A water tank has the shape of an inverted right - circular cone with its axis vertical and vertex lower most. Its semi - vertical angle is `tan^(-1)((1)/(2))`. Water is poured into it at a constant rate of 5 cubic meter per minute. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 10 m.

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript while providing a clear mathematical explanation.

### Step 1: Understand the Geometry of the Cone
The water tank is in the shape of an inverted right circular cone. The semi-vertical angle \( \alpha \) is given as \( \tan^{-1}\left(\frac{1}{2}\right) \). This means that the relationship between the radius \( r \) and height \( h \) of the cone can be established using the tangent function.

### Step 2: Establish the Relationship Between Radius and Height
From the definition of tangent, we have:
\[ \tan \alpha = \frac{r}{h} \]
Given \( \alpha = \tan^{-1}\left(\frac{1}{2}\right) \), we can write:
\[ \tan \alpha = \frac{1}{2} \implies \frac{r}{h} = \frac{1}{2} \]
This implies:
\[ r = \frac{1}{2} h \]

### Step 3: Volume of the Cone
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r = \frac{1}{2} h \) into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{1}{2} h\right)^2 h = \frac{1}{3} \pi \frac{1}{4} h^2 h = \frac{\pi}{12} h^3 \]

### Step 4: Differentiate the Volume with Respect to Time
We need to find the rate at which the height \( h \) is rising, \( \frac{dh}{dt} \). To do this, we differentiate the volume \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi}{12} h^3\right) = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt} \]

### Step 5: Set Up the Equation
We know that water is poured into the tank at a constant rate of \( \frac{dV}{dt} = 5 \) cubic meters per minute. Therefore, we can set up the equation:
\[ 5 = \frac{\pi}{4} h^2 \frac{dh}{dt} \]

### Step 6: Solve for \( \frac{dh}{dt} \)
Rearranging the equation to solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{5 \cdot 4}{\pi h^2} = \frac{20}{\pi h^2} \]

### Step 7: Substitute the Value of \( h \)
We need to find \( \frac{dh}{dt} \) when the depth of water \( h = 10 \) m:
\[ \frac{dh}{dt} = \frac{20}{\pi (10)^2} = \frac{20}{100\pi} = \frac{1}{5\pi} \]

### Final Answer
The rate at which the level of the water is rising when the depth is 10 m is:
\[ \frac{dh}{dt} = \frac{1}{5\pi} \text{ meters per minute} \]