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Let the function f(t) be the world population (in billions) as a function of time (measured in years). For simplicity, let t = years since 1800, and let f(0) = 1. Part 1: Using a Table of Data Below is a table for f(t). Use it to answer the following questions. year t f(t) (population in billions) f ' (t) f ‘’ (t) 1800 0 1.0 xxxxxxxxxxxx xxxxxxxxxxxx 1850 50 1.2 xxxxxxxxxxxx 1900 100 1.6 1910 110 1.8 1920 120 1.9 1930 130 2.1 1940 140 2.3 1950 150 2.6 1960 160 3.0 1970 170 3.7 1980 180 4.5 1990 190 5.3 2000 200 6.1 2010 210 7.0 2020 220 7.8 2030 230 8.6 2040 240 9.2 xxxxxxxxxxxx 2050 250 9.8 xxxxxxxxxxxx xxxxxxxxxxxx Questions: 1. In the column labeled f ‘ (t), estimate values of the first derivative (using the forwardbackward technique). Do not worry about the endpoints (years 1800 and 2050) 2. Explain what f ‘ (t) means in the context of this situation. 3. Approximately when was the fastest rate of growth? 4. Use f(150) and f ‘ (150) to create a linear approximation of f (t). Then use the linear approximation to estimate the population in the year 1953. 5. In the column labeled f ‘’ (t), estimate values of the second derivative (using the forwardbackward technique on f ‘ (t) ). 6. Approximately where is the inflection point? Explain the significance of this point, in the context of this situation. 7. Using the data, sketch a graph of f (t), f ‘ (t), and f ‘’ (t). Part 2: Using a Model There are many phenomena that have exponential growth, such as growth of a population with unlimited resources. ( ) kt f t e  k = the continuous growth rate. However, in many circumstances, resources (such as food or space) is limited. In this situation, in the beginning the growth is exponential. However, as the amount of food or space gets used up, the growth rate slows. This is the case with the world population. Let… M = the carrying capacity, which is the maximum population given the limited amount of food or space. k = the continual growth rate under unlimited conditions P0 = the initial population (at time t = 0) One model for this type of growth is called a Logistic function: ( ) 1 kt M f t Ae   , where 0 0 M P A P   Example: A population of bacteria has a continual growth rate of 3%. Currently, the population is 4 million cells. Due to the amount of available nutrition, the carrying capacity is 30 million cells. Let t = time in days. a) Write a logistic model for this population growth b) Make a graph, and estimate when the population is close to the carrying capacity Part (a) Solution: M = 30, k = 0.03, P0 = 4 0 0 30 4 6.5 4 M P A P      0.03 30 ( ) 1 1 6.5 kt t M f t Ae e       Part (a) Answer: the population of the bacteria can be modeled by the formula 0.03 30 ( ) 1 6.5 t f t e    Part (b) Solution: Using Mathematica, type: Plot[30/(1 + 6.5*Exp[-0.03*t]), {t, 0, 200}] The graph looks like this: In about 200 days, the population will be close to carrying capacity. The next several questions will help you determine the formula that will model the world population. Questions: 8. Define the logistic equation as a function in Mathematica. Keep the M, k, P0 and A as unknown constants for now. 9. Decide a good value for the carrying capacity. In other words, how many people can the Earth sustain? This will be the M in the formula. Feel free to research using the internet to find a reasonable number. Search something like “world population carrying capacity” 10. Let P0 = 1, since this is the value of f(0). Then, compute the value for A 11. Decide a good value for k. This number represents the percentage growth per year. One way to do this is to compute f ‘ (t) / f (t), using the values from the table above. 12. Use Mathematica to plot the function. Experiment with various values of k and M until you get a model that is similar to the population data (from part 1 of this project). 13. Use Mathematica to find f ‘ (t) 14. Use Mathematica to find f ‘’ (t). 15. Use Mathematica to solve an equation to find the inflection point.