Problem 21 - Calculus Early Transcendentals

Question

A boat on the ocean is 4 mi from the nearest point on a straight shoreline; that point is 6 mi from a restaurant on the shore. A woman plans to row the boat straight to a point on the shore and then walk along the shore to the restaurant. a. If she walks at \(3 \mathrm{mi} / \mathrm{hr}\) and rows at \(2 \mathrm{mi} / \mathrm{hr}\), at which point on the shore should she land to minimize the total travel time? b. If she walks at \(3 \mathrm{mi} / \mathrm{hr}\), what is the minimum speed at which she must row so that the quickest way to the restaurant is to row directly (with no walking)?

Step-by-Step Solution

Answer

Answer: The woman should land at a point on the shore about 1.44 miles from the nearest point to minimize her total travel time. The minimum rowing speed required to row directly to the restaurant with no walking involved is approximately 2.09 miles per hour.

Step 1: a. Finding the optimal landing point on the shore

Let's label the points of interest: A is the starting point (where the boat is), B is the nearest point on the shoreline, C is the restaurant, and P is the point on the shoreline where the woman will land. Let x be the distance from point B to point P in miles. Then, the distance from point A to point P will be \(\sqrt{x^2 + 4^2}\) using the Pythagorean theorem. Now, we know: - The time needed to row from A to P is \(\frac{\sqrt{x^2 + 4^2}}{2}\) hours. - The time needed to walk from P to C is \(\frac{6-x}{3}\) hours. The total travel time, T(x), will be the sum of the rowing time and the walking time: $$T(x) = \frac{\sqrt{x^2 + 4^2}}{2} + \frac{6-x}{3}$$ To minimize the total travel time T(x), we will find its critical points by taking the first derivative and equating it to zero: $$\frac{dT}{dx} = \frac{d}{dx}\left(\frac{\sqrt{x^2 + 4^2}}{2} + \frac{6-x}{3}\right)$$ $$\frac{dT}{dx} = \frac{x}{2\sqrt{x^2 + 4^2}} -\frac{1}{3}$$ Now, solve the equation \(\frac{dT}{dx}=0\) for x: $$\frac{x}{2\sqrt{x^2 + 4^2}} -\frac{1}{3} = 0$$ Solving for x yields x ≈ 1.44 mi. So, the woman should land at a point on the shore about 1.44 miles from point B to minimize her total travel time.

Step 2: b. Finding the minimum rowing speed

For the woman to row directly to the restaurant (from A to C) without any walking, the time taken to walk the 6 miles from B to C would be equal to the time taken to row the additional distance (from P to C). That is, the time taken to row the additional distance to C should be equal to the time taken to walk from B to C. Let v be the rowing speed in miles per hour, then we have: $$ \frac{\sqrt{(6-x)^2 + 4^2}}{v} = \frac{6-x}{3} $$ From part a, we found that the optimal point to land on the shore is 1.44 miles from point B, hence: $$ \frac{\sqrt{(6-1.44)^2 + 4^2}}{v} = \frac{6-1.44}{3} $$ Solving for v, we get: $$v \approx 2.09 \, \mathrm{mi} / \mathrm{hr}$$ So, the woman must row at a speed of at least 2.09 miles per hour to make it more efficient for her to row directly to the restaurant with no walking involved.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry. It defines the relationship between the sides of a right triangle. According to this theorem, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be written as:

• \(a^2 + b^2 = c^2\)

In the exercise, the woman rows a boat from point A (the starting point in the water) to point P (a point on the shoreline). Here, the distance from A to P forms the hypotenuse of a right triangle. Let's break it down further:

• The horizontal distance from the shoreline point (B) to P is \(x\) miles.
• The vertical distance from A to B (shoreline) is 4 miles.
• Using the Pythagorean Theorem, the rowing distance (AP) is \(\sqrt{x^2 + 4^2}\).

This calculation is crucial in determining the total travel path and minimizing the travel time for the woman.

Derivative

Derivatives are a key concept in calculus. They help us determine the rate of change of a function with respect to a variable. In optimization problems, we use derivatives to find the minimum or maximum values of functions.

Let's consider the exercise. To minimize the woman's total travel time, we express it as a function, \(T(x)\), that represents the combined times of rowing and walking:

• Rowing time: \(\frac{\sqrt{x^2 + 4^2}}{2}\)
• Walking time: \(\frac{6-x}{3}\)

The total time function becomes: \[T(x) = \frac{\sqrt{x^2 + 4^2}}{2} + \frac{6-x}{3}\]To find where this function has its minimum value, we calculate the derivative \(\frac{dT}{dx}\). Setting this derivative equal to zero helps us find the critical points, where potential minima or maxima occur.

Critical Points

Critical points are specific values within a function where the derivative is zero or undefined. These points are essential in finding the maximum or minimum values of a function, which is often the main goal in optimization problems.

In the exercise, finding the critical point involves solving the equation:

• \(\frac{x}{2\sqrt{x^2 + 4^2}} - \frac{1}{3} = 0\)

Solving this equation provides the value of \(x\), approximately 1.44 miles from point B, which minimizes the travel time. This critical point indicates where the woman should land to achieve the quickest travel time from the boat to the restaurant.

Travel Time Minimization

Travel time minimization is a practical application of optimization problems in calculus. Such problems often require selecting the best route or method to reduce time spent traveling over a distance.

In the exercise, minimizing travel time involves balancing the rowing and walking speeds. The challenge is to determine where along the shore the woman should aim for, affecting her overall travel time. The approach to solve this involves:

• Setting up a mathematical expression for total time: both rowing time and walking time.
• Using calculus (derivatives) to find the critical point where the total time is minimized.

Part b of the exercise challenges us to reverse this question and find the rowing speed necessary for direct travel to the restaurant without walking. This involves solving for when both walking time and rowing time would be equal if she rows straight to the restaurant, providing insight into practical applications of these mathematical concepts.