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Magazine Chapter 4: Problem 7
An F-22 aircraft is flying at \(500 \mathrm{mph}\) with an elevation of \(10,000 \mathrm{ft}\) on a straight-line path that will take it directly over an anti- aircraft gun. How fast must the gun be able to turn to accurately track the aircraft when the plane is: (a) 1 mile away? (b) \(1 / 5\) mile away? (c) Directly overhead?
Short Answer
Expert verified
Gun turns at 0.0174 rad/s (1 mile away), 0.935 rad/s (1/5 mile away), infinite directly overhead.
Step by step solution
01
Understand the Set-Up of the Problem
We model the scenario as a right triangle, where the hypotenuse is the line of sight from the gun to the plane. The vertical side is the elevation of the plane from the gun, and the horizontal side is the ground distance between the gun and the point on the ground vertically below the plane.
02
Define the Variables
Let \( h = 10,000 \text{ ft} \) be the height of the plane. \( x \) be the horizontal distance from the plane to the gun in feet. The angle \( \theta \) is the angle between the ground and the line of sight from the gun to the plane. We are looking for \( \frac{d\theta}{dt} \), the rate at which the gun must turn.
03
Relate Distance with the Angle
The tangent of angle \( \theta \), \( \tan{\theta} \), can be expressed as \( \frac{h}{x} \). Thus, \( \theta = \tan^{-1}{\left( \frac{h}{x} \right)} \).
04
Differentiate the Relation to Find the Rate of Change
Differentiate \( \theta = \tan^{-1}{\left( \frac{h}{x} \right)} \) with respect to time \( t \): \[ \frac{d\theta}{dt} = \frac{1}{1 + \left( \frac{h}{x} \right)^2} \cdot \left( -\frac{h}{x^2} \right) \cdot \frac{dx}{dt} \]
05
Express the Ground Speed in Appropriate Units
The speed of the plane \( \frac{dx}{dt} = 500 \text{ mph} \). Convert this to feet per second: \[ 500 \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1}{3600 \text{ s}} = 733.ar{3} \text{ ft/s} \].
06
Solve Part (a): When the Plane is 1 Mile Away
At 1 mile away, \( x = 5280 \text{ ft} \). Substitute into the differentiated equation:\[ \frac{d\theta}{dt} = \frac{1}{1 + \left( \frac{10000}{5280} \right)^2} \cdot \left( -\frac{10000}{(5280)^2} \right) \cdot 733.3 \]Solving gives: \[ \frac{d\theta}{dt} \approx -0.0174 \text{ rad/s} \] indicating clockwise rotation.
07
Solve Part (b): When the Plane is 1/5 Mile Away
At \( \frac{1}{5} \) mile away, \( x = 1056 \text{ ft} \). Substitute:\[ \frac{d\theta}{dt} = \frac{1}{1 + \left( \frac{10000}{1056} \right)^2} \cdot \left( -\frac{10000}{(1056)^2} \right) \cdot 733.3 \]Solving gives: \[ \frac{d\theta}{dt} \approx -0.935 \text{ rad/s} \] showing a faster rotation is needed.
08
Solve Part (c): When the Plane is Directly Overhead
When directly overhead, the horizontal distance \( x \) is 0, which mathematically leads to an undefined \( \theta \). Physically, it suggests rapid change to a new direction is needed, hence \( \frac{d\theta}{dt} \to \infty \).
09
Conclusion: Summarize the Results
Therefore, the gun needs to rotate at approximately \( 0.0174 \text{ rad/s} \) when the plane is 1 mile away, \( 0.935 \text{ rad/s} \) when 1/5 mile away, and an infinitely large rate when directly overhead.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Related Rates
Related rates problems involve finding the rate at which one quantity changes in relation to another quantity. They are critical in understanding how different variables interact over time. In this exercise, the related rates concept is applied to track how quickly the anti-aircraft gun must rotate to follow the plane's path. The gun needs to account for the changing distances and angles as the plane approaches.
The key steps when solving related rates problems include:
The key steps when solving related rates problems include:
- Identifying all relevant quantities and their relationships.
- Defining the rate of change for each variable with respect to time.
- Utilizing implicit differentiation to connect these rates.
Trigonometric Functions
Trigonometric functions are crucial tools when dealing with problems that involve angles and distances, such as this aircraft exercise. Here, the tangent function, \(\tan(\theta)\), relates the vertical height of the aircraft to its horizontal distance from the gun. This is captured in the equation \(\tan(\theta) = \frac{h}{x}\).
Trig functions are useful for:
Trig functions are useful for:
- Describing angles in terms of ratios of sides in right triangles.
- Providing a way to calculate angle measures.
- Offering derivatives that help in determining rates of change in related rates problems.
Differentiation
Differentiation is a fundamental concept in calculus, providing a tool for finding the rate at which a function changes at any point. In the aircraft problem, differentiation helps calculate the rate of change of the angle \(\theta\) with respect to time as the plane moves.
The differentiation process often involves:
The differentiation process often involves:
- Applying chain rules to compute the derivative of composite functions.
- Using implicit differentiation to handle equations involving multiple variables.
- Substituting known values to find specific rates of change.
Tangents and Angles
Tangents and angles are at the heart of this exercise. The tangent function specifically relates the angle \(\theta\) to the plane's height and ground distance. Understanding how tangents link angles to these distances is key to solving problems like the anti-aircraft scenario.
In problems involving tangents:
In problems involving tangents:
- The angle of interest is often between the line of sight and the horizontal ground.
- Changes in this angle dictate the rotation speed of an object, such as a gun turret.
- These changes are calculated using previously discussed trigonometric and differentiation techniques.
