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To solve this problem, let’s break down the information given and the relationships between variables.

### Problem Analysis
1. **Given Information:**
   - The kite is flying at a height of \( 70 \) feet.
   - The kite is moving horizontally away from the child at a rate of \( \frac{dx}{dt} = 25 \) ft/sec.
   - We need to find the rate at which the child must let out the string, \( \frac{dy}{dt} \), when the distance \( y = 250 \) feet.

2. **Variables and Relationships:**
   - Let \( x \) be the horizontal distance between the child and the kite.
   - Let \( y \) be the hypotenuse, or the string length, which is the direct distance between the child and the kite.

Since the kite is flying horizontally, we can use the Pythagorean theorem to relate \( x \), \( y \), and the fixed height of the kite (70 feet):
\[
y^2 = x^2 + 70^2
\]

### Differentiation
To find the relationship between \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), we differentiate both sides of the equation with respect to \( t \):
\[
2y \frac{dy}{dt} = 2x \frac{dx}{dt}
\]
This simplifies to:
\[
y \frac{dy}{dt} = x \frac{dx}{dt}
\]
or
\[
\frac{dy}{dt} = \frac{x}{y} \frac{dx}{dt}
\]

### Substitution
1. When \( y = 250 \) feet, we can use the Pythagorean theorem to find \( x \):
   \[
   250^2 = x^2 + 70^2
   \]
   \[
   x^2 = 250^2 - 70^2
   \]
   \[
   x = \sqrt{250^2 - 70^2} = \sqrt{62500 - 4900} = \sqrt{57600} = 240 \text{ feet}
   \]

2. Now we substitute \( x = 240 \), \( y = 250 \), and \( \frac{dx}{dt} = 25 \) ft/sec into the differentiated equation:
   \[
   \frac{dy}{dt} = \frac{240}{250} \cdot 25
   \]
   Simplifying,
   \[
   \frac{dy}{dt} = \frac{24}{25} \cdot 25 = 24 \text{ ft/sec}
   \]

### Answer
The rate at which the child must let out the string is **24 ft/sec**.

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Do you have any questions or would you like more details?

### Related Questions
1. What would the rate \( \frac{dy}{dt} \) be if the kite’s height changed?
2. How does the string length change if the kite moves at a different horizontal speed?
3. What happens to \( \frac{dy}{dt} \) if the distance \( y \) is greater than 250 feet?
4. How does the rate \( \frac{dy}{dt} \) vary with different heights of the kite?
5. Can we use the same differentiation approach if the kite’s path isn’t horizontal?

### Tip
Remember that in related rates problems, identifying and clearly defining each variable and its relationship is crucial to setting up the correct equations.

A child flies a kite at a height of 70 ft, the wind carrying the kite horizontally away from the child at a rate of 25 ft/sec. How fast must the child let out the string when the kite is 250 ft away from the child? Let x represent the horizontal distance through which the kite is blown and let y represent the distance between the child and the kite. Write the equation that relates these rates.

Learn how to solve a related rates problem involving a kite flying at a height of 70 ft, moving horizontally at 25 ft/sec. Determine the rate at which the string is let out when the kite is 250 ft away. This solution includes the Pythagorean theorem and differentiation.

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