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Calculus (Help please)

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A tank has the shape of an inverted circular cone with a base radius of 5 meters
and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters
per minute, find the rate at which the water level is rising when the water is 7
meters deep

  • Calculus (Help please) - ,

    is the answer 8/49pi= 0.052 meters per minute

  • Calculus (Help please) - ,

    I get dV/dt = pi/16 h^2 dh/dt
    giving
    2 = 49pi/16 dh/dt
    or
    dh/dt= 32/49pi

    One of us has a factor of 2 out of place. Could be me ...

  • Calculus (Help please) - ,

    I did not get that

    Let the height of the water be h m and the radius of the water level be r m
    by ratios:
    r/h = 5/20 = 1/4
    r = h/4

    V = (1/3)πr^2 h = (1/3)π(h^2/16)(h)
    = (1/48)π h^3
    dV/dt = (1/16)π h^2 dh/dt
    so when dV/dt = 2, h = 7
    2 = (1/16)π(49) dh/dt
    2(16)/(49π) = dh/dt
    = 32/(49π)
    = appr. .208 m/min

  • Calculus (Help please) - ,

    WELL I HAVE THIS

    v=1/3pi*r^2h
    v=1/3pi*r2(4r)
    v=4/3pi*r^3
    dv/dt=4pi*r^2
    2=4pi(7/4)^2
    2=49pi/4 * dr/dt
    dr/dt= 8/49pi = 0.052 meters per minute

  • Calculus (Help please) - ,

    You have found the rate at which the rate is changing.

    Unfortunately, your question asked for how fast the height is changing.

  • typo - Calculus (Help please) - ,

    should say:


    You have found the rate at which the radius is changing.

    Unfortunately, your question asked for how fast the height is changing.

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