We are dealing with a related rates problem, where we need to find the rate of change of the viewing angle, $\theta$, as you walk away from a theater screen. Here’s a step-by-step breakdown of the solution.

Step 1: Diagram and Variables

We can model the situation as a right triangle, where:

• The horizontal distance you are from the wall (where the screen is) is $x$ (which changes over time).
• The height of the screen ranges from $99$ ft (bottom of the screen) to $3030$ ft (top of the screen), so the height difference between the bottom and top is $3030 - 99 = 2931$ ft.
• The angle $\theta$ represents your viewing angle, which is the angle between your line of sight and the horizontal distance to the screen.

At the point when you are 4040 ft from the wall, we are asked to determine how $\theta$ is changing.

Step 2: Relating the Quantities

To relate the angle $\theta$ to the distance $x$, we use the fact that $\theta$ is the angle formed by the lines of sight to the top and bottom of the screen. Therefore, we can use the tangent function for each line of sight:

$\tan(\theta_{\text{top}}) = \frac{3030}{x}, \quad \tan(\theta_{\text{bottom}}) = \frac{99}{x}$

The overall viewing angle $\theta$ is the difference between these two angles:

$\theta = \theta_{\text{top}} - \theta_{\text{bottom}}$

Since the tangent function is involved, we can take the derivative with respect to time $t$ to find the rate of change of $\theta$.

Step 3: Implicit Differentiation

Differentiating both tangent functions with respect to $t$:

$\frac{d}{dt} \left( \tan(\theta_{\text{top}}) \right) = \frac{d}{dt} \left( \frac{3030}{x} \right)$ Using the chain rule, this becomes:

$\sec^2(\theta_{\text{top}}) \cdot \frac{d\theta_{\text{top}}}{dt} = - \frac{3030}{x^2} \cdot \frac{dx}{dt}$

Similarly, for the bottom:

$\sec^2(\theta_{\text{bottom}}) \cdot \frac{d\theta_{\text{bottom}}}{dt} = - \frac{99}{x^2} \cdot \frac{dx}{dt}$

The rate of change of $\theta$ is:

$\frac{d\theta}{dt} = \frac{d\theta_{\text{top}}}{dt} - \frac{d\theta_{\text{bottom}}}{dt}$

Step 4: Plugging in Values

Given that $\frac{dx}{dt} = 44$ ft/s and $x = 4040$ ft at the moment of interest:

• For the top of the screen: $\tan(\theta_{\text{top}}) = \frac{3030}{4040}$, so $\sec^2(\theta_{\text{top}})$ can be computed from this.
• For the bottom of the screen: $\tan(\theta_{\text{bottom}}) = \frac{99}{4040}$, and $\sec^2(\theta_{\text{bottom}})$ follows similarly.

Finally, using these values, you can compute $\frac{d\theta}{dt}$.

Would you like me to compute the exact values for you, or do you have any specific questions about the process?

5 Questions to Expand:

1. How do trigonometric identities help simplify problems like this?
2. Why is it important to apply the chain rule in related rates problems?
3. Can we approximate $\sec^2(\theta)$ when $\theta$ is small?
4. What happens if the speed of walking changes? How does that affect $\frac{d\theta}{dt}$?
5. How would this problem change if the screen were tilted?

Tip:

When solving related rates problems, it's crucial to carefully define all variables and relationships before differentiating to avoid errors.