From a cylindrical drum containing oil and kept vertical, the oil leaking at the rate of `10cm^(3)//s.` If the radius of the durm is 10 cm and height is 50 cm, then find the rate at which level of oil is changing when oil level is 20cm.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the problem
We have a cylindrical drum with a radius of 10 cm and a height of 50 cm. Oil is leaking out of the drum at a rate of 10 cm³/s. We need to find the rate at which the height of the oil level is changing when the height of the oil is 20 cm.

### Step 2: Write the formula for the volume of the 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 oil.

### Step 3: Differentiate the volume with respect to time
Since the volume of oil is changing as it leaks out, we differentiate the volume \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]
Here, \( \frac{dV}{dt} \) is the rate of change of volume, and \( \frac{dh}{dt} \) is the rate of change of height.

### Step 4: Substitute known values
We know:
- The radius \( r = 10 \) cm
- The rate of change of volume \( \frac{dV}{dt} = -10 \) cm³/s (negative because the volume is decreasing)
- We need to find \( \frac{dh}{dt} \) when the height \( h = 20 \) cm.

Substituting the known values into the differentiated volume equation:
\[ -10 = \pi (10^2) \frac{dh}{dt} \]
This simplifies to:
\[ -10 = 100\pi \frac{dh}{dt} \]

### Step 5: Solve for \( \frac{dh}{dt} \)
Now, we can solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{-10}{100\pi} = \frac{-1}{10\pi} \]

### Step 6: Conclusion
Thus, the rate at which the level of oil is changing when the oil level is 20 cm is:
\[ \frac{dh}{dt} = \frac{-1}{10\pi} \text{ cm/s} \]