calculus

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gas is pumped into a spherical balloon at the rate of 1 cubic feet per minute. How fast is the diameter of the balloon increasing when the balloo0n contains 36 cubic feet of gas?

  • calculus -

    V = (4/3)πr^3
    when V = 36 ...
    36 = (4/3)πr^3
    r^3 = 27/π
    r = 2.04835

    dV/dt = 4πr^2 dr/dt

    1 = 4π(2.04835^2) dr/dt
    dr/dt = .018966 ft^3/min

    since diameter = 2r
    d(diameter)/dt = .0379 ft^3 /min

  • calculus -

    first find dr/dt. The diameter changes twice as fast as the radius.
    v = (4/3) pi r^3
    dv/dt = 4 pi r^2 dr/dt
    so
    1 = 4 pi (r^2) dr/dt

    but r^3 = (3/4)(36)/pi
    so r = 2.05 ft
    so
    1 = 4 pi (4.2) dr/dt
    so
    dr/dt = .019 ft/min
    D = 2 r
    dD/dt = 2 dr/dt = .038 ft/min

    another way
    rate of volume increase = surface area * dr/dt
    1 = 4 pi r^2 * dr/dt
    same old equation

  • calculus -

    A spherical balloon is inflated so that its volume is increasing at the rate of 2.2 ft^3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.2 feet?

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