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.

Home » 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 seconds.

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.

Solution:

It is given that:
Volume $\mathrm{V}=\frac{4 \pi r^{3}}{\mathrm{a}}$ $\frac{d V}{d t}=\frac{4}{3} \pi 3 r^{2} \frac{d r}{d t}$ $\Rightarrow \frac{d V}{d t}=k$ (constant)
$\begin{array}{l} 4 \pi r^{2} \frac{d r}{d t}=k \\ \Rightarrow 4 \pi r^{2} d r=k d t \\ \Rightarrow \int 4 \pi r^{2} d r=\int k d t \\ \Rightarrow \frac{4 \pi r^{2}}{3}=k t+c \end{array}$
For $t=0, r=3$ and for $t=3, r=6$ , Therefore, we have,
$\begin{array}{l} \Rightarrow \frac{4 \pi(3)^{3}}{3}=0+c \\ \Rightarrow c=36 \pi \\ \frac{4 \pi(6)^{3}}{3}=k \cdot(3)+36 \pi \\ \Rightarrow k=84 \pi \end{array}$
Therefore after $t$ seconds the radius of the balloon will be,
$\begin{array}{l} \Rightarrow \frac{4 \pi r^{3}}{3}=84 \pi t+36 \pi \\ \Rightarrow 4 \pi r^{3}=252 \pi t+108 \pi \\ \Rightarrow r^{3}=\frac{252 \pi t+108 n}{4 \pi} \\ \Rightarrow r^{3}=63 t+27 \\ \Rightarrow r=\sqrt[3]{63 t+27} \end{array}$
As a result, the radius of the balloon as a function of time is
$\therefore r=(63 t+27)^{1 / 3}$