Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

An airplane is flying at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over an observer on the ground. How fast is the angle of elevation of the observer's line of sight increasing when the distance from her to the plane is 3 miles? Give your result in radians per minute.

Step-by-Step Solution

Answer

The angle of elevation is increasing at \( \frac{20}{9} \) radians per minute.

Step 1: Understand the Problem

The airplane is flying at an altitude of 2 miles and the observer is on the ground. When the plane is 3 miles away from the observer, we need to find how fast the angle of elevation is increasing.

Step 2: Define Variables

Let \( \theta \) be the angle of elevation, \( x \) be the horizontal distance from the observer to the point below the airplane on the ground, and \( y = 2 \) miles be the constant altitude of the airplane. We know the airplane is moving at 600 miles per hour, which equals 10 miles per minute.

Step 3: Set Up the Relationship

The distance from the observer to the plane forms a right triangle with legs \( x \) and \( y = 2 \), and hypotenuse 3 miles. We can write this using the tangent function: \( \tan(\theta) = \frac{y}{x} = \frac{2}{x} \).

Step 4: Differentiate with Respect to Time

Use implicit differentiation with respect to time \( t \). Differentiate both sides of \( \tan(\theta) = \frac{2}{x} \) to get \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{2}{x^2} \cdot \frac{dx}{dt} \).

Step 5: Solve for \( \frac{d\theta}{dt} \)

When the plane is 3 miles from the observer, apply the Pythagorean theorem: \( x^2 + 2^2 = 3^2 \Rightarrow x = \sqrt{5} \). Also, \( \frac{dx}{dt} = -10 \) miles per minute. Substitute these into the differentiated equation: \[ \sec^2(\theta) \frac{d\theta}{dt} = -\frac{2}{(\sqrt{5})^2}(-10) \] which simplifies to \( \sec^2(\theta) \frac{d\theta}{dt} = \frac{4}{5} \cdot 10 \).

Step 6: Calculate \( \sec(\theta) \)

Since \( \sec(\theta) = \frac{hypotenuse}{adjacent} = \frac{3}{\sqrt{5}} \), then \( \sec^2(\theta) = \left(\frac{3}{\sqrt{5}}\right)^2 = \frac{9}{5} \).

Step 7: Find \( \frac{d\theta}{dt} \)

Using \( \sec^2(\theta) = \frac{9}{5} \) in \( \sec^2(\theta) \frac{d\theta}{dt} = 4 \), solve for \( \frac{d\theta}{dt} \): \[ \frac{9}{5} \frac{d\theta}{dt} = 4 \Rightarrow \frac{d\theta}{dt} = \frac{4}{\frac{9}{5}} = \frac{20}{9} \] radians per minute.

Key Concepts

Angle of Elevation

The angle of elevation is a familiar term, often used when observing objects positioned above the observer's horizontal line of sight. Imagine you are standing on the ground and looking up at an airplane. The angle formed between your line of sight and the horizontal line is your angle of elevation. In this scenario, the airplane is maintaining a constant height of 2 miles.
The angle of elevation is crucial in determining how quickly this angle changes as the airplane approaches or recedes. In mathematical terms, this involves determining how fast the angle increases when the airplane is 3 miles away from the observer. To solve such problems, angles of elevation often come into play as part of related rates problems, which combine understanding of right triangles and trigonometric functions.
By visualizing this angle, it is easier to grasp how different variables are interconnected, helping to set up the needed equations for analysis.

Trigonometric Functions

Trigonometric functions are the mathematical principles used when dealing with angles and their relationships in triangles. In the exercise, we primarily use the tangent function. This function relates the angle of elevation \( \theta \) to the right triangle formed between the airplane, the observer, and the point directly below the airplane.
The relationship is given by \( \tan(\theta) = \frac{2}{x} \), where \(x\) is the distance along the ground from the observer to the point under the airplane. Trigonometric functions like sine, cosine, and tangent help establish these relationships, which describe how angles change relative to one another.
The concept of secant \( \sec(\theta) \) also arises naturally once we differentiate the tangent function. Here, \( \sec^2(\theta) \) captures how changes in \( \theta \) relate to changes in the observer's position and speed.

Differentiation

Differentiation is a core concept in calculus that measures how a function changes as its inputs change. In the context of this exercise, we want to see how small changes in time affect the angle of elevation. This involves differentiating with respect to time \( t \).
When using related rates, it's about connecting how different quantities change together. For instance, the differentiation of \( \tan(\theta) = \frac{2}{x} \) over time gives us \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{2}{x^2} \cdot \frac{dx}{dt} \). This equation lets us solve for \( \frac{d\theta}{dt} \), the rate at which the angle changes.
Understanding differentiation is key to mastering related rates problems, as it enables the calculation of rates of change and provides insight into how all parts of the problem interconnect.

Right Triangle Applications

Right triangle applications are often encountered in real-world scenarios like this exercise, where they form a basis for applying trigonometry and calculus. Here, the observer, the airplane, and the ground form a right triangle. The known lengths and changing distances create a helpful context for solving the problem.
This right triangle allows us to apply the Pythagorean theorem: \( x^2 + 2^2 = 3^2 \) in determining the horizontal distance \( x \) when the plane is 3 miles from the observer. This is fundamental in simplifying the correlation between distance, time, and angle.
Right triangles serve as the foundation for applying trigonometric relationships and further analyzing related rates. By recognizing this geometric setup, we can translate physical observations into mathematical equations that are easier to work with.