Calculus
posted by Mehak on .
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?

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 
T=3s