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15 April, 10:30

Water is leaking out of an inverted conical tank at a rate of 8200.08200.0 cm3/min cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 11.0 m11.0 m and the the diameter at the top is 4.5 m4.5 m. If the water level is rising at a rate of 16.0 cm/min16.0 cm/min when the height of the water is 3.0 m3.0 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute. Answer: cm3/min

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  1. 15 April, 10:54
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    I' = 197,474.47 cm^3/min

    the rate at which water is being pumped into the tank in cubic centimeters per minute is 197,474.47 cm^3/min

    Step-by-step explanation:

    Given;

    Tank radius r = d/2 = 4.5/2 = 2.25 m = 225 cm

    height = 11 m

    Change in height dh/dt = 16 cm/min

    The volume of a conical tank is;

    V = (1/3) πr^2 h ... 1

    The ratio of radius to height for the cone is

    r/h = 2.25/11

    r = 2.25/11 * h

    Substituting into equation 1.

    V = (1/3 * (2.25/11) ^2) πh^3

    the change in volume in tank is

    dV/dt = dV/dh. dh/dt

    dV/dt = ((2.25/11) ^2) πh^2. dh/dt ... 2

    And change in volume dV/dt is the aggregate rate at which water is pumped into the tank.

    dV/dt = inlet - outlet rate

    Let I' represent the rate of water inlet and O' represent the rate of water outlet.

    dV/dt = I' - O'

    Water outlet O' is given as 8200 cm^3/min

    dV/dt = I' - 8200

    Substituting into equation 2;

    I' - 8200 = ((2.25/11) ^2) πh^2. dh/dt

    I' = ((2.25/11) ^2) πh^2. dh/dt + 8200

    h = 3.0 m = 300 cm (water height)

    Substituting the given values;

    I' = ((2.25/11) ^2) * π*300^2 * 16 + 8200

    I' = 197,474.47 cm^3/min

    the rate at which water is being pumped into the tank in cubic centimeters per minute is 197,474.47 cm^3/min
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