Answer:
the surface area will increase 16 feet per minute
Step-by-step explanation:
Allow me to revise your question for a better understanding.
<em>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? </em>
My answer:
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.
