# calculus

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gas is pumped into a spherical balloon at the rate of 1 cubic feet per minute. How fast is the diameter of the balloon increasing when the balloo0n contains 36 cubic feet of gas?

• calculus -

V = (4/3)πr^3
when V = 36 ...
36 = (4/3)πr^3
r^3 = 27/π
r = 2.04835

dV/dt = 4πr^2 dr/dt

1 = 4π(2.04835^2) dr/dt
dr/dt = .018966 ft^3/min

since diameter = 2r
d(diameter)/dt = .0379 ft^3 /min

• calculus -

first find dr/dt. The diameter changes twice as fast as the radius.
v = (4/3) pi r^3
dv/dt = 4 pi r^2 dr/dt
so
1 = 4 pi (r^2) dr/dt

but r^3 = (3/4)(36)/pi
so r = 2.05 ft
so
1 = 4 pi (4.2) dr/dt
so
dr/dt = .019 ft/min
D = 2 r
dD/dt = 2 dr/dt = .038 ft/min

another way
rate of volume increase = surface area * dr/dt
1 = 4 pi r^2 * dr/dt
same old equation

• calculus -

A spherical balloon is inflated so that its volume is increasing at the rate of 2.2 ft^3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.2 feet?

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