A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
Find the rate (in rad/sec) at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall.


This is a related rates problem involving trigonometry and differentiation. The given variables are:

- The ladder length \( L = 25 \) feet (which remains constant),
- The base of the ladder is moving away from the wall at \( \frac{dx}{dt} = 2 \) feet per second,
- We need to find the rate at which the angle between the ladder and the wall is changing, denoted as \( \frac{d\theta}{dt} \), when the base is 24 feet away from the wall.

Let’s break this down.

### Step 1: Establish the relationship
Using trigonometry, the cosine of the angle \( \theta \) between the ladder and the wall can be described as:
\[
\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{L}
\]
Where:
- \( x \) is the distance of the base of the ladder from the wall,
- \( L = 25 \) is the length of the ladder.

Thus,
\[
\cos(\theta) = \frac{x}{25}
\]

### Step 2: Differentiate both sides with respect to time \( t \)
We now differentiate implicitly with respect to \( t \), using the chain rule. Remember that \( x \) is changing with respect to time.

\[
\frac{d}{dt} \left( \cos(\theta) \right) = \frac{d}{dt} \left( \frac{x}{25} \right)
\]

This gives:
\[
-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{25} \frac{dx}{dt}
\]

### Step 3: Plug in the known values
We know:
- \( \frac{dx}{dt} = 2 \) feet per second,
- When \( x = 24 \) feet.

To proceed, we need to find \( \sin(\theta) \) when \( x = 24 \). Using the Pythagorean theorem, the height of the ladder \( y \) can be calculated as:
\[
y^2 + x^2 = L^2
\]
\[
y^2 + 24^2 = 25^2
\]
\[
y^2 + 576 = 625
\]
\[
y^2 = 49
\]
\[
y = 7 \, \text{feet}
\]

Thus, \( \sin(\theta) \) is:
\[
\sin(\theta) = \frac{y}{L} = \frac{7}{25}
\]

### Step 4: Solve for \( \frac{d\theta}{dt} \)
Now, plug all the known values into the differentiated equation:
\[
-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{25} \times 2
\]
\[
-\frac{7}{25} \frac{d\theta}{dt} = \frac{2}{25}
\]

Solving for \( \frac{d\theta}{dt} \):
\[
\frac{d\theta}{dt} = -\frac{2}{7} \, \text{rad/sec}
\]

### Final Answer:
The angle between the ladder and the wall is decreasing at a rate of \( \frac{2}{7} \, \text{rad/sec} \) when the base of the ladder is 24 feet from the wall.

Would you like further explanation on any part? 

Here are 5 related questions to expand your understanding:
1. How does the rate change if the base of the ladder is moving faster?
2. How would the solution differ if the ladder's length changes over time?
3. What happens to \( \frac{d\theta}{dt} \) as \( x \) approaches 25 feet?
4. Can you derive the rate using sine instead of cosine?
5. What is the physical meaning of \( \frac{d\theta}{dt} \) being negative?

**Tip:** Always check the geometry of the problem before setting up related rates equations, especially if a triangle is involved!

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. Find the rate (in rad/sec) at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall.

This problem explores related rates in trigonometry, where a 25-foot ladder leans against a wall and its base moves at 2 feet per second. Learn how to calculate the rate at which the angle between the ladder and wall changes, applying concepts like implicit differentiation, the Pythagorean theorem, and trigonometric functions.

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