Asked by Anonymous
Water is flowing at a rate of 50 cubic meters per minute into a holding tank shaped like a cone, sitting vertex down. The tank's base diameter is 40 m and a height of 10 m.
Write an expression for the rate of change of the water level with respect to time, in terms of h (the water's height in the tank).
Write an expression for the rate of change of the water level with respect to time, in terms of h (the water's height in the tank).
Answers
Answered by
Steve
using similar triangles, the radius of the water surface at height h can be found
r/h = (40/2)/10 = 2
so, r = 2h
v = 1/3 π r^2 h
= π/3 (2h)^2 h
= 4π/3 h^3
dv/dt = 4π h^2 dh/dt
50 = 4π h^2 dh/dt
dh/dt = 25/(2πh^2)
r/h = (40/2)/10 = 2
so, r = 2h
v = 1/3 π r^2 h
= π/3 (2h)^2 h
= 4π/3 h^3
dv/dt = 4π h^2 dh/dt
50 = 4π h^2 dh/dt
dh/dt = 25/(2πh^2)
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