Question

The volume of 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 balloon after t seconds.

Answer 1

`V = 4/3 pir^3`                  ....(1)

Differentiating (1) w.r.t.t., we get

`(dV)/dt = 4pir^2 (dr)/dt`

Now, `(dV)/dt = a`

⇒ `4pir^2 = (dr)/dt = a`               ....(2)

Integrating (2) both sides, we get,

⇒ `int 4 pi r^2 dr = int a  dt`

⇒ `4pi r^3/3 = at + C`

When t = 0, r = 3

∴ `(4pi (3)^3)/3 = C`

⇒ C = 36π

∴ `(4 pir^3)/3 = at + 36 pi`

When t = 3, r = 6 then

`(4pi(6)^3)/3 = 3a + 36 pi`

⇒ `(4pi xx 36 xx 6)/3`

= 3a + 36π

⇒ 288π = 3a + 36π

⇒ 96π = a + 12π

⇒ a = 84π

Hence, `(4pir^3)/3 = 84pi t + 36 pi`

⇒ `(4r^3)/3 = 84t + 36`

⇒ `r^3/3 = 21t + 9`

⇒ `r^3 = 63t + 27`

⇒ `r = (63t + 27)^(1/3)`