Oil is leaking at the rate of 16 mL/s from a vertically kept cylindrical drum containing oil. If the radius of the drum is 7 cm and its height is 60 cm, find the rate at which the level of the oil is changing when the oil level is 18 cm

Explanation

To solve the problem step by step, we need to find the rate at which the height of the oil in the cylindrical drum is changing when the oil level is at 18 cm.

### Step 1: Identify the given information
- Rate of oil leaking, \( \frac{dV}{dt} = -16 \) mL/s (negative because the volume is decreasing)
- Radius of the cylinder, \( r = 7 \) cm
- Height of the cylinder, \( h = 60 \) cm (not directly needed for this calculation)
- Height of the oil when we need to find the rate of change, \( h = 18 \) 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 radius \( r \) is constant, we can differentiate the volume with respect to time \( t \):
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]

### Step 4: Substitute the known values into the differentiated equation
We know \( \frac{dV}{dt} = -16 \) mL/s and \( r = 7 \) cm. First, we need to convert the volume rate from mL to cm³ since \( 1 \text{ mL} = 1 \text{ cm}^3 \):
\[ \frac{dV}{dt} = -16 \text{ cm}^3/\text{s} \]
Now substituting the values into the equation:
\[ -16 = \pi (7^2) \frac{dh}{dt} \]
\[ -16 = \pi (49) \frac{dh}{dt} \]

### Step 5: Solve for \( \frac{dh}{dt} \)
Now we can isolate \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{-16}{49\pi} \]

### Step 6: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ \frac{dh}{dt} \approx \frac{-16}{49 \times 3.14} \approx \frac{-16}{153.86} \approx -0.104 \]
Thus, the rate at which the height of the oil is changing when the oil level is at 18 cm is approximately:
\[ \frac{dh}{dt} \approx -0.104 \text{ cm/s} \]

### Final Answer
The height of the oil is decreasing at a rate of approximately \( 0.104 \) cm/s when the oil level is at 18 cm.

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