QPaperGen

We are given that the radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/s. We want to find the rate at which the volume of the bubble is increasing when the radius is 1 cm.

Assuming the bubble is spherical, the volume is given by: \[V = \frac{4}{3}\pi r^3\]

Differentiating both sides with respect to time \(t\): \[\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\]

Given: \(\frac{dr}{dt} = \frac{1}{2}\) cm/s

At \(r = 1\) cm: \[\frac{dV}{dt} = 4\pi (1)^2 \cdot \frac{1}{2} = 2\pi \text{ cm}^3/\text{s}\]

Final Answer: The volume is increasing at a rate of \(2\pi\) cm\(^3\)/s when the radius is 1 cm.