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: Understanding the Concept.
This is a related-rates problem: two quantities, the horizontal distance $x$ and the cord length $L$, are both changing with time, but they stay linked by a fixed right triangle because the kite's height never changes as the wind blows it away only horizontally. Once we know that fixed height, we can connect the rate of change of $x$ to the rate of change of $L$ using the triangle's angle instead of differentiating the Pythagorean relation directly.

Step 2: Key Formula or Approach.
First find the fixed height using the 30-40-50 right triangle: since $50^2 - 30^2 = 2500-900=1600=40^2$, the height is $h=40$ m.
Let $\theta$ be the angle the cord makes with the ground, so $\cos\theta = \dfrac{x}{L}$ (adjacent over hypotenuse) at any instant. At the given instant, $\cos\theta = \dfrac{30}{50} = \dfrac{3}{5}$.

Step 3: Detailed Explanation.
Because $h$ is fixed, the rate at which the cord length grows is the horizontal rate scaled down by the cosine of the angle the cord makes with the ground:
\[ \frac{dL}{dt} = \cos\theta \cdot \frac{dx}{dt} \]
This matches the usual $\dfrac{dL}{dt}=\dfrac{x}{L}\dfrac{dx}{dt}$ relation since $\cos\theta = x/L$.
Substituting the known values:
\[ \frac{dL}{dt} = \frac{3}{5} \times 5 = 3 \]

Step 4: Final Answer.
Using the angle of the cord with the ground instead of direct implicit differentiation gives the same result: the cord is let out at 3 km per hour. \[ \boxed{3 \text{ km per hour}} \]