Question

The horizontal distance of a kite from the boy flying it is 30 m and 50 m of cord is out from the roll. If the wind moves the kite horizontally at the rate of 5 km per hour directly away from the boy, how fast is the cord being released?

• A. 3 km per hour
• B. 4 km per hour
• C. 5 km per hour
• D. 6 km per hour

Hint:

Find the fixed height using the 30-40-50 right triangle, then differentiate L squared = x squared + h squared with respect to time to relate dL/dt to dx/dt.

Answer 1

The Correct Option is A

Step 1: Find the constant height of the kite.
Let \(h\) be the height of the kite above the boy's hand, \(x\) be the horizontal distance from the boy, and \(L\) be the length of cord out. Since the wind only moves the kite horizontally, the height \(h\) stays fixed throughout. At the given instant, \(x=30\) m and \(L=50\) m, and these along with \(h\) form a right triangle with the cord as hypotenuse, so
\[ h^2 = L^2 - x^2 = 50^2 - 30^2 = 2500 - 900 = 1600 \]
\[ h = 40 \text{ m (constant for all time)} \]

Step 2: Write the relation between L and x at any instant.
Since \(h\) never changes, the right-triangle relation holds at every instant:
\[ L^2 = x^2 + h^2 = x^2 + 1600 \]

Step 3: Differentiate with respect to time.
Differentiating both sides with respect to time \(t\), remembering \(h\) is constant so its derivative is zero:
\[ 2L\frac{dL}{dt} = 2x\frac{dx}{dt} \]
\[ \frac{dL}{dt} = \frac{x}{L}\cdot\frac{dx}{dt} \]

Step 4: Substitute the known values.
At the instant in question, \(x=30\), \(L=50\), and \(\dfrac{dx}{dt}=5\) km per hour (the rate at which the kite moves away horizontally). Since \(x\) and \(L\) are used only as a ratio here, they can be substituted directly:
\[ \frac{dL}{dt} = \frac{30}{50}\times 5 = \frac{3}{5}\times 5 = 3 \]

Final Answer:
The cord is being released at a rate of 3 km per hour. \[ \boxed{3 \text{ km per hour}} \]