Jesse has constructed a huge cylindrical can with a diameter of 60 ft.

Question

Jesse has constructed a huge cylindrical can with a diameter of 60 ft. The can is being filled with water at a rate of 450 ft3 /min. How fast is the depth of the water increasing? (Hint: The volume of water in the cylinder is determined by πr2 h where r is the radius and h is the depth of the water

Answer 1

Hi Doyin,

Given that the flow in is 450 ft³/min and ithe rate of change of the can volume at a constant diameter is

dV/dt=d/dt(πr²h)=πr²dh/dt you can solve this for dh/dt which how fast the water level is rising

dV/dt=450 ft³/min

r=60/2 ft =30ft

dh/dt=450/(3.14*900)=.16 ft/mn or about 2"/min

Jim