Water is dripping out of a conical funnel of semi-vertical angle `45^@` at rate of `2(cm^3)/s`. Find the rate at which slant height of water is decreasing when the height of water is `sqrt(2)` cm.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a conical funnel with a semi-vertical angle of \(45^\circ\). Water is dripping out at a rate of \(2 \, \text{cm}^3/\text{s}\). We need to find the rate at which the slant height of the water is decreasing when the height of the water is \(\sqrt{2} \, \text{cm}\).

### Step 2: Set up the variables
Let:
- \(V\) = volume of water in the cone
- \(r\) = radius of the water surface
- \(h\) = height of the water
- \(L\) = slant height of the water

Given the semi-vertical angle is \(45^\circ\), we can relate \(r\) and \(h\):
\[ \tan(45^\circ) = \frac{h}{r} \implies h = r \]

### Step 3: Volume of the cone
The volume \(V\) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \(h = r\) into the volume formula:
\[ V = \frac{1}{3} \pi r^2 r = \frac{1}{3} \pi r^3 \]

### Step 4: Differentiate the volume with respect to time
We need to find \(\frac{dV}{dt}\):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{3} \pi r^3\right) = \pi r^2 \frac{dr}{dt} \]

### Step 5: Relate the rates of change
From the problem, we know:
\[ \frac{dV}{dt} = -2 \, \text{cm}^3/\text{s} \quad (\text{negative because the volume is decreasing}) \]
Thus:
\[ -2 = \pi r^2 \frac{dr}{dt} \]

### Step 6: Find \(r\) when \(h = \sqrt{2}\)
Since \(h = r\) and at this moment \(h = \sqrt{2}\):
\[ r = \sqrt{2} \]

### Step 7: Substitute \(r\) into the equation
Substituting \(r = \sqrt{2}\) into the equation:
\[ -2 = \pi (\sqrt{2})^2 \frac{dr}{dt} \implies -2 = 2\pi \frac{dr}{dt} \]
Thus,
\[ \frac{dr}{dt} = -\frac{1}{\pi} \, \text{cm/s} \]

### Step 8: Find the slant height \(L\)
The slant height \(L\) is given by:
\[ L = \sqrt{r^2 + h^2} = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \, \text{cm} \]

### Step 9: Differentiate the slant height with respect to time
Using the relationship:
\[ L = \sqrt{r^2 + h^2} \implies \frac{dL}{dt} = \frac{1}{2\sqrt{r^2 + h^2}}(2r\frac{dr}{dt} + 2h\frac{dh}{dt}) \]
Since \(h = r\), we have \(\frac{dh}{dt} = \frac{dr}{dt}\):
\[ \frac{dL}{dt} = \frac{1}{2\sqrt{4}}(2r\frac{dr}{dt} + 2h\frac{dr}{dt}) = \frac{1}{4}(2\sqrt{2}\frac{dr}{dt} + 2\sqrt{2}\frac{dr}{dt}) = \frac{1}{4}(4\sqrt{2}\frac{dr}{dt}) = \sqrt{2}\frac{dr}{dt} \]

### Step 10: Substitute \(\frac{dr}{dt}\)
Now substituting \(\frac{dr}{dt} = -\frac{1}{\pi}\):
\[ \frac{dL}{dt} = \sqrt{2}\left(-\frac{1}{\pi}\right) = -\frac{\sqrt{2}}{\pi} \, \text{cm/s} \]

### Final Answer
The rate at which the slant height of the water is decreasing when the height of the water is \(\sqrt{2} \, \text{cm}\) is:
\[ \frac{dL}{dt} = -\frac{\sqrt{2}}{\pi} \, \text{cm/s} \]