Water is being poured at the rate of `36" m"^(3)//"sec"` in a cylindrical vessel of base radius 3 metres. Find the rate at which water level is rising.

Answer

To solve the problem of finding the rate at which the water level is rising in a cylindrical vessel, we can follow these steps:

### Step 1: Understand the given information
We are given:
- The rate at which water is being poured into the cylindrical vessel: \( \frac{dV}{dt} = 36 \, \text{m}^3/\text{s} \)
- The radius of the base of the cylinder: \( r = 3 \, \text{m} \)

### Step 2: Write the formula for the volume of a cylinder
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height of the water in the cylinder.

### Step 3: Differentiate the volume with respect to time
To find the rate at which the height \( h \) is changing, we differentiate the volume \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]

### Step 4: Substitute the known values
We know that \( r = 3 \, \text{m} \), so we can substitute this into the equation:
\[ \frac{dV}{dt} = \pi (3^2) \frac{dh}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 9\pi \frac{dh}{dt} \]

### Step 5: Substitute the rate of volume change
Now, we substitute \( \frac{dV}{dt} = 36 \, \text{m}^3/\text{s} \) into the equation:
\[ 36 = 9\pi \frac{dh}{dt} \]

### Step 6: Solve for \( \frac{dh}{dt} \)
To find \( \frac{dh}{dt} \), we rearrange the equation:
\[ \frac{dh}{dt} = \frac{36}{9\pi} = \frac{4}{\pi} \]

### Conclusion
Thus, the rate at which the water level is rising is:
\[ \frac{dh}{dt} = \frac{4}{\pi} \, \text{m/s} \]