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 1.2 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? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

To solve this problem, we will use related rates in calculus, specifically focusing on the relationship between the angle \(\theta\) that the ladder makes with the ground, the distance of the bottom of the ladder from the wall (\(x\)), and the length of the ladder itself.

### Step 1: Relationship Setup
The ladder forms a right triangle with the wall and the ground. Let:
- \(x\) be the distance from the wall to the bottom of the ladder,
- \(y\) be the distance from the top of the ladder to the ground (height of the ladder on the wall),
- \(L\) be the length of the ladder (10 ft).

From the Pythagorean theorem:
\[
x^2 + y^2 = L^2 \implies x^2 + y^2 = 10^2 = 100.
\]

### Step 2: Find \(y\) When \(x = 6\) ft
Plug \(x = 6\) ft into the Pythagorean theorem to solve for \(y\):
\[
6^2 + y^2 = 100 \implies 36 + y^2 = 100 \implies y^2 = 64 \implies y = 8 \text{ ft}.
\]

### Step 3: Relate \(\theta\), \(x\), and \(y\)
We use the trigonometric relationship:
\[
\cos(\theta) = \frac{x}{L} = \frac{x}{10}.
\]

### Step 4: Differentiate with Respect to Time
Differentiate both sides with respect to time \(t\):
\[
-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}.
\]

Given that \(\frac{dx}{dt} = 1.2\) ft/s (the rate at which the bottom of the ladder slides away from the wall), we can now find \(\sin(\theta)\).

### Step 5: Find \(\sin(\theta)\)
When \(x = 6\) ft and \(y = 8\) ft, we have:
\[
\sin(\theta) = \frac{y}{L} = \frac{8}{10} = 0.8.
\]

### Step 6: Solve for \(\frac{d\theta}{dt}\)
Substitute the values into the differentiated equation:
\[
-0.8 \frac{d\theta}{dt} = \frac{1.2}{10}.
\]
Thus,
\[
\frac{d\theta}{dt} = -\frac{1.2}{10 \times 0.8} = -\frac{1.2}{8} = -0.15 \text{ rad/s}.
\]

The negative sign indicates that the angle \(\theta\) is decreasing. Therefore, the angle between the ladder and the ground is changing at a rate of \(0.15\) rad/s when the bottom of the ladder is 6 ft from the wall.

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### Related Questions:
1. How would the rate change if the ladder were 12 ft long instead of 10 ft?
2. What happens to the rate of change of the angle as the ladder becomes more horizontal?
3. How do we find the rate of change of the height of the ladder against the wall?
4. What if the rate of sliding away from the wall was changing with time?
5. How would the solution differ if the ladder length changed over time?

### Tip:
When solving related rates problems, identify the changing quantities, establish a relationship between them, and differentiate with respect to time to find the desired rate.

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 1.2 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?

Explore how to calculate the rate at which the angle between a 10 ft ladder and the ground is changing when the ladder slides away from a vertical wall at a rate of 1.2 ft/s. This problem applies the Pythagorean theorem, trigonometry, and related rates in calculus, providing step-by-step guidance for advanced high school students.

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