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:
[tex]\dfrac{dV}{dt} = 64\pi \dfrac{\text{ ft}^3}{\text{min}}[/tex] (1)
As we know that, the volume of a spherical balloon can be calculated by using the following formula:
[tex]V = \dfrac{4}{3}\pi r^3[/tex] where r is the radius
Substitute V into (1), we have:
[tex]\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{4}{3}\pi r^3)\\\\\dfrac{dV}{dt} =4\pi r^2\dfrac{dr}{dt}[/tex]
In this situation, the instant radius is: 2f
So we have:
[tex]64\pi = 4\pi (2)^2\dfrac{dr}{dt}\\\\\Rightarrow \dfrac{dr}{dt} = 4[/tex]
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[tex]r^{2}[/tex]
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[tex](4r)^{2}[/tex] = 16 (4π[tex]r^{2}[/tex] ) = 16 SA1
Hope it will find you well.
