Question

A spherical balloon is inflating with helium at a rate of 64 pi StartFraction ft cubed Over min EndFraction . How fast is the​ balloon's radius increasing at the instant the radius is 2 ​ft?

Answer 1

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:

$\dfrac{dV}{dt} = 64\pi \dfrac{\text{ ft}^3}{\text{min}}$

Volume of sphere is given by

$V = \dfrac{4}{3}\pi r^3$

where r is the radius of the balloon.

Instant radius, r = 2 ft

Rate of change of volume =

$\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{4}{3}\pi r^3)\\\\\dfrac{dV}{dt} =4\pi r^2\dfrac{dr}{dt}$

Putting values, we get,

$64\pi = 4\pi (2)^2\dfrac{dr}{dt}\\\\\Rightarrow \dfrac{dr}{dt} = 4$

Thus, the radius of the balloon is increasing at a rate of 4 feet per minute.