China Population Trends Practice Math Ia

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  • Topic: Polynomial, Differential equation, Logistic function
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Population Trends in China

The goal of this mathematical study is to explore the different functions that best model the population of China from the years 1950 to 1995.

Year | 1950| 1955| 1960| 1965| 1970| 1975| 1980| 1985| 1990| 1995| Population in Millions| 554.8| 609.0| 657.5| 729.2| 830.7| 927.8| 998.9| 1070.0| 1155.3| 1220.5|

Using the Chinese population data from 1950 to 1995, let us construct a graph using technology. Before graphing the data though, we must first determine the relevant variables, which are, the year and the population (in millions) of each coinciding year. The parameters are strictly confined to the data for the years 1950 and 1995 in the sense that the data cannot fall below the population number for the year 1950 and cannot fall above the data for the year 1995.

Upon reviewing the graph, we notice that the data appears to increase, but not precisely in a straight line. Going through the trend lines on excel, we find that the trend line that fits the best is the polynomial trend line, which is displayed in the graph down below. If we were to analytically develop one model function to determine if the polynomial trend line is indeed the most accurate fit, I would propose creating a system of equations.

Before jumping to far ahead, we need to make it clear the equation we are going to be analyzing. We will use the equation given to us by the polynomial trend line which is: y= ax2 + bx +c and the reason that we are using this equation is because of the fact that the R2 value is 0.9955. The closer the R2 value is to 1 the better it will fit the graph.

We will rearrange the equation y= ax2 + bx +c so that we can solve for the unknowns which are, the letters a, b and c. To do this, we need to add data to the equation and create three matrices. In order to continue on, let us first add the known values to the equation.

Given y= ax2 + bx +c , we know that the y-values are China’s population in millions and the x-values are the years at which the population is measured. We will use three points to solve this, one from the beginning, one from the middle, and one from the end in order to create the matrices that will then be used to find values for a, b, and c.

So we take the equation y= ax2 + bx +c and plug 1950 (first year) in as the x-value and 554.8 as the y-value and we will do the same with the next two points that will be used.
y= ax2 + bx +c
554.8=a (19502) + b (1950) + c
830.7=a (19702) + b (1970) + c
1120.5=a (19952) + b (1995) + c
Now that we have created our system of equations we can split them up into matrices. Looking at the equations we have come up with, we notice that we have a column of y-values (bold), a column of x-values (blue), and a column of unknowns (purple)

554.8=a (19502) + b (1950) + c
830.7=a (19702) + b (1970) + c
1120.5=a (19952) + b (1995) + c

We can call the y-values matrix ‘B’ or [B] which is going to be a 3x1 matrix, the x-values matrix ‘A’ or [A] which will be a 3x3 matrix, and the unknowns matrix ‘C’ or [C] which is another 3x1 matrix.

[A] = [B] = [C] =

So then we get an equation that looks like this: [B]=[A]*[C]. We can divide [A] from one side of the equation to isolate [C] so, [A]-1*[B]=[C]. We multiply by the inverse of [A]. Multiplying [A]-1*[B] we get the [C] to equal.

Year | 1950| 1955| 1960| 1965| 1970| 1975| 1980| 1985| 1990| 1995| New Population in Millions| 554.8| 620.8| 688.8| 758.7| 830.7| 904.6| 980.6| 1058.6| 1138.5| 1220.5|

Having plugged the original years into the found function we receive the data in the data table above and get a graph that looks like the data is close to the original graph.

We are next shown an equation where the population at time is modeled by: P (t)=

Using a calculator we can run a logistical test on the original data where we get K to equal 1950, L to equal 4.34, and M to...
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