Problem 38 - Essential Calculus Early Transcendentals

Question

A woman at a point \(A\) on the shore of a circular lake with radius 2 mi wants to arrive at the point \(C\) diametrically opposite \(A\) on the other side of the lake in the shortest possible time. She can walk at the rate of 4 \(\mathrm{mi} / \mathrm{h}\) and row a boat at 2 \(\mathrm{mi} / \mathrm{h} .\) How should she proceed?

Step-by-Step Solution

Answer

She should walk \(\frac{2}{\sqrt{5}}\) miles around the lake and then row across to minimize time, taking about 1.43 hours.

Step 1: Determine the Route Options

The woman at point \(A\) can either row straight across the lake to point \(C\), or walk around the lake circumference. Walking along the circumference will be slower because it's twice the distance of rowing straight across. We'll explore the combination of walking and rowing for optimal time.

Step 2: Calculate Time to Row Straight Across

Rowing straight across the diameter of the circular lake, the distance is 4 miles (since the diameter is twice the radius). At a rowing speed of 2 miles per hour, the time to row across is \( \frac{4}{2} = 2 \) hours.

Step 3: Calculate Time to Walk Around the Lake

Walking the entire circumference of the lake involves a distance of \(2\pi\) miles (since the circumference is \(2 \times \pi \times 2\)). At a walking speed of 4 miles per hour, the time is \( \frac{2\pi}{4} = \frac{\pi}{2} \) hours. Since \(\pi\) is approximately 3.14, the time to walk all the way around is about 1.57 hours.

Step 4: Combine Rowing and Walking

To minimize time, she can walk a portion along the lake's circumference to a point \(B\), then row across to \(C\). Define \(x\) as the distance walked along the circumference from \(A\) to \(B\). The time for walking is \( \frac{x}{4} \). The distance rowing is \(\sqrt{4^2-x^2}\) due to the Pythagorean theorem between points \(B\) and \(C\), making the rowing time \( \frac{\sqrt{4-x^2}}{2} \). Total time \(T(x) = \frac{x}{4} + \frac{\sqrt{4-x^2}}{2} \).

Step 5: Optimize Total Time Function

To find the minimum time, take the derivative of \(T(x)\) with respect to \(x\) and set it to zero: \(T'(x) = \frac{1}{4} - \frac{x}{2\sqrt{4-x^2}} = 0\). Solving, \(\frac{1}{4} = \frac{x}{2\sqrt{4-x^2}}\). Multiply both sides by \(2\sqrt{4-x^2}\) to get \(\sqrt{4-x^2} = 2x\). Squaring both sides results in \(4-x^2 = 4x^2\). Thus, \(5x^2 = 4\), giving \(x^2 = \frac{4}{5}\), and \(x = \frac{2}{\sqrt{5}}\).

Step 6: Calculate Total Minimum Time

Substitute \(x = \frac{2}{\sqrt{5}}\) back into the total time formula \(T(x)\). Walk time is \(\frac{2}{4\sqrt{5}}\) and rowing time is \(\frac{\sqrt{4 - \frac{4}{5}}}{2}\). The total time is approximately 1.43 hours.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry that helps us relate the lengths of sides in a right triangle. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
This theorem is essential in various optimization problems, particularly when dealing with distances.In the exercise, the woman chooses to both walk along the circumference and row across the lake. To determine her route, we consider a right triangle composed of her walking path, rowing distance, and the diameter of the lake.

• Let point A be where she starts and point C be directly across the lake from A.
• The walking portion starts from A and ends at some point B on the circumference.
• The rowing distance can then be seen as the hypotenuse of the right triangle, calculated using the Pythagorean theorem.

This gives us the equation: \[4^2 = x^2 + \text{rowing distance}^2\]Solving this equation helps find the optimal trade-off between walking and rowing to minimize her travel time.

Derivative test

The derivative test is a standard mathematical method used to find the extrema (minimum or maximum) of functions. By taking the derivative of a function and setting it to zero, we effectively find points where the function's rate of change is zero—possible candidates for minima or maxima.In the context of the exercise, after establishing the time equation for walking and rowing, we obtain:\[T(x) = \frac{x}{4} + \frac{\sqrt{4-x^2}}{2}\]Here, \(T(x)\) represents the total travel time as a function of \(x\), which is the distance walked. To find the optimal \(x\), we calculate the derivative \(T'(x)\) and set it equal to zero:

• The derivative \(T'(x)\) is computed.
• Setting \(T'(x) = 0\), we solve for \(x\).
• This involves some algebraic manipulation, including solving for \(x\) using the resulting equation.

Finding where \(T'(x) = 0\) helps identify the walking distance that minimizes travel time by effectively balancing the time spent walking against the time spent rowing.

Distance-rate-time calculation

Distance, rate, and time are interrelated and are used in conjunction to solve many real-world problems. The basic relationship can be given by the formula:\[ \text{Distance} = \text{Rate} \times \text{Time} \]In this exercise, understanding this relationship is crucial for calculating the woman's travel time through different modalities of movement (walking and rowing).

• For rowing, the distance is the diameter of the lake, 4 miles, with a rate of 2 miles per hour, giving a time of 2 hours for crossing directly.
• If she chooses to walk around the circuit completely, the distance is the circumference \(2\pi\), and at 4 miles per hour, the total time is about 1.57 hours.
• If she uses a mix of these methods, the function \(T(x)\) considers part of the distance at 4 miles per hour and the rest at 2 miles per hour.

Mastering this calculation allows for the adjustment of rates and distances to minimize the time taken across different travel segments, showcasing the practical blend of theory and problem-solving using these principles.