Post a New Question


posted by .

A spherical balloon a has a diameter 2 m when it is 1500m high . It is observed that the diameter increases at a constant rate of 4cm/min as it continues to rise. At what rate is the volume increasing when the diameter is 4m . At what rate is the surface area imcreasing then?

  • Calculus -

    r = 1 meter
    dr/dt = .02m/min
    This is the same idea as your cone problem.

    The rate of change of the volume is the rate of change of the surface area times the added thickness of the surface, dr/dt
    In other words:
    dV/dt = 4 pi r^2 dr/dt

    surface area = A = 4 pi r^2
    dA/dr = 8 pi r
    dA/dt = 8 pi r dr/dt

  • Calculus -


Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question