Question

Suppose you park your car at a trailhead in a national park and begin a 2 -hr hike to a lake at 7 A.M. on a Friday morning. On Sunday morning, you leave the lake at 7 A.M. and start the 2 -hr hike back to your car. Assume the lake is 3 mi from your car. Let \(f(t)\) be your distance from the car \(t\) hours after 7 A.M. on Friday morning and let \(g(t)\) be your distance from the car \(t\) hours after 7 A.M. on Sunday morning. a. Evaluate \(f(0), f(2), g(0),\) and \(g(2)\). b. Let \(h(t)=f(t)-g(t) .\) Find \(h(0)\) and \(h(2)\). c. Use the Intermediate Value Theorem to show that there is some point along the trail that you will pass at exactly the same time of morning on both days.

Short answer

If so, prove it using the Intermediate Value Theorem. Answer: Yes, there is a point along the trail that the hiker passes at exactly the same time on both days. The Intermediate Value Theorem can be used to prove this, as h(0) < 0, h(2) > 0, and h(t) is continuous on the interval [0, 2]. Therefore, there must be a point (c) in the interval [0, 2] such that h(c) = 0.

Step-by-step solution

Find the distance functions f(t) and g(t) for Friday and Sunday

On Friday, the hiker starts at a distance of 0 miles and travels 3 miles in the 2 hours. So the distance function for Friday is: \(f(t) = \frac{3}{2}t\) On Sunday, the hiker starts at a distance of 3 miles and travels back to the car, decreasing by 3 miles in the 2 hours. So the distance function for Sunday is: \(g(t) = 3 - \frac{3}{2}t\) Now, we'll evaluate these functions at the requested points.

Evaluate f(0), f(2), g(0), and g(2)

\(f(0) = \frac{3}{2}(0) = 0\) \(f(2) = \frac{3}{2}(2) = 3\) \(g(0) = 3 - \frac{3}{2}(0) = 3\) \(g(2) = 3 - \frac{3}{2}(2) = 0\) b) Finding \(h(t)\)

Compute h(t) = f(t) - g(t)

Now, we'll find \(h(t)\) by subtracting \(g(t)\) from \(f(t)\) \(h(t) = f(t) - g(t) = \frac{3}{2}t - (3 - \frac{3}{2}t)\)

Simplify h(t) and evaluate it at t=0 and t=2

Simplify h(t) by combining the terms: \(h(t) = \frac{3}{2}t + \frac{3}{2}t - 3 = 3t - 3\) Now, evaluate h(t) at t = 0 and t = 2 \(h(0) = 3(0) - 3 = -3\) \(h(2) = 3(2) - 3 = 3\) c) The Intermediate Value Theorem

Analyze the properties of h(t)

By looking at h(0) and h(2), we noticed that h(0) is negative and h(2) is positive. It means that the function h(t) changes its value while hiking at some time between Friday and Sunday.

Apply the Intermediate Value Theorem

According to the Intermediate Value Theorem, since h(t) is continuous on the interval [0, 2] (as it is a linear function), and h(0) < 0 while h(2) > 0, there must be a point (c) in the interval [0, 2] such that h(c) = 0. Therefore, there is a point along the trail that the hiker will pass at exactly the same time of morning on both days.

Key concepts

Understanding the Distance Function

When we describe a distance function in mathematics, we're talking about how faraway something is from a particular point, depending on some variable like time. In real-world terms, imagine tracking your steps on a hike using a smartphone app. As time passes, the app shows you how far you've gone from your starting point.

In our exercise, we have two distance functions, \(f(t)\) and \(g(t)\). For \(f(t)\), the distance from the car increases as time increases, since you're moving away from the starting point. On the other hand, \(g(t)\) represents the return trip, where the distance from the car decreases as you approach it. These two functions represent the distance traveled at various times on the hike.

Understanding the behaviour of these distance functions is crucial because it shapes the foundation for problems involving rates, movement, and change over time, which are common in calculus and physics.

The Importance of Continuity

Continuity is a fundamental concept in calculus, and it refers to a function that is smooth and unbroken. A continual function doesn't suddenly leap in value or have gaps or holes. It's like drawing a line without lifting your pencil from the paper.

In our scenario, we're dealing with a linear distance function for each of the hikes. Because these functions are linear, they're also continuous over their domains. Why does this matter? Well, continuity is a prerequisite for the Intermediate Value Theorem (IVT), which is essential in proving that there's a point on the trail crossed at the same time on two separate days.

Why Continuity Matters for the IVT

The Intermediate Value Theorem tells us that if you have a continuous function on a closed interval, and it takes on different values at the endpoints of the interval, then it must take on every value in between at some point. The continuous nature of our distance functions ensures that at some time between the hikes, the distances from the car matched.

Function Evaluation and Its Significance

Function evaluation might seem straightforward—plug in a number, get out a result—but it's a powerful tool in understanding functions behaviour over time. By evaluating \(f(t)\) and \(g(t)\) at specific times, we're able to gather information about where the hiker was on the trail.

The exercise leads us to evaluate these functions at \(t=0\) and \(t=2\) hours. For instance, \(f(0)\) gives us the distance from the car at the start of the hike on Friday, which is 0 miles. On the other end, \(f(2)\) gives us the distance from the car after 2 hours, which is 3 miles. A similar process is applied to \(g(t)\), which gives us the location on the return trip.

Knowing how to evaluate functions at certain points enables us to comprehend the hiker's journey, and when we subtract one function from another, \(h(t) = f(t) - g(t)\), we're combining the information of both trips to find points of intersection—in this case, where the hiker is at the same distance from the car at the same time during both the outbound and return journeys.