Question

A spherical balloon is inflating with helium at a rate of 6464piπ StartFraction ft cubed Over min EndFraction ft3 min. How fast is the​ balloon's radius increasing at the instant the radius is 22 ​ft? How fast is the surface area​ increasing?

Answer 1

Answer:

the surface area​ will increase 16  feet per minute

Step-by-step explanation:

Allow me to revise your question for a better understanding.

A spherical balloon is inflating with helium at a rate of 64π Start Fraction ft cubed Over min End Fraction ft3 min. How fast is the​ balloon's radius increasing at the instant the radius is 2 ​ft? How fast is the surface area​ increasing?

My answer:  

Given that:

$\dfrac{dV}{dt} = 64\pi \dfrac{\text{ ft}^3}{\text{min}}$  (1)

As we know that, the volume of a  spherical balloon can be calculated by using the following formula:

$V = \dfrac{4}{3}\pi r^3$ where r is the radius

Substitute V into (1), we have:

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

In this situation, the instant radius is: 2f

So we have:

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

The radius of the balloon is increasing at a rate of 4 feet per minute

The the surface area​ has the formula as:

SA1 = 4π$r^{2}$

If the radius of the balloon is increasing at a rate of 4 feet per minute, so the

the surface area​ will increase 16  feet per minute  because:

SA2 : 4π$(4r)^{2}$  = 16 (4π$r^{2}$ ) = 16 SA1

Hope it will find you well.