Water is leaking out of an inverted conical tank at a rate of 700.0 cubic centimeters per min at
the same time that water is being pumped into the tank at a constant rate. The tank has height
10.0 meters and the diameter at the top is 3.5 meters. If the water level is rising at a rate of 26.0
centimeters per minute when the height of the water is 4.5 meters, find the rate at which water is
being pumped into the tank in cubic centimeters per minute.
The volume of a conical tank is
V=
π 2
r h
3
Using cm as the unit of length, the container has a radius of 175 cm, and a
€ height of 1000 cm. The radius of the cone at a height h is in proportion:
r=
175
h = 0.175h
1000
To make the volume a simple function of height h, replace r2 to yield
V=
€
π
(0.175) 2 h 3
3
The change in volume with time is dV/dt:
€
dV
dh
= π (0.175h) 2
dt
dt
The net change dV/dt is known to be
€
dV
= pump − 700 cc/min
dt
where pump is the rate at which water is being pumped into the tank in cc/min. Equate the two
€ expressions for dV/dt and evaluate at h=450 cm and dh/dt= 26 cm/min to find:
pump = 700 cc/min + π (0.175⋅ 450 cm) 2 ⋅ 26 cm/min
pump = 507252.36 cc/min
€