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The image shows a related rates problem involving a ladder leaning against a wall.

### Problem Summary:
- A ladder is 18 feet long and rests against a vertical wall.
- The base of the ladder slides away from the wall at a rate of 3 feet per second.
- We need to determine at what rate the angle between the ladder and the ground is decreasing when the base of the ladder is 8 feet from the wall.

The solution provided shows that the angle is decreasing at approximately \(-0.185 \, \text{rad/s}\).

### Steps to Solve the Problem:
1. **Variables**:
   - Let \( \theta \) be the angle between the ladder and the ground.
   - Let \( x \) be the distance from the wall to the base of the ladder.
   - Let \( y \) be the height of the ladder on the wall (the vertical distance from the top of the ladder to the ground).

2. **Relations**:
   - By the Pythagorean theorem: \( x^2 + y^2 = 18^2 \), since the length of the ladder is 18 feet.

3. **Rates**:
   - We are given \( \frac{dx}{dt} = 3 \, \text{ft/s} \).
   - We are asked to find \( \frac{d\theta}{dt} \) when \( x = 8 \, \text{ft} \).

4. **Relating \( \theta \) to \( x \)**:
   - Using trigonometry: \( \cos(\theta) = \frac{x}{18} \).
   - Differentiate this with respect to time \( t \):
     \[
     -\sin(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{18} \cdot \frac{dx}{dt}
     \]
   - This gives:
     \[
     \frac{d\theta}{dt} = -\frac{1}{18 \sin(\theta)} \cdot \frac{dx}{dt}
     \]

5. **Find \( \sin(\theta) \) when \( x = 8 \, \text{ft} \)**:
   - From the Pythagorean theorem: \( y = \sqrt{18^2 - 8^2} = \sqrt{324 - 64} = \sqrt{260} \approx 16.12 \, \text{ft} \).
   - So, \( \sin(\theta) = \frac{y}{18} \approx \frac{16.12}{18} \approx 0.8956 \).

6. **Calculate \( \frac{d\theta}{dt} \)**:
   \[
   \frac{d\theta}{dt} = -\frac{1}{18 \cdot 0.8956} \cdot 3 \approx -0.185 \, \text{rad/s}
   \]

Thus, the rate at which the angle is decreasing is \(-0.185 \, \text{rad/s}\), as shown in the image.

Do you want more details or have any questions about the steps?

### Relative Questions:
1. How does the Pythagorean theorem help in related rates problems involving triangles?
2. Why do we use trigonometric functions to relate angles and distances in this problem?
3. What would change in the problem if the ladder were shorter or longer than 18 feet?
4. How does the sign of \( \frac{d\theta}{dt} \) indicate whether the angle is increasing or decreasing?
5. How do you find the rate of change of a different quantity, like the height of the ladder on the wall?

### Tip:
In related rates problems, always focus on identifying the relationship between variables before differentiating with respect to time.

A 18-ft-long ladder rests against a vertical wall. If the base of the ladder slides away from the wall at 3 ft/s, at what rate is the angle between the ladder and the ground decreasing when the base of the ladder is 8 ft from the wall?

This problem explores how to calculate the rate at which the angle between a ladder and the ground is decreasing as the base of the ladder slides away from a wall. The ladder is 18 feet long, and the base moves at a rate of 3 ft/s. When the base is 8 feet from the wall, the angle decreases at a rate of -0.185 rad/s. The solution uses trigonometry, the Pythagorean theorem, and calculus to find the answer.

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