Post a New Question

Calculus (Help please)

posted by .

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.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question