Question

Déjà Vu At \(8 : 00\) A.M. on Saturday, a man begins running up the side of a mountain to his weekend campsite (see figure). On Sunday morning at \(8 : 00\) A.M. he runs back down the mountain. It takes him 20 minutes to run up, but only 10 minutes to run down. At some point on the way down, he realizes that he passed the same place at exactly the same time on Saturday. Prove that he is correct. [Hint: Let \(s(t)\) and \(r(t)\) be the position functions for the runs up and down, and apply the Intermediate Value Theorem to the function \(f(t)=s(t)-r(t) . ]\)

Short answer

Yes, he is correct. By creating a function to track the difference in his positions on both days, and applying the Intermediate Value Theorem, we can see that there must exist a time \(t'\) when he is in the exact same position during both runs.

Step-by-step solution

Define the Continuous Functions

Define the position function \(s(t)\) as the man's position at time \(t\) on Saturday (his way up), and \(r(t)\) as his position at time \(t\) on Sunday (his way down). We know \(s(0) = r(0)\) because he starts at the same point on both days, and also \(s(20) = r(10)\) because he ends at the same point on both days.

Formulate the Difference Function

Consider the difference function \(f(t) = s(t) - r(t)\). We want to prove that there's a time \(t'\) such that \(f(t') = 0\), meaning that \(s(t') = r(t')\), which translates to him being at the same place at the same time on both days. At the start, we have \(f(0) = s(0) - r(0) = 0\), and after he has run up but before he runs down, we have \(f(20) = s(20) - r(20)\). Since he takes longer to run up than down, \(f(20) < 0\).

Apply the Intermediate Value Theorem

Apply the Intermediate Value Theorem on the function \(f(t)\). Since \(f(0)\) and \(f(20)\) have different signs and \(f(t)\) is continuous (the man never teleports), the Intermediate Value Theorem ensures there's a time \(t'\) in the interval \([0, 20]\) such that \(f(t') = 0\). That is, for some time \(t'\), the man is at the same place on his way down as he was on Saturday at the same time.

Key concepts

Understanding Continuous Functions

When we talk about continuous functions in mathematics, especially in calculus, we are referring to functions that graph as an uninterrupted, unbroken line or curve. A function is considered continuous at a point if the following three conditions are met: the function has a value at that point, the limit of the function as it approaches the point exists, and the value of the function at that point equals the limit.

In simpler terms, you can think of it as being able to draw the graph of the function without lifting your pencil. Continuous functions are vital in calculus because they allow us to use tools like the Intermediate Value Theorem (IVT), which is essential in solving many real-world problems, like the one posed in the exercise.

The Intermediate Value Theorem states that for any function that is continuous over a closed interval, if the function takes two values at each end of the interval, it must take any value between those two values at some point within the interval. So, if our continuous function starts with a positive value and ends with a negative value, somewhere in between there must be a point where it crosses the x-axis (reaches zero). This core concept is the foundation for proving the scenario in the given problem.

Position Functions in Motion

In the realm of calculus, specifically when analyzing motion, position functions play a crucial role. These functions, usually denoted as 's(t)' for position with respect to time 't', give us the location of a moving object at any given time. What makes the position function particularly useful is its ability to be differentiated into a velocity function and integrated into a displacement function, giving us a wealth of information about the object's motion.

In our textbook exercise, the runner's positions going up and down the mountain are defined by two separate position functions, namely 's(t)' for his ascent and 'r(t)' for his descent. By examining these functions, we can start analyzing his journey. By comparing these two functions, we can deduce a lot about his movement — including where and when he was at the same place during both the ascent and descent, as he experiences a 'déjà vu' moment. This provides a practical application of position functions to solve real-life problems.

Calculus Applications in Real Life

Calculus is more than just an academic subject; it has numerous real-life applications. One of the most profound applications is the analysis of motion — whether it's physics, engineering, economics, or even sports. Calculus enables us to understand the changes and patterns over time through its fundamental operations, differentiation, and integration, which can tell us the rate at which things change and the accumulation of changes over time, respectively.

The usage of the Intermediate Value Theorem (IVT) in the provided exercise is a prime example of applying calculus to everyday scenarios. It helps us prove that a runner, following a continuous path up and down a mountain, must have been at some point at the same position at the same time on two consecutive days. This is not only an elegant illustration of calculus in motion but also demonstrates the power of mathematical concepts to make logical and precise determinations about events in our daily lives.