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$$ \begin{array}{|c|c|c|c|} \hline \text { Year } & \text { Wolves } & \text { Year } & \text { Wolves } \\\ \hline 1985 & 15 & 1993 & 40 \\ \hline 1986 & 16 & 1994 & 57 \\ \hline 1987 & 18 & 1995 & 83 \\ \hline 1988 & 28 & 1996 & 99 \\ \hline 1989 & 31 & 1997 & 145 \\ \hline 1990 & 34 & 1998 & 178 \\ \hline 1991 & 40 & 1999 & 197 \\ \hline 1992 & 45 & 2000 & 266 \\ \hline \end{array} $$ 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’s 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

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.

Nearly linear or exponential data: One of the two tables below shows data that are better approximated with a linear function, and the other shows data that are better approximated with an exponential function. Make plots to identify which is which, and then use the appropriate regression to find models for both. $$ \begin{aligned} &\begin{array}{|c|c|} \hline t & f(t) \\ \hline 1 & 3.62 \\ \hline 2 & 23.01 \\ \hline 3 & 44.26 \\ \hline 4 & 62.17 \\ \hline 5 & 83.25 \\ \hline \end{array}\\\ &\begin{array}{|c|c|} \hline t & g(t) \\ \hline 1 & 3.62 \\ \hline 2 & 5.63 \\ \hline 3 & 8.83 \\ \hline 4 & 13.62 \\ \hline 5 & 21.22 \\ \hline \end{array} \end{aligned} $$

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 natural logarithm of the data points. Does this plot make it look reasonable to approximate the original data with an exponential function? b. Find the regression line for the natural logarithm of the data and add its graph to the plot in part a. c. Construct an exponential model for the original subscribership data using the logarithm as a link.

Atmospheric pressure: The table below gives a measurement of atmospheric pressure, in grams per square centimeter, at the given altitude, in kilometers.17 $$ \begin{array}{|c|c|} \hline \text { Altitude } & \text { Atmospheric pressure } \\ \hline 5 & 569 \\ \hline 10 & 313 \\ \hline 15 & 172 \\ \hline 20 & 95 \\ \hline 25 & 52 \\ \hline \end{array} $$ (For comparison, 1 kilometer is about 0.6 mile, and 1 gram per square centimeter is about 2 pounds per square foot.) a. Plot the data on atmospheric pressure. b. Make an exponential model for the data on atmospheric pressure. c. What is the atmospheric pressure at an altitude of 30 kilometers? d. Find the atmospheric pressure on Earth’s surface. This is termed standard atmospheric pressure. e. At what altitude is the atmospheric pressure equal to 25% of standard atmospheric pressure?

Grains of wheat on a chess board: A children's fairy tale tells of a clever elf who extracted from a king the promise to give him one grain of wheat on a chess board square today, two grains on an adjacent square tomorrow, four grains on an adjacent square the next day, and so on, doubling the number of grains each day until all 64 squares of the chess board were used. How many grains of wheat did the hapless king contract to place on the 64th square? There are about \(1.1\) million grains of wheat in a bushel. Assume that a bushel of wheat sells for \(\$ 4.25\). What was the value of the wheat on the 64th square?

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