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Math Help - Curve in parameter form

  1. #1
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    Curve in parameter form

    A curve is given in this form

    x=t^3-3t, y=t^2 and  -2\leq t \leq 2

    Eliminate the t-variable so you get a funktion in the form x=f(y) and draw the curve as accurately as you can!

    Good luck!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lynch-mob View Post
    A curve is given in this form

    x=t^3-3t, y=t^2 and  -2\leq t \leq 2

    Eliminate the t-variable so you get a funktion in the form x=f(y) and draw the curve as accurately as you can!

    Good luck!
    Well, y = t^2 so t = \pm \sqrt{y}

    For the + branch:
    x = y^{3/2} - 3y^{1/2}

    and for the - branch:
    x = -(y^{3/2} - 3y^{1/2})

    Sketch both of these and you'll have your graph.

    -Dan
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by lynch-mob View Post
    A curve is given in this form

    x=t^3-3t, y=t^2 and  -2\leq t \leq 2

    Eliminate the t-variable so you get a funktion in the form x=f(y) and draw the curve as accurately as you can!

    Good luck!
    Just to be more general, given

    y=f(\xi) and x=g(\xi) where \xi is some parameter,

    Then converting to rectangular coordinates we need to get y as a function of x so we see that

    x=g(\xi)\Rightarrow{\xi=g^{-1}(x)}

    So now making a substituion

    y=f(\xi)\Rightarrow{y=f(g^{-1}(x))}
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  4. #4
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    Interval values

    How do I convert the interval values. I mean should I change it to an interval between to x-values or between two y-values. And how do I do it?
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  5. #5
    MHF Contributor kalagota's Avatar
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    yes, you have to.. just check what will the values of x or y when you substitute the values a \leq t \leq b to x(t) and/or y(t)

    in your case, -2 \leq x \leq 2
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  6. #6
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    little problem

    Thanks. A little problem. Ive made a table with t,x and y. I have 9 different values for t and get nine different values for x and so on. But the problem is y values. My calculator tells me the graph will look like an S. But the table values for y tells me something else. When t=-2 then x=-2 and y=+4 but it should be y=-4. This goes on and on. I suppose it has something to do with t=\pm\sqrt{y}

    How do I figure this out?
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  7. #7
    MHF Contributor kalagota's Avatar
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    look at what topsquark posted..
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  8. #8
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    is this it?

    Is this the correct graph? And by the way. Is there a way to draw a graph on a calculator when the function is given as x=f(y)?

    Thanks man!
    Attached Thumbnails Attached Thumbnails Curve in parameter form-graph.jpg  
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lynch-mob View Post
    Is this the correct graph? And by the way. Is there a way to draw a graph on a calculator when the function is given as x=f(y)?

    Thanks man!
    Edit: I just graphed the original parametric function (and the pair of implicit functions that I solved for) and got a different graph than you did. See if you can find out what went wrong with yours.

    As to graphing on a calculator, no most calculators will not graph an implicit function. You might find this graphing program to be useful, however. It will graph implicit functions, as well as a host of other graphing methods.

    -Dan
    Attached Thumbnails Attached Thumbnails Curve in parameter form-parametric.jpg  
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  10. #10
    MHF Contributor kalagota's Avatar
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    i don't have a graphing software here, but i think it is correct..

    yes there is.. note that x = f(y) \Longleftrightarrow f^{-1}(x) = y
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