Problema Solution
A spherical balloon is inflated so that its volume is increasing at the rate of 3.9 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.2 feet?
Answer provided by our tutors
let
dV/dt = 3.9 ft^2/min
d = 1.2 ft
r = d/2
r = 1.2/2 = 0.6 ft
we need to find: d(2r)/dt = 2*(dr/dt)
V = (4/3)*pi*r^3
dV/dt = 4*pi*(r^2)*(dr/dt)
4*pi*(r^2)*(dr/dt) = 3.9
dr/dt = 3.9/(4*pi*(r^2))
dr/dt = 3.9/(4*3.14*0.6^2)
dr/dt = 0.86 ft/min
2(dr/dt) = 2*0.86 ft/min = 1.72 ft/min
the diameter of the balloon is increasing with a rate of 1.72 ft/min.