Homework Help: Calculus

Posted by Mehak on Sunday, March 13, 2011 at 11:22pm.

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 - Damon, Monday, March 14, 2011 at 4:56am
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 - Abba, Friday, July 29, 2011 at 9:00am
T=3s

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