Problem 35 - Calculus Early Transcendentals

Question

Starting at the same time and place, Abe and Bob race, running at velocities \(u(t)=4 /(t+1) \mathrm{mi} / \mathrm{hr}\) and \(v(t)=4 e^{-t / 2} \mathrm{mi} / \mathrm{hr},\) respectively, for \(t \geq 0\) a. Who is ahead after \(t=5 \mathrm{hr}\) ? After \(t=10 \mathrm{hr} ?\) b. Find and graph the position functions of both runners. Which runner can run only a finite distance in an unlimited amount of time?

Step-by-Step Solution

Answer

Answer: At t=5 hours and t=10 hours, Abe is ahead. In the long run, Abe's position goes to infinity, while Bob's position approaches a constant value, meaning Bob can run only a finite distance in an unlimited amount of time.

Step 1: Integrate the velocity functions

To find the position function for each runner, we need to integrate their velocity functions. Let's denote the position functions for Abe and Bob as \(s_A(t)\) and \(s_B(t)\) respectively. Integrate their velocity functions with respect to \(t\): $$s_A(t) = \int u(t) dt = \int \frac{4}{t+1} dt$$ $$s_B(t) = \int v(t) dt = \int 4e^{-t/2} dt$$

Step 2: Determine the position functions

Now, we determine the position functions by solving the integrals: $$s_A(t) = 4\int \frac{1}{t+1} dt = 4\ln{(t+1)} + C_A$$ $$s_B(t) = -8e^{-t/2} + C_B$$ Since the runners start at the same location (i.e., their positions are equal when \(t=0\)), we can find the constants \(C_A\) and \(C_B\) as follows: $$s_A(0) = 4\ln(1) + C_A = C_A$$ $$s_B(0) = -8e^0 + C_B = -8 + C_B$$ To make their position functions equal at \(t=0\), we have \(C_A = C_B - 8\). Thus, their position functions are: $$s_A(t) = 4\ln{(t+1)} + (C_B - 8)$$ $$s_B(t) = -8e^{-t/2} + C_B$$

Step 3: Determine who is ahead at specific times

Now, we compare the positions of Abe and Bob at the given times of \(t=5\) hours and \(t=10\) hours: At \(t=5\) hours: $$s_A(5) = 4\ln{(5+1)} + (C_B - 8) = 4\ln{6} + (C_B - 8)$$ $$s_B(5) = -8e^{-5/2} + C_B$$ We don't need to calculate the exact values, since we can see that \(s_A(5) > s_B(5)\), so Abe is ahead at \(t=5\) hours. At \(t=10\) hours: $$s_A(10) = 4\ln{(10+1)} + (C_B - 8) = 4\ln{11} + (C_B - 8)$$ $$s_B(10) = -8e^{-10/2} + C_B$$ Again, we can see that \(s_A(10) > s_B(10)\), so Abe is ahead at \(t=10\) hours.

Step 4: Determine the long-term position

As \(t\) approaches infinity, we analyze the position functions to determine which runner can run only a finite distance in an unlimited amount of time: $$\lim_{t \to \infty} s_A(t) = \lim_{t \to \infty} (4\ln{(t+1)} + (C_B - 8)) = \infty$$ $$\lim_{t \to \infty} s_B(t) = \lim_{t \to \infty} (-8e^{-t/2} + C_B) = C_B$$ As \(t\) goes to infinity, Abe's position goes to infinity, whereas Bob's position approaches a constant value \(C_B\). This shows that Bob can run only a finite distance in an unlimited amount of time.

Step 5: Graph the position functions

To graph the position functions, we can plot the functions \(s_A(t) = 4\ln{(t+1)} + (C_B - 8)\) and \(s_B(t) = -8e^{-t/2} + C_B\) separately. The graph of \(s_A(t)\) will show a logarithmic growth with time, starting at \(C_B-8\) and increasing without bound as \(t\) increases. The graph of \(s_B(t)\) will show an exponential decay with time, starting at \(C_B\) and decreasing asymptotically to the value \(C_B\).

Key Concepts

Velocity Functions

Velocity functions represent how the speed of an object changes with time. In this exercise, Abe and Bob each have a unique velocity function that determines their speed over time, evaluated in miles per hour.

Abe's velocity function is defined as:

• \( u(t) = \frac{4}{t+1} \) mi/hr

Bob's velocity function is given by:

• \( v(t) = 4e^{-t/2} \) mi/hr

The way these functions behave over time is crucial. Abe's function is a decreasing rational function, meaning his speed decreases the longer he runs. As time passes, the denominator \( t+1 \) increases, causing the overall value of \( u(t) \) to decrease.

On the other hand, Bob's velocity is modeled by an exponentially decreasing function. This means his speed drops off quickly, and the factor \( e^{-t/2} \) causes his velocity to decrease at a base exponential rate (e is the base of natural logarithms, approximately 2.718). Understanding how these velocity functions behave is the first step to understanding the overall problem.

Integration

Integration is the process of finding the position from a velocity function. Essentially, when you integrate a velocity function, you are accumulating all the small distance changes over time to find the total distance (or position).

For Abe, integrating his velocity function \( u(t) = \frac{4}{t+1} \) involves calculating:

• \( s_A(t) = \int \frac{4}{t+1} \, dt = 4\ln{(t+1)} + C_A \)

For Bob, the integration of his exponential velocity function \( v(t) = 4e^{-t/2} \) results in:

• \( s_B(t) = \int 4e^{-t/2} \, dt = -8e^{-t/2} + C_B \)

These integrals give us the position functions for Abe and Bob, with \( C_A \) and \( C_B \) representing integration constants determined by initial conditions. Integration transforms these continuous arrays of velocities into a comprehensible form that tells us where each runner is along a track relative to their starting point.

Position Functions

Position functions provide a snapshot of where an object is at any given time, relative to a starting point. They are integrals of velocity functions and are dependent on time, \( t \).

For Abe, the position function is:

• \( s_A(t) = 4\ln{(t+1)} + (C_B - 8) \)

For Bob, it is:

• \( s_B(t) = -8e^{-t/2} + C_B \)

Initially, both positions were adjusted with constants \( C_A \) and \( C_B \) so that they both start from the same point (time \( t=0 \)). These functions showcase the behavior of each runner over time.

Abe's position function grows logarithmically due to the \( \ln(t+1) \) element, meaning it continues to grow indefinitely as time stretches. Bob's function, however, approaches a constant value (the integration constant \( C_B \)) because of the exponential decay, highlighting a limit to how far Bob can run.

These functions help in determining who is ahead at different times as well as identifying any inherent limitations in their capacity to cover distance.

Infinite Distance

The concept of infinite distance refers to whether a moving object can continue without stopping over an infinite amount of time. In this scenario, it is determined by the nature of the position functions derived from their velocity.

For Abe, his position function:

• \( \lim_{t \to \infty} s_A(t) = \infty \)

This indicates that over an unlimited amount of time, Abe will continue to cover more ground without an inherent stopping point. The logarithmic nature of his function guarantees infinite expansion as time grows.

For Bob, his position function approaches a finite limit:

• \( \lim_{t \to \infty} s_B(t) = C_B \)

This means Bob can only run a finite maximum distance since his position asymptotically approaches the constant \( C_B \). His exponential decay velocity indicates an upper bound on his position.

These concepts show the limitations of each runner and are crucial in understanding how different mathematical models translate into real-world limitations.