Water pours out of a conical tank of height 10 feet and radius 4 feet at a rate of 10 cubic feet per minute. How fast is the water level changing when it is 5 feet high? Use V= (1/3)*pi*r^(2)*h


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Oddman answered
The radius of a cone is proportional to height. In this case, it looks like the constant of proportionality is (4 ft/(10 ft)) = 0.4, so the equation for the volume is
  v = (π/3)(0.4h)^2*h = .16π/3*h^3
The rate of change of volume with respect to time is
  dv/dt = (.16π/3)*3*h^2 = .16π*h^2*dh/dt
Solving for dh/dt, we get
  (dv/dt)/(.16π*h^2) = dh/dt
Using the given values for dv/dt and h, we get
  (-10 ft^3/min)/(.16π*(5 ft)^2) = dh/dt
  -5/(2π) ft/min = dh/dt
  The water level is dropping at the rate of 5/(2π) ≈ 0.795775 ft/min.

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