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Air is being pumped into a spherical balloon so that its volume increases at a rate of 80cm^3/s. How fast is the surface area of the balloon increasing when its radius is 7cm?

  • calculus -

    V = (4/3)pi(r^3)
    dV/dt = 4pi(r^2)dr/dt
    when t=7
    80 = 4pi(49)dr/dt
    dr/dt = 80/(196pi)

    SA = 4pi(r^2)
    d(SA)/dt = 8pi(r)dr/dt
    = 8pi(7)*80/(196pi)


  • calculus -


  • calculus -

    22.85 cubic cm per second

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