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?

Question

Ezra Adams · Accepted Answer

# -40 m^2 "/ min" # A square of diagonal #l# has area #A = l/(sqrt 2) * l/(sqrt 2) = l^2/2# Thus #dot A = l dot l# here #dot A = 5*(-8) = -40 m^2 "/ min" #

Isaiah Allen · Answer

undefined 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.