Chapter 3: Problem 16
An aircraft spotter observes a plane flying at a constant altitude of 4000 feet toward a point directly above her head. She notes that when the angle of elevation is \(\frac{1}{2}\) radian it is increasing at a rate of \(\frac{1}{10}\) radian per second. What is the speed of the airplane?
Short Answer
Expert verified
The speed of the airplane is approximately 58.78 feet/second.
Step by step solution
01
Identify Given Values
We begin by noting the values given in the problem. The altitude of the plane is 4000 feet, the angle of elevation \( \theta \) is \( \frac{1}{2} \) radian, and the rate of change of the angle of elevation is \( \frac{d\theta}{dt} = \frac{1}{10} \) radian/second.
02
Understanding Relationships
The relationship between the altitude of the plane \(h\), the horizontal distance \(x\) from the spotter to the point under the plane, and the angle of elevation \(\theta\) is given by the tangent function: \( \tan(\theta) = \frac{h}{x} \).
03
Differentiate the Tan Function
Differentiate \( \tan(\theta) = \frac{h}{x} \) with respect to time \(t\). By using the chain rule, we get \( \sec^2(\theta) \frac{d\theta}{dt} = -\frac{h}{x^2} \frac{dx}{dt} \), where \( \frac{dx}{dt} \) is the speed of the airplane.
04
Substitute Known Values
We know \( h = 4000 \) feet and that \( \theta = \frac{1}{2} \) radian. So, \( \sec^2(\theta) = 1 + \tan^2(\theta) \) becomes \( 1 + \left(\frac{4000}{x}\right)^2 \) when substituting \( \tan(\theta) \). Substitute \( \theta = \frac{1}{2} \) rad into the equation, and also plug in \( \frac{d\theta}{dt} = \frac{1}{10} \) and solve for \( \frac{dx}{dt} \).
05
Calculate the Horizontal Distance
Using \( \tan(\theta) = \frac{h}{x} \), substitute \( \theta = \frac{1}{2} \) and \( h = 4000 \) to find \( x \). Thus, \( \tan(\frac{1}{2}) = \frac{4000}{x} \) leads to \( x = \frac{4000}{\tan(\frac{1}{2})} \).
06
Solve for Plane's Speed
Now, calculate \( \frac{dx}{dt} \) from the equation derived in the differentiation step, which is \( \sec^2(\frac{1}{2}) \cdot \frac{1}{10} = -\frac{4000}{x^2} \cdot \frac{dx}{dt} \). Now, compute and solve for \( \frac{dx}{dt} \) using the calculated \( x \) and plugging back into the equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation Technique
Differentiation is an essential calculus technique used to find the rate at which quantities change. In the given exercise, we applied differentiation to solve a problem involving a moving aircraft. This involved identifying given values, understanding mathematical relationships, and applying derivative rules to find an unknown rate, specifically the speed of the airplane.
In this context, the differentiation was applied to the trigonometric function, the tangent function, because it described the relationship between the angle of elevation and the horizontal distance under the plane. Having been given the rate of change of an angle, we required the derivative of the tangent function to relate various rates of change, particularly focusing on how the change in angle (dθ/dt) translates to a change in horizontal distance (dx/dt).
The chain rule, a powerful differentiation rule, was key here. It allowed us to differentiate a function that changes with respect to another changing variable—in this case, time. Mastering differentiation techniques such as this is crucial for solving related rates problems effectively.
In this context, the differentiation was applied to the trigonometric function, the tangent function, because it described the relationship between the angle of elevation and the horizontal distance under the plane. Having been given the rate of change of an angle, we required the derivative of the tangent function to relate various rates of change, particularly focusing on how the change in angle (dθ/dt) translates to a change in horizontal distance (dx/dt).
The chain rule, a powerful differentiation rule, was key here. It allowed us to differentiate a function that changes with respect to another changing variable—in this case, time. Mastering differentiation techniques such as this is crucial for solving related rates problems effectively.
