A cylindrical water trough has a diameter of 6 feet and a height of 2 feet. It is being filled at the rate of 4 ft^3/min. How fast is the water level rising when the water is 1 foot deep?

Volume of Cylinder V = πr²h

diameter = 6 feet, radius = 6/2 = 3 feet

dV/dt = (dV/dh) * (dh/dt) by change rule.

V = πr²h

dV/dh = πr² = π*3² = 9π ft²

dV/dt = 2 ft ³/min

dV/dt = (dV/dh) * (dh/dt)

2 ft³/min = 9π ft² * (dh/dt)

9π ft² * (dh/dt) = 2 ft ³/min

(dh/dt) = (2 ft³/min) / (9π ft ²)

(dh/dt) = (2 / 9π) ft/min

So the water level is rising at (2/9 π) ft/min.