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To solve for \(\frac{dy}{dt}\) in this related rates problem, let’s break down the information and set up the necessary equations.

### Given Information:
1. The height of the kite above the child is \(70\) feet (constant).
2. The horizontal distance \(x\) is changing as the kite moves horizontally away from the child.
3. The rate at which the horizontal distance \(x\) is increasing (i.e., \(\frac{dx}{dt}\)) is \(25\) ft/sec.
4. We need to find \(\frac{dy}{dt}\), the rate at which the distance \(y\) (the length of the string) is changing, when the kite is \(250\) feet away from the child.

### Step-by-Step Solution

1. **Set up the relationship between \(x\), \(y\), and the height of the kite:**
   Since the kite is flying horizontally and maintaining a constant height, this forms a right triangle, where:
   - \(x\) is the horizontal distance from the child to the kite.
   - \(y\) is the hypotenuse (distance between the child and the kite).
   - \(70\) feet is the vertical side.

   Using the Pythagorean theorem:
   \[
   y^2 = x^2 + 70^2
   \]

2. **Differentiate both sides with respect to time \(t\):**
   Applying implicit differentiation:
   \[
   2y \cdot \frac{dy}{dt} = 2x \cdot \frac{dx}{dt}
   \]

3. **Solve for \(\frac{dy}{dt}\):**
   Divide both sides by \(2y\):
   \[
   \frac{dy}{dt} = \frac{x}{y} \cdot \frac{dx}{dt}
   \]

4. **Substitute known values:**
   - When \(y = 250\) feet (distance from the child to the kite),
   - \(dx/dt = 25\) ft/sec,
   - Use the Pythagorean theorem to find \(x\) when \(y = 250\):
     \[
     x^2 + 70^2 = 250^2
     \]
     \[
     x^2 + 4900 = 62500
     \]
     \[
     x^2 = 57600
     \]
     \[
     x = 240 \text{ feet}
     \]

5. **Calculate \(\frac{dy}{dt}\):**
   Substitute \(x = 240\), \(y = 250\), and \(\frac{dx}{dt} = 25\):
   \[
   \frac{dy}{dt} = \frac{240}{250} \cdot 25
   \]
   Simplify:
   \[
   \frac{dy}{dt} = 0.96 \cdot 25
   \]
   \[
   \frac{dy}{dt} = 24 \text{ ft/sec}
   \]

### Answer
The rate at which the child must let out the string when the kite is \(250\) feet away is:
\[
\frac{dy}{dt} = 24 \text{ ft/sec}
\]

Would you like further details or have any questions?

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### Follow-Up Questions:
1. How would the solution change if the height of the kite varied?
2. What if the horizontal distance were increasing at a different rate?
3. How would you solve this problem if given only angles and not distances?
4. How does implicit differentiation help in solving related rates problems?
5. Can you generalize this solution for any height \(h\) and distance \(y\)?

### Tip:
Always start related rates problems by identifying all given rates and writing an equation that links the variables involved before differentiating.

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?

Learn how to solve a related rates problem involving a kite flown at a height of 70 ft, moving horizontally at 25 ft/sec. Calculate the rate of the string being let out when the kite is 250 ft from the child. Suitable for Grades 10-12.

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