Answer :
Let 's' be the length side of a square
A is its area
From pythagorean theorem, we know the diagonal, y, is:
[tex]\begin{gathered} y^2=s^2+s^2 \\ y^2=2s^2 \end{gathered}[/tex]Also, we know the area of a square is:
[tex]A=s^2[/tex]We can substitute this equation in previous one to get:
[tex]\begin{gathered} y^2=2s^2 \\ y^2=2A \end{gathered}[/tex]We can implicitly differentiate with respect to time (t) to get:
[tex]\begin{gathered} y^2=2A \\ 2y(\frac{dy}{dt})=2(\frac{dA}{dt}) \end{gathered}[/tex]dA/dt is rate of change of Area with time, which is 32.
dy/dt is rate of change of diagonal [what we want to find]
At the instant diagonals are 12, y = 12
Thus, we have:
[tex]\begin{gathered} 2y(\frac{dy}{dt})=2(\frac{dA}{dt}) \\ 2(12)(\frac{dy}{dt})=2(32) \\ \frac{dy}{dt}=\frac{2(32)}{2(12)} \\ \frac{dy}{dt}=2.67\text{ m/s} \end{gathered}[/tex]