To solve this problem, we need to use related rates. The setup involves a rotating beam of light sweeping across a shoreline. We can approach it geometrically by understanding that the light beam's position relative to the shoreline changes as it rotates, and we're asked to determine how fast the beam is moving along the shoreline when the light is 1 km from point P.

Step-by-Step Solution

1. Visualize the problem:

   • The lighthouse is 4 km away from point P (the closest point on the shoreline).
   • The light rotates at 7 revolutions per minute.

2. Define the variables:

   • Let $\theta$ be the angle between the line from the lighthouse to the point P and the beam of light at some time $t$.
   • Let $x$ be the distance of the beam's position along the shoreline from point P at time $t$.

   By trigonometry, we know: $\tan(\theta) = \frac{x}{4}$

3. Differentiate with respect to time: We need to find how fast $x$ is changing with respect to time. So, differentiate both sides of the equation $\tan(\theta) = \frac{x}{4}$ with respect to time $t$: $\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{4} \cdot \frac{dx}{dt}$ Here, $\frac{dx}{dt}$ is the speed of the beam along the shoreline that we are looking for, and $\frac{d\theta}{dt}$ is the angular speed of the light.

4. Angular speed: The lighthouse makes 7 revolutions per minute, so its angular speed is: $\frac{d\theta}{dt} = 7 \times 2\pi = 14\pi \text{ radians per minute}$

5. Evaluate when the light is 1 km away: When the beam of light is 1 km away from point P, $x = 1$. At this moment, we can find $\theta$ using: $\tan(\theta) = \frac{1}{4} \quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{1}{4}\right)$ Now, find $\sec^2(\theta)$. Recall that: $\sec^2(\theta) = 1 + \tan^2(\theta)$ Substituting $\tan(\theta) = \frac{1}{4}$: $\sec^2(\theta) = 1 + \left(\frac{1}{4}\right)^2 = 1 + \frac{1}{16} = \frac{17}{16}$

6. Solve for $\frac{dx}{dt}$: Now substitute everything into the differentiated equation: $\frac{17}{16} \cdot 14\pi = \frac{1}{4} \cdot \frac{dx}{dt}$ Solving for $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{17 \cdot 14\pi}{4} \times \frac{16}{17} = 14\pi \times 4$ $\frac{dx}{dt} = 56\pi \, \text{km/min}$

7. Final answer: Now approximate $\pi \approx 3.1416$: $\frac{dx}{dt} \approx 56 \times 3.1416 = 175.9 \, \text{km/min}$

So, the beam of light is moving along the shoreline at approximately 175.9 km/min when it is 1 km away from point P.

Let me know if you want any details or have further questions!

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Here are 5 related questions:

1. What is the speed of the light beam along the shoreline when it is 2 km from point P?
2. How would the answer change if the lighthouse were 5 km away from point P instead of 4 km?
3. What happens to the speed of the beam as the distance from P increases?
4. Can we calculate the angular speed in degrees per minute instead of radians per minute?
5. How does the problem change if the rotation speed of the lighthouse increases to 10 revolutions per minute?

Tip: The further the beam is from the lighthouse, the faster it moves along the shoreline due to the nature of angular motion and its relationship to linear velocity!