Water is drpping out from a conical funnel at the uniform rate of `2cm^(3)//s` through a tiny hole at the vertex at the bottom. When the slant height of the water is 5 cm, find the rate of decrease of the slant higher of the water. Given that `alpha` is semi-vertical angle of the cone.

Answer

To solve the problem, we need to find the rate of decrease of the slant height \( l \) of the water in the conical funnel when the slant height is 5 cm. We are given that water is draining from the funnel at a rate of \( \frac{dV}{dt} = -2 \, \text{cm}^3/\text{s} \).

### Step 1: Understand the relationship between volume, height, and radius of the cone
The volume \( V \) of a cone is given by the formula:

\[ V = \frac{1}{3} \pi r^2 h \]

where \( r \) is the radius of the base and \( h \) is the height of the cone.

### Step 2: Relate radius and height to the slant height
For a cone, the radius \( r \), height \( h \), and slant height \( l \) are related by the following relationships, considering the semi-vertical angle \( \alpha \):

\[ r = l \sin \alpha \]
\[ h = l \cos \alpha \]

### Step 3: Substitute \( r \) and \( h \) into the volume formula
Substituting the expressions for \( r \) and \( h \) into the volume formula gives:

\[ V = \frac{1}{3} \pi (l \sin \alpha)^2 (l \cos \alpha) = \frac{1}{3} \pi l^2 \sin^2 \alpha (l \cos \alpha) = \frac{1}{3} \pi l^3 \sin^2 \alpha \cos \alpha \]

### Step 4: Differentiate the volume with respect to time
Now differentiate \( V \) with respect to time \( t \):

\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi l^3 \sin^2 \alpha \cos \alpha \right) \]

Using the chain rule, we have:

\[ \frac{dV}{dt} = \frac{1}{3} \pi \sin^2 \alpha \cos \alpha \cdot 3l^2 \frac{dl}{dt} \]

This simplifies to:

\[ \frac{dV}{dt} = \pi \sin^2 \alpha \cos \alpha \cdot l^2 \frac{dl}{dt} \]

### Step 5: Substitute known values and solve for \( \frac{dl}{dt} \)
We know \( \frac{dV}{dt} = -2 \, \text{cm}^3/\text{s} \) and \( l = 5 \, \text{cm} \). Substituting these values into the equation gives:

\[ -2 = \pi \sin^2 \alpha \cos \alpha \cdot (5^2) \frac{dl}{dt} \]

This simplifies to:

\[ -2 = 25\pi \sin^2 \alpha \cos \alpha \cdot \frac{dl}{dt} \]

Now, solving for \( \frac{dl}{dt} \):

\[ \frac{dl}{dt} = \frac{-2}{25\pi \sin^2 \alpha \cos \alpha} \]

### Step 6: Find the rate of decrease of the slant height
The rate of decrease of the slant height is given by the absolute value of \( \frac{dl}{dt} \):

\[ \text{Rate of decrease of slant height} = -\frac{dl}{dt} = \frac{2}{25\pi \sin^2 \alpha \cos \alpha} \]

### Final Answer
Thus, the rate of decrease of the slant height of the water when the slant height is 5 cm is:

\[ \frac{2}{25\pi \sin^2 \alpha \cos \alpha} \, \text{cm/s} \]