Math Problem Statement

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.

Solution

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 QQ (which is 300 m from point PP, the point on the shore closest to the lighthouse).

To calculate the speed of the beam at QQ, 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 ω\omega is: ω=3×2πradians per minute=6πradians per minute\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 PP is 600 m (vertical distance). The distance from PP to QQ is 300 m.

    At any given time, the position of the light on the shore, measured from PP, is denoted as x(t)x(t). The distance from the lighthouse to the point where the beam strikes the shore is: r=6002+x2r = \sqrt{600^2 + x^2}

    Using trigonometry, the angle θ\theta between the beam and the shore satisfies: tan(θ)=x600\tan(\theta) = \frac{x}{600} Taking the derivative of both sides with respect to time tt, we get: sec2(θ)dθdt=1600dxdt\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{600} \cdot \frac{dx}{dt} But dθdt=ω\frac{d\theta}{dt} = \omega, so the equation becomes: dxdt=600ωsec2(θ)\frac{dx}{dt} = 600 \cdot \omega \cdot \sec^2(\theta)

  3. At point QQ (where x=300x = 300): tan(θ)=300600=0.5θ=tan1(0.5)\tan(\theta) = \frac{300}{600} = 0.5 \quad \Rightarrow \quad \theta = \tan^{-1}(0.5) Using sec2(θ)=1+tan2(θ)\sec^2(\theta) = 1 + \tan^2(\theta): sec2(θ)=1+0.52=1.25\sec^2(\theta) = 1 + 0.5^2 = 1.25

    Now, the speed of the beam at QQ is: dxdt=6006π1.25\frac{dx}{dt} = 600 \cdot 6\pi \cdot 1.25 dxdt=4500πmeters per minute\frac{dx}{dt} = 4500\pi \, \text{meters per minute} Converting to meters per second (since 1minute=60seconds1 \, \text{minute} = 60 \, \text{seconds}): dxdt=4500π60235.62meters per second\frac{dx}{dt} = \frac{4500\pi}{60} \approx 235.62 \, \text{meters per second} Rounding to the nearest integer gives 236meters per second\approx 236 \, \text{meters per second}.

The speed of the beam along the shore at point QQ is approximately 236 meters per second.


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=100x = 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 PP?
  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.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Related Rates
Angular Velocity
Linear Velocity

Formulas

Angular velocity formula: ω = θ/t
Relationship between linear and angular speed: v = rω
Derivative of the tangent function for related rates

Theorems

Pythagorean Theorem
Trigonometric Identities

Suitable Grade Level

Grades 11-12