Math Problem Statement

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.)

Solution

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 PP 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 PP.

Step 1: Setup Variables

Let:

  • xx be the distance of the beam of light from point PP along the shoreline.
  • θ\theta be the angle between the line from the lighthouse to point PP 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 PP, so the distance between the lighthouse and point PP is constant and equals 5 km.
  • We are looking for dxdt\frac{dx}{dt} when x=1 kmx = 1 \text{ km}.

Step 2: Use Trigonometry to Relate Variables

From trigonometry, we have: tan(θ)=x5\tan(\theta) = \frac{x}{5} Differentiating both sides with respect to tt, we get: sec2(θ)dθdt=15dxdt\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot \frac{dx}{dt} We need to solve for dxdt\frac{dx}{dt}, so rearranging the equation: dxdt=5sec2(θ)dθdt\frac{dx}{dt} = 5 \cdot \sec^2(\theta) \cdot \frac{d\theta}{dt}

Step 3: Determine dθdt\frac{d\theta}{dt}

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

Step 4: Calculate sec2(θ)\sec^2(\theta)

When x=1 kmx = 1 \text{ km}, we can find θ\theta using the tangent relationship: tan(θ)=15    θ=tan1(15)\tan(\theta) = \frac{1}{5} \implies \theta = \tan^{-1}\left(\frac{1}{5}\right) Now, sec2(θ)=1+tan2(θ)\sec^2(\theta) = 1 + \tan^2(\theta): sec2(θ)=1+(15)2=1+0.04=1.04\sec^2(\theta) = 1 + \left(\frac{1}{5}\right)^2 = 1 + 0.04 = 1.04

Step 5: Calculate dxdt\frac{dx}{dt}

Substitute the values of sec2(θ)=1.04\sec^2(\theta) = 1.04 and dθdt=6π\frac{d\theta}{dt} = 6\pi into the related rates equation: dxdt=51.046π\frac{dx}{dt} = 5 \cdot 1.04 \cdot 6\pi dxdt51.0418.849698.14 km/min\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 PP.

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 PP 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 PP?
  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 PP?

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

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Math Problem Analysis

Mathematical Concepts

Related Rates
Trigonometry
Differentiation

Formulas

tan(θ) = x / 5
sec^2(θ) · dθ/dt = 1/5 · dx/dt
dx/dt = 5 · sec^2(θ) · dθ/dt

Theorems

Trigonometric Functions
Implicit Differentiation

Suitable Grade Level

Grades 10-12 (Advanced Calculus and Trigonometry)