# Parallelogram...

• Aug 5th 2009, 10:46 PM
PTL
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...
• Aug 5th 2009, 10:59 PM
red_dog
Let $\displaystyle O(0,0), \ A(1,2t), \ B(2t,1), C(1+2t, 1+2t)$

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

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

Therefore $\displaystyle f(t)=|1-4t^2|$
• Aug 6th 2009, 03:08 AM
alunw
Another way to do it is using the cross product of v and w. That also works with 3D coordinates.
• Aug 6th 2009, 06:49 AM
Soroban
Hello, PTL!

Yet another approach . . .

Did you make a sketch?

Quote:

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

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

Find a formula for $\displaystyle f(t).$

Code:

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

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

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

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

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

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

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

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

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