Water is dripping out from a conical funnel of semi-vertical angle $\frac{\pi}{4}$ at the uniform rate of $2 \text{ cm}^2/\text{sec}$ in the surface area,through a tiny hole at the vertex of the bottom. When the slant height of the cone is $4 \text{ cm}$,find the rate of decrease of the slant height of water.

(N/A) Let $s$ be the curved surface area of the conical water level. The formula for the curved surface area is $s = \pi r l$.
Given the semi-vertical angle $\alpha = \frac{\pi}{4}$,we have $r = l \sin(\alpha) = l \sin(\frac{\pi}{4}) = \frac{l}{\sqrt{2}}$.
Substituting $r$ into the surface area formula: $s = \pi (\frac{l}{\sqrt{2}}) l = \frac{\pi}{\sqrt{2}} l^2$.
Differentiating with respect to time $t$: $\frac{ds}{dt} = \frac{\pi}{\sqrt{2}} \cdot 2l \cdot \frac{dl}{dt} = \sqrt{2} \pi l \frac{dl}{dt}$.
Given $\frac{ds}{dt} = -2 \text{ cm}^2/\text{sec}$ (since the area is decreasing),we have $-2 = \sqrt{2} \pi l \frac{dl}{dt}$.
At $l = 4 \text{ cm}$,$-2 = \sqrt{2} \pi (4) \frac{dl}{dt}$.
Solving for $\frac{dl}{dt}$: $\frac{dl}{dt} = \frac{-2}{4 \sqrt{2} \pi} = -\frac{1}{2 \sqrt{2} \pi} = -\frac{\sqrt{2}}{4 \pi} \text{ cm/sec}$.
Thus,the rate of decrease of the slant height is $\frac{\sqrt{2}}{4 \pi} \text{ cm/sec}$.