QPaperGen

Let \(x\) be the horizontal distance from the wall, and \(y\) be the height on the wall. \[\]
Using Pythagoras: \(x^2 + y^2 = 25\)\[\]

Differentiate:

\[\frac{dy}{dt} = \frac{d}{dt}\left(\sqrt{25 - x^2}\right) = \frac{-x}{\sqrt{25 - x^2}} \cdot \frac{dx}{dt}\]

Given \(\frac{dx}{dt} = 2,\text{cm/s} = 0.02,\text{m/s}\) and \(x = 4\):

\[\frac{dy}{dt} = \frac{-4}{\sqrt{25 - 16}} \cdot 0.02 = \frac{-4}{3} \cdot 0.02 = -\frac{0.08}{3} \approx -0.0267,\text{m/s}\]

So, the top of the ladder is sliding down at approximately \(0.0267,\text{m/s}\).