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This problem describes a rotating lighthouse with a beam of light that sweeps across the shore. The lighthouse is 600 m from the shore, and the beam sweeps across the point \( Q \) (which is 300 m from point \( P \), the point on the shore closest to the lighthouse). 

To calculate the speed of the beam at \( Q \), we need to relate the angular velocity of the beam to the linear speed along the shore. 

### Steps:
1. **Angular velocity:**
   The beam revolves 3 times per minute. This means the angular velocity \( \omega \) is:
   \[
   \omega = 3 \times 2\pi \, \text{radians per minute} = 6\pi \, \text{radians per minute}
   \]

2. **Right triangle setup:**
   The distance from the lighthouse to point \( P \) is 600 m (vertical distance). The distance from \( P \) to \( Q \) is 300 m. 
   
   At any given time, the position of the light on the shore, measured from \( P \), is denoted as \( x(t) \). The distance from the lighthouse to the point where the beam strikes the shore is:
   \[
   r = \sqrt{600^2 + x^2}
   \]
   
   Using trigonometry, the angle \( \theta \) between the beam and the shore satisfies:
   \[
   \tan(\theta) = \frac{x}{600}
   \]
   Taking the derivative of both sides with respect to time \( t \), we get:
   \[
   \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{600} \cdot \frac{dx}{dt}
   \]
   But \( \frac{d\theta}{dt} = \omega \), so the equation becomes:
   \[
   \frac{dx}{dt} = 600 \cdot \omega \cdot \sec^2(\theta)
   \]
   
3. **At point \( Q \) (where \( x = 300 \)):**
   \[
   \tan(\theta) = \frac{300}{600} = 0.5 \quad \Rightarrow \quad \theta = \tan^{-1}(0.5)
   \]
   Using \( \sec^2(\theta) = 1 + \tan^2(\theta) \):
   \[
   \sec^2(\theta) = 1 + 0.5^2 = 1.25
   \]
   
   Now, the speed of the beam at \( Q \) is:
   \[
   \frac{dx}{dt} = 600 \cdot 6\pi \cdot 1.25
   \]
   \[
   \frac{dx}{dt} = 4500\pi \, \text{meters per minute}
   \]
   Converting to meters per second (since \( 1 \, \text{minute} = 60 \, \text{seconds} \)):
   \[
   \frac{dx}{dt} = \frac{4500\pi}{60} \approx 235.62 \, \text{meters per second}
   \]
   Rounding to the nearest integer gives \( \approx 236 \, \text{meters per second} \).

The speed of the beam along the shore at point \( Q \) is approximately **236 meters per second**.

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Would you like further details on any step, or do you have any questions?

Here are 5 related questions to expand your understanding:
1. What is the beam's speed at the point where \( x = 100 \) meters?
2. How would the speed change if the lighthouse were further from the shore?
3. How does the speed vary as the beam moves farther away from point \( P \)?
4. What would happen to the beam's speed if it revolved 4 times per minute instead of 3?
5. How would the problem change if the height of the lighthouse were not neglected?

**Tip:** When solving related rates problems, always start by finding an equation that connects the variables and differentiate with respect to time.

A lighthouse stands 600 m off a straight shore and the focused beam of its light revolves three times each minute. As shown in the figure, P is the point on shore closest to the lighthouse, and Q is a point on the shore 300 m from P. What is the speed of the beam along the shore when it strikes the point Q? Describe how the speed of the beam along the shore varies with the distance between P and Q. Neglect the height of the lighthouse.

Learn how to calculate the speed of a lighthouse beam at a point along the shore 300 meters from the closest shore point. This problem covers related rates, angular velocity, and trigonometric identities. Suitable for high school students in Grades 11-12.

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