A lighthouse is located on a small island 5 km away from the nearest point P on a straight shoreline and its light makes two revolutions per minute. How fast (in km/min) is the beam of light moving along the shoreline when it is 1 km from P?

This problem is a classic example of related rates in calculus. We'll break it down step by step.

### Step 1: Analyze the problem geometrically
- The lighthouse is 5 km away from point \( P \), which is the closest point on the shoreline to the lighthouse.
- The beam of light rotates at 2 revolutions per minute. So, the angular speed \( \omega \) is:
  \[
  \omega = 2 \times 2\pi \, \text{radians/min} = 4\pi \, \text{radians/min}.
  \]

We are asked to find the speed at which the beam of light is moving along the shoreline when the beam is 1 km from point \( P \).

### Step 2: Set up the relationship
We can use trigonometry to describe the position of the beam along the shoreline. Let:
- \( x \) be the distance of the point of light from \( P \) along the shoreline.
- \( \theta \) be the angle between the line from the lighthouse to \( P \) and the current position of the light beam.

Using the geometry of the situation, we have:
\[
\tan(\theta) = \frac{x}{5}.
\]
Differentiating both sides with respect to time \( t \) gives:
\[
\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot \frac{dx}{dt}.
\]
We know that \( \frac{d\theta}{dt} = \omega = 4\pi \), so the equation becomes:
\[
\sec^2(\theta) \cdot 4\pi = \frac{1}{5} \cdot \frac{dx}{dt}.
\]
Solving for \( \frac{dx}{dt} \), the speed of the beam of light along the shoreline:
\[
\frac{dx}{dt} = 5 \cdot 4\pi \cdot \sec^2(\theta).
\]

### Step 3: Find \( \sec^2(\theta) \)
When the beam is 1 km from \( P \), \( x = 1 \). From the relationship \( \tan(\theta) = \frac{x}{5} \), we have:
\[
\tan(\theta) = \frac{1}{5}.
\]
Thus, using the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \):
\[
\sec^2(\theta) = 1 + \left(\frac{1}{5}\right)^2 = 1 + \frac{1}{25} = \frac{26}{25}.
\]

### Step 4: Calculate the speed
Now we substitute \( \sec^2(\theta) = \frac{26}{25} \) into the expression for \( \frac{dx}{dt} \):
\[
\frac{dx}{dt} = 5 \cdot 4\pi \cdot \frac{26}{25} = 4\pi \cdot \frac{26}{5} = \frac{104\pi}{5} \, \text{km/min}.
\]
Thus, the speed of the light beam along the shoreline when it is 1 km from \( P \) is:
\[
\frac{104\pi}{5} \, \text{km/min} \approx 65.34 \, \text{km/min}.
\]

### Final Answer:
The beam of light is moving along the shoreline at approximately \( 65.34 \, \text{km/min} \) when it is 1 km from point \( P \).

Would you like more details or have any questions?

Here are 5 questions to further expand on this topic:
1. What would the speed of the light beam be if it were 2 km from point \( P \)?
2. How would the speed change if the lighthouse were 10 km from the shoreline instead of 5 km?
3. Can you explain the role of trigonometric identities in related rates problems?
4. How would the speed of the beam change if the light rotated at 3 revolutions per minute instead of 2?
5. What is the significance of the secant function in this problem?

**Tip**: Always make sure to relate your variables clearly when working with related rates problems—visualizing the geometry can help guide the setup.

Explore a detailed solution to a related rates problem involving a lighthouse beam 5 km from the shore, rotating at 2 revolutions per minute. Learn how calculus and trigonometry determine the speed of the beam when it is 1 km from point P, using angular speed and differentiation techniques. Suitable for advanced high school students.

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