at a time of t minutes,
let the radius of the water level be r cm
let the height of the water level be h cm
by ratios:
r/h = 1.75/6
1.75h = 6r
r = 1.75h/6 = 7h/24
V = (1/3)π r^2 h
= (1/3)π(49h^2/576)h
= (49π/1728) h^3
dV/dt = (49π/576) h^2 dh/dt
without leakage ... for the given data
dV/dt = (49π/576)(1.5^2)(.27)
= .1623565... m^3/min
now we must compensate for the leakage of
6700 cm^3/min
= .0067 m^3/min (behold the beauty of the metric system)
so the actual rate pumped in is
.1623565.. + .0067 = .16906 m^3/min
or 169056.5 cm^3/min
check my arithmetic
Water is leaking out of an inverted conical tank at a rate of 6700 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 meters and the diameter at the top is 3.5 meters. If the water level is rising at a rate of 27 centimeters per minute when the height of the water is 1.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
1 answer