Calculus - Rates of Change

A water tank has a shape of an inverted cone with a base radius of 2m and a height of 4 m. If water is being pumped into the tank at a rate of 2m3/min, then find the rate at which the water level is rising when the water is 3m deep.

Attempted solution:
V= 1/3πr^2h
V= 1/3π(2h/4)^2 h
V= 1/3π (4h^2/16) h
V= 4/48 π h^3
Dv/Dt = 4/48 π h^3 dh/dt
2= 4/48 π 2^3dh/dt
2(48)/4π x 8 = dh/dt

dh/dt = 3π m/min

Is this the correct solution, or did I go wrong?

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  1. if the point of the cone is up, then
    volume= PI*2^2 *4 when full
    so the volume of the cone filling up is
    volume=full-PI* radius of top^2*distance to top^2, and if h is the distance from the bottom to the top of the water,
    V = full- PI ((4=h)/2)^2* (4-h)
    = full-PI/4 * (4-h)^3 check that
    then figure out
    dV/dt from this function, you know dv/dt, solve for dh/dt

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    bobpursley
  2. I do not understand that.

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  3. You were ok to here:

    V= 4/48 π h^3
    V = π/12 h^3

    But that means

    dV/dt = π/12 * 3h^2 dh/dt
    = π/4 h^2 dh/dt

    Now plug in dV/dt=2, h=3, and you get

    π/12 * 9 dh/dt = 2
    dh/dt = 8/3π ft/min

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