Calculus I

The volume of a sphere is changing at a rate of 8pi cm/sec. What is the rate of change of its surface area when the radius is 1? (the volume of a sphere is given by V=4/3pir^3 and its surface area, by A=4pi r ^2).

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  1. V = (4/3)π r^3
    dV/dt = 4π r^2 dr/dt

    given: dV/dt = 8π , r = 1
    8π = 4π(1) dr/dt
    dr/dt = 8π/4π = 2

    A = 4πr^2 ----> (did you notice that surface area equals the derivative of volume ??)

    dA/dt = 8πr dr/dt
    when r = 1
    dA/dt = 8π(1)(1) = 8π

    the surface area is changing at 8π cm/sec

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    posted by Reiny

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