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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.