A hypothetical square shrinks so that the length of its diagonals are changing at a rate of −8 m/min. At what rate is the area of the square changing when the diagonals are 5 m each?

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Answer 1 · 2016-08-05T18:55:47

$- 40 m^{2} / min$ $-40 m^2 "/ min"$

A square of diagonal $l$ $l$ has area $A = \frac{l}{\sqrt{2}} \cdot \frac{l}{\sqrt{2}} = \frac{l^{2}}{2}$ $A = l/(sqrt 2) * l/(sqrt 2) = l^2/2$

Thus $. A = l . l$ $dot A = l dot l$

here

$. A = 5 \cdot (- 8) = - 40 m^{2} / min$ $dot A = 5*(-8) = -40 m^2 "/ min"$