Problem 326 - Calculus Volume 1

Question

Consider the construction of a pen to enclose an area. You can run at a speed of \(6 \mathrm{mph}\) and swim at a speed of \(3 \mathrm{mph}\) and are located on the shore, 4 miles east of an island that is 1 mile north of the shoreline. How far should you run west to minimize the time needed to reach the island?

Step-by-Step Solution

Answer

Run west to approximately 1.2 miles to minimize time reaching the island.

Step 1: Understanding the Problem

You are on a shoreline, 4 miles east of a point directly south of an island located 1 mile north of the shoreline. You need to find the point on the shore where you should start swimming to the island, in order to minimize the total time needed to reach the island.

Step 2: Setting Up the Function for Time

Let the point where you start swimming be \(x\) miles west from your starting point. The distance run on the shoreline is \(x\) miles, while the swimming distance is the hypotenuse of a right triangle with sides \(4 - x\) miles and \(1\) mile. The time function can be expressed as: \[ T(x) = \frac{x}{6} + \frac{\sqrt{(4-x)^2 + 1^2}}{3} \] Here, \( \frac{x}{6} \) is the time running and \( \frac{\sqrt{(4-x)^2 + 1}}{3} \) is the time spent swimming.

Step 3: Finding Critical Points

Find the derivative of the time function to locate the critical points. First, differentiate the time function: \[ T'(x) = \frac{1}{6} - \frac{2(4-x)}{3 \sqrt{(4-x)^2 + 1}} \] Set \( T'(x) = 0 \) to find critical points. This equation will yield values of \( x \) where the time is minimized.

Step 4: Solving the Derivative Equation

Solve \( \frac{1}{6} = \frac{2(4-x)}{3 \sqrt{(4-x)^2 + 1}} \) to find the optimal \( x \). Simplify the equation:\[ \sqrt{(4-x)^2 + 1} = 12(4-x) \] Expand and rearrange into a quadratic equation, then solve for \( x \).

Step 5: Comparing Boundary Values

Since \( x \) is constrained between 0 and 4 (you can run west between 0 and 4 miles), evaluate \( T(x) \) for \( x = 0 \), \( x = 4 \), and the critical point found. Compare these times to determine the smallest value, which represents the minimum time.

Key Concepts

Calculus Applications

Calculus is a powerful tool used to solve a variety of optimization problems, such as finding the most efficient path to take. In this exercise, you need to minimize the time it takes to reach an island by deciding on a combination of running and swimming.
By applying calculus, specifically derivatives, we can streamline our decision-making process.
Optimization in calculus involves determining the best way to allocate resources or minimize effort to achieve a desired outcome. Whether you are managing a budget or planning a route, calculus helps in pinpointing the best solutions.
Calculus allows us to use mathematical functions and their derivatives to find key values that represent optimal choices.
In this context, the time taken to reach the island is expressed as a function of the point where you transition from running to swimming. By optimizing this function, we find the point that results in the least amount of time.

Derivative-Based Optimization

Derivative-based optimization is the process of using derivatives to find the minimum or maximum values of a function. This principle is central to solving optimization problems in calculus.
To determine the minimum time needed in the given problem, we start by defining a function of time in terms of distance run before swimming, represented by the variable \( x \).
The derivative \( T'(x) \) indicates how the time changes with small variations in \( x \). A critical point is any \( x \) where the derivative \( T'(x) = 0 \). It signals a possible minimum or maximum value.
Differentiating the function by taking the derivative enables us to explore how the time function behaves, allowing us to identify when the time is minimized.

Critical Points Analysis

Critical points analysis involves identifying points where the derivative of a function is zero or undefined, as these points can represent potential minima or maxima.
In the exercise, by setting the derivative equal to zero \( T'(x) = 0 \), we determine where the function's rate of change switches from decreasing to increasing. This provides candidates for the minimum time.
Critical points occur when changes to the variable, in this case, \( x \), cause no immediate increase or decrease in the value of the function \( T(x) \). Skipping this step means missing out on optimizing the outcome.
After calculating possible critical points, we must also consider boundary values, since they can sometimes yield the minimum or maximum results, especially when they fall within the feasible range. Comparing the values at these points helps determine the optimal decision.

Calculus in Real-World Scenarios

Calculus finds numerous applications in real-world scenarios, and this problem is a classic example. Whether navigating on land or water, we often seek the most efficient routes.
In the context of this exercise, determining the optimal mix of running and swimming involves real-life considerations like speed and distance. Here, calculus breaks down what could seem an overwhelming task into manageable parts.
By constructing a time function and analyzing derivative results, calculus helps in making informed decisions, such as how far to run before transitioning to swimming for the quickest route to the island.
Such applications of calculus are not limited to just crossing water but extend to various fields, including economics, engineering, and environmental science. Whether minimizing cost, maximizing output, or optimizing resource use, calculus methods guide effective goal achievement.