The oil is leaking from a drum at a rate of `16 cm^(3)//sec`. If the radius and height of drum are 7 cm and 60 cm respectively, find the rate of change of the height of oil when height of oil in drum is 18 cm.

Answer

To solve the problem, we need to find the rate of change of the height of oil in a cylindrical drum as oil is leaking out. We will use the concept of related rates in calculus.

### Step 1: Identify the variables
Let:
- \( V \) be the volume of oil in the drum (in cm³).
- \( h \) be the height of the oil in the drum (in cm).
- \( r \) be the radius of the drum (in cm), which is constant at 7 cm.

### Step 2: Write the formula for the volume of a cylinder
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
Since the radius \( r \) is constant (7 cm), we can substitute it into the formula:
\[ V = \pi (7^2) h = 49\pi h \]

### Step 3: Differentiate the volume with respect to time
We need to find the rate of change of height \( h \) with respect to time \( t \). We will differentiate both sides of the volume equation with respect to \( t \):
\[ \frac{dV}{dt} = 49\pi \frac{dh}{dt} \]

### Step 4: Substitute known values
We know:
- The rate at which oil is leaking (the rate of change of volume) is \( \frac{dV}{dt} = -16 \, \text{cm}^3/\text{s} \) (negative because the volume is decreasing).
- We need to find \( \frac{dh}{dt} \) when the height \( h = 18 \, \text{cm} \).

Substituting into the differentiated equation:
\[ -16 = 49\pi \frac{dh}{dt} \]

### Step 5: Solve for \( \frac{dh}{dt} \)
Rearranging the equation to solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{-16}{49\pi} \]

### Step 6: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ \frac{dh}{dt} = \frac{-16}{49 \times 3.14} \approx \frac{-16}{153.86} \approx -0.104 \]
Thus,
\[ \frac{dh}{dt} \approx -0.104 \, \text{cm/s} \]

### Final Answer
The rate of change of the height of oil when the height of oil in the drum is 18 cm is approximately \(-0.104 \, \text{cm/s}\).

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