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Suppose we pump water into an inverted right-circular cone tank at the rate of 6 cubic feet per minute. The tank has the height 9 ft and radius on the top is 8 ft. What is the rate at which the water level is rising when the water is 3 ft deep? Leave the answer approximate. You may use the formula for the volume of the right-circular cone of radius r and height h: V=1/3pi*r^2*h

  • math -

    At a time of t minutes, let the
    radius of the water level be r
    let the height of the water be h
    by ratio:
    r/h = 8/9
    r = 8h/9

    V = (1/3)π r^2 h
    = (1/3)π(64h^2/81)h
    = (64π/243) h^3
    dV/dt = (64π/81) h^2 dh/dt
    6 = (64π/81) (9) dh/dt
    I will let you solve for dh/dt

  • math -

    thanks so much!

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