First let's find the Volume of the trough, I assume it's an inverted triangle where the sides are equal since the diagram is not shown. All units match so I will work with numbers only.
Volume of a triangle prism is given by:
V = (b * h )/2 * l
Length or l is a constant at 9, so we can sub it in the equation.
V = 9/2 * (b * h)
Substitute out base
Base and height of the filled trough at any time is related by trigonometry. We're interested in height so we will substitute out base with height.
This part would be a lot easier with a diagram, but unfortunately I can't upload one. Using properties of isosceles right triangle and tan:
tan(45 degrees) = h/(b/2)
b = 2h
V = 9/2 *(2h * h) = 9h^2
Take derivative of volume with respect to time to find equation for rate of filling the trough.
dV/dt = 2 * 9 *h dh/dt = 18h dh/dt
re-arranging for dh/dt
dh/dt = dV/dt /(18h)
Subbing in dV/dt = 2
dh/dt = 2/(18 h) = 1/(9h)
Find h at 2 min
V = 9h^2
V = t * rate = 2 * 2 = 4
4 = 9h^2
re-arranging for h:
h = sqrt(4/9) = 2/3
Subbing in h to rate equation
dh/dt = 1/(9h) = 1/(9 * 2/3) = 1/6
The rate of change for height is 1/6 cubic feet/min at 2 min.