Question

The volume of a spherical ball increase at the rate of 4π cc/s . Then the rate of increase of the surface area when the volume is 288π cc ? a)4π/3 b)3π/2 c)6π d)6

Answer 1

A

Explanation:

"volume of a spherical ball increase at the rate of `4pi` cc/s"

`(dv)/(dt)=4pi`

Now, you are required to find the rate of the increase of the surface area when the volume is `288pi`

`(dA)/(dt)=?`

Since the volume of the sphere is `288pi`, we can find the radius
using the volume of a sphere is `4/3pir^3`

`4/3pir^3=288pi`
`r^3=288times3/4`
`r^3=216`
`r=6`

Now `(dv)/(dt)=(dv)/(dr)times(dr)/(dt)`

Differentiate the volume of the sphere gives us:
`(dv)/(dr)=4/3pitimes3r^2=4pir^2`

To find `(dr)/(dt)`,
`(dv)/(dt)=(dv)/(dr)times(dr)/(dt)`
`4pi=4pir^2times(dr)/(dt)`
`(dr)/(dt)=1/r^2`

Now,
`(dA)/(dt)=(dA)/(dr)times(dr)/(dt)`

Remember, the surface area of the sphere is `4pir^2`
so:

`A=4pir^2`

`(dA)/(dr)=8pir`

We already found that `r=6` and `(dr)/(dt)=1/r^2`.
Putting this all together,

`(dA)/(dt)=(dA)/(dr)times(dr)/(dt)`

`(dA)/(dt)=8pir times 1/r^2`

`(dA)/(dt)=(8pi)/r`

`(dA)/(dt)=(8pi)/6=4/3pi`