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

We can solve this problem using related rates, a powerful technique in calculus.

### Step-by-Step Solution:

1. **Set up the problem:**
   - The man is walking along a straight path, and the searchlight is 10 ft away from the path.
   - We need to find how fast the angle of rotation of the searchlight is changing when the man is 8 ft from the point on the path closest to the searchlight.
   - Let:
     - \( x \) be the distance of the man from the point on the path closest to the searchlight (which changes over time).
     - \( \theta \) be the angle that the searchlight makes with the perpendicular from its position to the path (the angle we are solving for).
   - Given:
     - \( \frac{dx}{dt} = 5 \, \text{ft/s} \) (man's speed),
     - \( x = 8 \, \text{ft} \) (man's distance at the moment of interest),
     - The distance from the searchlight to the path is 10 ft.

2. **Establish a relationship between the variables:**
   - The situation forms a right triangle where:
     - One leg is the distance from the man to the point on the path closest to the searchlight (\( x \)),
     - The other leg is the fixed distance from the searchlight to the path (10 ft),
     - The angle \( \theta \) is the angle between the searchlight's beam and the perpendicular to the path.
   - Using trigonometry, we can relate \( \theta \) and \( x \) by:
     \[
     \tan(\theta) = \frac{x}{10}
     \]
   - Differentiating both sides with respect to time \( t \), using implicit differentiation:
     \[
     \sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}
     \]
   - Now, substitute \( \sec^2(\theta) = 1 + \tan^2(\theta) \), and we have:
     \[
     \left( 1 + \tan^2(\theta) \right) \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}
     \]

3. **Substitute known values:**
   - First, calculate \( \tan(\theta) \) when \( x = 8 \, \text{ft} \):
     \[
     \tan(\theta) = \frac{8}{10} = 0.8
     \]
   - Therefore:
     \[
     \sec^2(\theta) = 1 + (0.8)^2 = 1 + 0.64 = 1.64
     \]
   - Now, substitute into the differentiated equation:
     \[
     1.64 \frac{d\theta}{dt} = \frac{1}{10} \cdot 5
     \]
     \[
     1.64 \frac{d\theta}{dt} = 0.5
     \]
   - Solve for \( \frac{d\theta}{dt} \):
     \[
     \frac{d\theta}{dt} = \frac{0.5}{1.64} \approx 0.305 \, \text{rad/s}
     \]

### Conclusion:
The searchlight is rotating at approximately \( 0.305 \, \text{rad/s} \) when the man is 8 ft from the point on the path closest to the searchlight.

Would you like more details on the solution? Here are some related questions for practice:

1. What would happen to the rotation rate if the man increased his speed to 7 ft/s?
2. How would the rotation rate change when the man is 15 ft from the closest point on the path?
3. What is the rotation rate when the man is exactly at the point closest to the searchlight?
4. How does the rotation rate change as the man moves farther away from the searchlight?
5. Can you generalize the relationship between the rotation rate and the distance of the man from the closest point?

**Tip:** When solving related rates problems, always begin by identifying the relationship between the variables and applying implicit differentiation!

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?

Solve a related rates problem where a man walks along a path at 5 ft/s and a searchlight, 10 ft away from the path, rotates to follow him. Learn how to calculate the rate of rotation (0.305 rad/s) when the man is 8 ft from the closest point to the searchlight. Involves trigonometry and calculus.

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