Asked by Ram
A balloon (in a shape of a sphere) is inflated with helium at a constant rate of 125 cm^3/s. Calculate the rate of increase of the diameter when the volume is 2000 cm^3.
Answers
Answered by
Reiny
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
when V = 2000
(4/3)πr^3 = 2000
r^3 = 1500/π
r = (1500/π)^(1/3)
= 7.8159... I stored in my calculator's memory
so in
dV/dt = 4π r^2 dr/dt
125 = 4π (7.8159...)^2 dr/dt
dr/dt = .1628618
rate of change of diameter = .3225536 cm/s
dV/dt = 4π r^2 dr/dt
when V = 2000
(4/3)πr^3 = 2000
r^3 = 1500/π
r = (1500/π)^(1/3)
= 7.8159... I stored in my calculator's memory
so in
dV/dt = 4π r^2 dr/dt
125 = 4π (7.8159...)^2 dr/dt
dr/dt = .1628618
rate of change of diameter = .3225536 cm/s