Rate of change of radius (in feet per second)

Question

Helium is pumped into a spherical balloon at a rate of 2 cubic feet per second. How fast is the radius increasing after 2 minutes?

Note: The volume of a sphere is given by:V=(4/3)pir^3

Answer 1

V =(4/3)pi(r^3)

V' = 4pi(r^2)r' = 2 ft^3/sec

at time t = 2 minutes = 120 seconds, V = 2(120) = 240 ft^3

240 = (4/3)pi(r^3)

r^3 = (3/4pi)240 = 180/pi

r = cube root of 180/pi ft = about 3.855 feet radius after 2 minutes

V' = 4pi(r^2)r' = 2 ft^3/sec

2pi(r^2)r' = 1

r' = 1/2pi(r^2)

= 1/2pi(3.855)^2

r' = 0.010709 feet per minute

in one second the rate of change e radius is about 1/2 a foot per second

it gets smaller each second, closer and closer to zero with more time

since V' remains 2, it has very little effect on the radius, with larger Volume and more time

= about .129 inch per minute

= a little more than 1/8 of an inch per minute rate of change in radius after 2 minutes = r'(t=2)

Answer 2

v = 4πr³/3; dv/dt = 4πr²dr/dt; dr/dt = (dv/dt)/(4πr²)

What will be radius after 2 minutes: 2·120 = 4πr³/3; r = (180/π)^1/3_.

So, dr/dt = (dv/dt)/[4π(180/π)^2/3] = (dv/dt)/(4π^1/3/180^2/3) = 2/(4π^1/3/180^2/3 = 10.88 ft/s