- #1
gillgill
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am i doing this right?
A cylindrical tank with a radius of 5 ft and a height of 20ft is filled with a certain liquid chemical. A hole is punched in the bottom. At that moment the chemical drains ut of the tank at the rate of 2ft^3/min. At what rate is the height of liquid in the tank changing?
V = 2(pi)rh
dV/dt = [2(pi)r]dh/dt
2 = [2(pi)5]dh/dt
dh/dt = 1/5(pi)
The height of liquid in the tank is changing at the rate is 1/5(pi) /min
A cylindrical tank with a radius of 5 ft and a height of 20ft is filled with a certain liquid chemical. A hole is punched in the bottom. At that moment the chemical drains ut of the tank at the rate of 2ft^3/min. At what rate is the height of liquid in the tank changing?
V = 2(pi)rh
dV/dt = [2(pi)r]dh/dt
2 = [2(pi)5]dh/dt
dh/dt = 1/5(pi)
The height of liquid in the tank is changing at the rate is 1/5(pi) /min