Solution:

By ratio and proportion on the triangle, at any given height, the base is same as the height (\(b = h\)). Thus at a base of the triangle at \(h = 0.5 ft\) is also \(0.5 ft\)

\(Vol = A_{triangle}(10)\)

\(Vol = \frac{1}{2}bh (10)\)

*As the water increases, the value of b and h are also changing, thus, b and h are not constants.*

\(Vol = 5 h^2\), Since \(b = h\)

\(\frac{dV}{dt} = 10 h \frac{dh}{dt}\)

\(3 = 10 (0.5) \frac{dh}{dt}\)

\(\frac{dh}{dt} = 0.6 \text{ft per min}\)