Acylindrical 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 ft³/min

dV/dt =  (dV/dh) * (dh/dt)

2 ft³/min = 9π ft²   * (dh/dt)

9π ft²   * (dh/dt) = 2 ft³/min

(dh/dt) = (2 ft³/min)/(9π ft²)   

(dh/dt) = (2/9π)  ft/min

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

B. 2 1/2 feet. multiply 2.5 inches by 12 to get thirty inches then divide by 12 to get 2.5 feet