A lighthouse is located on a small island 5 km away from the nearest point P on a straight shoreline and its light makes three revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P? (Round your answer to one decimal place.)

This is a related rates problem that involves understanding the geometry of the situation and how the speed of the beam of light along the shoreline changes with time.

### Given:
- The lighthouse is 5 km away from point \( P \) on a straight shoreline.
- The light makes 3 revolutions per minute.
- We need to find how fast the beam of light is moving along the shoreline when the beam is 1 km away from point \( P \).

### Step 1: Setup Variables

Let:
- \( x \) be the distance of the beam of light from point \( P \) along the shoreline.
- \( \theta \) be the angle between the line from the lighthouse to point \( P \) and the line from the lighthouse to the beam of light.

From the problem's geometry, we know that:
- The lighthouse is located 5 km from point \( P \), so the distance between the lighthouse and point \( P \) is constant and equals 5 km.
- We are looking for \( \frac{dx}{dt} \) when \( x = 1 \text{ km} \).

### Step 2: Use Trigonometry to Relate Variables

From trigonometry, we have:
\[
\tan(\theta) = \frac{x}{5}
\]
Differentiating both sides with respect to \( t \), we get:
\[
\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot \frac{dx}{dt}
\]
We need to solve for \( \frac{dx}{dt} \), so rearranging the equation:
\[
\frac{dx}{dt} = 5 \cdot \sec^2(\theta) \cdot \frac{d\theta}{dt}
\]

### Step 3: Determine \( \frac{d\theta}{dt} \)

The lighthouse's light makes 3 revolutions per minute, which is equivalent to an angular speed of:
\[
\frac{d\theta}{dt} = 3 \times 2\pi = 6\pi \text{ radians per minute}
\]

### Step 4: Calculate \( \sec^2(\theta) \)

When \( x = 1 \text{ km} \), we can find \( \theta \) using the tangent relationship:
\[
\tan(\theta) = \frac{1}{5} \implies \theta = \tan^{-1}\left(\frac{1}{5}\right)
\]
Now, \( \sec^2(\theta) = 1 + \tan^2(\theta) \):
\[
\sec^2(\theta) = 1 + \left(\frac{1}{5}\right)^2 = 1 + 0.04 = 1.04
\]

### Step 5: Calculate \( \frac{dx}{dt} \)

Substitute the values of \( \sec^2(\theta) = 1.04 \) and \( \frac{d\theta}{dt} = 6\pi \) into the related rates equation:
\[
\frac{dx}{dt} = 5 \cdot 1.04 \cdot 6\pi
\]
\[
\frac{dx}{dt} \approx 5 \cdot 1.04 \cdot 18.8496 \approx 98.14 \text{ km/min}
\]

Thus, the beam of light is moving at approximately **98.1 km/min** when it is 1 km from point \( P \).

Would you like further explanation or any more details on this solution?

Here are 5 related questions you might consider:
1. How does the speed of the beam change as the distance from \( P \) increases?
2. What would happen to the speed of the beam if the lighthouse was closer to or farther from the shore?
3. How fast would the beam be moving along the shoreline when it is exactly at point \( P \)?
4. How would the speed change if the lighthouse rotated faster or slower?
5. Can you find a general formula for the speed of the beam as a function of distance from \( P \)?

**Tip:** Related rates problems often involve implicit differentiation, so keep track of how each variable is changing with respect to time!

Solve a related rates problem involving a lighthouse 5 km from the shore. Learn how to calculate the speed of a beam of light along the shoreline when it is 1 km from the nearest point using trigonometry and differentiation. Perfect for Grades 10-12.

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