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20 June, 11:54

A spherical balloon is inflating with helium at a rate of 72 ft2/min. How fast is the balloon's radius increasing at the instant the radius is 3 ft? How fast is the surface area increasing?

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  1. 20 June, 12:30
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    The volume of the balloon is given by:

    V = 4πr³/3

    V = volume, r = radius

    Differentiate both sides with respect to time t:

    dV/dt = 4πr² (dr/dt)

    Isolate dr/dt:

    dr/dt = (dV/dt) / (4πr²)

    Given values:

    dV/dt = 72ft³/min

    r = 3ft

    Plug in and solve for dr/dt:

    dr/dt = 72 / (4π (3) ²)

    dr/dt = 0.64ft/min

    The radius is increasing at a rate of 0.64ft/min

    The surface area of the balloon is given by:

    A = 4πr²

    A = surface area, r = radius

    Differentiate both sides with respect to time t:

    dA/dt = 8πr (dr/dt)

    Given values:

    r = 3ft

    dr/dt = 0.64ft/min

    Plug in and solve for dA/dt:

    dA/dt = 8π (3) (0.64)

    dA/dt = 48.25ft²/min

    The surface area is changing at a rate of 48.25ft²/min
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