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A spherical balloon is inflated so that its volume is increasing at the rate of 5ft^3/min. How fast is the diameter of the balloon increasing when the radius is 3ft?

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Rate of change of the diameter = ??? ft/min

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  • Ian H
    Lv 7
    1 month ago

    Without calculus:

    V = (4π/3)(D/2)^3 = (π/6)D^3

    When D = 6, V = 36π

    In 1/1000 of a minute V becomes 36π + 5/1000

    D^3 = 6V/π = 216 + 3/(100π) ~ 216.0095493

    D ~ 6.000088418...

    Change in diameter in 1/1000 of a minute ~ 0.000088418

    Estimate of dD/dt is ~ 0.088 ft/min

    dV/dt = dV/dD * dD/dt = 5 ft^3/min

    V = (4π/3)r^3 = (π/6)D^3

    dV/dD = πD^2/2

    dD/dt = 5/(dV/dD) = 10/(πD^2), (in general)

    When diameter D = 6 ft, dD/dt = 10/(36π)

    dD/dt = 5/(18π) ft/min (answer in exact form)

    dD/dt ~ 0.088419 ft/min, or, about 1.06 inches per minute

  • 1 month ago

    V = (4/3)πr³

    so, V = (4/3)π(d/2)³ => (1/6)πd³

    Hence, dV/dt = (1/2)πd².dd/dt

    At the point when r = 3, d = 6 so,

    5 = (1/2)π(6)².dd/dt

    so, dd/dt = 5/18π => 0.088 feet/min

    or, about 1 inch/min

    :)>

  • 1 month ago

    Sphere V = ⁴/₃πr³

      A = 4πr²

    d = 2r

    V = ⁴/₃πr³ = ⁴/₃π(d/2)³ = (1/6)πd³

    dV/dd = (1/2)πd²

    dV/dt = 5 ft³/min

    dd/dt = dV/dt / dV/dd = (1/2)πd² / 5 = (1/10)πd²

    dd/dt  (d is 6) =  (1/10)π6² = 21.6π ft/min

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