Problem 328 - Calculus Volume 1

Question

For the following problems, 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 \(\theta = \cos^{-1}(\dfrac{1}{3})\).

Step 1: Understand the Problem

The problem involves finding the optimal path a lifeguard should take to minimize the time to reach a person drowning at point C, directly across a circular pool. The lifeguard can swim with speed \(v\) and run around the pool at speed \(w = 3v\).

Step 2: Define Variables and Equations

Let the diameter of the pool be 40 m, making the radius 20 m. Define \(\theta\) as the angle between the line from the initial point to the center and the lifeguard's swimming path. The distance to swim is \(20 \cdot \sin(\theta)\), and the distance to run is the perimeter segment \(20 \cdot (\pi - \theta)\).

Step 3: Compute Time for Each Path Segment

Time to swim is given by \(t_s = \dfrac{20 \sin(\theta)}{v}\). Time to run is \(t_r = \dfrac{20 (\pi - \theta)}{3v}\). Total time \(T(\theta) = t_s + t_r = \dfrac{20 \sin(\theta)}{v} + \dfrac{20 (\pi - \theta)}{3v}\).

Step 4: Simplify and Differentiate Total Time Function

Simplify \(T(\theta)\) to \(\dfrac{20}{v} \left( \sin(\theta) + \dfrac{\pi - \theta}{3} \right)\). Differentiate with respect to \(\theta\): \(T'(\theta) = \dfrac{20}{v} \left( \cos(\theta) - \dfrac{1}{3} \right)\).

Step 5: Find Critical Points

Set the derivative equal to zero: \(\cos(\theta) = \dfrac{1}{3}\). Solve for \(\theta\): \(\theta = \cos^{-1}(\dfrac{1}{3})\).

Step 6: Confirm the Result is a Minimum

Check the second derivative \(T''(\theta) = -\dfrac{20}{v} \sin(\theta)\). Since \(\sin(\theta) > 0\) in the first quadrant, \(T''(\theta) < 0\) indicating a minimum.

Key Concepts

Critical Points

In calculus, a critical point is where the derivative of a function is zero or undefined. It's an important concept for determining where your function reaches local maxima, minima, or saddle points. The lifeguard problem involves finding the critical point of the total time function, which shows how much time it takes to swim and run to the other side of the pool. By differentiating this function, the goal is to find an angle \( \theta \) that minimizes the time, allowing us to determine the optimal path.- Setting the derivative to zero \( T'(\theta) = 0 \)- Solving this helps find the critical angle \( \theta \)Understanding critical points helps us solve optimization problems by examining when functions change direction or behavior.

Trigonometric Functions

Trigonometric functions are crucial in problems involving angles and circular motion, like this pool problem. They help describe the relationship between the sides and angles of triangles, and also solve real-world scenarios involving circular paths.In this exercise, the sine and cosine functions are used extensively:- \( \sin(\theta) \) helps calculate the distance the lifeguard swims directly across the pool.- \( \cos(\theta) \) comes in handy to determine the rate of change for the optimal angle, since it appears in the derivative used to find critical points.Understanding the roles trigonometric functions play is crucial in simplifying \( T(\theta) \) , leading us toward the minimum time solution.

Differentiation

Differentiation is a fundamental operation in calculus used to find rates of change. It plays a key role in optimizing functions, like finding the least time to reach from one point to another. For the lifeguard example, we need to differentiate the total time function \( T(\theta) \). The derivative \( T'(\theta) \)shows how the function's value changes with respect to angle \( \theta \).- Simplify the total time: \( T(\theta) = \dfrac{20}{v} \left( \sin(\theta) + \dfrac{\pi - \theta}{3} \right) \)- Differentiate it with respect to \( \theta \): \( T'(\theta) = \dfrac{20}{v} \left( \cos(\theta) - \dfrac{1}{3} \right) \)Differentiation allows us to find the optimal angle by setting \( T'(\theta) = 0 \), ensuring we identify points where the time can be minimized.

Problem Solving

Problem-solving in calculus combines logical thinking, creative reasoning, and mathematical techniques to arrive at a solution.For this problem:- Understand the situation and constraints: Identify the need to reach point C across the pool.- Break down the problem: Recognize that the lifeguard can swim and run - each with different speeds, translating into different parts of the function.- Develop a solution strategy: Formulate equations and express them using known variables like \( v \) and \( \theta \).- Analyze and solve: Differentiate, find critical points, then confirm solutions represent a minimum.Mastering problem-solving in calculus involves comprehending the problem, simplifying the solution path, and verifying the correctness of the approach.

Minimization Problem

A minimization problem aims to find the minimum value of a function within a given set of constraints. Here, it involves determining the least time for the lifeguard to reach the opposite side of the pool.The steps to solve the minimization aspect of this problem are:- Define the objective function: In this case, it's the total time \( T(\theta) \) for swimming and running.- Differentiate the function and find derivative(s): Use differentiation to explore where the function's rate of change is zero.- Solve for zeroes of the derivative: Find \( \theta \) where \( T'(\theta) = 0 \); ensure it represents a minimum by checking the second derivative \( T''(\theta) \).- Check for contextual validity: Ensure the solution fits within the real-world context – here, the angles must be realistic for the swimming path around the circular pool.Minimization problems require balancing constraints and objectives, ultimately reaching the most efficient solution.