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Chapter 4: Problem 16

$$ \begin{array}{|c|c|c|c|} \hline \text { Year } & \text { Wolves } & \text { Year } & \text { Wolves } \\\ \hline 1985 & 15 & 1993 & 40 \\ \hline 1986 & 16 & 1994 & 57 \\ \hline 1987 & 18 & 1995 & 83 \\ \hline 1988 & 28 & 1996 & 99 \\ \hline 1989 & 31 & 1997 & 145 \\ \hline 1990 & 34 & 1998 & 178 \\ \hline 1991 & 40 & 1999 & 197 \\ \hline 1992 & 45 & 2000 & 266 \\ \hline \end{array} $$ Gray wolves in Wisconsin: Gray wolves were among the first mammals protected under the Endangered Species Act in the \(1970 \mathrm{~s}\). Wolves recolonized in Wisconsin beginning in 1980 . Their population has grown reliably since 1985 as follows: \({ }^{21}\) a. Explain why an exponential model may be appropriate. b. Are these data exactly exponential? Explain. c. Find an exponential model for these data. d. Plot the data and the exponential model. e. Comment on your graph in part d. Which data points are below or above the number predicted by the exponential model?

Short Answer

Expert verified
The wolves' growth resembles exponential growth. It's not perfectly exponential, but a model like \( P(t) = 15e^{0.2t} \) describes it well. The initial numbers are below the exponential model, but later numbers sometimes exceed the predictions.

Step by step solution

01

Understanding Exponential Models

An exponential model is appropriate for situations where a quantity increases by a constant rate, or factor, over equal intervals of time. In this scenario, the wolf population in Wisconsin is experiencing sustained growth over the period from 1985 to 2000. This pattern suggests that the population might be growing at a consistently increasing rate, making an exponential model potential a suitable fit.
02

Evaluating Data for Exponential Growth

To check if the data shows exponential growth, examine if the ratio of population sizes between consecutive years remains approximately constant. Calculate the ratios for each year pair: e.g., from 1985 to 1986 (16/15), from 1986 to 1987 (18/16), and so on. If these ratios are similar, it indicates exponential growth characteristics.
03

Calculating Yearly Growth Ratios

Compute the ratios between the population of wolves for consecutive years:- 1985 to 1986: \(\frac{16}{15} = 1.067\)- 1986 to 1987: \(\frac{18}{16} = 1.125\)- 1987 to 1988: \(\frac{28}{18} = 1.556\)- 1988 to 1989: \(\frac{31}{28} = 1.107\)- Continue this for all pairs of consecutive years to identify if the growth ratios are close to constant.
04

Observing Ratio Consistency

The calculated ratios are not exactly constant but show an overall increase trend. This suggests that while the population is not perfectly following an exponential pattern, it still exhibits characteristics of exponential growth since the annual change in population is growing larger over time.
05

Determining an Exponential Growth Model

Use the exponential model formula \( P(t) = P_0e^{rt} \), where \( P_0 \) is the initial amount, \( r \) is the growth rate, and \( t \) is time. Use year 1985 as \( t=0 \), with wolf population 15.Estimated \( r \) using the trend in ratios can be found using a fitting technique (or tool) such as regression analysis, resulting typically in an equation like \( P(t) = 15e^{0.2t} \), where \( e \) makes the base Euler's number.
06

Plotting Data and Model

Plot the original data points (1985 - 2000) on a graph and superimpose the exponential model derived in the previous step. The x-axis represents years, and the y-axis represents the wolf population, allowing visualization of both real data and the model for comparison.
07

Analyzing the Graph

Examine which data points lie above or below the expected numbers from the exponential model plot. Discuss the trend: generally, some initial data points might fall below the model curve, indicating slower growth initially, while later points align or surpass the model prediction due to acceleration in population growth.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Growth
Population growth refers to the increase in the number of individuals in a population. In the context of the gray wolves in Wisconsin, it offers an interesting case study. Ever since the gray wolves were recaptured in the region in the 1980s, their population has grown. As seen from the recorded figures from 1985 to 2000, the number of wolves significantly increased from 15 to 266. This process of population growth often originates from factors like improved habitat conditions, conservation efforts, and reduced threats. The important part of studying population growth is understanding whether the growth is constant, rapid, or follows particular patterns, like the exponential type. Studying these patterns helps in creating effective strategies for conservation or management plans. Understanding population growth deeply allows us to comprehend both the biological and environmental aspects influencing it.
Mathematical Modeling
Mathematical modeling is a powerful tool that helps to describe real-world systems using mathematical equations and concepts. When it comes to population studies, such as the wolf population, models help predict future trends based on past data. The concept involves the representation of biological scenarios using numbers and formulas. For the wolf population, creating a model begins with observing the data: the number of wolves year after year. Mathematical models can range from simple arithmetic to complex calculus-based structures, depending on the situation. In this case, by examining the population data from 1985 to 2000, a model can be crafted to reasonably predict future population sizes. Mathematical modeling helps researchers and policymakers make informed decisions by understanding the potential future scenarios.
Exponential Functions
Exponential functions are instrumental in understanding scenarios where quantities grow or decay at rates proportional to their current value. In simpler terms, it鈥檚 about growth that accelerates over time. For populations like the gray wolves, an exponential function is often used because their numbers appear to grow at a rate that increases each year. The formula for an exponential growth model is given by \[ P(t) = P_0 e^{r t} \] Here, \( P(t) \) denotes the population at time \( t \), \( P_0 \) is the initial population size, \( r \) is the growth rate, and \( e \) is Euler's number, a constant approximately equal to 2.71828.In our specific case, using 1985 as the starting point and analyzing the data, we can estimate the growth rate \( r \) and create a model that forecasts the wolf population in the subsequent years. Such a model can then be plotted against actual data to visualize the accuracy and adaptability of exponential functions in predicting real-world phenomena.

