Problem 66 - Essential Calculus Early Transcendentals

Question

Under certain circumstances a rumor spreads according to the equation $$p(t)=\frac{1}{1+a e^{-k t}}$$ where \(p(t)\) is the proportion of the population that knows the rumor at time \(t\) and \(a\) and \(k\) are positive constants. (a) Find \(\lim _{t \rightarrow \infty} p(t)\) . (b) Find the rate of spread of the rumor. (c) Graph \(p\) for the case \(a=10, k=0.5\) with \(t\) measured in hours. Use the graph to estimate how long it will take for 80\(\%\) of the population to hear the rumor.

Step-by-Step Solution

Answer

(a) 1; (b) \(\frac{a k e^{-kt}}{(1 + a e^{-kt})^2}\); (c) Approximately 6.93 hours for 80% to hear the rumor.

Step 1: Determine the Limit

To find \(\lim_{t \to \infty} p(t)\), we observe the behavior of the function as \(t\) goes to infinity. The function is given by \(p(t) = \frac{1}{1 + a e^{-kt}}\). As \(t \to \infty\), the term \(e^{-kt} \to 0\) since \(k\) is positive. Therefore, the equation simplifies to \[\lim_{t \to \infty} p(t) = \frac{1}{1 + a \cdot 0} = \frac{1}{1} = 1.\]Thus, the limit is \(1\).

Step 2: Find the Rate of Spread

The rate of spread of the rumor can be determined by differentiating \(p(t)\) with respect to \(t\). Using the chain rule, we have:\[\frac{dp}{dt} = \frac{d}{dt}\left(\frac{1}{1 + a e^{-kt}}\right) = \frac{a k e^{-kt}}{(1 + a e^{-kt})^2}.\]This expression represents the instantaneous rate at which the rumor spreads at any given time \(t\).

Step 3: Graph the Function

Using the provided values \(a = 10\) and \(k = 0.5\), the function becomes:\[p(t) = \frac{1}{1 + 10 e^{-0.5t}}\]To graph this function, plot \(p(t)\) on the y-axis and \(t\) on the x-axis. Notice how the graph approaches \(p(t) = 1\) as \(t\) increases. By estimating from the graph, determine the time \(t\) when \(p(t) = 0.8\).

Step 4: Estimate Time to Reach 80%

From the graph in Step 3, observe where \(p(t) = 0.8\) on the y-axis. This will give an approximate value of \(t\) when 80% of the population knows the rumor. Solving the equation analytically, \[0.8 = \frac{1}{1 + 10 e^{-0.5t}},\]yields the value of \(t\) by rearranging and solving. After solving, it turns out that such a time \(t\) is approximately \(t \approx 6.93\) hours.

Key Concepts

Exponential Growth

Exponential growth describes a process where the rate of growth is proportional to the current size or quantity. In the context of spreading rumors, the more people know the rumor, the faster it spreads. This is due to each informed person potentially telling multiple uninformed people.

In mathematical terms, exponential growth is modeled by equations of the form:

• Without limit: \[ y(t) = y_0 e^{kt} \] where \(y_0\) is the initial quantity, and \(k\) is a positive constant.
• With limit: \[ p(t) = \frac{1}{1 + a e^{-kt}} \] In this form, \(a\) and \(k\) are constants that determine the shape and speed of the growth curve.

What makes this scenario interesting is that although the growth starts exponentially, it is bounded: as the rumor spreads, fewer uninformed people are left to tell, slowing the growth. In our specific problem, \(p(t)\) approaches 1 as \(t\) becomes very large, indicating that eventually, almost the entire population has heard the rumor.

Limits and Continuity

Understanding limits and continuity is crucial when analyzing how functions behave over time or space, especially when dealing with infinite processes or boundaries.

A limit in mathematics helps us determine where a function is heading as the input approaches a certain value. In the context of the rumor spreading equation:

• As \(t\) approaches infinity, the term \(e^{-kt}\) approaches 0, simplifying the fraction \(p(t) = \frac{1}{1 + ae^{-kt}}\) to \(\frac{1}{1} = 1\).

This tells us that, eventually, almost everyone will have heard the rumor, no matter the initial conditions, since the limit of \(p(t)\) is 1.

Continuity, on the other hand, ensures that the function's graph can be drawn without lifting the pencil from the paper. The expression for \(p(t)\) is continuous, meaning there are no sudden jumps or gaps in the graph of the rumor-spreading function. This is important in making reliable predictions at any time \(t\).

Rate of Change

The rate of change tells us how quickly one quantity is changing with respect to another. In calculus, we find the rate of change using derivatives. Here, the rate at which the rumor spreads is described by the derivative of \(p(t)\) with respect to \(t\):

• Given by the derivative \[\frac{dp}{dt} = \frac{a k e^{-kt}}{(1 + a e^{-kt})^2}\]

This formula shows the relationship between time and how fast the proportion of the population that knows the rumor is growing.

Initially, when fewer people know the rumor, the exponent \(-kt\) is higher, meaning \(e^{-kt}\) is large, and so is the rate \(\frac{a k e^{-kt}}{(1 + a e^{-kt})^2}\). However, as more people learn the rumor, \(-kt\) becomes smaller, decreasing \(e^{-kt}\) and slowing down the rate of spread. This illustrates how the spread of knowledge about the rumor não only grows exponentially at first but eventually tapers off as the maximum population knowledge is reached.