# 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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To find the rate at which the area of the square is changing when the diagonals are both 5 m each, we use the chain rule and the fact that the diagonals are related to the side length of the square by (d = \sqrt{2} s). Let (A) be the area of the square, (s) be the side length, and (d) be the length of a diagonal. We have the equation (A = s^2). Differentiating implicitly with respect to time, we get (\frac{{dA}}{{dt}} = 2s\frac{{ds}}{{dt}}). We are given that (\frac{{dd}}{{dt}} = -8) m/min when (d = 5). Solving for (s) using the relationship between (s) and (d), we find (s = \frac{d}{\sqrt{2}} = \frac{5}{\sqrt{2}}). Substituting the values into our equation for (\frac{{dA}}{{dt}}), we get (\frac{{dA}}{{dt}} = 2\left(\frac{5}{\sqrt{2}}\right)(-8) = -40\sqrt{2}) square meters per minute. Therefore, the area of the square is changing at a rate of approximately (−56.57 , \text{m}^2/\text{min}) when the diagonals are 5 m each.

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