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To solve this problem, we need to use related rates. We are given:

1. A 10 ft ladder, which forms a right triangle with the wall and the ground.
2. The ladder's bottom moves away from the wall at a rate of \( \frac{dx}{dt} = 0.8 \) ft/s.
3. We are asked to find the rate at which the angle \( \theta \) between the ladder and the ground changes, \( \frac{d\theta}{dt} \), when the bottom of the ladder is 6 ft away from the wall (i.e., \( x = 6 \)).

### Step-by-Step Solution:

#### Step 1: Set up the relationship between the variables
Using the Pythagorean theorem:
\[
x^2 + y^2 = 10^2
\]
Since the ladder is 10 ft, \( x^2 + y^2 = 100 \).

#### Step 2: Differentiate with respect to \( t \)
To find \( \frac{d\theta}{dt} \), we use trigonometry. From the triangle, we have:
\[
\cos \theta = \frac{x}{10}
\]
Now differentiate both sides with respect to \( t \):
\[
-\sin \theta \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot \frac{dx}{dt}
\]
Thus,
\[
\frac{d\theta}{dt} = -\frac{1}{10} \cdot \frac{\frac{dx}{dt}}{\sin \theta}
\]

#### Step 3: Find \( \sin \theta \) when \( x = 6 \)
When \( x = 6 \):
\[
x^2 + y^2 = 100 \Rightarrow 6^2 + y^2 = 100 \Rightarrow 36 + y^2 = 100 \Rightarrow y^2 = 64 \Rightarrow y = 8
\]
Now, \( \sin \theta = \frac{y}{10} = \frac{8}{10} = 0.8 \).

#### Step 4: Substitute values to find \( \frac{d\theta}{dt} \)
Now we have:
\[
\frac{d\theta}{dt} = -\frac{1}{10} \cdot \frac{0.8}{0.8} = -\frac{1}{10} = -0.1 \text{ rad/s}
\]

### Final Answer
The rate at which the angle between the ladder and the ground is changing when the bottom of the ladder is 6 ft from the wall is:
\[
\frac{d\theta}{dt} = -0.1 \text{ rad/s}
\]

Would you like further details on this solution or have any questions?

Here are some related questions that build on this problem:

1. What would happen to \( \frac{d\theta}{dt} \) if the ladder length were different?
2. How would the answer change if the rate at which the ladder moves were doubled?
3. What would be \( \frac{d\theta}{dt} \) if \( x \) were initially at 8 feet instead of 6?
4. How does the angle between the ladder and the ground change as the ladder moves further from the wall?
5. What if we wanted to find the rate of change of \( y \) instead of \( \theta \)?

**Tip:** In related rates problems, identifying and setting up the relationships between variables early simplifies the differentiation process.

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.8 ft/s, how fast (in rad/s) is the angle (in radians) between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall?

Learn how to solve a classic related rates problem involving a 10 ft ladder sliding away from a wall. Find out how fast the angle between the ladder and the ground is changing when the ladder's base is 6 ft from the wall. This problem covers key calculus concepts such as differentiation, trigonometry, and the Pythagorean theorem, making it ideal for Grade 12 and college-level students.

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