Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

You must cross a river that is 50 meters wide and reach a point on the opposite bank that is \(1 \mathrm{~km}\) up stream. You can travel \(6 \mathrm{~km}\) per hour along the river bank and \(1 \mathrm{~km}\) per hour in the river. Describe a path that will minimize the amount of time required for your trip. Neglect the flow of water in the river.

Step-by-Step Solution

Answer

Travel directly across the river and then move upstream along the bank.

Step 1: Understanding the Problem

You need to cross a river, which is 50 meters wide, and reach a point on the opposite bank that is 1 km upstream. You have different speeds: 6 km/h along the bank and 1 km/h in the water. The goal is to find the path that minimizes the total time needed for this journey.

Step 2: Convert Distances and Speeds

Convert the river's width from meters to kilometers for uniformity. 50 meters is 0.05 km. You will travel across these 0.05 km in the river and then move 1 km along the bank.

Step 3: Setup the Variables

Let \( x \) be the distance you travel in the river at a 90-degree angle to the bank before landing. The remaining horizontal distance on the bank will be \( 1 - x \) km.

Step 4: Time Across the River

The time taken to cross the river's width (0.05 km) is determined by the river speed: \( t_1 = \frac{0.05}{1} = 0.05 \) hours.

Step 5: Time Along the River Your Path

Calculate the time to travel along the bank. Since you travel \( 1 - x \) km at 6 km/h, the time is \( t_2 = \frac{1 - x}{6} \) hours.

Step 6: Express Total Time

The total time taken is the sum of crossing and traveling along the river bank: \( T = t_1 + t_2 = 0.05 + \frac{1 - x}{6} \).

Step 7: Optimize the Time

To minimize time, take the derivative of \( T \) with respect to \( x \) and set it to zero. The derivative \( \frac{dT}{dx} = -\frac{1}{6} \). Since it's constant, not dependent on \( x \), we should minimize \( x \) to 0.

Step 8: Check Minimum Condition

Since \( \frac{d^2T}{dx^2} = 0 \) (second derivative is zero), it confirms constant speed in the river and along will always result in a linear path along the vector sum direction.

Step 9: Conclusion of Path

Since derivative equals zero (indicating constant path characteristics), the optimal path is straight across to the point directly opposite the starting point, then travel 1 km upstream on land at 6 km/h.

Key Concepts

Time Minimization

Time minimization in calculus refers to finding the quickest way to complete a given task by optimizing variables affecting time. In this context, it involves using mathematical tools to determine the shortest possible time for crossing a river and reaching a point upstream.

To minimize time, we need to consider both segments of your journey: crossing the river and moving along the bank. This requires balancing the speeds: you cross the river at a slower speed of 1 km/h and move along the bank at a faster speed of 6 km/h. Minimizing time means finding the optimal point to land on the bank so you can take advantage of higher speed along the bank.

The strategy involves calculating the time for each segment and then summing these times. With calculus, particularly derivatives, we can adjust the variables to find the minimum total time required, ensuring the most efficient path is chosen. Understanding time minimization allows us to apply this principle to various real-world scenarios beyond the river crossing problem.

River Crossing Problem

The river crossing problem is a classic optimization challenge in calculus. It involves determining the most efficient path across a river to minimize travel time. This problem is intriguing because it combines linear movement across a river with follow-up travel along a bank, each having different conditions and speeds.

In this scenario, you start on one bank, needing to cross a 50-meter wide river to reach a point that is 1 km upstream. Your challenge is to determine how far downstream or upstream to aim for on the opposite bank, balancing the trade-off between a shorter, slower river crossing and a longer, faster walk along the river bank.

The solution involves setting up an equation that expresses the total journey time as a combination of time spent in the river and time spent traveling along the bank. The aim is to land at a strategic point on the opposite bank that minimizes the total time consumed. This practical aspect of calculus in optimizing pathways underscores the relevance of mathematical analysis in real-world navigation and decision-making.

Vector Analysis

Vector analysis is a mathematical framework widely used in optimization problems like the river crossing problem. It focuses on magnitude and direction, crucial in dealing with paths and movements through space.

When solving the river crossing problem, we employ vector analysis to visualize the journey's path. The river crossing journey can be broken into vectors representing movements across the river and along the bank. This separation helps in calculating each segment's time based on their respective speeds.

In this exercise, the vectors allow us to compute the path that simultaneously accounts for both the river crossing and bank travel, essentially optimizing the journey. By interpreting movements as vectors, calculus principles such as derivatives are applied to optimize the route, ensuring that the chosen path results in the shortest travel time from start to finish. This approach highlights vector analysis's power in addressing physical problems involving spatial dynamics and emphasizes its importance as a conceptual tool in applied mathematics.