Problem 56 - Calculus Early Transcendentals

Question

A man is floating in a rowboat 1 mile from the (straight) shoreline of a large lake. A town is located on the shoreline 1 mile from the point on the shoreline closest to the man. As suggested in Figure \(\mathrm{Ex}-56,\) he intends to row in a straight line to some point \(P\) on the shoreline and then walk the remaining distance to the town. To what point should he row in order to reach his destination in the least time if (a) he can walk \(5 \mathrm{mi} / \mathrm{h}\) and row \(3 \mathrm{mi} / \mathrm{h}\) (b) he can walk \(5 \mathrm{mi} / \mathrm{h}\) and row \(4 \mathrm{mi} / \mathrm{h} ?\)

Step-by-Step Solution

Answer

Row to \( x = \frac{3}{4} \) miles for (a) and \( x = \frac{4}{3} \) miles for (b).

Step 1: Define the problem

The man needs to row to point \( P \) on the shoreline and then walk to a town located 1 mile along the shoreline from the closest point to where he starts. We need to find the point \( P \) where rowing to it and walking from it takes the least time. This involves considering both rowing and walking distances, plus their respective speeds.

Step 2: Set up the variables

Let \( x \) be the distance along the shoreline from the closest point on the shore to the point \( P \) where the man lands. Therefore, the rowing distance forms a right triangle with one leg of length 1 mile (distance from the man to the shoreline) and the other leg of length \( x \) miles.

Step 3: Express distances

The hypotenuse of the triangle for the rowing path is \( \sqrt{x^2 + 1} \) (using the Pythagorean theorem). The walking distance from \( P \) to the town is \( 1 - x \).

Step 4: Formulate time function for case (a)

For case (a), where the rowing speed is 3 mi/h and walking speed is 5 mi/h, the total time \( T(x) \) is given by:\[T(x) = \frac{\sqrt{x^2 + 1}}{3} + \frac{1 - x}{5}\]

Step 5: Differentiate the time function for case (a)

To minimize \( T(x) \), differentiate it with respect to \( x \):\[T'(x) = \frac{x}{3\sqrt{x^2 + 1}} - \frac{1}{5}\]Set \( T'(x) = 0 \) to find critical points.

Step 6: Solve for critical point in case (a)

Set the derivative equal to zero and solve for \( x \):\[\frac{x}{3\sqrt{x^2 + 1}} = \frac{1}{5}\]Cross-multiply and solve the resulting equation to find:\[ 5x = 3\sqrt{x^2 + 1} \Rightarrow 25x^2 = 9(x^2 + 1) \Rightarrow 16x^2 = 9 \Rightarrow x = \frac{3}{4} \]

Step 7: Perform similar steps for case (b)

For case (b), where rowing speed is 4 mi/h, repeat the process:\( T(x) = \frac{\sqrt{x^2+1}}{4} + \frac{1-x}{5} \)Now,\[T'(x) = \frac{x}{4\sqrt{x^2 + 1}} - \frac{1}{5}\]Set \( T'(x) = 0 \), and solve:\[\frac{x}{4\sqrt{x^2+1}} = \frac{1}{5}\]Cross-multiply and solve for \( x \):Solve:\[ 5x = 4\sqrt{x^2 + 1} \Rightarrow 25x^2 = 16(x^2 + 1) \Rightarrow 9x^2 = 16 \Rightarrow x = \frac{4}{3} \]

Step 8: Verify solution of critical points

Verify that these values of \( x \) given local minima by checking second derivatives or evaluating behavior at boundaries (0 and 1 mile). Since the functions will be minimized as both are near the center due to their symmetry and rates.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a cornerstone in geometry and plays a critical role in problems involving right triangles. In this exercise, the man forms a right triangle between his starting point, the point he reaches on the shoreline, and the point right across from him on the shoreline. The theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Mathematically, it is expressed as:

• \[ c^2 = a^2 + b^2 \]

Here, the distance the man rows forms the hypotenuse since it's the direct line to the shoreline. Given one leg is perpendicular (1 mile to the shoreline) and the other is parallel (distance along the shore \( x \)), the hypotenuse is \( \sqrt{x^2 + 1} \).

Understanding how the theorem sets up the problem geometrically will help in optimizing the man's route to reduce time spent.

Differentiation

Differentiation is a key calculus tool used to determine how a function changes at any given point. In the context of the problem, it is used to find the rate at which the total travel time changes with respect to \( x \), the point where the man reaches the shoreline.

This is done by forming a time function, \( T(x) \), which incorporates both rowing and walking times. The derivative of this function, \( T'(x) \), tells us how changes to \( x \) affect total travel time:

• For scenario (a), \( T'(x) = \frac{x}{3\sqrt{x^2 + 1}} - \frac{1}{5} \)
• For scenario (b), \( T'(x) = \frac{x}{4\sqrt{x^2 + 1}} - \frac{1}{5} \)

Setting the derivative to zero helps in finding critical points, indicating where the travel time is potentially minimized.

Mastery of differentiation makes optimization problems more manageable and provides insights into how variables change with respect to each other.

Critical Points

Critical points in calculus mark where a function's slope is zero or undefined. These points are crucial for finding local maxima or minima of a function, and they play an essential role in optimization.

For this problem, once the time function \( T(x) \) is derived, setting the derivative \( T'(x) \) to zero allows us to solve for \( x \):

• For part (a): Solve \( 5x = 3\sqrt{x^2 + 1} \) to find \( x = \frac{3}{4} \)
• For part (b): Solve \( 5x = 4\sqrt{x^2 + 1} \) to find \( x = \frac{4}{3} \)

These values are the critical points where the travel time could be minimized.

Further verification, such as using the second derivative test, is necessary to confirm whether these points indeed provide a minimum time for travel from the boat to the town.

Rates of Change

Rates of change describe how a quantity changes as another quantity changes. This concept is especially important in calculus when assessing speed, growth, or decline in various scenarios.

In this exercise, the man’s travel times are determined by different rates of change: rowing speed and walking speed. Rowing a diagonal distance versus walking a fixed shoreline distance at different speeds requires calculating how long each part of the journey takes.

• Rowing speed changes the time it takes to cover any given distance. Faster rowing impacts the optimal landing point \( P \) along the shore.
• Walking speed affects how much distance can efficiently be covered after reaching the shore.

Understanding rates of change is vital in deciding the best strategy for the man’s journey, as they dictate the time optimization constraints when combined with the distances determined by the Pythagorean Theorem. Integrating these rates into the time function ensures that each part of the journey is precisely evaluated for the shortest travel time.