/*! 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 24 Commercial air traffic Two comme... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 24

Commercial air traffic Two commercial airplanes are flying at an altitude of 40,000 ft along straight-line courses that intersect at right angles. Plane \(A\) is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nautical mile is 2000 yd). Plane \(B\) is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when \(A\) is 5 nautical miles from the intersection point and \(B\) is 12 nautical miles from the intersection point?

Short Answer

Expert verified
The distance between the planes is decreasing at approximately 614 knots.

Step by step solution

01

Understanding the Problem

To find how the distance between the planes is changing, we need to consider the movement of both planes approaching the intersection at right angles. Plane A is approaching at 442 knots and Plane B at 481 knots.
02

Identify Given Values

Let the distance of Plane A from the intersection be represented by \( x \), and the distance of Plane B from the intersection be \( y \). When \( x = 5 \) nautical miles and \( y = 12 \) nautical miles, Plane A is approaching at a rate of \( \frac{dx}{dt} = -442 \) knots, and Plane B at \( \frac{dy}{dt} = -481 \) knots.
03

Set Up the Distance Equation

Using the Pythagorean theorem, the distance between the planes, \( z \), is given by \[ z = \sqrt{x^2 + y^2} \]. We aim to find \( \frac{dz}{dt} \), the rate of change of \( z \).
04

Differentiate the Distance Equation

Differentiate both sides of \( z = \sqrt{x^2 + y^2} \) with respect to time \( t \): \[ \frac{dz}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \].
05

Plug In Given Distances and Rates

When \( x = 5 \) and \( y = 12 \), \[ \frac{dz}{dt} = \frac{1}{\sqrt{5^2 + 12^2}} \cdot (5)(-442) + (12)(-481) \]. Calculate the values to determine \( \frac{dz}{dt} \).
06

Calculate Distance and Solve

First, find \( \sqrt{25 + 144} = \sqrt{169} = 13 \). Then, substitute into the equation: \[ \frac{dz}{dt} = \frac{1}{13} \cdot (-2210 - 5772) = \frac{-7982}{13} \approx -614 \text{ knots} \].

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Key Concepts

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

Understanding Rate of Change
The concept of "rate of change" is crucial in calculus, especially when dealing with motion problems. In the context of airplanes flying, the rate of change refers to how quickly the distance between two planes decreases or increases per unit of time.
Here are the main key points to grasp this concept:
  • The rate of change is often represented by derivatives in calculus. This is because a derivative signifies how a particular quantity changes with respect to another.
  • In the problem at hand, we are interested in how the distance between planes changes over time, labeled as \(\frac{dz}{dt}\).
  • This rate of change depends on how each plane moves, which is described by their respective speeds: Plane A at 442 knots and Plane B at 481 knots.
Understanding this helps us solve complex motion problems involving multiple moving parts.
Applying the Pythagorean Theorem
The Pythagorean theorem is an essential mathematical tool, especially for problems that involve right-angled intersections. In the problem where two planes approach an intersection at right angles, this theorem becomes highly relevant.
Here is how it applies:
  • The theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: \(a^2 + b^2 = c^2\).
  • In our exercise with planes A and B, their positions relative to the intersection form a right triangle, where the hypotenuse is the distance between the planes.
  • By letting \(x\) be the distance from Plane A to the intersection and \(y\) for Plane B, the distance between the planes can be expressed with the formula: \(z = \sqrt{x^2 + y^2}\).
This theorem allows us to link the planes' straight-line distances to the intersection with the overall distance between them.
Differentiation in Motion Problems
Differentiation is a fundamental calculus concept used to find how a quantity changes. In motion problems, like the airplanes problem, differentiation helps us determine the rate at which distances change.
Key aspects include:
  • Differentiation involves applying derivatives to functions. For instance, to find the rate of change of distance \(z\), we differentiate the distance formula \(z = \sqrt{x^2 + y^2}\) with respect to time \(t\).
  • This gives us the formula: \(\frac{dz}{dt} = \frac{1}{\sqrt{x^2 + y^2}} \, (x \frac{dx}{dt} + y \frac{dy}{dt})\).
  • By substituting known values, such as the current distances and speeds of the planes (\(\frac{dx}{dt} = -442\) knots and \(\frac{dy}{dt} = -481\) knots), we can calculate the exact rate at which the distance between the planes changes.
Differentiation, thus, translates tangible speed into a clear rate of spatial change.
Converting Nautical Miles to Knots
Understanding the units of nautical miles and knots is essential when dealing with aviation and maritime scenarios.
Here's how these units interact:
  • A "nautical mile" is a unit of measurement used in aviation and maritime contexts, measuring length and equaling 2000 yards or approximately 1.15 miles.
  • "Knots" is a unit of speed, representing one nautical mile per hour. It is widely used in marine and aviation fields to denote speed.
  • In our problem, plane speeds are given in knots, which correlates directly to the distance in nautical miles they cover per hour. This simplifies calculations when determining how fast the planes are closing in on each other or changing distances.
Converting between these units is generally straightforward, but understanding their definitions is crucial for solving real-world problems efficiently.

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Most popular questions from this chapter

A balloon and a bicycle \(A\) balloon is rising vertically above a level, straight road at a constant rate of 1 \(\mathrm{ft} / \mathrm{sec} .\) Just when the balloon is 65 \(\mathrm{ft} /\) above the ground, a bicycle moving at a constant rate of 17 \(\mathrm{ft} / \mathrm{sec}\) passes under it. How fast is the distance \(s(t)\) between the bicycle and balloon increasing 3 sec later?

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x+\tan \left(\frac{y}{x}\right)=2, \quad P\left(1, \frac{\pi}{4}\right) \end{equation}

Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=\sin 2 x, \quad x_{0}=\pi / 2$$

Cardiac output In the late 1860 \(\mathrm{s}\) , 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 \(\mathrm{L} / \mathrm{min} .\) $$\begin{array}{c}{\text { Your cardiac output can be calculated with the formula }} \\ {y=\frac{Q}{D}}\end{array}$$ 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 \(D=97-56=41 \mathrm{ml} / \mathrm{L},\) $$y=\frac{233 \mathrm{ml} / \min }{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min},$$ 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?
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