Question

The volume of a spherical balloon is increasing at the rate of 25 cm³/sec. Find the rate of change of its surface area at the instant when radius is 5 cm ?

Answer 1

\[\text { Let r be the radius and V be the volume of the sphere at any  time t.}\hspace{0.167em}\text { Then,}\]

\[V=\frac{4}{3}\pi r^3 \]

\[ \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\]

\[ \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}\]

\[ \Rightarrow \frac{dr}{dt} = \frac{25}{4\pi \left( 5 \right)^2} \left[ \because r = 5 \text { cm and }\frac{dV}{dt} = 25 {cm}^3 /\sec \right]\]

\[ \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi}cm/\sec\]

\[\text { Now , let S be the surface area of the sphere at any time t.Then},\]

\[S=4\pi r^2 \]

\[\Rightarrow\frac{dS}{dt}=8\pi r\frac{dr}{dt}\]

\[\Rightarrow\frac{dS}{dt}=8\pi\left( 5 \right)\times\frac{1}{4\pi}\left[ \because r = 5 \text { cm and }\frac{dr}{dt} = \frac{1}{4\pi} cm/\sec \right]\]

\[\Rightarrow\frac{dS}{dt} {=10 cm}^2 /sec\]