QPaperGen

Let the side length of the cube be \(x\), and its volume be \(V\). \[\]
Since \(V = x^3\), we differentiate with respect to time \(t\):\[\]

\[\frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} = 3x^2 \cdot \frac{dx}{dt}\]

Given \(\frac{dx}{dt} = 3\,\text{cm/s}\), we substitute:

\[\frac{dV}{dt} = 3x^2 \cdot 3 = 9x^2\]

When \(x = 10\,\text{cm}\), \[\]we find:

\(\frac{dV}{dt} = 9(10)^2 = 900\,\text{cm}^3/\text{s}\) \[\]

Hence, the volume of the cube is increasing at the rate of \(900\,\text{cm}^3/\text{s}\) when the edge is 10
cm long.