]y=t^3[/tex]
    [tex]x=t^2[/tex]

    The [tex]2-[/tex] is added for the final values, but that could also be done by chaning the range for [tex]t[/tex].
     
  6. Feb 16, 2004 #5

    HallsofIvy

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    NateTG is, as usual, exactly right.

    There exist many "parametrizations" for any given curve. Given that y2= x3, the simplest to find is to set
    y2= x3= t and solve for x and y separately.

    If you don't like fractions (and who does?) you might think instead
    ( )2= ( )3 and not that those will be "obviously" equal if you put t3 in the first parenthesis and t2 in the second. That is: x= t2 and y= t3 just as NateTG gave.

    That's a perfectly good parametrization- in particular, when t= 1, it gives the point (1, 1) and when t= 2, it gives the point (4,8).

    Once you have that, you could replace the "t" by any formula (in particular "2- t") and still have a parametrization for the same curve. With x= (2-t)2 and y= (2-t)3, when t= 1, we get (1,1) and when t= 0, we get (4,8), just the reverse of what we had before.
     
  7. Feb 16, 2004 #6
    Oooh, thank you both very much NateTG and HallsofIvy! That made perfect sense! :-)
     
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