/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A revolving searchlight \(\frac{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 29

A revolving searchlight \(\frac{1}{2}\) mile from a straight shoreline makes I revolution per minute. How fast is the light moving along the shore as it passes over a shore point 1 mile from the shore point nearest to the scarchlight?

Short Answer

Expert verified
Therefore, when the light shines over a point along the shore 1 mile from the nearest point, it is moving along the shore at a speed of \( 2\sqrt{5}\pi \) miles/minute.

Step by step solution

01

- Define the Variables

Let \( x \) be the distance the light is moving along the shore with respect to the nearest shore point, and \( \theta \) the angle the shoreline and the beam make as the light rotates. We know that \( x = \frac{1}{2}\tan(\theta) \) from the right triangle created with the shoreline and the beam. We also know that \( \frac{d\theta}{dt} = \frac{2\pi}{60} \) rad/sec because the light makes one revolution per minute.
02

- Differentiate with respect to time

Differentiating both sides of the equation \( x = \frac{1}{2}\tan(\theta) \) with respect to time \( t \) will result in \( \frac{dx}{dt} = \frac{1}{2}\sec^2(\theta) * \frac{d\theta}{dt} \). This equation shows us how the light's position \( x \) changes over time with respect to the angle \( \theta \).
03

- Plug in given value and solve

We know that at the point where \( x = 1 \) mile, \( \theta = \arctan(2) \) from the right triangle created. Plugging this value of \( \theta \) and the given value of \( \frac{d\theta}{dt} \) into our equation gives \( \frac{dx}{dt} = \frac{1}{2} * \sec^2(\arctan(2)) * \frac{2\pi}{60} = \frac{\sqrt{5}\pi}{30} \) miles/sec.
04

- Convert Units

To get the speed in miles per minute we simply multiply our result by 60. So the light is moving along the shore at a speed of \( 2\sqrt{5}\pi \) miles/minute.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Trigonometric Functions
One of the fundamental concepts frequently encountered in related rates calculus problems is the use of trigonometric functions.

Trigonometric functions such as sine (\textbf{sin}), cosine (\textbf{cos}), and tangent (\textbf{tan}) are essential when solving problems involving right triangles. In our searchlight example, the tangent function relates the opposite side of the right triangle (distance light moves along the shore) to the adjacent side (distance from the searchlight to the shore). Here, the equation is given by \( x = \frac{1}{2}\tan(\theta) \).

When the rate of change of the angle \( \theta \) is considered, the secant function (\textbf{sec}), which is the reciprocal of \textbf{cos}, becomes significant, particularly when we differentiate the tangent (\textbf{tan}) with respect to time.

Understanding the basic identities and properties of trigonometric functions allows students to transition smoothly between these functions and their derivatives, which is crucial in solving related rates problems.
Differentiation
Differentiation is a core mathematical operation in calculus that helps us understand how a quantity changes instantaneously. It is especially important in related rates problems. In the provided solution,

we differentiate the function \( x = \frac{1}{2}\tan(\theta) \) with respect to time to find out how the distance \( x \) along the shore changes as the light rotates and \( \theta \) varies over time.

The derivative of \( \tan(\theta) \) concerning \( \theta \) is \( \sec^2(\theta) \), hence why we get \( \frac{dx}{dt} = \frac{1}{2}\sec^2(\theta) \frac{d\theta}{dt} \).

The rate of change of the angle, \( \frac{d\theta}{dt} \) is given, and this rate is constant because of the uniform rotational speed of the light. By understanding differentiation, students can relate rates of change between different quantities in the problem.
Right Triangle Geometry
The geometry of right triangles is essential in solving many related rates problems, including our searchlight example. Right triangles have one angle precisely equal to \(90^\circ\) (a right angle), and the classic trigonometric functions are defined based on the ratios of the sides of these triangles.

In the given problem, we conceive a right triangle with the searchlight at the triangle's right angle, the shore point nearest the searchlight forming the adjacent side, and the path along the shore the opposite side鈥攁llowing us to use the \( \tan(\theta) \) function that relates these sides.

Understanding Similar Triangles

By considering different positions of the light along the shore, we deal with multiple similar right triangles, which scale proportionally and allow the use of trigonometric functions to represent the related variables.

The key relation coming from right triangle geometry in our exercise is \( x = \frac{1}{2}\tan(\theta) \), which enables us to connect the geometric properties of the triangle with the trigonometric function and subsequently with the rates of change through differentiation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sketch the graph of the function using the approach presented in this section. $$f(x)=3+2 \cot x+\csc ^{2} x, \quad x \in\left(0, \frac{1}{2} \pi\right)$$

An athlete is running around a circular track of radius 50 meters at the rate of 5 meters per second. A spectator is 200 meters from the center of the track. How fast is the distance between the two changing when the runner is approaching the spectator and the distance between them is 200 meters?

Sketch the graph of a function \(f\) that satisfies the given conditions. Indicate whether the graph of \(f\) has any horizontal or vertical asymptotes, and whether the graph has any vertical tangents or vertical cusps. If you find that no function can satisfy all the conditions, explain your reasoning. $$\begin{aligned}&f(0)=0, f(3)=f(-3)=0 ; f(x) \rightarrow-\infty \quad \text { as } \quad x \rightarrow 1,\\\&f(x) \rightarrow-\infty \text { as } x \rightarrow-1, f(x) \rightarrow \operatorname{las} x \rightarrow \infty, f(x) \rightarrow 1\\\&\text { as } x \rightarrow-\infty ; f^{\prime \prime}(x)<0 \text { for all } x \neq\pm 1\end{aligned}$$

The cost of erecting a small office building is 1,000,000 dollar for the first story, 1,100,000 dollar for the second, 1,200,000 dollar for the third, and so on. Other expenses (lot, basement, etc.) are \(5.5,000,000 .\) Assume that the annual rent is 200,000 dollar per story. How many stories will provide the greatest return on investment?

Set \(f(x)=x^{k}-a, k\) a positive integer, \(a>0 .\) The number \(a^{1 / k}\) is a root of the equation \(f(x)=0\). (a) Show that if \(x_{1}\) is any initial estimate for \(a^{1 / k}\), then the Newton-Raphson method gives the iteration formula $$x_{n+1}=\frac{1}{k}\left[(k-1) x_{n}+\frac{a}{x_{n}^{k-1}}\right]$$. Note that for \(k=2\) this formula reduces to the formula given in Exercise 13. (b) Use the formula in part (a) to approximate \(\sqrt[3]{23} .\) Begin at \(x_{1}=3\) and calculate \(x_{4}\) rounded off to five decimal places. Evaluate \(f\left(x_{4}\right)\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.