Answer:
The radius of the balloon is increasing at a rate of 4 feet per minute.
Step-by-step explanation:
We are given the following in the question:
[tex]\dfrac{dV}{dt} = 64\pi \dfrac{\text{ ft}^3}{\text{min}}[/tex]
Volume of sphere is given by
[tex]V = \dfrac{4}{3}\pi r^3[/tex]
where r is the radius of the balloon.
Instant radius, r = 2 ft
Rate of change of volume =
[tex]\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{4}{3}\pi r^3)\\\\\dfrac{dV}{dt} =4\pi r^2\dfrac{dr}{dt}[/tex]
Putting values, we get,
[tex]64\pi = 4\pi (2)^2\dfrac{dr}{dt}\\\\\Rightarrow \dfrac{dr}{dt} = 4[/tex]
Thus, the radius of the balloon is increasing at a rate of 4 feet per minute.
