Asked by Jeff
A conical tank (with its vertex down) is 8 feet tall and 6 feet across its diameter. If water is flowing into the tank at the rate of 2 feet3/min, find the rate at which the water level is rising at the instant when the water depth is 2.5 feet.
Answers
Answered by
drwls
Water depth = y
Surface water radius r = 3 y/8 ft
Water volume V = (1/3)*pi*(3y/8)^2*y
= (3 pi/64)y^3
dV/dt = 2.0 ft^3/min = (9*pi*y^2/64)*dy/dt = pi*r^2*(dy/dt)
When y = 2.5 ft
dy/dt = 4.53 ft/min
Surface water radius r = 3 y/8 ft
Water volume V = (1/3)*pi*(3y/8)^2*y
= (3 pi/64)y^3
dV/dt = 2.0 ft^3/min = (9*pi*y^2/64)*dy/dt = pi*r^2*(dy/dt)
When y = 2.5 ft
dy/dt = 4.53 ft/min
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