Problem 34 - Calculus - AP Edition

Question

A beacon that makes one revolution every \(10 \mathrm{s}\) is located on a ship anchored 4 kilometers from a straight shoreline. How fast is the beam moving along the shoreline when it makes an angle of \(45^{\circ}\) with the shore?

Step-by-Step Solution

Answer

The beam moves at a speed of \( \frac{8\pi}{5} \text{ km/s} \) along the shoreline.

Step 1: Understand the problem setup

We have a beacon on a ship which is 4 kilometers away from a straight shoreline. The beacon rotates, creating a light beam moving along the line of sight on the shore. We are asked to find out how fast the beam of light moves along the shoreline when the angle between the line to the ship and the line of sight is \(45^\circ\).

Step 2: Convert optimal information to math

Define the angle \( \theta \) between the beam and the shore. The beam makes one revolution every \(10\) seconds, corresponding to a rate of change of \( \theta \) with respect to time, \( \frac{d\theta}{dt} = \frac{2\pi}{10} \) radians/second.

Step 3: Relate the angle and distance along the shoreline

The position \( x \) of the point where the beam hits the shore, in relation to the angle \(\theta\), is \( x = 4 \tan(\theta) \). We need to find \( \frac{dx}{dt} \) when \( \theta = 45^\circ \).

Step 4: Differentiate with respect to time

Use implicit differentiation to find the rate of change \( \frac{dx}{dt} \):\[ \frac{dx}{dt} = \frac{d}{dt} (4 \tan(\theta)) = 4 \sec^2(\theta) \cdot \frac{d\theta}{dt} \]

Step 5: Calculate the value when \( \theta = 45^\circ \)

Substitute \( \theta = 45^\circ \) and \( \frac{d\theta}{dt} = \frac{2\pi}{10} \) into the equation:\[ \sec(45^\circ) = \sqrt{2} \]\[ \frac{dx}{dt} = 4 (\sqrt{2})^2 \left(\frac{2\pi}{10}\right) = 8 \left(\frac{2\pi}{10}\right) = \frac{16\pi}{10} \]

Step 6: Simplify the result

Simplify the calculation result:\[ \frac{16\pi}{10} = \frac{8\pi}{5} \]Therefore, the speed at which the beam is moving along the shoreline is \( \frac{8\pi}{5} \text{ km/s} \).

Key Concepts

Trigonometric Functions

Trigonometric functions are essential tools in solving problems involving angles and lengths in triangles. In this exercise, the tangent function is used to relate the angle of elevation \( \theta \) with the distance \( x \) along the shoreline. Specifically, the relationship is given by \( x = 4 \tan(\theta) \). This function helps us model how the distance changes as the beam rotates. Understanding the tangent function involves knowing that it is the ratio of the opposite side to the adjacent side of a right-angled triangle. In context, it connects the angle \( \theta \) and the physical distance on the shore. The trigonometric identity \( \sec(\theta) = \frac{1}{\cos(\theta)} \) is also used in implicit differentiation, as we'll see. To solve problems like this, you should be comfortable using:

• Definitions of trigonometric functions
• Trigonometric identities
• Relationships between sides and angles in triangles

Implicit Differentiation

Implicit differentiation is a technique used when a function is not easily solved for one variable in terms of another. In this problem, implicit differentiation helps find the rate at which the beam of light is moving along the shoreline, \( \frac{dx}{dt} \).Rather than solving for \( x \) explicitly in terms of \( \theta \), we observe that \( x = 4 \tan(\theta) \). By implicitly differentiating both sides with respect to time \( t \), we find \[ \frac{dx}{dt} = 4 \sec^2(\theta) \cdot \frac{d\theta}{dt} \]. This involves several steps:

• Differentiating \( \tan(\theta) \) to get \( \sec^2(\theta) \)
• Applying the chain rule, which involves differentiating inner functions
• Substituting known values for \( \theta \) and \( \frac{d\theta}{dt} \)

Implicit differentiation is very useful when dealing with related rates, where changes in one quantity affect another.

Angle of Elevation

The angle of elevation is a key concept in this exercise. It refers to the angle between the line of sight from the observer (or the light beam from the beacon) and the horizontal line, here represented by the shoreline. In this problem, the angle of elevation \( \theta \) is given as \( 45^\circ \). At this angle, certain trigonometric values become simpler: \( \tan(45^\circ) = 1 \) and \( \sec(45^\circ) = \sqrt{2} \). This makes calculations more straightforward.The understanding of angles of elevation is crucial in:

• Calculating and connecting different components of triangles
• Determining distances or changes in position
• Solving navigation problems like this one

Being comfortable with angles of elevation enables better problem solving in scenarios involving height and distances.