Problem 18 - Calculus Early Transcendentals

Question

A man wishes to get from an initial point on the shore of a circular lake with radius 1 mi to a point on the shore directly opposite (on the other end of the diameter). He plans to swim from the initial point to another point on the shore and then walk along the shore to the terminal point. a. If he swims at \(2 \mathrm{mi} / \mathrm{hr}\) and walks at \(4 \mathrm{mi} / \mathrm{hr}\), what are the minimum and maximum times for the trip? b. If he swims at \(2 \mathrm{mi} / \mathrm{hr}\) and walks at \(1.5 \mathrm{mi} / \mathrm{hr},\) what are the minimum and maximum times for the trip? c. If he swims at \(2 \mathrm{mi} / \mathrm{hr}\), what is the minimum walking speed for which it is quickest to walk the entire distance?

Step-by-Step Solution

Answer

Answer: The minimum time is 1 hour, and the maximum time is \(\frac{\pi}{2}\) hours.

Step 1: Calculate Minimum Distance (Direct Path Across the Lake)

He will swim directly across the lake (diameter) from point A to point B which is 2 mi. Distance: \(2 \times 1 = 2 \mathrm{mi}\)

Step 2: Calculate Maximum Distance (Walk Around the Lake)

He will walk all around the lake for the maximum distance. Circumference: \(2 \times \pi \times 1 = 2\pi \mathrm{mi}\)

Step 3: Calculate Minimum Time (Direct Path Across the Lake)

He swims at 2 mi/hr. Time: \(\frac{2 \mathrm{mi}}{2 \mathrm{mi/hr}} = 1 \mathrm{hr}\)

Step 4: Calculate Maximum Time (Walk Around the Lake)

He walks at 4 mi/hr. Time: \(\frac{2\pi \mathrm{mi}}{4 \mathrm{mi/hr}} = \frac{\pi}{2} \mathrm{hr}\) Minimum time is 1 hour, and maximum time is \(\frac{\pi}{2}\) hours. b) Swims at 2 mi/hr and walks at 1.5 mi/hr

Step 5: Calculate Maximum Time (Walk Around the Lake)

He walks at 1.5 mi/hr. Time: \(\frac{2\pi \mathrm{mi}}{1.5 \mathrm{mi/hr}} = \frac{4\pi}{3} \mathrm{hr}\) Minimum time is 1 hour, and maximum time is \(\frac{4\pi}{3}\) hours. c) Swims at 2 mi/hr

Step 6: Calculate Swimming Time

Swimming directly across from A to B. 𝑇\(_{swimming}\) = \(\frac{2 \mathrm{mi}}{2 \mathrm{mi/hr}} = 1 \mathrm{hr}\)

Step 7: Calculate Walking Time

Walking around the lake. 𝑇\(_{walking}\) = \(\frac{2\pi \mathrm{mi}}{x \mathrm{mi/hr}} = \frac{2\pi}{x} \mathrm{hr}\)

Step 8: Set Swimming Time Equal to Walking Time and Solve for 𝑥.

1 = \(\frac{2\pi}{x}\) x = \(2\pi\) The minimum walking speed for which it is quickest to walk the entire distance is 2π mi/hr.

Key Concepts

Calculus

Calculus is a fundamental branch of mathematics that deals with rates of change and the accumulation of quantities. In this exercise, calculus helps us optimize or find the extreme values of time based on different rates of swimming and walking. The problem involves assessing both a direct swim across the lake and walking along its perimeter.
This task essentially deals with the concept of "optimization," where calculus tools, like differentiation, could typically find the minimum and maximum time taken to travel these distances. However, in this exercise, calculations were straightforward and did not require differentiation or integration, focusing primarily on arithmetic functions. Calculus enables us to analyze similar problems by using functions to calculate the best possible outcomes, known as critical points.

• Optimization through calculus involves finding either the maximum or minimum value of a function.
• Key calculus tools for such tasks include derivatives and integrals, which help discover these critical points.

Even though calculus wasn't explicitly used here, understanding its concepts allows predicting and analyzing similar problems efficiently.

Rate Problems

Rate problems involve calculating and understanding different speeds or rates, working together to achieve the best outcome. Here, the problem's objective was to find the minimum and maximum time for crossing a lake considering both swimming and walking as the methods of transport. Different swimming and walking speeds were used in scenarios (a, b, and c) to calculate these times. By understanding the rates involved, one can determine the best combination of actions to achieve an optimum outcome.

• The term 'rate' refers to how fast or slow an action occurs.
• In this exercise, different rates or speeds are used for swimming and walking.

Rate problems often need one to set up equations that link distance, speed, and time, comparing different actions to identify the optimal choice. This highlights the importance of strategy when calculating time, acknowledging how altering rates affects overall outcomes.

Distance-Speed-Time Problems

Distance-speed-time problems are a common element of applied mathematics involving calculating one of these factors when provided with the other two. They are based on the basic formula:
\[\text{Time} = \frac{\text{Distance}}{\text{Speed}}\]In the textbook exercise, the man has two transportation methods: swimming and walking. Each of them involves different speeds, demanding the use of this formula to determine the time needed for each scenario.
For example, when swimming directly across, we find the time using his swim speed and distance across the lake, whereas walking was calculated based on the lake's perimeter and the walk speed.

• This problem illustrates how varying conditions (different swim and walk rates) affect overall travel time.
• The direct application of the formula allows getting solutions without complicated calculations.

Understanding distance-speed-time relations lets you solve diverse real-life problems easily, informing you how modifying one variable can impact others.