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• Anthony Lu

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IB Math HL: Exploration Project Modeling Europe’s Populations Using the Logistic Model with Carrying Capacity Personal Code: fwj500 Word Count: 2464

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Introduction Logistic equations are often taught in calculus classes and used to model populations, such as bacteria or animals. In this investigation, I will focus on a specific type of logistic equation, which models populations with a carrying capacity. Although in math class we studied these models by solving math problems involving the population of bacteria, I was interested in determining whether the world population could be modeled with these equations, and whether a logistic model that takes carrying capacity into account can model populations with a reasonable degree of accuracy. However, I soon noticed that my original plan to model the population of the world using a logistic equation with a carrying capacity would not work because the population of the world has been growing at an approximately constant or increasing rate. Because my plan was to find the carrying capacity of a human population, I decided that I needed to model populations with a declining growth rates, and decided to model the population of Europe since 1960. My goal in this exploration is to figure out whether the population of Europe will reach a carrying capacity and stop growing at a specific population value. However, another goal of this investigation is to identify limitations of a carrying capacity model. A limitation that I thought I might encounter is that this model does not take into account major events that can influence population trends, such as epidemics or mass migrations in or out of a region.

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Rationale The aim of this investigation is to determine a carrying capacity Europe, because the trend in the population of these regions is that the population is growing at a slower rate. I can also determine an equation to model the population of Europe based on the assumption that there is a carrying capacity. The methodology will be solving a differential equation and using the method of partial fraction decomposition, as well as finding a general solution based on given initial values for the differential equation. First, after gathering population data I will solve a differential equation to find a logistic model for the population based two specific data points. Once I find one logistic model for the population, I will use two different data points to determine another equation to model the population. I will do this two times so that I have three equations in total to compare to determine whether different samples from a data set affects the final derived equation for the model. I will also compare these equations to an equation derived from using data points that begin in 1980 instead of 1960, and determine whether omitting data points can significantly affect the final equation derived from the same model. Through the derivation of multiple models, I can assess the reliability of this model and the extent to which this population data leads to a common model, as well as uncovering limitations in applying mathematical models to a real life situation, which often assume ideal conditions.

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First Logistic Population Model The following is the data for the population of Europe for certain years since 1960 (Info) Population of Europe since 1960 Year 1960 1961 1970 1971 1980 1981 1990 1991 2000 2001 2010 2011

Population (Millions) 605.619 611.682 657.221 661.377 693.859 696.802 721.086 723.079 726.407 726.453 735.394 736.316

Growth Rate 6.063 4.155 2.943 1.993 0.046 0.921 dp The difference in population between consecutive years is equal to in the logistic dt equation, or the rate of change of population in any given year.

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This graph, created from the Logger Pro graphing software, has the general shape of a graph with an asymptote. I was interested in exploring possible best-fit lines, such as an exponential model or logarithmic model. Using the Logger Pro software, the two best-fit lines I attempted ended up having the equations of y  49.52 ln(kx) , where k=60660 +/- 67000, and y  144.9*0.953x  749.4 , with relatively low uncertainty values in the constants.

The logarithmic graph is the best fit line that starts at a much lower y intercept and has a higher y value than the exponential graph as x increases (towards the right side of the graph, the curve is above the exponential graph). The logarithmic graph has high uncertainty in the constants, and the exponential graph fits the data points much better than the logarithmic graph, which suggests that the logistic equation, which is similar to the exponential equation, is quite likely to fit the data set. The model for a logistic equation is

dp dp Mp p  kp(  kp (1  ) , or ) , where M is the dt dt M M

carrying capacity and P is the population at a given year (Powell). In this model, P will represent the population during a specific year, while t will represent the number of years that have passed since 1960. I will be using two data points to solve for k and M in this logistic equation. However, as we can see from the table, the year 2000 is an outlier, with a very low rate of population growth. The rate is slower than the rate in 2010, which is a limitation for this model because the model I will use assumes that the rate of population growth is slowing down. I was interested in exploring the impact of this outlier on the final logistic equation derived because it will help determine the extent that assumptions of mathematical models must be met to reach an accurate conclusion.

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For the first equation that I will derive, the two data points that I will use are the years 1980 and 1990. The population of these two years are 693.859 and 721.086, respectively, based on the table, and the rates of population growth for these two years are 2.943 and 1.993. I will substitute these values into the logistic differential equation to solve for k and M through a system of two equations. M  693.859 ) M M  721.086 1.993  k *721.086*( ) M 2.943  k *693.859*(

Dividing the first equation by the second equation eliminates k and by moving constants to the left, 1.535 

M  693.859 M  721.086

1.535M 1106.588  M  693.859 M  771.456 , so the carrying capacity of Europe is approximately 771 million people based on these two specific data points. This value is interesting because the population statistics I used estimated that in the near future, a population of approximately 740 million to 745 million will be the maximum population of Europe before the population for Europe begins to decrease (Info). This demonstrates that using math to solve a population model, such as the logistic model that assumes ideal conditions, often does not take into account the dynamic nature of real life issues such as human populations. Factors such as an influx of migrations are not taken into account due to the assumptions of the model.

