Question

The surface area of a spherical bubble is increasing at the rate of 2 cm²/s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?

Answer 1

\[\text { Let r be the radius,S be the surface area and V be the volume of the sphere at any time t. Then },\]

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

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

\[ \Rightarrow \frac{dr}{dt} = \frac{1}{8\pi r}\frac{dS}{dt}\]

\[ \Rightarrow \frac{dr}{dt} = \frac{2}{8\pi \times 6}\]

\[ \Rightarrow \frac{dr}{dt} = \frac{1}{24\pi} cm/\sec\]

\[\text { Now,} \]

\[\text { Volume of sphere } = \frac{4}{3}\pi r^3 \]

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

\[ \Rightarrow \frac{dV}{dt} = \frac{4\pi \left( 6 \right)^2}{24\pi}\]

\[ \Rightarrow \frac{dV}{dt} = \text{6 cm}^3 /\sec\]