Question

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Q: Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6cm and a height of 12cm, at the rate of 3 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep? Give an exact answer showing all work and include units in your answer. Show your work

Answer 1

h = 9 cm
r = (6/12)9 = 4.5 cm

dV = pi r^2 dh
so
dV/dt = pi r^2 dh/dt

-3 cm^3/min = pi (4.5)^2 dh/dt
so
dh/dt = -3/[ pi(4.5)^2]
about .047 cm/min

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