Pre-calculas

A conical tank of radius R=19 feet and height of H=16 feet is being filled with water at a rate of 9ft 3 /min .


(a) Express the height h of the water in the tank, in feet, as a function of time t in minutes

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  1. assuming the tank is pointed-end down, the water forms a cone, so

    v = π/3 r^2 h

    Now, using similar triangles, (r/h) = 9/16, so

    v = (π/3)(9/16 h)^2 h = 27π/256 h^3
    h = 3/8 ∛(2πv)

    since this is precal, not calculus, I'm not sure what tools you have available to convert that to a function of time

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  2. Is the point at the bottom or the top?

    whichever way you draw it
    Volume of a cone = (1/3)pi R^2 x
    where r is the radius of the surface and x is distance from the tip
    r/x = 9.5/ 16 = .594
    so
    r = .594 x

    v = (1/3) pi (.594x)^2 x
    v = .369 x^3
    dv/dx = 1.11 x^2
    dv/dt = 1.11 x^2 dx/dt

    but dv/dt = 9 ft^3/min
    x^2 dx/dt = 9/1.11 = 8.11
    x^2 dx = 8.11 dt
    (1/3) x^3 = 8.11 t
    x^3 = 24.3 t
    x = (24.3 t)^(1/3)

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