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Jesse has constructed a huge cylindrical can with a diameter of 60 ft. The can is being filled with water at a rate of 450 ft3/min. How fast is the depth of the water increasing? (Hint: The volume of water in the cylinder is determined by πr2h where r is the radius and h is the depth of the water )

  • Calculus -

    the rate of change of volume is the surface area times the rate of change of height

    450 ft^3/min = surface area * dh/dt
    surface area = pi r^2

    450 = pi (30^2) dh/dt

    dh/dt = .159 ft/min

    You could do this by saying
    V = pi r^2 h
    dV/dh = pi r^2
    dV/dh*dh/dt = pi r^2 dh/dt
    chain rule
    dV/dt = pi r^2 dh/dt

    but most of us just look at the lake rising an inch and seeing that the added volume is the area times one inch

  • Calculus -

    Thanks Damon, that really clears it up for me

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