Question

Air is pumped into a balloon such that its volume increases at the rate of 75cm^3 per second. It is assumed that the balloon is spherical all the time. Find in term of pi the rate at which the radius of the balloon is increasing when the radius is 5cm. Given the balloon was initially empty show that one minute after the pumping begins the readius is increasing at the rate of 1/12pi r^-1/3 per second. Help please!!

Answer 1

surface area = 4 pi r^2
so
dv/dt = 4 pi r^2 dr/dt

dv/dt = 75 cm^3/s
so
dr/dt = 75/(4 pi r^2)
when r = 5
dr/dt = 3/(4 pi) = .75/pi cm/s

after 60 seconds volume = 75*60 = 4500 cm^3
(4/3)pi r^3 = 4500
pi r^3 = 3375
r = 15/pi^(1/3)
r^2 = 225/pi^(2/3)
4 pi r^2 = 900 pi^(1/3)
dr/dt = 75/(4 pi r^2)
= (75/900)pi^-1/3)
= (1/12) pi^(-1/3)
So I disagree

Answer 2

So did I - thank you so much!! I have spent hours on this one and I am sure you are right.

Answer 3

Well, let's hope we are right :)