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Chapter 3: Problem 38

A lighthouse is located on a small island 3 \(\mathrm{km}\) away from the nearest point \(\mathrm{P}\) on a straight shoreline and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 \(\mathrm{km}\) from P?

Short Answer

Expert verified
The beam moves at 80蟺 km/minute along the shore when 1 km from P.

Step by step solution

01

Understand the Problem

We have a lighthouse 3 km away from the coastline at point P, and a beacon that makes four revolutions per minute. We need to find the speed of the light beam along the shore when it is 1 km from point P.
02

Setup the Geometry

Imagine the lighthouse at point L and the point on the shore where the light hits as point Q. The distance LQ changes as the light rotates. Let x be the distance PQ along the shore, and 胃 the angle LPQ.
03

Relate the Geometry to the Angle

From the right triangle LPQ, we have the relationship: \[ \tan(\theta) = \frac{x}{3} \]Differentiate both sides with respect to time (t).
04

Differentiate the Tangent Equation

Differentiating \( \tan(\theta) = \frac{x}{3} \) gives:\[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{3} \cdot \frac{dx}{dt} \]
05

Calculate \( \sec^2(\theta) \)

When x = 1 km, use the tangent relationship to find \( \theta \):\[ \theta = \tan^{-1}\left( \frac{1}{3} \right) \]Calculate \( \sec^2(\theta) \):\[ \sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left( \frac{1}{3} \right)^2 = \frac{10}{9} \]
06

Calculate \( \frac{d\theta}{dt} \)

The light makes four revolutions per minute, so \( \theta = 4 \times 2\pi \text{ radians per minute} \). Then,\[ \frac{d\theta}{dt} = 8\pi \text{ radians per minute} \]
07

Solve for \( \frac{dx}{dt} \)

Substitute \( \sec^2(\theta) = \frac{10}{9} \) and \( \frac{d\theta}{dt} = 8\pi \) into the differentiated equation:\[ \frac{10}{9} \cdot 8\pi = \frac{1}{3} \cdot \frac{dx}{dt} \]Solving for \( \frac{dx}{dt} \) gives\[ \frac{dx}{dt} = 80\pi \text{ km/minute} \]
08

Summary

The beam of light is moving along the shoreline at a speed of \( 80\pi \text{ km/minute} \) when it is 1 km from point P.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
In the realm of mathematics, calculus is a powerful tool used to describe change and motion. It focuses on two main concepts: derivatives and integrals. Derivatives describe how a quantity changes over time, which is incredibly useful for solving problems involving motion, like our lighthouse problem. Using calculus to find rates allows us to understand how different quantities relate as they change. This is precisely what's happening in our exercise where we deal with the rotating beam of light and its effect on distances.
Differentiation
Differentiation is a key concept in calculus, and it's all about finding derivatives. A derivative represents the rate at which a quantity changes with respect to another. In practical terms, it tells us the slope of a function at any given point.
  • In the lighthouse problem, we differentiated the equation \( \tan(\theta) = \frac{x}{3} \) to relate \( \theta \) and \( x \) to time, \( t \).
  • This differentiation gives us \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{3} \cdot \frac{dx}{dt} \), which links the angle's change over time to the light's movement along the shore.
Understanding differentiation allows you to manipulate and solve for one rate if you know the other, a crucial skill in related rates problems.
Trigonometry
Trigonometry is the study of triangles, and it helps relate angles to side lengths. In the context of our problem, trigonometry provides the mathematical relationships needed to solve it, mainly through the tangent function.
  • The triangle formed by the lighthouse, the point on the shoreline, and the distance between them uses the tangent function: \( \tan(\theta) = \frac{x}{3} \).
  • This relationship is crucial because it connects the linear distance \( x \) along the shore to the rotational angle \( \theta \).
Trigonometry simplifies the process of studying non-linear relationships and is a foundational tool for many physics and engineering problems.
Radians
Radians provide an alternative way of measuring angles that is often more natural in calculus and trigonometry. Unlike degrees, which divide a circle into 360 parts, radians are based on the circle's radius. One complete revolution is \( 2\pi \) radians, as we used in solving our problem.
  • The lighthouse's light making four revolutions per minute translates to \( 4 \times 2\pi = 8\pi \) radians per minute.
  • Using radians simplifies the math involved when calculating derivatives and integrating functions, especially when dealing with rotations and oscillations like a lighthouse beam.
Embracing radians as a measurement allows for streamlined calculations and is essential for solving many advanced mathematics and physics problems.

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Most popular questions from this chapter

The frequency of vibrations of a vibrating violin string is given by $$\mathrm{f}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\rho}}$$ where \(L\) is the length of the string, T is its tension, and \(\rho\) is its linear density. [See Chapter 11 in D. E. Hall, Musical Acoustics, 3 \(\mathrm{d}\) ed. (Pacific Grove, CA: Brooks/Cole, \(2002 ) . ]\) (a) Find the rate of change of the frequency with respect to (i) the length (when T and \(\rho\) are constant), (ii) the tension (when L and \(\rho\) are constant), and (iii) the linear density (when L and T are constant). (b) The pitch of a note (how high or low the note sounds) is determined by the frequency f. (The higher the frequency, the higher the pitch.) Use the signs of the derivatives in part (a) to determine what happens to the pitch of a note (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates, (ii) when the tension is increased by turning a tuning peg. (iii) when the linear density is increased by switching to another string.

(a) If \(\$ 3000\) is invested at 5\(\%\) interest, find the value of the investment at the end of 5 years if the interest is compounded ( i ) annually, (ii) semiannually, (iil) monthly, (iv) weekly, (v) daily, and (vi) continuously. (b) If \(\mathrm{A}(\mathrm{t})\) is the amount of the investment at time t for the case of continuous compounding, write a differential equation and an initial condition satisfied by \(\mathrm{A}(\mathrm{t})\) .

A plane flies horizontally at an altitude of 5 \(\mathrm{km}\) and passes directly over a tracking telescope on the ground. When the angle of elevation is \(\pi / 3,\) this angle is decreasing at a rate of \(\pi / 6 \mathrm{rad} / \mathrm{min}\) . How fast is the plane traveling at that time?

One side of a right triangle is known to be 20 \(\mathrm{cm}\) long and the opposite angle is measured as \(30^{\circ},\) with a possible error of \(\pm 1^{\circ} .\) $$ \begin{array}{l}{\text { (a) Use differentials to estimate the error in computing the }} \\ {\text { length of the hypotenuse. }} \\ {\text { (b) What is the percentage error? }}\end{array} $$

$$ \begin{array}{l}{\text { Suppose that we don't have a formula for } g(\mathrm{x}) \text { but we know }} \\ {\text { that } g(2)=-4 \text { and } g^{\prime}(\mathrm{x})=\sqrt{\mathrm{x}^{2}+5} \text { for all } \mathrm{x} \text { . }}\end{array} $$ $$ \begin{array}{l}{\text { (a) Use a linear approximation to estimate } g(1.95)} \\ {\text { and } g(2.05) .} \\ {\text { (b) Are your estimates in part (a) too large or too small? }} \\ {\text { Explain. }}\end{array} $$
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