A spherical balloon is being inflated at the rate of $35 \mathrm{cc} / \mathrm{min}$. The rate of increase in the surface area (in $\mathrm{cm}^2 / \mathrm{min}$.) of the balloon when its diameter is $14 \mathrm{~cm}$, is :

Question

Aicharya · Accepted Answer

Volume of sphere $\mathrm{V}=\frac{4}{3} \pi r^3$ $\frac{d \mathrm{~V}}{d t}=\frac{4}{3} \cdot \pi \cdot 3 r^2 \cdot \frac{d r}{d t}$ $35=4 \pi r^2 \cdot \frac{d r}{d t}$ or $\frac{d r}{d t}=\frac{35}{4 \pi r^2}$ Surface area of sphere $=\mathrm{S}=4 \pi r^2$ $\frac{d \mathrm{~S}}{d t}=4 \pi 	imes 2 r 	imes \frac{d r}{d t}=8 \pi r \cdot \frac{d r}{d t}$ $\frac{d \mathrm{~S}}{d t}=\frac{70}{r}$ Now, diameter $=14 \mathrm{~cm}, r=7$ $	herefore \quad \frac{d \mathrm{~S}}{d t}=10$

A spherical balloon is being inflated at the rate of $35 \mathrm{cc} / \mathrm{min}$. The rate of increase…

Solution