Class 12

Math

Calculus

Differential Equations

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.

So, $dtdV =k$, where $k$ is a constant.

Now, Volume of a sphere, $V=34 πr_{3}$

So, $dtdV =34 ∗3πr_{2}dtdr $

$⇒dtdV =4πr_{2}dtdr $

As,$dtdV =k$.So,

$⇒kdt=4πr_{2}dr$

Now, integrating both sides,

$⇒kt+c=34 πr_{3}+c_{1}$->(1)

$⇒kt+C=34 πr_{3}$, where $C=c−c_{1}$

So, at $(r,t)=(3,0)$

$⇒C=34 π∗(3)_{3}⇒C=36π$

At$(r,t)=(6,3)$

$⇒k(3)+36π=34 π(6)_{3}⇒3k=288π−36π⇒k=84π$

Putting values of $C$ and $k$ in (1),

$84πt+36π=34 πr_{3}⇒84t+36=34 r_{3}$

$⇒r_{3}=(63t+27)⇒r=(63t+27)_{31}$

So, at any time $t$ radius of the baloon will be $(63t+27)_{31}$.