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

the surface area​ will increase 16 feet per minute

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:

`(dV)/(dt) = 64\pi \frac{\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 = (4)/(3)\pi r^3` where r is the radius

Substitute V into (1), we have:

`(dV)/(dt) = (d)/(dt)((4)/(3)\pi r^3)\\\\(dV)/(dt) =4\pi r^2(dr)/(dt)`

In this situation, the instant radius is: 2f

So we have:

`64\pi = 4\pi (2)^2(dr)/(dt)\\\\\Rightarrow (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.