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.
Answer Details
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) \(48\pi cm^{2}s^{-1}\).