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Most popular questions from this chapter

Dispersion models: Animal populations move about and disperse. A number of models for this dispersion have been proposed, and many of them involve the logarithm. For example, in \(1965 \mathrm{O}\). H. Paris \(^{30}\) released a large number of pill bugs and after 12 hours recorded the number \(n\) of individuals that could be found within \(r\) meters from the point of release. He reported that the most satisfactory model for this dispersion was $$ n=-0.772+0.297 \log r+\frac{6.991}{r} $$ a. Make a graph of \(n\) against \(r\) for the circle around the release point with radius 15 meters. b. How many pill bugs were to be found within 2 meters from the release point? c. How far from the release point would you expect to find only a single individual?

Magazine sales: The following table shows the income from sales of a certain magazine, measured in thousands of dollars, at the start of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Income } \\ \hline 2001 & 7.76 \\ \hline 2002 & 8.82 \\ \hline 2003 & 9.88 \\ \hline 2004 & 10.94 \\ \hline 2005 & 12.00 \\ \hline 2006 & 13.08 \\ \hline 2007 & 14.26 \\ \hline 2008 & 15.54 \\ \hline \end{array} $$ Over an initial period the sales grew at a constant rate, and over the rest of the time the sales grew at a constant percentage rate. Calculate differences and ratios to determine what these time periods are, and find the growth rate or percentage growth rate, as appropriate.

Headway on four-lane highways: When traffic is flowing on a highway, the headway is the average time between vehicles. On four-lane highways, the probability \(P\) that the headway is at least \(t\) seconds is given to a good degree of accuracy \({ }^{6}\) by $$ P=e^{-q t}, $$ a. On a four-lane highway carrying an average of 500 vehicles per hour in one direction, what is the probability that the headway is at least 15 seconds? (Note: 500 vehicles per hour is \(\frac{500}{3600}=0.14\) vehicle per second.) b. On a four-lane highway carrying an average of 500 vehicles per hour, what is the decay factor for the probability that headways are at least \(t\) seconds? Reminder: An important law of exponents tells us that \(a^{b c}=\left(a^{b}\right)^{c}\). where \(q\) is the average number of vehicles per second traveling one way on the highway.

Cell phones: The following table shows the number, in millions, of cell phone subscribers in the United States at the end of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Subscribers (millions) } \\ \hline 2001 & 128.4 \\ \hline 2002 & 140.8 \\ \hline 2003 & 158.7 \\ \hline 2004 & 182.1 \\ \hline 2005 & 207.9 \\ \hline \end{array} $$ a. Plot the data points. Does this plot make it look reasonable to approximate the data with an exponential function? b. Use exponential regression to construct an exponential model for the subscriber data. c. Add the graph of the exponential model to the plot in part a. d. What was the yearly percentage growth rate from the end of 2001 through the end of 2005 for cell phone subscribership? e. In 2005 an executive had a plan that could make money for the company, provided that there would be at least 250 million cell phone subscribers by the end of 2007. Solely on the basis of an exponential model for the data in the table, would it be reasonable for the executive to implement the plan?

Magnitude and distance: Astronomers measure brightness of stars using both the absolute magnitude, a measure of the true brightness of the star, and the apparent magnitude, a measure of the brightness of a star as it appears from Earth.18 The difference between apparent and absolute magnitude should yield information about the distance to the star. The table on the following page gives magnitude difference m and distance d, measured in light-years, for several stars. $$ \begin{array}{|l|c|c|} \hline \text { Star } & \begin{array}{c} \text { Magnitude } \\ \text { difference } m \end{array} & \text { Distance } d \\ \hline \text { Algol } & 2.56 & 105 \\ \hline \text { Aldebaran } & 1.56 & 68 \\ \hline \text { Capella } & 0.66 & 45 \\ \hline \text { Canopus } & 2.38 & 98 \\ \hline \text { Pollux } & 0.13 & 35 \\ \hline \text { Regulus } & 2.05 & 84 \\ \hline \end{array} $$ a. Plot d against m and determine whether it is reasonable to model distance as an exponential function of magnitude difference. b. Give an exponential model for the data. c. If one star shows a magnitude difference 1 greater than the magnitude difference that a second star shows, how do their distances from Earth compare? d. Alphecca shows a magnitude difference of 1.83. How far is Alphecca from Earth? e. Alderamin is 52 light-years from Earth and has an apparent magnitude of 2.47. Find the absolute magnitude of Alderamin.
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