QPaperGen

Differentiate w.r.t. Time (\(t\)):\[\frac{dA}{dt} = \frac{d}{dt} \left( \frac{\sqrt{3}}{4} s^2 \right) = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt}\]\[\frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}\]

Substitute the given values: \(s = 15 \text{ cm}\) and \(\frac{ds}{dt} = 3 \text{ cm/s}\).\[\frac{dA}{dt} = \frac{\sqrt{3}}{2} (15) (3)\]\[\frac{dA}{dt} = \frac{45\sqrt{3}}{2}\]The rate of increase of the area is \(\mathbf{\frac{45\sqrt{3}}{2} \text{ cm}^2\text{/s}}\).