1. A man walks along a straight path at a speed of 5 ft/s. A searchlight is closed on the ground 10 ft form the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 8 ft from the point on the path closest to the searchlight?

To find the rate at which the searchlight is rotating, we’ll use related rates. Let’s define the situation mathematically:

1. Let \( x \) be the distance of the man from the point on the path closest to the searchlight (in this case, 8 feet at the moment we’re examining).
2. Let \( \theta \) be the angle that the line from the searchlight to the man makes with the perpendicular from the searchlight to the path.
3. The distance between the searchlight and the path is fixed at 10 feet.

Given:
- The man’s speed along the path is \( \frac{dx}{dt} = 5 \, \text{ft/s} \).
- We want to find the rate of rotation of the searchlight, \( \frac{d\theta}{dt} \), when \( x = 8 \, \text{ft} \).

### Step 1: Set up the relationship

From the right triangle formed by the path, the line to the searchlight, and the line from the searchlight to the man:
\[
\tan(\theta) = \frac{x}{10}
\]

### Step 2: Differentiate with respect to time \( t \)

Differentiating both sides with respect to \( t \):
\[
\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot \frac{dx}{dt}
\]

### Step 3: Substitute known values

We know:
- \( \frac{dx}{dt} = 5 \, \text{ft/s} \)
- \( x = 8 \, \text{ft} \)

First, find \( \theta \) when \( x = 8 \):
\[
\tan(\theta) = \frac{8}{10} = 0.8
\]
Thus, \( \theta = \tan^{-1}(0.8) \).

Next, calculate \( \sec^2(\theta) \):
\[
\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + (0.8)^2 = 1 + 0.64 = 1.64
\]

### Step 4: Solve for \( \frac{d\theta}{dt} \)

Now, plug in the values:
\[
1.64 \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot 5
\]
\[
\frac{d\theta}{dt} = \frac{5}{10 \cdot 1.64} = \frac{5}{16.4} \approx 0.305 \, \text{rad/s}
\]

### Answer

The rate at which the searchlight is rotating when the man is 8 feet from the closest point on the path is approximately:
\[
\frac{d\theta}{dt} \approx 0.305 \, \text{rad/s}
\]

Would you like a detailed breakdown of any part? Here are five related questions for further exploration:

1. What happens to the rate of rotation as the man moves further away from the closest point on the path?
2. How would this problem change if the searchlight were a variable distance from the path?
3. What if the man’s speed along the path changed over time?
4. How is the rate of change in angle affected if the searchlight were at a higher elevation above the path?
5. How can this concept of related rates apply to scenarios involving other angles in motion?

**Tip:** In related rates problems, focus on setting up the geometric relationships clearly before differentiating with respect to time.

A man walks along a straight path at a speed of 5 ft/s. A searchlight is closed on the ground 10 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 8 ft from the point on the path closest to the searchlight?

Explore how to calculate the rate at which a searchlight rotates when a man walks along a straight path at 5 ft/s, and the searchlight is 10 ft from the path. The problem uses related rates, involving trigonometric relationships and differentiation. Ideal for Pre-Calculus and Calculus students.

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