An aeroplane is flying horizontally at a height of `2/3k m` with a velocity of 15 km/h. Find the rate at which it is receding from a fixed point on the ground which it passed over 2 min ago.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
The aeroplane is flying horizontally at a height of \( \frac{2}{3} \) km with a velocity of 15 km/h. We need to find the rate at which it is receding from a fixed point on the ground, which it passed over 2 minutes ago.

### Step 2: Convert time into hours
Since the velocity is given in km/h, we need to convert the time from minutes to hours.
- 2 minutes = \( \frac{2}{60} \) hours = \( \frac{1}{30} \) hours.

### Step 3: Calculate the horizontal distance traveled in 2 minutes
Using the speed of the aeroplane:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 15 \text{ km/h} \times \frac{1}{30} \text{ h} = \frac{15}{30} = \frac{1}{2} \text{ km}. \]
So, the horizontal distance \( x \) from the fixed point is \( \frac{1}{2} \) km.

### Step 4: Set up the relationship using the Pythagorean theorem
Let \( l \) be the distance from the aeroplane to the fixed point on the ground. The height \( h \) of the aeroplane is \( \frac{2}{3} \) km, and the horizontal distance \( x \) is \( \frac{1}{2} \) km. According to the Pythagorean theorem:
\[ l^2 = h^2 + x^2. \]
Substituting the values:
\[ l^2 = \left(\frac{2}{3}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{4}{9} + \frac{1}{4}. \]
To add these fractions, we need a common denominator, which is 36:
\[ l^2 = \frac{4 \times 4}{36} + \frac{1 \times 9}{36} = \frac{16}{36} + \frac{9}{36} = \frac{25}{36}. \]
Thus,
\[ l = \sqrt{\frac{25}{36}} = \frac{5}{6} \text{ km}. \]

### Step 5: Differentiate the equation
We need to find the rate of change of \( l \) with respect to time \( t \). Differentiating \( l^2 = h^2 + x^2 \) with respect to \( t \):
\[ \frac{d}{dt}(l^2) = \frac{d}{dt}(h^2) + \frac{d}{dt}(x^2). \]
Since \( h \) is constant (the height of the aeroplane does not change), \( \frac{d}{dt}(h^2) = 0 \):
\[ 2l \frac{dl}{dt} = 2x \frac{dx}{dt}. \]
We can simplify this to:
\[ l \frac{dl}{dt} = x \frac{dx}{dt}. \]

### Step 6: Substitute known values
We know:
- \( l = \frac{5}{6} \) km,
- \( x = \frac{1}{2} \) km,
- \( \frac{dx}{dt} = 15 \) km/h (the speed of the aeroplane).

Substituting these values into the equation:
\[ \frac{5}{6} \frac{dl}{dt} = \frac{1}{2} \times 15. \]
Calculating the right side:
\[ \frac{5}{6} \frac{dl}{dt} = \frac{15}{2}. \]
To isolate \( \frac{dl}{dt} \):
\[ \frac{dl}{dt} = \frac{15}{2} \cdot \frac{6}{5} = \frac{90}{10} = 9 \text{ km/h}. \]

### Final Answer
The rate at which the aeroplane is receding from the fixed point on the ground is \( 9 \) km/h.

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