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Thread: Calculus III Word Problem

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    Calculus III Word Problem

    I have absolutely no clue. Can someone at least get me started?

    Thanks a Million!!!

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  2. #2
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    Vectors

    Hello qbkr21
    Quote Originally Posted by qbkr21 View Post
    I have absolutely no clue. Can someone at least get me started?

    Thanks a Million!!!

    For the geometric argument, consider an arbitrary point in the plane, O, as origin. Then take two lines, $\displaystyle l$ and $\displaystyle m$, through O parallel to $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ respectively (effectively setting up a system of non-perpendicular* coordinate axes). Then consider point C whose displacement from O is represented by $\displaystyle \vec{c}$. From C, draw a line parallel to $\displaystyle l$, meeting $\displaystyle m$ at P. Then $\displaystyle \vec{PC} = s\vec{a}$ for some scalar $\displaystyle s$, and $\displaystyle \vec{OP} = t\vec{b}$ for some scalar $\displaystyle t$, and then:

    $\displaystyle \vec{OC} = \vec{OP} + \vec{PC}$

    $\displaystyle \Rightarrow \vec{c} = s\vec{a} + t\vec{b}$

    For the components argument, let $\displaystyle \vec{a} = a_1\vec{i} + a_2\vec{j}$, and $\displaystyle \vec{b}$ and $\displaystyle \vec{c}$ similarly. Then consider separate components and show that equations

    $\displaystyle c_1 = sa_1 + tb_1$

    $\displaystyle c_2 = sa_2 + tb_2$

    have a solution for some $\displaystyle s$ and $\displaystyle t$.

    Grandad

    * PS. Not necessarily perpendicular is what I mean. Instead of being covered by a grid of unit squares, as in Cartesian geometry, the plane is spanned by parallelograms with sides $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$.
    Last edited by Grandad; Jan 21st 2009 at 11:59 PM. Reason: PS added
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