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?

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  

Given that:

 (1)

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

where r is the radius

Substitute V into (1), we have:

In this situation, the instant radius is: 2f

So we have:

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

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  = 16 (4π ) = 16 SA1

Hope it will find you well.

answer:     true

step-by-step explanation: