/*! 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 A highway patrol plane flies 3 k... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The car's speed along the highway is 200 km/h.

Step by step solution

01

Understand the Relationship

The problem involves a right triangle where the plane's altitude, the distance along the road, and the line-of-sight distance form the sides of the triangle. The plane is flying at a constant altitude of 3 km, and we need to find the car's speed along the road (horizontal distance) when certain distances and speed rates are given.
02

Set Up the Geometry

We denote the altitude of the plane as \( a = 3 \) km, the horizontal distance from the plane to the car as \( x \) km, and the line-of-sight distance as \( s \) km. According to the Pythagorean theorem, we have:\[s^2 = x^2 + a^2\]When \( s = 5 \) km, we have \( 5^2 = x^2 + 3^2 \). Calculate \( x \).
03

Solve for x

Using the equation from the previous step:\[25 = x^2 + 9\]Subtract 9 from both sides:\[16 = x^2 \implies x = 4 \text{ km}\]Therefore, the horizontal distance at the instant consideration is 4 km.
04

Apply the Chain Rule to Derivatives

The rate of change of the line-of-sight distance \( \frac{ds}{dt} = -160 \) km/h (negative because \( s \) is decreasing). Also, note the plane's speed is constant at \( \frac{dx}{dt} \) (which is actually what we want). By differentiating the Pythagorean relation \( s^2 = x^2 + a^2 \) with respect to time \( t \), we get:\[2s \frac{ds}{dt} = 2x \frac{dx}{dt}\]
05

Substitute Known Values

Plug in the known values \( s = 5 \), \( \frac{ds}{dt} = -160 \), \( x = 4 \):\[2(5)(-160) = 2(4)(\frac{dx}{dt})\]This gives:\[-1600 = 8\frac{dx}{dt}\]
06

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

To find \( \frac{dx}{dt} \), rearrange the equation:\[\frac{dx}{dt} = \frac{-1600}{8}\]Calculate:\[\frac{dx}{dt} = -200 \text{ km/h}\]The negative sign indicates the car is moving towards the plane, but in terms of speed, the car travels at 200 km/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 connects the three sides of a right triangle. It states that the square of the hypotenuse, the triangle's longest side, is equal to the sum of the squares of the other two sides. In equation form, this is expressed as:\[c^2 = a^2 + b^2\]where \(c\) represents the hypotenuse, and \(a\) and \(b\) are the triangle's other sides.
This theorem is extremely useful in real-world applications, especially when dealing with problems involving distance in multi-dimensional spaces. In our exercise, the line-of-sight distance forms the hypotenuse of a right triangle, while the altitude of the plane and the horizontal distance from the car to the plane form the other two sides. By using the Pythagorean theorem in our solution, we were able to calculate the horizontal distance \(x\) when the line-of-sight distance \(s\) was 5 km and the altitude \(a\) was 3 km.
Understanding this relation helps in visualizing and solving problems that can be broken down into geometric interpretations, making it easier to handle complex distance and area calculations.
Differentiation
Differentiation is a core concept in calculus that deals with understanding how functions change. It is about finding the derivative, which represents the rate at which a quantity changes with respect to another quantity. In simpler terms, it's a measure of how something is changing at any given point.
In the context of the problem, we're interested in how the line-of-sight distance and the horizontal distance from the plane to the car are changing over time. By differentiating the Pythagorean relationship \(s^2 = x^2 + a^2\) with respect to time, we determine the relationship between these rates of change. This process involves applying the chain rule, a differentiation technique used to work with composed functions:\[2s \frac{ds}{dt} = 2x \frac{dx}{dt}\]
Here, \(\frac{ds}{dt}\) is the rate at which the line-of-sight distance is changing (given as \(-160\, \text{km/h}\)), and \(\frac{dx}{dt}\) is the rate at which the horizontal distance is changing, which is our unknown. This method allows us to solve for the unknown rate, providing insight into how one change affects the other.
Physics in Calculus
Calculus is a powerful tool in physics, where it often describes how physical quantities change over time or space. The essence of physics in calculus is about unpacking how dynamics, motion, and forces interact in continuously changing systems.
In this airplane-car exercise, calculus helps us describe the motion of the car relative to the plane. By analyzing rates of change, we identify how fast the car moves along the road, leveraging known rates of change for the line-of-sight distance. Calculus aids in translating these dynamic relationships into mathematical expressions, accounting for changes in variables over time.
This approach illustrates how differential calculus provides a framework for understanding continuous variations in physics. Whether it's an accelerating car, a descending plane, or any moving object, calculus allows us to predict and interpret motion, offering insights that are critical for real-world applications such as navigation and engineering.

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Most popular questions from this chapter

(Continuation of Exercise \(65 .)\) The quotient $$ \frac{f(x+h)-f(x-h)}{2 h} $$ may have a limit as \(h \rightarrow 0\) when \(f\) has no derivative at \(x .\) As a case in point, take \(f(x)=|x|\) and calculate $$ \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} $$ As you will see, the limit exists even though \(f(x)=|x|\) has no derivative at \(x=0 .\) Moral: Before using a centered difference quotient, be sure the derivative exists.

What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions. Find \(d y / d x\) if \(y=x^{3 / 2}\) by using the Chain Rule with \(y\) as a composite of a. \(y=u^{3}\) and \(u=\sqrt{x}\) b. \(y=\sqrt{u}\) and \(u=x^{3}\)

The line that is normal to the curve \(x^{2}+2 x y-3 y^{2}=0\) at (1,1) intersects the curve at what other point?

Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains along the wall of the crater, activity was later confined to a single vent in the crater's floor, which at one point shot lava \(580 \mathrm{m}\) straight into the air (a Hawaiian record). What was the lava's exit velocity in meters per second? In kilometers per hour? (Hint: If \(v_{0}\) is the exit velocity of a particle of lava, its height \(t\) seconds later will be \(s=v_{0} t-4.9 t^{2} \mathrm{m} .\) Begin by finding the time at which \(d s / d t=0 .\) Neglect air resistance.)

If \(r=\sin (f(t)), f(0)=\pi / 3,\) and \(f^{\prime}(0)=4,\) then what is \(d r / d t\) at \(t=0 ?\)

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