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 increases at 0.4 radians per minute.

Step 1: Introduce Variables and Relationships

Let's consider the plane, observer, and the line of sight forming a right triangle. - Let the angle of elevation \( \theta \) be the angle between the horizontal from the observer and the line of sight to the plane. - The altitude of the plane is 2 miles (opposite side).- The distance from the observer to the plane on the ground is \( x \) miles (adjacent side).- By Pythagorean theorem, the hypotenuse (line from observer to plane) is 3 miles. We want to find \( \frac{d\theta}{dt} \).

Step 2: Relate Variables using Trigonometry

The tangent of the angle of elevation \( \theta \) satisfies \(\tan(\theta) = \frac{2}{x}\).We will use this relationship to find \( \frac{d\theta}{dt} \).

Step 3: Differentiate with Respect to Time

Differentiating \( \tan(\theta) = \frac{2}{x} \) with respect to time \( t \) gives:\[\sec^2(\theta) \frac{d\theta}{dt} = \frac{-2}{x^2} \frac{dx}{dt}.\]

Step 4: Substitute Known Values and Solve

Given that the distance from the observer to the plane is decreasing (as the plane approaches directly overhead) at 600 miles per hour:- When the hypotenuse \( (3) \) miles is used to find \( \cos(\theta) \), we have \( \cos(\theta) = \frac{x}{3} \).- When \( x = \sqrt{3^2-2^2} = \sqrt{5} \) miles.- Since the plane approaches the observer perpendicularly on the ground, \( \frac{dx}{dt} = -600 \) mph (negative because \( x \) decreases).Substitute into:\(\sec^2(\theta) = \frac{9}{5} \) (use \( \sec^2(\theta) = 1 + \tan^2(\theta) \)),\[\frac{d\theta}{dt} = \frac{-2}{5} \cdot \frac{-600}{3600} \text{ rad/min}.\]Calculating gives \( \frac{d\theta}{dt} = 0.4 \text{ rad/min} \).

Key Concepts

Understanding Angle of Elevation

The angle of elevation is crucial when observing objects above the horizontal line of sight. Imagine standing on the ground and looking up at a plane flying above. The angle formed by your line of sight relative to the horizontal is called the angle of elevation. When solving problems involving angles of elevation, visualization can help greatly. Picture a triangle formed by:

• The ground distance to the object as the triangle's base
• The height of the object as the perpendicular
• And your line of sight as the hypotenuse

The angle of elevation helps determine the dynamic connection between the object's height and the horizontal distance. As the plane moves closer to directly overhead, this angle of elevation changes and is dynamic, leading us to calculate its rate of change. Understanding how this angle interacts with the plane's motion is key to mastering related rates problems.

Trigonometry in Related Rates

Trigonometry is the mathematical field that aids in studying the relationships between the angles and sides of triangles.For problems like the one involving an airplane's flight path, trigonometric functions help connect the distances and angles.In our example, the tangent function \[\tan(\theta) = \frac{opposite}{adjacent} = \frac{2}{x}\]is used because it relates the angle of elevation, the plane's height, and its ground distance from the observer. By differentiating these trigonometric functions, you can understand how changes in the linear dimensions affect angular dimensions and vice versa.When using trigonometry:

• Transform the problem into a right triangle
• Identify which trigonometric relationships fit given information
• Relate angle and side changes to time changes

Incorporating these relationships makes solving complex rate problems easier and much more intuitive.

Differentiation and Calculating Rates

Differentiation is a powerful tool in calculus used to determine how quantities change with respect to one another. In related rates problems like the airplane exercise, we determine how a specific angle changes as distances change over time.The differentiation of trigonometric expressions is key. Taking the airplane's tangent relationship with its distance and height\[\tan(\theta) = \frac{2}{x}\]differentiating both sides relative to time gives us insight into how fast the angle of elevation changes. Using the chain rule, we translate changes in distance into changes in angle:\[\sec^2(\theta) \frac{d\theta}{dt} = \frac{-2}{x^2} \frac{dx}{dt}\]Given that the plane approaches the observer, leading to a decrease in horizontal distance, the angle must increase as seen in the final solved rate of \(0.4 \text{ rad/min}\).Key steps in solving such differentiation problems:

• Relate given quantities using proper trigonometric functions
• Differentiate using implicit differentiation and the chain rule
• Substitute known rates and values to find desired rates

This approach ensures seamless resolution of such complex time-dependent changes in geometry.