Gas is being pumped into a spherical balloon

Solution:

Let r be the radius and V be the volume of the balloon at time t.

It is given that, $\frac{d V}{d t}=30 \mathrm{~cm}^{3} / \mathrm{min}$

Now,

Volume of the spherical balloon, $V=\frac{4}{3} \pi r^{3}$

$V=\frac{4}{3} \pi r^{3}$

Differentiating both sides with respect to $t$, we get

$\frac{d V}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)$

$\Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi \times 3 r^{2} \frac{d r}{d t}$

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

When $r=15 \mathrm{~cm}$ and $\frac{d V}{d t}=30 \mathrm{~cm}^{3} / \mathrm{min}$, we get

$30=4 \pi \times(15)^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{30}{4 \pi \times(15)^{2}}$

 

$\Rightarrow \frac{d r}{d t}=\frac{1}{30 \pi} \mathrm{cm} / \mathrm{min}$

Thus, the radius of the balloon is increasing at the rate of $\frac{1}{30 \pi} \mathrm{cm} / \mathrm{min}$.

Gas is being pumped into a spherical balloon at the rate of $30 \mathrm{~cm}^{3} / \mathrm{min}$. The rate at which the radius increases when it reaches the value $15 \mathrm{~cm}$, is $\frac{1}{30 \pi} \mathrm{cm} / \mathrm{min}$