# The volume of spherical balloon being inflated changes at a constant rate.

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 seconds.

Solution:

Let the rate of change of the volume of the balloon be k (where k is a constant).

$\Rightarrow \frac{d v}{d t}=k$

$\Rightarrow \frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=k$                    $\left[\right.$ Volume of sphere $\left.=\frac{4}{3} \pi r^{3}\right]$

$\Rightarrow \frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=k$

$\Rightarrow 4 \pi r^{2} d r=k d t$

Integrating both sides, we get:

$4 \pi \int r^{2} d r=k \int d t$

$\Rightarrow 4 \pi \cdot \frac{r^{3}}{3}=k t+\mathrm{C}$

$\Rightarrow 4 \pi r^{3}=3(k t+\mathrm{C})$                   ...(1)

Now, at $t=0, r=3:$

$\Rightarrow 4 \pi \times 3^{3}=3(k \times 0+C)$

$\Rightarrow 108 \pi=3 C$

$\Rightarrow C=36 \pi$

At $t=3, r=6$

$\Rightarrow 4 \pi \times 6^{3}=3(k \times 3+C)$

$\Rightarrow 864 \pi=3(3 k+36 \pi)$

$\Rightarrow 3 k=-288 \pi-36 \pi=252 \pi$

$\Rightarrow k=84 \pi$

Substituting the values of k and C in equation (1), we get:

$4 \pi r^{3}=3[84 \pi t+36 \pi]$

$\Rightarrow 4 \pi r^{3}=4 \pi(63 t+27)$

$\Rightarrow r^{3}=63 t+27$

$\Rightarrow r=(63 t+27)^{\frac{1}{3}}$

Thus, the radius of the balloon after $t$ seconds is $(63 t+27)^{\frac{1}{3}}$.