SOLUTION: A chemical is poured into a tank the shape of a cone with vertex down at the rate of 1/5 m^3/min (meters cuber per minute). if the radius of the cone is 6m and its height is 10 m,

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Question 245091: A chemical is poured into a tank the shape of a cone with vertex down at the rate of 1/5 m^3/min (meters cuber per minute). if the radius of the cone is 6m and its height is 10 m, how fast is the liquid rising at the instant when the depth of the liquid is 2m in the center?
Answer by Alan3354(69443) About Me  (Show Source):
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A chemical is poured into a tank the shape of a cone with vertex down at the rate of 1/5 m^3/min (meters cuber per minute). if the radius of the cone is 6m and its height is 10 m, how fast is the liquid rising at the instant when the depth of the liquid is 2m in the center?
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Vol+=+pi%2Ar%5E2%2Ah%2F3
Eliminate r by solving for it in terms of h
r = 0.6h (that wasn't hard)
V+=+pi%2A%280.6h%29%5E2%2Ah%2F3
V+=+pi%2A0.12h%5E3
Differentiate wrt time
dV/dt = 3*pi*0.12h^2*dh/dt = 0.2
dh/dt = 0.2/(3*pi*0.12*4) at h=2
dh/dt =~ 0.0442097 meter/min