Problem 67 - Calculus

Question

An ant crawls at 1 foot/second along a rubber band whose original length is 2 feet. The band is being stretched at 1 foot/second by pulling the other end. At what time \(T,\) if ever, does the ant reach the other end? One approach: The band's length at time \(t\) is \(t+2\). Let \(y(t)\) be the fraction of that length which the ant has covered, and explain (a) \(y^{\prime}=1 /(t+2)\) (b) \(y=\ln (t+2)-\ln 2\) (c) \(T=2 e-2\).

Step-by-Step Solution

Answer

The ant reaches the end at time \( T = 2e - 2 \).

Step 1: Understand the Problem

We need to determine when the ant reaches the end of a rubber band that is being stretched while the ant crawls along it. The band is initially 2 feet long and is being stretched at 1 foot per second. The ant crawls at a speed of 1 foot per second along this band.

Step 2: Define the Variables

Let the length of the rubber band at time \(t\) be \(L(t) = t + 2\). Let \(y(t)\) be the fraction of the rubber band's length that the ant has covered at time \(t\). Thus, the location of the ant at any time \(t\) is \(y(t) \times L(t)\).

Step 3: Establish the Differential Equation for y(t)

The ant moves at 1 foot/second relative to the rubber band, so its rate of progress along the band fraction-wise is \(y'(t) = \frac{1}{L(t)}\). Substituting \(L(t) = t + 2\), we have:\[ y'(t) = \frac{1}{t+2} \]

Step 4: Solve the Differential Equation

Integrating both sides with respect to \(t\) gives:\[ y(t) = \int \frac{1}{t+2} \ dt = \ln(t+2) + C \]Using the initial condition at \(t = 0\), where the ant has not yet moved, we find:\[ y(0) = 0 = \ln(2) + C \quad \Rightarrow \quad C = -\ln(2) \]Thus, the solution is:\[ y(t) = \ln(t+2) - \ln(2) \]

Step 5: Determine When the Ant Reaches the End

The ant reaches the end when \(y(t) = 1\). Setting \(y(t) = 1\), we solve for \(t\):\[ 1 = \ln(t+2) - \ln(2) \]This implies:\[ \ln\left(\frac{t+2}{2}\right) = 1 \]\[ \frac{t+2}{2} = e \]\[ t+2 = 2e \]\[ t = 2e - 2 \]

Step 6: Final Step: Conclusion

The ant reaches the end of the rubber band at time \(T = 2e - 2\).

Key Concepts

Differential Equations

Differential equations play a crucial role in many aspects of calculus problem solving. They are equations that involve a function and its derivatives. In this exercise, we set up a differential equation to track the ant's progress along the rubber band.
This type of problem is typical for illustrating how real-world situations can be modeled with differential equations.

• The function is about the fraction of the rubber band length that the ant has traversed, denoted as \( y(t) \).


• The derivative \( y'(t) \) describes the rate of change of \( y(t) \) with respect to time \( t \).

By solving this differential equation, we can determine how the ant moves over time. It essentially tells us how quickly the ant covers the growing length of the rubber band. Here, the rate of change is given by \( \frac{1}{t+2} \), which accurately reflects how the speed of the ant as a function of the band length changes. Handling differential equations is about finding functions that satisfy given conditions and rates of change, which we thoroughly explored in this example.

Exponential Functions

Understanding exponential functions is key when you encounter them in calculus and differential equations. Here, the expression \( \ln(t+2) \) comes from integrating \( \frac{1}{t+2} \). This integral shows up because the natural logarithm is the antiderivative of \( \frac{1}{x} \). Exponential functions are critical as they grow or decay at a rapid rate, modeling many natural processes. In this exercise, by transforming the problem into an integral of an exponential function,

• we reveal how the speed in which the ant covers distances transforms with time


• and how these concepts of growth apply to distances expanding over time.

The connection here between the logarithmic function and \( e \), Euler's number, comes out notably when reversing the logarithmic transformation to find \( t \), the time at which the ant reaches the end of the rubber band. It highlights the reciprocal nature of exponential and logarithmic functions.

Integration

Integration is one of the fundamental concepts of calculus. It's used to find areas, volumes, central points, and other useful things. In the exercise, we integrated to solve the differential equation for \( y(t) \).When integrating \( \frac{1}{t+2} \),

• we find that it results in \( \ln(t+2) \), highlighting the relationship between differentiation and integration as inverse processes.

Using an initial condition or boundary, such as \( y(0) = 0 \) in this scenario, allows us to solve for the constant and complete the function that characterizes the ant’s movement. This process of using integration reinforces its utility as a tool for rediscovering relationships between varying quantities over time or space, and in solving a wide array of real-world scenarios. This entire approach illustrates the power and versatility of integration in calculus, not only in finding specific functions but also in describing dynamic systems.