Question

A spherical balloon is being inflated at a rate of 8 cm³s-1. Find the rate of change of the surface area when the radius is 10 cm.​

Answer 1

1.6

Explanation:

Firstly, we know the rate of change of volume
`v'=8`.

The equation for surface area is:
`A = 4\pi r^2`.

The derivative (rate of change) is then:
`A'=8\pi r*r'`

The equation of the volume of a sphere is:
`V=(4)/(3)\pi r^3`.

The derivative (rate of change) is then:
`V' = 4\pi r^2 * r'`.

This means that
`r'=(V')/(4\pi r^2)`.

We can then plug that into our equation for
`A'` and get:

`A'=8\pi r*(V')/(4\pi r^2)=2*(V')/(r)`

After that, we can just plug in:

`2*(8)/(10)=1.6`