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Math Help - paramatised curve and self-intersection points.

  1. #1
    Newbie paronga's Avatar
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    paramatised curve and self-intersection points. Need Extra help!

    So i have the curve:

    \gamma(t) =(t^2-t+1, t^2-t)

    And i need to show that it has one self-intersection point.

    To me, it is obviously at 0 and 1, as at these points \gamma is equal to zero.

    Now, the thing is, i think i must have missed this lecture. I can't find any reference in my differential geometry book, and my google fu is failing me.

    Can anyone show me the method on how to show self-intersection points?
    I'll solve the question once i know how.

    Thus far, i've been unable to figure out my own way. These damn parametric equations upset me so.

    so far i've though maybe if
    \gamma(t) = \gamma(u)
    where
    (t^2-t+1, t^3-t) = (u^2-u+1, u^3-u)
    but yeh.

    Halp?
    Last edited by paronga; August 22nd 2010 at 09:22 PM.
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  2. #2
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    Hello, paronga!

    Your idea is a good one!


    So i have the curve: . \gamma(t) \:=\:(t^2-t+1, t^2-t)

    And i need to show that it has one self-intersection point.


    So far, i've thought maybe if : . \gamma(t) = \gamma(u)

    where: . (t^2-t+1, t^3-t) \:=\: (u^2-u+1, u^3-u) . Yes!

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


    From [1], we have: . t^2-u^2 - (t - u) \:=\:0

    . . (t-u)(t+u) - (t - u) \;=\;0 \quad\Rightarrow\quad (t-u)(t+u-1) \:=\:0


    We have: . t-u \:=\:0 \quad\Rightarrow\quad \rlap{//////}t \:=\:u
    . . . . . t + u - 1 \:=\:0 \quad\Rightarrow\quad u \:=\:1 - t\;\;[3]


    From [2], we have: . t^3 - u^3 -t+u\:=\:0

    . . (t-u)(t^2+tu+u^2) - (t-u)\:=\:0

    . . (t-u)(t^2+tu+u^2-1) \:=\:0

    . . t^2 + tu + u^2-1 \:=\:0\;\;[4]


    Substitute [3] into [4]: . t^2 + t(1-t) + (1-t)^2-1 \:=\:0

    . . which simplifies to: . t^2-t\:=\:0

    . . . . . . . . . . . . . . . t(t-1)\:=\:0 \quad\Rightarrow\quad t \:=\:0,\,1


    It turns out that: . \gamma(0) \:=\:\gamma(1) \:=\:(1,0)

    Therefore, there is one self-intersection point.
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  3. #3
    Newbie paronga's Avatar
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    huuzar!!
    I see where i was going wrong.

    I thought it was something like that but my simplification was wrong.
    I thank you very much.

    I did have another question. Maybe i should just post it in this thread instead of making a new one?
    It's on the same problem set for the same subject. My weekend if just me doing problems

    It's to do with showing a vector function in planar. Once again, i can do it say visually.

    Anyway, i've got the unit tangent vector to the curve:

    \gamma(t) = \frac{1+t^2}{t} , t+1 , \frac{1-t}{t}

    with

    \gamma\prime (t) = \frac{t^2-1}{t^2} , 1 , \frac{1}{t^2}

    Which means the unit tangent vector is

    T(t) = \frac{\gamma\prime(t)}{|\gamma\prime(t)|}

    where (for simplification)

    |\gamma\prime(t)| = \frac{\sqrt{2t^4-2t^2+2}}{t^2}

    finally

    T(t) = \frac{t^2-1}{|\gamma\prime(t)|}, \frac{t^2}{|\gamma\prime(t)|} , \frac{1}{|\gamma\prime(t)|}

    So now i have this vector, which i believe to be normal to the whole original curve. how do i show that this is in a plane? or is also normal to a plane. Hmm

    The original question says to show that \gamma(t) is planar.
    Perhaps i'm on the wrong path?
    Any halp?
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