Problem 42 - Thomas Calculus

Question

Quickest route Jane is 2 \(\mathrm{mi}\) offshore in a boat and wishes to reach a coastal village 6 \(\mathrm{mi}\) down a straight shoreline from the point nearest the boat. She can row 2 \(\mathrm{mph}\) and can walk 5 \(\mathrm{mph}\) . Where should she land her boat to reach the village in the least amount of time?

Step-by-Step Solution

Answer

Jane should land approximately 0.873 miles along the shoreline from the nearest point.

Step 1: Understanding the Problem

Jane is on a boat 2 miles from the shore and wants to reach a village 6 miles along the shore from the nearest point. She can row and walk at different speeds. We need to find the landing point on the shore that minimizes her total travel time.

Step 2: Identify Variables

Let's denote the landing point on the shore as point \(x\). This point is \(x\) miles from the point directly closest to Jane. Therefore, she will row a distance of \(\sqrt{x^2 + 2^2}\) miles to reach point \(x\), and walk \(6 - x\) miles to the village.

Step 3: Establish Time Equations

The time taken to row to point \(x\) is \(\frac{\sqrt{x^2 + 4}}{2}\), since her rowing speed is 2 mph. The time taken to walk from point \(x\) to the village is \(\frac{6 - x}{5}\), since her walking speed is 5 mph. Therefore, the total travel time is \(T(x) = \frac{\sqrt{x^2 + 4}}{2} + \frac{6 - x}{5}\).

Step 4: Differentiate the Time Function

To minimize \(T(x)\), we find the derivative \(T'(x)\) and set it to zero. Differentiating, we find: \[ T'(x) = \frac{x}{2\sqrt{x^2 + 4}} - \frac{1}{5} \] Set \(T'(x) = 0\) to find critical points.

Step 5: Solve for x

Solving \(\frac{x}{2\sqrt{x^2 + 4}} = \frac{1}{5}\), we get \(5x = 2 \sqrt{x^2 + 4}\). Squaring both sides, we obtain \(25x^2 = 4(x^2 + 4)\), simplifying to \(21x^2 = 16\). Thus, \(x = \frac{4}{\sqrt{21}}\), approximately \(0.873\) miles from the nearest point to Jane.

Step 6: Verify and Conclude

We verify the second derivative or check the endpoints to ensure this is indeed the minimum point. Given the constraints, the location \(x = \frac{4}{\sqrt{21}}\) provides the minimum time. Thus, Jane should land her boat approximately 0.873 miles along the shoreline from the nearest point.

Key Concepts

Distance Minimization

In calculus optimization problems, distance minimization is a key concept. Here, Jane's problem can be broken down into minimizing the distance she travels to reach her destination in the shortest possible time. Total travel distance comprises two components: a rowing distance and a walking distance. However, since Jane's rowing and walking speeds differ, the focus here is on minimizing travel time rather than physical distance.

The goal is to identify the optimal landing point along the shoreline where Jane should dock her boat to minimize this travel time. By considering various potential landing points, we calculate the respective distances for rowing and walking. These distances ultimately affect her total travel time, which becomes the focal point of our minimization task.

Derivatives

Derivatives play a crucial role in solving optimization problems like Jane's. By leveraging derivatives, we can establish how a function changes at any given point. In this instance, we are dealing with a travel time function, denoted as \( T(x) \), which combines both rowing and walking times as dependent variables of distance \( x \).

To optimize the function, we take its derivative, \( T'(x) \), with respect to the distance \( x \). The derivative provides insights into the slope of the time function. When the derivative is zero, it signifies that the function has reached a critical point where the slope is neither increasing nor decreasing. We explore these critical points to determine optimal solutions for Jane's journey.

Critical Points

Critical points are fundamental when examining calculus optimization problems. These are the points where the derivative of a function equals zero or is undefined. In the context of Jane’s travel problem, critical points help us pinpoint potential solutions for minimizing travel time.

Once we have the derivative \( T'(x) = \frac{x}{2\sqrt{x^2 + 4}} - \frac{1}{5} \), setting it to zero helps identify these critical points. Solving this equation results in \( x = \frac{4}{\sqrt{21}} \). This value represents a candidate for Jane's optimal landing point. It's crucial to confirm this as a minimum by checking either the second derivative or evaluating endpoint behaviors to ensure it's not a maximum or a saddle point.

Travel Time Calculation

Travel time calculation is the essence of solving Jane's problem. The total travel time involves both rowing and walking components. These are calculated separately based on her speeds: 2 mph for rowing and 5 mph for walking.

For rowing to the point \( x \), the time equation is \( \frac{\sqrt{x^2 + 4}}{2} \), while for walking from this landing point to the village, the time is \( \frac{6-x}{5} \). These two components add up to give the total travel time, expressed as \( T(x) = \frac{\sqrt{x^2 + 4}}{2} + \frac{6-x}{5} \).

Understanding and accurately calculating these elements is vital for determining the most efficient route for reaching the village from Jane's current position in the least amount of time possible.