The radius of a sphere is increasing at a rate \(3cm s^{-1}\). Find the rate of increase in the surface area, when the radius is 2cm.

The surface area, \(A\), of a sphere is given by \(A = 4\pi r^2\), where \(r\) is the radius. We are given that \(\frac{dr}{dt} = 3\text{ cm/s}\) and \(r = 2\text{ cm}\). We want to find \(\frac{dA}{dt}\) at this point. Taking the derivative of the surface area formula with respect to time, we get: \begin{align*} \frac{dA}{dt} &= \frac{d}{dt}(4\pi r^2) \\ &= 8\pi r \frac{dr}{dt} \\ \end{align*} Substituting the given values, we get: \begin{align*} \frac{dA}{dt} &= 8\pi (2\text{ cm})(3\text{ cm/s}) \\ &= 48\pi\text{ cm}^2/\text{s} \\ \end{align*} Therefore, the rate of increase in the surface area when the radius is 2cm is \(48\pi\text{ cm}^2/\text{s}\). So, the answer is (d) \(48\pi cm^{2}s^{-1}\).