Water is dripping out from a conical funnel of semi-vertical angle `pi/4` at the uniform rate of `2cm^2//sec` in the surface through a tiny hole at the vertex of the bottom. When the slant height of the water level is 4 cm, find the rate of decrease of the slant heights of the water.

Answer

To solve the problem, we need to find the rate of decrease of the slant height of the water in a conical funnel when water is dripping out at a given rate. We will follow these steps:

### Step 1: Understand the Geometry of the Cone
Given that the semi-vertical angle \( \alpha = \frac{\pi}{4} \), we can relate the radius \( r \), height \( h \), and slant height \( l \) of the cone using trigonometric relationships.

### Step 2: Relate the Radius and Slant Height
From the semi-vertical angle, we have:
\[ \tan(\alpha) = \frac{r}{h} \]
Since \( \alpha = \frac{\pi}{4} \), we know that \( \tan\left(\frac{\pi}{4}\right) = 1 \). Thus, we can write:
\[ r = h \]
Also, using the Pythagorean theorem, we have:
\[ l = \sqrt{r^2 + h^2} \]
Substituting \( r = h \):
\[ l = \sqrt{h^2 + h^2} = \sqrt{2h^2} = h\sqrt{2} \]

### Step 3: Express the Surface Area
The surface area \( S \) of the cone is given by:
\[ S = \pi r l \]
Substituting \( r = h \) and \( l = h\sqrt{2} \):
\[ S = \pi h (h\sqrt{2}) = \pi \sqrt{2} h^2 \]

### Step 4: Differentiate the Surface Area with Respect to Time
We know that the rate of change of the surface area \( \frac{dS}{dt} \) is given as \( -2 \, \text{cm}^2/\text{s} \) (negative because the area is decreasing). Therefore, we differentiate:
\[ \frac{dS}{dt} = \frac{d}{dt}(\pi \sqrt{2} h^2) = 2\pi \sqrt{2} h \frac{dh}{dt} \]
Setting this equal to the rate of change of the surface area:
\[ 2\pi \sqrt{2} h \frac{dh}{dt} = -2 \]

### Step 5: Solve for \( \frac{dh}{dt} \)
Rearranging gives:
\[ \frac{dh}{dt} = \frac{-2}{2\pi \sqrt{2} h} = \frac{-1}{\pi \sqrt{2} h} \]

### Step 6: Substitute the Given Slant Height
We need to find \( \frac{dh}{dt} \) when the slant height \( l = 4 \, \text{cm} \). From our earlier relationship:
\[ l = h\sqrt{2} \implies h = \frac{l}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{cm} \]
Now substituting \( h \) into the equation for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{-1}{\pi \sqrt{2} (2\sqrt{2})} = \frac{-1}{4\pi} \]

### Final Result
Thus, the rate of decrease of the slant height \( \frac{dl}{dt} \) is:
\[ \frac{dl}{dt} = \frac{d}{dt}(h\sqrt{2}) = \sqrt{2} \frac{dh}{dt} = \sqrt{2} \left(\frac{-1}{4\pi}\right) = \frac{-\sqrt{2}}{4\pi} \, \text{cm/s} \]

### Summary
The rate of decrease of the slant height of the water is:
\[ \frac{dl}{dt} = \frac{-\sqrt{2}}{4\pi} \, \text{cm/s} \]