A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate (in rad/s) is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

We are given that a kite is flying 100 ft above the ground and moves horizontally at a speed of 11 ft/s. The problem asks us to find the rate at which the angle between the kite string and the horizontal is decreasing (in radians per second) when 200 ft of string have been let out.

### Step-by-step solution:

1. **Identify the variables:**
   - Let \(\theta\) be the angle between the string and the horizontal.
   - The length of the string is \( s = 200 \, \text{ft} \).
   - The height of the kite is \( h = 100 \, \text{ft} \).
   - The horizontal distance between the kite and the point on the ground where the string is anchored is \( x \).
   - The kite’s horizontal speed is given as \( \frac{dx}{dt} = 11 \, \text{ft/s} \).

2. **Relating the variables:**
   - From the geometry of the situation, we have a right triangle where:
     \[
     s^2 = x^2 + h^2
     \]
     Given that \( h = 100 \, \text{ft} \) and \( s = 200 \, \text{ft} \), this can be used to find \( x \) when 200 ft of string is let out:
     \[
     200^2 = x^2 + 100^2
     \]
     Solving for \( x \):
     \[
     40000 = x^2 + 10000
     \]
     \[
     x^2 = 30000
     \]
     \[
     x = \sqrt{30000} = 100\sqrt{3} \, \text{ft}
     \]

3. **Using trigonometry to relate \(\theta\) and \(x\):**
   The angle \(\theta\) can be expressed using the trigonometric relationship:
   \[
   \sin(\theta) = \frac{h}{s} = \frac{100}{200} = \frac{1}{2}
   \]
   Therefore:
   \[
   \cos(\theta) = \frac{x}{s} = \frac{100\sqrt{3}}{200} = \frac{\sqrt{3}}{2}
   \]

4. **Differentiate the equation with respect to time \( t \):**
   Using the equation \( x = s \cos(\theta) \), differentiate both sides with respect to \( t \):
   \[
   \frac{dx}{dt} = -s \sin(\theta) \frac{d\theta}{dt}
   \]
   Plugging in known values:
   \[
   11 = -200 \times \frac{1}{2} \times \frac{d\theta}{dt}
   \]
   \[
   11 = -100 \frac{d\theta}{dt}
   \]
   Solving for \( \frac{d\theta}{dt} \):
   \[
   \frac{d\theta}{dt} = -\frac{11}{100} \, \text{rad/s}
   \]
   So, the angle \(\theta\) is decreasing at a rate of \( \frac{11}{100} \, \text{rad/s} \).

### Final answer:
The rate at which the angle between the string and the horizontal is decreasing is \( \frac{11}{100} \, \text{rad/s} \).

Would you like further details or have any other questions?

Here are 5 related questions:
1. How would the solution change if the height of the kite were different?
2. What would happen if the kite's speed increased to 15 ft/s?
3. How would you approach this problem if the string were not straight?
4. Can you find the rate of change of the string length at a different point in time?
5. How does the horizontal distance change as the string length increases?

**Tip:** Always check units during differentiation steps to ensure consistency, especially when working with rates of change!

This problem explores how to find the rate at which the angle between a kite string and the horizontal is decreasing, using trigonometry and related rates. The kite is 100 ft above the ground, moving horizontally at 11 ft/s, and 200 ft of string is let out. Detailed solution included for Grades 10-12.

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