Water is leaking out of an inverted conical tank at a rate of 1100.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 15.0 meters and the diameter at the top is 5.0 meters. If the water level is rising at a rate of 300 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
3 answers
Can you double check if the water going out is 1100 litres per minute (1.1 m³/min.) or 1100 cc (0.0011 m³/min)?
It's cubic centimeter per minute
Assume "inverted" conical tank is wider at the top, with a diameter of 5m.
At the base, r=0, h=0.
At the top, r=2.5, h=15
radius at height h:
r(h)=h/6
Area of surface of water, Aw
=πr(h)^2
=πh²/36
Equate input (Q m³/min.) and output (x):
dh/dt = (Q-x)/Aw
At h=2.5m,
dh/dt=300cm/min = 3m/min.
x=0.0011 m³/min.
Therefore
Q=(dh/dt)*Aw+x
=3 m/min. * πh²/36
=0.521π+0.0011 m³/min
=1.637 m³/min.
Please check my calculations.
At the base, r=0, h=0.
At the top, r=2.5, h=15
radius at height h:
r(h)=h/6
Area of surface of water, Aw
=πr(h)^2
=πh²/36
Equate input (Q m³/min.) and output (x):
dh/dt = (Q-x)/Aw
At h=2.5m,
dh/dt=300cm/min = 3m/min.
x=0.0011 m³/min.
Therefore
Q=(dh/dt)*Aw+x
=3 m/min. * πh²/36
=0.521π+0.0011 m³/min
=1.637 m³/min.
Please check my calculations.