Homework SourseA conical tank with an upper radius of 4m and a height of 5m
A conical tank with an upper radius of 4m and a height of 5m drains into a cylindrical take with a radius of 4m and a height of 5m. If the water level in the conical tank drops at a rate of 0.5m/min, at what rate does the water level in the cylindrical tank rise when the water level in the conical tank is 3m and at 1m?
Solution
V = (1/3) * pi * r^2 * h
v = (1/3) * pi * (16/25) * h^3 [ r = (4/5) * h]
V = (16/75) * pi * h^3
dV/dt = (16/25) * pi * h^2 * dh/dt
dh/dt = -0.5 m/min, h= 3
dV/dt = (16/25) * pi * 9 * -0.5
dV/dt = (-144/50) * pi
dV/dt = (-72/25) * pi
The water is flowing out of the conical tank at a rate of (-72/25) * pi m^3/min.
Volume for cylinder:
V = pi * r^2 * h
r is unchanging, so let\'s find dV/dt and dh/dt
dV/dt = pi * r^2 * dh/dt
dV/dt = (72/25) * pi
Solve for dh/dt
(72 * pi / 25) / (pi * r^2) = dh/dt
72 / (25 * r^2) = dh/dt
r = 4
72 / (25 * 16) = dh/dt
9 / (25 * 2) = dh/dt
9 / 50 = dh/dt
The height in the cylindrical tank is rising at 9/50 m/min
