/*! 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 42 Highway patrol A highway patrol ... [FREE SOLUTION] | 91Ó°ÊÓ

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Highway patrol A highway patrol plane flies 3 mi above a level, straight road at a steady 120 \(\mathrm{mi} / \mathrm{h}\) . The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 \(\mathrm{mi}\) , the line-of- sight distance is decreasing at the rate of 160 \(\mathrm{mi} / \mathrm{h} .\) Find the car's speed along the highway.

Short Answer

Expert verified
The car's speed is 200 mi/h.

Step by step solution

01

Understand the Problem

We are given a right triangle with the plane's altitude as one leg (3 miles) and the distance the car has traveled along the highway as the other leg. The hypotenuse is the line-of-sight distance from the plane to the car (5 miles). We need to find the car's speed given that the rate of change of this line-of-sight distance is -160 mi/h.
02

Define Variables

Let \( x \) be the distance the car has traveled along the highway from the point directly below the plane. The plane's height is 3 miles, and the line-of-sight distance (hypotenuse) is \( s = 5 \) miles at that instant.
03

Use Pythagorean Theorem

According to the Pythagorean theorem, the relation is \( s^2 = x^2 + 3^2 \). At our moment of interest, this becomes \( 5^2 = x^2 + 3^2 \).
04

Solve for x

Substitute known values into the equation: \( 25 = x^2 + 9 \), then solve for \( x \): \( x^2 = 16 \), hence \( x = 4 \) miles.
05

Differentiate with Respect to Time

To find the car's speed, take the derivative of \( s^2 = x^2 + 3^2 \) with respect to time \( t \). This gives \( 2s\frac{ds}{dt} = 2x\frac{dx}{dt} \).
06

Substitute Known Rates and Solve

We know \( s = 5 \) miles, \( \frac{ds}{dt} = -160 \) mi/h, and previously found \( x = 4 \) miles. Substituting into the equation: \( 2(5)(-160) = 2(4)\frac{dx}{dt} \). Solve for \( \frac{dx}{dt} \): \(-1600 = 8\frac{dx}{dt} \), thus \( \frac{dx}{dt} = -200 \) mi/h.
07

Interpret the Result

The negative sign indicates the car is approaching the point directly below the plane. Therefore, the car's speed along the highway is 200 mi/h.

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Key Concepts

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

Pythagorean theorem
The Pythagorean theorem is a fundamental principle in geometry that relates the sides of a right triangle. Specifically, it states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In mathematical terms, if a triangle has sides of length \(a\) and \(b\), and hypotenuse length \(c\), the theorem can be described as:\[ c^2 = a^2 + b^2 \]This theorem not only helps in calculating distances, widths, and heights in real-world problems but is also essential in solving exercises involving related rates of change. For instance, in the problem of the highway patrol plane and the oncoming car, we use the Pythagorean theorem to understand the relationship between the line-of-sight, the plane's altitude, and the car's distance along the highway. Thus, it forms the basis for linking these variables before determining the rates of their change.
derivative
Derivatives are a core concept in calculus, representing the rate of change of a function concerning its variable. Think of it as a measure of how a quantity changes as something else changes. In mathematical notation, if \(f(x)\) represents a function, its derivative \(f'(x)\) can be seen as:\[ f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \]In the context of the exercise about the highway patrol plane, derivatives play a crucial role in determining how fast the distance between the car and the point directly below the plane changes over time. By taking the derivative of the equation derived from the Pythagorean theorem, we formulate expressions that relate these changes. This is crucial for calculating quantities such as the speed of the car, using information about the change in the line-of-sight distance.
right triangle
A right triangle is a triangle where one of the angles is exactly 90 degrees, known as a right angle. Because of this distinct angle, right triangles have special properties and mathematical relations, like the Pythagorean theorem. In any right triangle, the side opposite the right angle is the hypotenuse, and the other two sides are called legs.
  • The hypotenuse is the longest side.
  • The legs form the right angle.
In the exercise involving the highway patrol, the right triangle is formed with one leg being the altitude of the plane, and the other being the distance the car has traveled along the highway. The hypotenuse is the line-of-sight distance from the plane to the car. Understanding the structure of right triangles is essential for solving various geometric and real-world problems, especially when it involves calculating unknown distances or verifying relationships between different parts of the triangle.
instantaneous rate of change
The instantaneous rate of change refers to how quickly a variable is changing at a specific moment in time. It is a concept widely used in calculus to describe the behavior of functions and is essentially represented by a derivative. If you think about speed, it indicates how fast something is moving at a particular instant. This concept is crucial in related rates problems, like the example of the highway patrol plane. Here, the instantaneous rate of change of the line-of-sight distance, which is given as \(-160\ \text{mi/h}\), helps determine how fast the distance from the plane to the car is decreasing. By calculating the derivative, we convert this change into the car's speed along the highway. Thus, understanding instantaneous rates provides a snapshot of how a system is evolving at a precise time, allowing us to link and calculate various connected rates in dynamic situations.

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Most popular questions from this chapter

Ships Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

Find the derivatives of the functions in Exercises \(19-40\) $$ y=(2 x-5)^{-1}\left(x^{2}-5 x\right)^{6} $$

Cardiac output In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min.}\) . At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min}\) . If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min.}\) Your cardiac output can be calculated with the formula $$y=\frac{Q}{D}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{ml} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L}\) $$y=\frac{233 \mathrm{ml} / \mathrm{min}}{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

In Exercises \(41-58\) find \(d y / d t\) $$ y=\cos ^{4}\left(\sec ^{2} 3 t\right) $$

Radians versus degrees: degree mode derivatives What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$ f(h)=\frac{\sin h}{h} $$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$ \lim _{h \rightarrow 0} \frac{\cos h-1}{h} $$ c. Now go back to the derivation of the formula for the derivative of sin \(x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos \(x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of sin \(x\) and \(\cos x ?\)

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