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Math Help - Parallelogram...

  1. #1
    PTL
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    Parallelogram...

    Let f(t) be the area of the parallelogram with vertices v=(1,2t), w = (2t,1), (0,0), and v+w = (2t+1,2t+1). Find a formula for f(t).


    I'm assuming one can't simply 'stand up' the parallelogram so that it's a rectangle...
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let O(0,0), \ A(1,2t), \ B(2t,1), C(1+2t, 1+2t)

    Area[OACB]=2\cdot Area[OAB]=2\cdot\frac{1}{2}|\Delta| where

    \Delta=\begin{vmatrix}0 & 0 & 1\\1 & 2t & 1\\2t & 1 & 1\end{vmatrix}=1-4t^2

    Therefore f(t)=|1-4t^2|
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  3. #3
    Member alunw's Avatar
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    Another way to do it is using the cross product of v and w. That also works with 3D coordinates.
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  4. #4
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    Hello, PTL!

    Yet another approach . . .

    Did you make a sketch?


    Let f(t) be the area of the parallelogram with vertices:

    \vec v\,=\,(1,2t),\;\vec w\,=\,(2t,1),\;(0,0), and \overrightarrow{v+w} \:=\:(2t+1,2t+1).

    Find a formula for f(t).
    Code:
            |
       2t+1 + - - - - - - - - - - o v+w
            |                 *  *:
            |             *     * :
            |     v   *        *  :
         2t + - - o           *   :
            |    *           *    :
           1+   *           o w   :
            |  *        *   :     :
            | *     *       :     :
            |*  *           :     :
        - - o - - + - - - - + - - * - -
          (0,0)   1        2t    2t+1
    We have a rhombus.

    If d_1,\:d_2 are diagonals of a rhombus,

    . . the area of the rhombus is: . A \:=\:\tfrac{1}{2}d_1d_2


    The long diagonal d_1 goes from (0,0) to (2t+1,2t+1)

    . . d_1 \:=\:\sqrt{(2t+1)^2 + (2t+1)^2} \;=\;\sqrt{2}\,|2t+1|


    The short diagonal d_2 goes from (1,2t) to (2t-1)

    . . d_2 \;=\;\sqrt{(2t-1)^2 + (1-2t)^2} \;=\;\sqrt{2}\,|2t-1|


    Hence: . A \;=\;\tfrac{1}{2}\bigg[\sqrt{2}\,|2t+1|\bigg]\,\bigg[\sqrt{2}\,|2t-1|\bigg] \;=\;|4t^2-1|


    Therefore: . A \;=\;\begin{Bmatrix}4t^2-1 & &\text{for } |t| > \frac{1}{2} \\ \\[-4mm] 0 && \text{for }t \:=\:\frac{1}{2} \\ \\[-4mm] 1-4t^2 && \text{for }|t| < \frac{1}{2} \end{Bmatrix}

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