Question

A revolving beacon light is located on an island and is 2 miles away from the nearest point \(P\) of the straight shoreline of the mainland. The beacon throws a spot of light that moves along the shoreline as the beacon revolves. If the speed of the spot of light on the shoreline is \(5 \pi\) miles per minute when the spot is 1 mile from \(P\), how fast is the beacon revolving?

Short answer

The beacon is revolving at \( 2\pi \) radians per minute.

Step-by-step solution

Set Up the Problem

Visualize the scenario as a right triangle where the beacon (point on the island) is one vertex, point \( P \) on the shoreline is another vertex, and the spot of light (\( x \) miles away from \( P \) along the shore) is the third vertex of the triangle. The distance from the beacon to \( P \) is 2 miles (the hypotenuse), and \( x \) is the horizontal leg of the triangle.

Express the Relationship Using Trigonometry

Let \( \theta \) be the angle at the beacon between the line connecting the beacon to \( P \) and the rotating light (hypotenuse). The position of the spot of light along the shoreline is given by \( x = 2\tan \theta \).

Differentiate with Respect to Time

Differentiate \( x = 2\tan \theta \) with respect to time \( t \), yielding \( \frac{dx}{dt} = 2 \sec^2 \theta \frac{d\theta}{dt} \).

Solve for \( \frac{d\theta}{dt} \)

Given that \( \frac{dx}{dt} = 5\pi \) miles per minute when \( x = 1 \), first find \( \theta \) using \( x=2\tan\theta \Rightarrow 1=2\tan\theta \Rightarrow \tan\theta=\frac{1}{2} \). Thus, \( \theta = \tan^{-1}(\frac{1}{2}) \). Use the trigonometric identity \( \sec^2\theta = 1 + \tan^2\theta \) to find \( \sec^2\theta = 1 + \left(\frac{1}{2}\right)^2 = \frac{5}{4} \). Substitute \( \sec^2\theta \) into \( \frac{dx}{dt} = 2 \sec^2 \theta \frac{d\theta}{dt} \) with \( \frac{dx}{dt} = 5\pi \) to solve for \( \frac{d\theta}{dt} \): \( 5\pi = 2 \cdot \frac{5}{4} \cdot \frac{d\theta}{dt} \).

Calculate \( \frac{d\theta}{dt} \)

Simplify \( 5\pi = \frac{5}{2} \frac{d\theta}{dt} \) to find \( \frac{d\theta}{dt} = \frac{5\pi}{5/2} = 2\pi \) radians per minute.

Key concepts

Trigonometry in Calculus

The problem involving the revolving beacon introduces us to the fascinating interplay between trigonometry and calculus. When you visualize the scenario involving the beacon and the shoreline, you are faced with a right triangle. Trigonometry becomes essential here as it helps us describe the relationship between the angle of rotation of the beacon and the position of the light spot along the shoreline.
The key trigonometric function used in this problem is the tangent. The formula we use is \(x = 2\tan \theta\), where \(x\) is the distance from point \(P\) along the shoreline. This expression allows us to relate the angle \(\theta\) that the beacon makes to one of the triangle's sides. Understanding and properly applying trigonometric functions is crucial in solving such problems as they help us build the mathematical model needed for further calculus operations.

Differentiation with Respect to Time

In calculus, differentiation with respect to time is a powerful method to determine how quantities change over time. In our beacon example, we want to understand how the change in the angle \(\theta\) affects the movement of the light spot along the shoreline.
We begin by differentiating the trigonometric expression \(x = 2\tan \theta\) with respect to time \(t\). This process, known as implicit differentiation, gives us \(\frac{dx}{dt} = 2 \sec^2 \theta \frac{d\theta}{dt}\). Here, \(\sec^2 \theta\) is the derivative of \(\tan \theta\), and \(\frac{d\theta}{dt}\) represents the rate at which the beacon is rotating. This setup helps us connect the rate of angular change \(\frac{d\theta}{dt}\) with the rate of light spot movement \(\frac{dx}{dt}\). By finding these derivatives, we can solve for changes over time, locally representing dynamic situations in calculus.

Related Rates

Related rates problems, like the one involving the beacon, achieve their complexity because they deal with how two or more related quantities change over time. The task involves using calculus to determine unknown rates of change when others are known.
In this exercise, the known rate is the speed of the spot of light along the shoreline, \(\frac{dx}{dt} = 5\pi\), and we want to find how fast the angle \(\theta\) is changing, \(\frac{d\theta}{dt}\). The established relationship \(\frac{dx}{dt} = 2 \sec^2 \theta \frac{d\theta}{dt}\) allows us to substitute known values and solve for \(\frac{d\theta}{dt}\).

• This interconnected calculation highlights why understanding the geometry of the situation and the relevant rates is vital. For many students, wrapping their heads around the idea that these rates are proportionally related through time is one of the most crucial steps in mastering related rates problems.

Geometry in Calculus

Geometry plays a fundamental role in this calculus problem by providing the framework for modeling the situation. Here, the points form a right triangle that is pivotal to solving the problem using calculus and trigonometry.

• Visualizing geometric shapes helps translate real-world scenarios into mathematical equations. Without grasping this spatial relationship, applying calculus effectively could be challenging.
• The hypotenuse of the right triangle is constant at 2 miles, while one leg of the triangle changes with \(x\), the spot of light on the shoreline. This geometric understanding empowers us to apply differential calculus formulas correctly.
• Such integration of geometry and calculus allows students to solve real-world problems, highlighting how mathematics describes physical phenomena accurately.