To solve this problem, we will use related rates involving trigonometric functions. Here’s the step-by-step solution:

1. Define the variables:

   • Let $x$ be the distance from the bottom of the ladder to the wall (horizontal distance).
   • Let $y$ be the height of the ladder on the wall (vertical distance).
   • Let $\theta$ be the angle between the ladder and the ground.
   • Given: $\frac{dx}{dt} = 0.5 \text{ ft/s}$, $x = 6 \text{ ft}$, and the ladder length is 10 ft (constant).

2. Apply the Pythagorean theorem: Since the ladder length is constant, we have: $x^2 + y^2 = 10^2 \implies y^2 = 100 - x^2.$ When $x = 6 \text{ ft}$, we find $y$: $y^2 = 100 - 6^2 = 100 - 36 = 64 \implies y = 8 \text{ ft}.$

3. Relate the angle $\theta$ with $x$ and $y$: Using the trigonometric relationship: $\cos(\theta) = \frac{y}{10}.$

4. Differentiate with respect to time $t$: Using the chain rule: $\frac{d}{dt}[\cos(\theta)] = \frac{d}{dt}\left(\frac{y}{10}\right),$ $-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{10} \frac{dy}{dt}.$

5. Find $\sin(\theta)$: $\sin(\theta) = \frac{x}{10} = \frac{6}{10} = 0.6.$

6. Find $\frac{dy}{dt}$ using the derivative of the Pythagorean theorem: Differentiating $x^2 + y^2 = 100$ with respect to time: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0.$ Substituting $x = 6$, $y = 8$, and $\frac{dx}{dt} = 0.5 \text{ ft/s}$: $2(6)(0.5) + 2(8) \frac{dy}{dt} = 0 \implies 6 + 16 \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{6}{16} = -0.375 \text{ ft/s}.$

7. Substitute $\sin(\theta)$ and $\frac{dy}{dt}$ to find $\frac{d\theta}{dt}$: $-0.6 \frac{d\theta}{dt} = \frac{-0.375}{10} \implies \frac{d\theta}{dt} = \frac{0.375}{10 \times 0.6} = \frac{0.375}{6} = 0.0625 \text{ rad/s}.$

Therefore, the rate of change of the angle when the bottom of the ladder is 6 ft from the wall is approximately $0.06 \text{ rad/s}$.

Would you like a more detailed explanation or have any other questions?

Here are five related questions to consider:

1. How does changing the length of the ladder affect the rate of change of the angle?
2. What happens to the rate of change of the angle as the bottom of the ladder gets closer to the wall?
3. How would the problem change if the ladder was sliding at a different rate, such as 1 ft/s?
4. Can we find the rate of change of the height of the ladder at the same moment?
5. What would happen if the ladder were not straight but had a bend in it?

Tip: When dealing with related rates problems, always differentiate equations with respect to time and relate variables using given relationships or physical constraints.