Question

At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 4 square meters and the radius is increasing at the rate of 1/6 meters per minute?

Answer 1

We know the volume of a sphere relative to its radius is given by:
`V(r) = 4/3 pi r^3`

We are given that
Surface area at the time in question is `4 m^2`
which implies since `S = 4 pi r^2` that the radius at that time is
`r = 1/sqrt(pi)`

We are asked for `(dV)/(dt)`

`(d V)/(dt) = (d V)/(dr) * (dr)/(dt)`

`(d V)/(dr) = 4 pi r^2` (using the derivative ofour formula for the Volume)
and we are told that `(dr)/(dt) = 1/6 m/sec`

So at the time indicated
`(d V)/(dt) = 4 pi r^2 * 1/6`

and with `r= 1/sqrt(pi)`

`(d V)/(dt) = 2/3 m/sec`