Problem 328 - Calculus Volume 1

Question

Consider a lifeguard at a circular pool with diameter \(40 \mathrm{~m}\). He must reach someone who is drowning on the exact opposite side of the pool, at position \(C\). The lifeguard swims with a speed \(v\) and runs around the pool at speed \(w=3 v\). Find at what angle \(\theta\) the lifeguard should swim to reach the drowning person in the least amount of time.

Step-by-Step Solution

Answer

The lifeguard should swim at an angle \(\theta = \cos^{-1}(\frac{1}{6})\).

Step 1: Understand the Scenario

The problem involves a circular pool with a diameter of 40 meters, so the radius is 20 meters. The lifeguard needs to reach a person on the opposite side of the pool by swimming and then running. The angle \(\theta\) is the angle at which the lifeguard should swim toward the opposite side to minimize time.

Step 2: Define the Distances and Speed

The lifeguard swims from point \(A\) on the pool's edge to point \(B\), covering a distance \(d_{AB}\), and then runs from \(B\) to \(C\), covering \(d_{BC}\), which is the arc of the circle. The distance AB is \(2r\sin(\theta)\), and the arc length BC is \(r\theta\). The swimming speed is \(v\), and the running speed is \(w = 3v\).

Step 3: Express Time in Terms of θ

The time to swim (\(T_{swim}\)) is given by \( \frac{2r\sin(\theta)}{v} \), and the time to run (\(T_{run}\)) is \( \frac{r\theta}{3v} \). The total time \(T\) is the sum of these times, \(T = \frac{2r\sin(\theta)}{v} + \frac{r\theta}{3v}\).

Step 4: Differentiate Total Time Expression

Differentiate the expression for total time \(T\) with respect to \(\theta\): \(\frac{dT}{d\theta} = \frac{2r\cos(\theta)}{v} - \frac{r}{3v}\). Set this derivative to zero to find the critical points: \(\frac{2r\cos(\theta)}{v} = \frac{r}{3v}\).

Step 5: Solve for θ

Solving \(2\cos(\theta) = \frac{1}{3}\) gives \(\cos(\theta) = \frac{1}{6}\). Thus, \(\theta = \cos^{-1}\left(\frac{1}{6}\right)\) is the angle at which the lifeguard should swim to minimize the time.

Key Concepts

Calculus

Calculus plays a pivotal role in solving optimization problems like those faced by our lifeguard. Calculus is a branch of mathematics that deals with change and motion, precisely underlining processes such as finding extrema (maximum or minimum values). For this problem, calculus helps us determine the angle \( \theta \) at which the time taken is minimized.

Understanding calculus allows us to find the critical points of a function by taking its derivative. A critical point is a moment where the function's slope, or rate of change, is zero or undefined. In the context of finding the optimal angle \( \theta \), the calculus process involves identifying a critical point where the total time \( T \) is minimized.

The differentiation process reveals where the function is increasing or decreasing. By setting the derivative equal to zero, as we did when calculating \( \frac{dT}{d\theta} = 0 \), we can solve for \( \theta \) to find the point that minimizes travel time for the lifeguard. Using this method ensures the lifeguard swims and runs in the most time-efficient way possible.

Trigonometry

Trigonometry involves the study of angles and their relationships in triangles, and it is crucial in understanding the pool scenario. The lifeguard's path is dictated by the angle \( \theta \), influencing the distances swum and run around the pool.

Key trigonometric functions include sine and cosine, which relate a triangle's angles to the proportions of its sides. In this exercise, we use the sine function to express the swimming distance: \( 2r\sin(\theta) \). Similarly, by employing the cosine function, we calculate \( \cos(\theta) \) to find the angle when solving for the critical point.

The equations involving sine and cosine functions are directly tied to the circular nature of the pool. These functions allow us to effectively break down the geometrical space into understandable lengths and paths, helping us optimize the lifeguard's movement.

Derivatives

Derivatives are a fundamental aspect of calculus. They represent the rate of change of a function with respect to one of its variables. In our problem, differentiating the time function \( T \) with respect to \( \theta \) helps us determine how changes in the angle affect the total time.

• The derivative \( \frac{dT}{d\theta} \) tells us how fast the overall time \( T \) changes as \( \theta \) changes.
• Setting the derivative to zero, \( \frac{2r\cos(\theta)}{v} - \frac{r}{3v} = 0 \), helps us find critical points where time is minimized.
• The point where \( \cos(\theta) = \frac{1}{6} \) was determined by solving the derivative equation for \( \theta \).

By analyzing these points, we ensure that the lifeguard reaches the drowning person as quickly as possible, demonstrating the power of derivatives in finding optimal solutions to real-world problems.