Question

The volume of a sphere is changing at a constant rate of `pi/3 \ cm^3s^-1`. How fats is the surface area changing when the volume is `(9pi)/2`?

Answer 1

`(dA)/dt =(4pi)/9 \ cm^2s^-1`

Explanation:

Let us set up the following variables:

`{(r, "Radius of sphere at time t","(cm)"), (A, "Surface area of sphere at time t", "(cm"^2")"), (V, "Volume of sphere at time t", "(cm"^3")"), (t, "time", "(sec)") :}`

Our aim is to find `(dA)/dt` when `V=(9pi)/2` and `(dV)/dt=pi/3`.

The standard formula for Area & Volume of a sphere are:

`V=4/3pir^3 \ \ \ \ .... [1]`
`A=4pir^2 \ \ \ \ \ \ .... [2]`

When `V=(9pi)/2 => 4/3pir^3 =(9pi)/2`

`:. r^3 =9/2*3/4`
`:. r =3/2`

Differentiating [1] and [2] wrt `r` we get;

`(dV)/(dr)=4pir^2` and `(dA)/(dr) = 8pir`

And from the chain rule we get:

`(dA)/dt =(dA)/(dr) * (dr)/(dV)* (dV)/(dt)`
`\ \ \ \ \ \ \=8pir * 1/(4pir^2) * (dV)/(dt)`
`\ \ \ \ \ \ \=2/r * (dV)/(dt)`

So when `V=(9pi)/2`, `(dV)/dt=pi/3` and `r =3/2`, then:

`(dA)/dt =2/(3/2) * pi/3`
`\ \ \ \ \ \ \=(4pi)/9`

Answer 2

`4/9 pi` cm^2/s

Explanation:

`V = 4/3 pi*r^3`, where V =volume of sphere and r =radius.

given that,
`(dV)/(dt) =pi/3` and `V =9/2 pi`

`4/3 pi *r^3 = 9/2 pi`

`r^3 = 9/2 pi * 3/(4 pi)`

`r^3 = 27/8`, `r =3/2`

`V=4/3 pi*r^3,` then `(dV)/(dr) = 4 pi*r^2`

when `r =3/2`, `(dV)/(dr) = 4 pi*(3/2)^2=9 pi`

`(dr)/(dt) = (dV)/(dt)/((dV)/(dr))`

`= (pi/3)/(9 pi)=1/27`

Area pf sphere, `A =4 pi r^2`

`(dA)/(dr) = 8 pi r`
when `r =3/2`, `(dA)/(dr) = 8 pi r =8 pi (3/2)=12 pi`

therefore,
`(dA)/(dt) = (dA)/(dr)*(dr)/(dt)`

`=12 pi*1/27 = 4/9 pi` `(cm)^2/s`