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The problem involves finding the rate of change of the angle of elevation (denoted as \( \theta \)) of an elevator as it rises at a constant speed. Let’s break it down step by step.

### Given:
- The observer is 24 m above the ground and is 20 m horizontally from the elevator shaft.
- The elevator rises at a constant rate of 3 m/s.
- The question asks for the rate of change of the angle of elevation when the elevator is 14 m above the ground.

We can solve this using related rates in calculus, considering the relationship between the height of the elevator and the angle of elevation \( \theta \).

### Step 1: Define the Variables
- Let \( y(t) \) represent the height of the elevator above the ground.
- Let \( x = 20 \) m be the constant horizontal distance between the observer and the elevator.
- Let \( \theta(t) \) be the angle of elevation of the observer’s line of sight to the elevator.

The relationship between \( y \), \( x \), and \( \theta \) is given by:
\[
\tan(\theta) = \frac{y - 24}{x}
\]
This equation accounts for the observer's 24 m height.

### Step 2: Differentiate with Respect to Time
We want to find \( \frac{d\theta}{dt} \), the rate of change of \( \theta \), when \( y = 14 \) m. Differentiate both sides of the tangent equation with respect to time \( t \):

\[
\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{x} \cdot \frac{dy}{dt}
\]

### Step 3: Substitute Known Values
We know:
- \( x = 20 \) m,
- \( \frac{dy}{dt} = 3 \) m/s (the rate at which the elevator is rising),
- \( y = 14 \) m.

First, we calculate \( \theta \) when \( y = 14 \):
\[
\tan(\theta) = \frac{14 - 24}{20} = \frac{-10}{20} = -\frac{1}{2}
\]
Thus, \( \theta = \tan^{-1}\left(-\frac{1}{2}\right) \).

Now, use the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \) to find \( \sec^2(\theta) \):
\[
\sec^2(\theta) = 1 + \left(-\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4}
\]

### Step 4: Solve for \( \frac{d\theta}{dt} \)
Substitute the known values into the differentiated equation:
\[
\frac{5}{4} \cdot \frac{d\theta}{dt} = \frac{1}{20} \cdot 3
\]
\[
\frac{d\theta}{dt} = \frac{3}{20} \cdot \frac{4}{5} = \frac{12}{100} = 0.12 \text{ rad/s}
\]

### Final Answer:
When the elevator is 14 m above the ground, the rate of change of the angle of elevation is \( \boxed{0.12} \) rad/s.

Would you like a deeper explanation of any steps or details? Here are a few related questions:

1. How would the rate change if the observer's height were different?
2. How does the horizontal distance impact the rate of change of \( \theta \)?
3. What if the elevator speed increased to 5 m/s? How would that affect the result?
4. How would you calculate the rate of change of \( \theta \) when the elevator is higher, say 44 m?
5. What mathematical principle allows us to relate rates of change in this way?

**Tip:** When solving related rates problems, always start by identifying the geometric or physical relationship between the variables before differentiating.

An observer is 24 m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 20 m horizontally from the observer. The angle of elevation of the elevator is the angle that the observer's line of sight makes with the horizontal. Assuming that the elevator rises at a rate of 3 m/s, what is the rate of change of the angle of elevation when the elevator is 14 m above the ground?

Solve a related rates problem involving the rate of change of the angle of elevation of an elevator rising at 3 m/s. Use trigonometric identities and calculus to find the solution when the elevator is 14 m above the ground. Suitable for high school calculus students.

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