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

Let the rate of change of the volume of the balloon be gsuch that,

dv dt =g

Volume of the sphere is V= 4 3 π r 3 . So,

d dt ( 4 3 π r 3 )=g 4 3 π⋅3 r 2 dr dt =g 4π r 2 dr=gdt

Integrate both sides, we get,

∫ 4π r 2 dr = ∫ gdt 4π ∫ r 2 dr =g ∫ dt 4π( r 3 3 )=gt+C 4π r 3 =3( gt+C ) (1)

Given that at time t=0, radius is r=3.

4π( 3 3 )=3( g×0+C ) C=36π

Also at time t=3,radius is r=6.

4π ( 6 ) 3 =3( g×3+36π ) 864π 3 =3g+36π 288π−36π 3 =g g=84π

Substitute values of gand Cin equation (1).

4π r 3 =3( 84πt+36π ) r 3 =3( 21t+9 ) r= ( 63t+27 ) 1 3

Therefore, the radius of the balloon after t seconds is ( 63t+27 ) 1 3 .