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Substituting this value for M back into the differential equation gives k  0.042 . Thus, by substituting the values for k and M, the differential equation to model the population of Europe is dp p  0.042 p(1  ) dt 771.456

By integrating this equation, I can find an equation for the population of Europe. First, the variables must be separated, as this is a separable differential equation. dp p 0.042 p (1  ) 771.456

 dt

18368dp  dt p(771.456  p)

After partial fraction decomposition of the left hand side of the equation, the left hand side can be integrated. To decompose

fractions,

18368dp into two more easily integrated p(771.456  p)

18368dp must be rewritten as p(771.456  p)

18368 A B   p(771.456  p) p 771.456  p Multiplying both sides by p(771.456  p) results in 18368  A(771.456  p)  Bp

Substituting the value p  0 solves for A, and A  23.810 , while substituting the value p  771.456 solves for B, and B  23.810 . The differential equation can be rewritten as

23.810 23.810  dp  dt p 771.456  p

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Integrating the equation, and combining the integration constants into one constant, 23.810 ln | p | 23.810 ln | 771.456  p | t  C

ln |

p t |  C  0.042t  C 771.456  p 23.810

Ce0.042t 

p 771.456  p

To solve for C, the first data point, (0, 605.619), which represents the population of 605.619 million in 1960 (since the value of t is measured in the number of years passed since 1960), will be substituted into the equation.

605.619 771.456  605.619 C  3.652

Ce0 

After solving for C, solving for P from this linear equation determines the final equation. p (1  771.456e0.042t )  2817.275e0.042t p

2817.275e0.042t 771.456  0.042 t 1  3.652e 1  0.274e 0.042t

This seems like a logical equation because the carrying capacity, which is the value in the numerator, is reasonable. However, to reach this final equation, I used two specific data points that were space apart by 10 (the data points were the points at t=20 and t=30) and solved for the values of k and M. I will use other pairs of data points out of the data set from the table and compare the other equations I derive. I will attempt to use two points that are separated by a greater distance and also attempt to incorporate the outlier data point (t=40, where the rate of growth was 0.046, a very small value). I will compare constants such as carrying capacity and the power that the exponential in the denominator is raised to (the value of r in the ert term in the denominator), which is the Malthusian parameter and an indicator of the rate of maximum population growth

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(Weisstein). This is important to investigate because it will help me identify whether analysis of a data set through this particular model, regardless of which data points specifically are selected, will generally lead to a common final result. This will reveal how much ideal models can account for uncertainty in real life situations. General Solution and Other Derived Equations for the Model I will derive a general formula for solving the equation from the logistic differential equation, assuming that the values for k and M can be solved for or are known (Powell). dp p  kp (1  ) dt M

M dp *  dt k p( M  p) By partial fraction decomposition,

1 A B   p( M  p) p M  p

1  A( M  p)  Bp .

Substituting p=M gives the solution B 

A

1 while substituting p=0 gives the solution M

1 . The equation will be rewritten as M

1 1 1 (  )dp  dt . After integration, k p Mp

1 *(ln | p |  ln | M  p |)  k

ln |

p | Mp  t C k

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p | kt  C Mp p Ce kt  Mp ln |

p

MCe kt M  kt 1 1  Ce 1  e  kt C

Substituting the data point at t=0 and population p=p0, where p0 is the initial population, into the second equation returns the equation C 

p0 , so substituting this into the M  p0

final equation returns p

M , which is the formula I will use for the solution of this equation using M  p0  kt 1 e p0

different data points or different data sets.

Returning to the model of the population of Europe, I will use two data points that are further apart for this other model. The data points I will use are the points (10, 657.221) with a rate of growth of 4.155 and (50, 735.394) with a rate of growth of 0.921. The logistic equation with these two data points, using the previously derived formula, is p

754.706 1  0.246e 0.049t

I will also compare these equations to the model obtained when using the data points (30, 721.1) with a rate of change of 1.993 and (40,726.407) with a rate of change of 0.046 to derive a final logistic equation. The final equation using these data points is p

725.532 1  0.200e 0.037 t

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771.456 Comparing these two equations to the original equation, 1  0.274e 0.042t , the important

constants in these values are very similar. First, the carrying capacity values, which equal the values on the numerators, suggest that using different data points can lead to different derived equations for the model, being approximately 725, 755, and 771. The Malthusian Parameter values also were not completely consistent, as the values were 0.037, -0.042, and -0.049. In the equation involving the outlier data point of t=40, the Malthusian Parameter was slightly different from the other values (equal to -0.037), and the coefficient of the exponential in the denominator was also slightly different, being equal to 0.274, compared to 0.246 and 0.200 when the outlier data point was not used. Thus, while the constants in the logistic equations derived differed, this may somewhat be accounted for by the fact that one of the models was based on an outlier. It would be interesting to explore whether there is more agreement in the models when outlier data points are not used at all. Any limitation in the methodology to derive these logistic equations would apply to all the three derived models. These limitations could arise from unforeseen factors resulting from the dynamic nature of human populations, such as migrations or changing cultural attitudes towards having children. Thus, this model is only accurate to the extent that it predicts the future population if current population trends continue. The first logistic equation derived from the data points (20, 693.859) and (30, 721.086), y

771.5 1  0.274e 0.042 x , appears to fit the data set the best out of the derived equations.

