Question

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Answer 1

Let r be the radius and V be the volume of the spherical balloon at any time t. Then rate of change in volume of spherical balloon is `"dV"/"dt"` which is a constant.

∴ `"dV"/"dt" = "k"`' , where k' is a constant.

But V = `4/3 pi"r"^3`

∴ `"dV"/"dt" = (4pi)/3 xx 3"r"^2 "dr"/"dt" = 4 pi"r"^2 "dr"/"dt"`

∴ `4 pi"r"^2 "dr"/"dt"` = k'

∴ `"r"^2 "dr"/"dt" = (k')/(4pi)` = k, where k = `(k')/(4pi)`

∴ r² dr = k dt

Integrating both sides, we get

`int "r"^2 "dr" = "k" int "dt"`

∴ `"r"^3/3 = "kt" + c`

Initially the radius is 3 units

i.e. r = 3, when t = 0

∴ `3^3/3 = k(0) + "c"`

∴ c = 9

∴ `"x"^3/3 = "kt" + 9`

Also, r = 6, when t = 3

∴ `(6)^3/3 = "k"(3) + 9`

∴ 3k = 72 - 9 = 63

∴ k = 21

∴ `"r"^3/3 = 21"t" + 9`

∴ r³ = 63t + 27

∴ r = `(63"t" + 27)^(1/3)`

Hence, the radius of the spherical balloon after t seconds is `(63"t" + 27)^(1/3)` units.