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Math Help - Intersection of parametric curve in 2d plane, and winplot q's

  1. #1
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    Intersection of parametric curve in 2d plane, and winplot q's

    The attachment is a set of equations that define a parametric curve in the xy plane. My instructions are to find the lowest point on the curve. I can see how my solution my solution is correct, but when I have winplot give me the derivative curve, which is an inverse parabola with vertex (-3,1). It's intercepts are nowhere in the -1/2 neighborhood on either axis. So what am i seeing in the derivative curve? What does its vertex represent?

    So while dinking about with this thing, i became curious how i would find the point of intersection on the curve (just the original curve) . I know how to find the intersection of two parametric curves, but not sure I've learned the
    process of finding it with one. Can it be done?

    Thanx!
    Attached Thumbnails Attached Thumbnails Intersection of parametric curve in 2d plane, and winplot q's-capture-parametric-10-27.jpg  
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  2. #2
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    Hello, boardguy67!

    I can help with one of your questions . . .


    The set of equations defines a parametric curve in the xy-plane.

    . . \begin{array}{ccc}x &=& t^3 - 3t \\ y &=& t^2 + t + 1 \end{array}

    My instructions are to find the lowest point on the curve.

    Your point is correct, but labeled wrong on your graph.
    It is at the very bottom of the loop, where there is a horizontal tangent.



    So while dinking about with this thing,
    i became curious how i would find the point of intersection on the curve.
    Let s be a value of the parameter.
    . . We have a point: . P(s^3-3s,\;s^2+s+1)

    Let t be another value of the parameter (s \neq t).
    . . We have another point: . Q(t^3-3t,\;t^2+t+1)

    When will P = Q ?

    We have: . \begin{array}{ccccc}s^3-3s &=& t^3-3t & [1] \\ s^2+s+1 &=& t^2+t+1 & [2] \end{array}


    From [1]: . s^3-t^3 \:=\: 3s - 3t \quad\Rightarrow\quad (s-t)(s^2+st + t^2) \:=\: 3(s-t)

    . . Since s\neq t, we have: . s^2 + st + t^2 \:=\:3 .[1]


    From [2]: . s^2-t^2 + s - t \:=\:0 \quad\Rightarrow\quad (s-t)(s+t) + (s-t) \:=\:0 \quad\Rightarrow\quad (s-t)(s+t+1) \:=\:0

    . . Since s \neq t, we have: . s + t + 1 \:=\:0 \quad\Rightarrow\quad t \:=\:\text{-}s-1 .[2]


    Substitute into [1]: . s^2 + s(\text{-}s-1) + (\text{-}s-1)^2 \:=\:3 \quad\Rightarrow\quad s^2 - s^2 - s + s^2 + 2s + 1 \:=\:3

    . . . s^2 + s - 2 \:=\:0 \quad\Rightarrow\quad (s-1)(s+2) \:=\:0 \quad\Rightarrow\quad s \:=\:1,\:-2

    Substitute into [2]: . t \:=\:-2,\:1


    If s = 1\!:\;\begin{Bmatrix}x &=& 1^3 - 3(1) &=& \text{-}2 \\ y &=& 1^2 + 1 + 1 &=& 3 \end{Bmatrix}

    The curve intersects itself at (\text{-}2,\:3).

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  3. #3
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    That is awesome. Great detail Tank you very much. From the sounds of it, it's not terribly likely i can return the favor in mathematical question, but if you ever need a skateboard, i'll give you a smokin discount.



    Be well.
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