A kite is 300 m high and there are 500 m of cord out. If the wind moves the kite horizontally at the rate of`5(km)/(hr) `directly away from the person who is flying it, find the rate at which the cord is being paid?

Options

A. 4
B. 8
C. 3
D. cannot be determined

Answer

To solve the problem step by step, we will use the Pythagorean theorem and the concept of related rates.

### Step 1: Understand the scenario
We have a kite flying at a height of 300 m, and the length of the cord is 500 m. The kite is moving horizontally at a rate of 5 km/hr away from the person flying it.

### Step 2: Define the variables
- Let \( y \) be the height of the kite, which is constant at 300 m.
- Let \( x \) be the horizontal distance from the person to the point directly below the kite.
- Let \( r \) be the length of the cord, which is 500 m.

### Step 3: Apply the Pythagorean theorem
According to the Pythagorean theorem:
\[ r^2 = x^2 + y^2 \]
Substituting the known values:
\[ 500^2 = x^2 + 300^2 \]

### Step 4: Calculate \( x \)
Calculating the squares:
\[ 250000 = x^2 + 90000 \]
Subtracting 90000 from both sides:
\[ x^2 = 250000 - 90000 = 160000 \]
Taking the square root:
\[ x = \sqrt{160000} = 400 \text{ m} \]

### Step 5: Differentiate with respect to time
Now we differentiate the equation \( r^2 = x^2 + y^2 \) with respect to time \( t \):
\[ \frac{d}{dt}(r^2) = \frac{d}{dt}(x^2 + y^2) \]
Using the chain rule:
\[ 2r \frac{dr}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]
Since the height \( y \) is constant, \( \frac{dy}{dt} = 0 \):
\[ 2r \frac{dr}{dt} = 2x \frac{dx}{dt} \]

### Step 6: Substitute known values
We know:
- \( r = 500 \) m
- \( x = 400 \) m
- \( \frac{dx}{dt} = 5 \) km/hr (which we need to convert to m/hr: \( 5 \times 1000 = 5000 \) m/hr)

Substituting these values into the differentiated equation:
\[ 2 \times 500 \times \frac{dr}{dt} = 2 \times 400 \times 5000 \]
Simplifying:
\[ 1000 \frac{dr}{dt} = 4000000 \]
Dividing both sides by 1000:
\[ \frac{dr}{dt} = 4000 \text{ m/hr} \]

### Step 7: Convert to km/hr
To convert \( \frac{dr}{dt} \) from m/hr to km/hr:
\[ \frac{dr}{dt} = \frac{4000}{1000} = 4 \text{ km/hr} \]

### Final Answer
The rate at which the cord is being paid out is \( 4 \text{ km/hr} \).

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