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Thus, while this method has a medium degree of uncertainty, the equations derived show some agreement over the values of the constants in the logistic model.

Finally, I will apply this model to the population with a different starting year, and compare the equation to the previously derived equations to determine how much changing the data set (in this case, omitting points) affects the final equation derived. Specifically, I will use data points for the population beginning in 1980 instead of 1960.

In this new data set, the values for x are shifted down by twenty, because the starting time of t=0 using a starting year of 1980 corresponds to the time value of t=20 using the population data beginning in 1960. Based on this new data set, the previous points (30, 721.086) and (50, 735.394) translate to the points (10, 721.086) and (30, 735.394). Using the points (10, 721.086) with a rate of change of 1.993 and (30, 735.394) with a rate of change of 0.921, the population modeling equation becomes

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p

738.305 . While the value for carrying capacity, 738.305, is close to the values 1  0.064e 0.092t

derived using the data set beginning in 1960, the constants in the denominator are quite different from previously derived values. Thus, omitting data points significantly affects the derived equation for the logistic model. Conclusion Based on the modeling of the population over time, the population of Europe fits the logistic model. The carrying capacity is similar to the projected maximum population of Europe, but there is some uncertainty because different carrying capacity values and Malthusian Parameters were derived from using different data points to determine the capacity. However, a major issue with these models is that they may not take into account events that will change the rate of population growth of these regions. A limitation of this investigation is that it only uses one model, the logistic model, for these populations. Although the logistic model with a carrying capacity was justified because of the declining rate of population growth in the regions modelled, an extension would be to test how well other modelling equations with different sets of assumptions fit populations that fit the assumptions of those models, and assess limitations of those models. Examples of such extensions include using the single species model to model the world population based on the birth and death rates in the world, or the SinkoStreifer model, which assumes a constant birth rate while taking into account the mortality rate with respect to the population size (Population Size). Also, it would be interesting to explore whether other fixed data sets, such as other populations, will return approximately the same equation when other models are applied to the data sets, regardless of what part of the data set was selected to determine the model.

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Works Cited [email protected]. "EUROPE Population 1958." EUROPE Population 1958 by UN Statistics since 1950 to 2100. N.p., n.d. Web. 26 Jan. 2017. . "Population Size." Population Parameters: Estimation for Ecological Models (2008): 49101. Web. 16 Mar. 2017. . Powell, James. "Solution of the Logistic Equation." Solution of the Logistic Equation. USU Math, 31 July 2000. Web. 13 Mar. 2017. . Weisstein, Eric W. "Logistic Equation." Logistic Equation -- from Wolfram MathWorld. MathWorld, n.d. Web. 16 Mar. 2017. .

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Appendix: Population of Europe since 1960 Year 1960 1961 1970 1971 1980 1981 1990 1991 2000 2001 2010 2011

Population (Millions) 605.619 611.682 (growth: 6.063) 657.221 661.377 (growth: 4.155) 693.859 696.802 (growth: 2.943) 721.086 723.079 (growth: 1.993) 726.407 726.453 (growth: 0.046) 735.394 736.316 (growth: 0.921)

Note that all these derivations are based off the equation

can be rewritten as

dp p  kp (1  ) , which dt M

dp Mp  kp( ) dt M

Derivation of values of k and M using the data points (10, 657.221) with a rate of change of 4.155 and (50, 735.394) with a rate of change of 0.921 M  657.221 ) M M  735.394 0.921  735.394k ( ) M 4.155  657.221k (

Dividing the two equations to cancel k and moving all the constants to the left side gives 5.048 

M  657.221 M  735.394

M  754.706 k  0.049

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M  p0 754.706  605.619   0.246 p0 605.619

Derivation of the values of k and M using the data points (30, 721.1) with a rate of change of 1.993 and (40,726.407) with a rate of change of 0.046:

M  721.086 ) M M  726.407 0.046  726.407 k ( ) M M  721.086 43.646  M  726.407 M  726.532 k  0.037

1.993  721.086k (

M  p0 726.532  605.619   0.200 p0 605.619

Derivation of the values of k and M using the data points (10, 721.086) with a rate of change of 1.993 and (30, 735.394) with a rate of change of 0.921 from data set with population of Europe beginning in the year 1980

M  721.086 ) M M  735.394 0.921  735.394k ( ) M M  721.086 2.122  M  735.394 M  748.146 k  0.074

1.993  721.086k (

M  p0 738.305  693.859   0.078 p0 693.859