Tangent Function Relationship
In trigonometry, the tangent function is particularly useful when dealing with right triangles and angles of elevation. For mathematical problems involving observation angles, such as in our exercise, the tangent of the angle gives the ratio of the opposite side (altitude) to the adjacent side (horizontal distance). In mathematical terms:
\[ \tan(\theta) = \frac{h}{x} \]
Here, \( \theta \) is the angle of elevation, \( h \) is the altitude of the plane (4000 feet), and \( x \) is the horizontal distance from the observer directly underneath the aircraft. This tangent relationship is foundational in finding changing rates because, with an increasing angle, both the tangent value and the physical relationships change.
Understanding this relationship helps us link the observed angle with the geometric dimensions of the situation. It informs us how changes in the observed angle translate into changes in the physical layout, which is crucial for solving related rates problems involving motion.
\[ \tan(\theta) = \frac{h}{x} \]
Here, \( \theta \) is the angle of elevation, \( h \) is the altitude of the plane (4000 feet), and \( x \) is the horizontal distance from the observer directly underneath the aircraft. This tangent relationship is foundational in finding changing rates because, with an increasing angle, both the tangent value and the physical relationships change.
Understanding this relationship helps us link the observed angle with the geometric dimensions of the situation. It informs us how changes in the observed angle translate into changes in the physical layout, which is crucial for solving related rates problems involving motion.
Rate of Change Applications
In related rates problems, like aircraft speed questions, understanding how quantities change relative to each other in real-time is key. The concept of rate of change is not just abstract—it's highly applicable in everyday contexts, driving decisions in physics, engineering, and even economics.
For our airplane scenario, we were interested in how fast the horizontal distance, \( x \), was changing, known as \( \frac{dx}{dt} \). Given the rate of change of the angle \( \frac{d\theta}{dt} \), we wanted to find \( \frac{dx}{dt} \) using the tangent relationship mentioned earlier.
By substituting known values into our derived equation \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{h}{x^2} \cdot \frac{dx}{dt} \), this relationship allowed us to compute the speed of the airplane, illustrating how calculus directly applies to motion and measurement problems.
Such applications of calculus enable us to solve problems across a breathtaking range of disciplines and scenarios, determining rates of change thoughtfully and effectively.
For our airplane scenario, we were interested in how fast the horizontal distance, \( x \), was changing, known as \( \frac{dx}{dt} \). Given the rate of change of the angle \( \frac{d\theta}{dt} \), we wanted to find \( \frac{dx}{dt} \) using the tangent relationship mentioned earlier.
By substituting known values into our derived equation \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{h}{x^2} \cdot \frac{dx}{dt} \), this relationship allowed us to compute the speed of the airplane, illustrating how calculus directly applies to motion and measurement problems.
Such applications of calculus enable us to solve problems across a breathtaking range of disciplines and scenarios, determining rates of change thoughtfully and effectively.
Calculus Problem Solving
Solving calculus problems systematically is an invaluable skill. The key is in dividing the problem into manageable steps and using mathematical principles accordingly. In the given problem, a clear convergence of trigonometry and calculus was necessary.
We started by identifying the relationships among distance, altitude, and angle through the tangent function. Next, we used calculus to differentiate this relationship with respect to time, highlighting the chain rule as our tool for handling composite functions.
Lastly, substituting the known values and solving for the unknown involved algebraic manipulation and attention to detail. This logical sequence, some might call a calculus problem solving routine, not only helps in handling complex related rates problems but also builds foundational skills for tackling future mathematical challenges.
Developing structured problem solving approaches is important in streamlining your solutions in calculus and offers clarity in thought. Always keep practice consistent and break down problems carefully for easier management and more successful outcomes.
We started by identifying the relationships among distance, altitude, and angle through the tangent function. Next, we used calculus to differentiate this relationship with respect to time, highlighting the chain rule as our tool for handling composite functions.
Lastly, substituting the known values and solving for the unknown involved algebraic manipulation and attention to detail. This logical sequence, some might call a calculus problem solving routine, not only helps in handling complex related rates problems but also builds foundational skills for tackling future mathematical challenges.
Developing structured problem solving approaches is important in streamlining your solutions in calculus and offers clarity in thought. Always keep practice consistent and break down problems carefully for easier management and more successful outcomes.
