I've GOT to assume that this is a calculus problem, so it's OK for me to use it to solve the problem.
Volume of a sphere = 4/3 π R³
Rate of change of the volume = d(Vol)/dt = (4/3 π) (3R² · dR/dt)
We know the rate of change of the volume ... 4 ft³ / sec
4 ft³/sec = (4/3 π) (3R²) dR/dt
After 3 minutes (180 seconds), the volume of the balloon is
(4 ft³/sec) · (180 sec) = 720 ft³
What is the balloon's radius at that time ?
Volume = 4/3 π R³
720 ft³ = 4/3 π R³
R³ = 720 ft³ / (4/3 π)
R = ∛ (720 ft³ / 4/3π)
R = 5.56 ft
Plug this back into the equation with the rates of change:
4 ft³/sec = (4/3 π) (3R²) dR/dt
4 ft³/sec = (4/3 π) (3) · (5.56ft)² · dR/dt
dR/dt = 4 / (4/3 π · 3 · 5.56²)
I get <em>dR/dt = 0.000103 ft/sec</em>. Seems weird. You'd better check this.
"At what time is the radius increasing at a rate of 140 feet per second?" Well, think about it. That's a huge huge rate of expansion. So it has to be right at the beginning, when the first tiny sniff of gas enters the balloon, and the volume jumps from zero to a-little-something.
Start back at our equation for dR/dt :
4 ft³/sec = (4/3 π) (3R²) dR/dt
We want to find out what ' R ' is when dR/dt is 140 ft/sec. Then, when we know 'R', we can calculate the volume, and we'll know how long the gas has been blowing in, because we know that it comes in at 4ft³/sec.
4 ft³/sec = (4/3 π) (3R²) dR/dt
4 ft³/sec = (4/3 π) (3R²) (140 ft/sec)
3 R² = 4 / (4/3 π · 140)
R² = 4 / (4 π · 140)
R = √ (1 / 140π)
R = 0.0477 ft.
Volume = 4/3 π R³ ft³
Time = (Volume/4 ft³/sec) seconds
Time = (4/3 π R³ / 4) seconds
Time = π R³ / 3 seconds
Time = π (0.0476)³/3 sec
I get <em>Time = 0.000114 second</em> . You'd REALLY better check it.
