Question

when the area of an expanding square is increasing three times as fast as its side is increasing the side is

Answer 1

Let A be the area and s be the side of the square.
From your given condition we have
\[\frac{dA}{dt}=3\frac{ds}{dt}\]
But we know that A=s^2 so
\[\frac{dA}{dt}=2s \frac{ds}{dt}\]
Chugging this back to the first equation, we have:
\[2s\frac{ds}{dt}=3\frac{ds}{dt}\\
2s=3\\
s=\frac{3}{2}\]