Problem 4 - Advanced Engineering Mathematics

Question

(a) We have \(d P / d t=P(a-b P)\) with \(P(0)=3.929\) million. Using separation of variables we obtain \\[ \begin{aligned} P(t) &=\frac{3.929 a}{3.929 b+(a-3.929 b) e^{-a t}}=\frac{a / b}{1+(a / 3.929 b-1) e^{-a t}} \\ &=\frac{c}{1+(c / 3.929-1) e^{-a t}} \end{aligned} \\] where \(c=a / b .\) At \(t=60(1850)\) the population is 23.192 million, so \\[ 23.192=\frac{c}{1+(c / 3.929-1) e^{-60 a}} \\] or \(c=23.192+23.192(c / 3.929-1) e^{-60 a} .\) At \(t=120(1910)\) \\[ 91.972=\frac{c}{1+(c / 3.929-1) e^{-120 a}} \\] or \(c=91.972+91.972(c / 3.929-1)\left(e^{-60 a}\right)^{2}\). Combining the two equations for \(c\) we get \\[ \left(\frac{(c-23.192) / 23.192}{c / 3.929-1}\right)^{2}\left(\frac{c}{3.929}-1\right)=\frac{c-91.972}{91.972} \\] or \\[ 91.972(3.929)(c-23.192)^{2}=(23.192)^{2}(c-91.972)(c-3.929) \\] The solution of this quadratic equation is \(c=197.274 .\) This in turn gives \(a=0.0313 .\) Therefore, \\[ P(t)=\frac{197.274}{1+49.21 e^{-0.0313 t}} \\] $$\begin{array}{|c|rrrr|} \hline & {}{\underline{\phantom{xx}}} {\text { Census }} & {}{\underline{\phantom{xx}}} {\text { Predicted }} & {}{\underline{\phantom{xx}}} {} \\ \text { Year } & \text { Population } & \text { Population } & \text {} & \text { Error } \\ \hline 1790 & 3.929 & 3.929 & 0.000 & 0.00 \\ 1800 & 5.308 & 5.334 & -0.026 & -0.49 \\ 1810 & 7.240 & 7.222 & 0.018 & 0.24 \\ 1820 & 9.638 & 9.746 & -0.108 & -1.12 \\ 1830 & 12.866 & 13.090 & -0.224 & -1.74 \\ 1840 & 17.069 & 17.475 & -0.406 & -2.38 \\ 1850 & 23.192 & 23.143 & 0.049 & 0.21 \\ 1860 & 31.433 & 30.341 & 1.092 & 3.47 \\ 1870 & 38.558 & 39.272 & -0.714 & -1.85 \\ 1880 & 50.156 & 50.044 & 0.112 & 0.22 \\ 1890 & 62.948 & 62.600 & 0.348 & 0.55 \\ 1900 & 75.996 & 76.666 & -0.670 & -0.88 \\ 1910 & 91.972 & 91.739 & 0.233 & 0.25 \\ 1920 & 105.711 & 107.143 & -1.432 & -1.35 \\ 1930 & 122.775 & 122.140 & 0.635 & 0.52 \\ 1940 & 131.669 & 136.068 & -4.399 & -3.34 \\ 1950 & 150.697 & 148.445 & 2.252 & 1.49 \\ \hline \end{array}$$ The model predicts a population of 159.0 million for 1960 and 167.8 million for \(1970 .\) The census populations for these years were 179.3 and \(203.3,\) respectively. The percentage errors are 12.8 and \(21.2,\) respectively.

Step-by-Step Solution

Answer

The population model predicts incorrect populations for 1960 and 1970, with significant errors of 12.8% and 21.2%.

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{dP}{dt} = P(a - bP) \), which represents a population model. The initial condition is \( P(0) = 3.929 \) million.

Step 2: Apply Separation of Variables

To solve the equation, use separation of variables to express it as \( \frac{dP}{P(a-bP)} = dt \). This can be integrated directly.

Step 3: Find the General Solution

The general solution is provided as \( P(t) = \frac{c}{1 + (c/3.929 - 1)e^{-at}} \), where \( c = a/b \). The formula represents the population \( P(t) \) at any time \( t \).

Step 4: Determine Constants Using Initial Conditions

Given \( P(60) = 23.192 \) million and \( P(120) = 91.972 \) million, calculate \( c \) and \( a \) by creating equations: \( c = 23.192 + 23.192(c/3.929 - 1)e^{-60a} \) and \( c = 91.972 + 91.972(c/3.929 - 1)(e^{-60a})^2 \).

Step 5: Solve Simultaneous Equations for Parameters

Utilize \( 91.972(3.929)(c - 23.192)^2 = (23.192)^2(c - 91.972)(c - 3.929) \) to solve for \( c \) and then determine \( a \). This results in \( c = 197.274 \) and \( a = 0.0313 \).

Step 6: Formulate the Population Model

Substitute \( c \) and \( a \) back into the solution to get \( P(t) = \frac{197.274}{1 + 49.21e^{-0.0313t}} \), which can now predict population at any year.

Step 7: Predict Population and Calculate Error

Using the model, predict populations for census years and compare to actual data. Compute errors to assess model accuracy. Particularly notice predictions for 1960 and 1970 differ significantly from the census, with errors of 12.8% and 21.2% respectively.

Key Concepts

Separation of Variables

One of the fundamental methods in solving differential equations is separation of variables. This method allows you to rearrange equations so that each variable appears on a separate side. In our case, we have a differential equation describing population growth:

• \( \frac{dP}{dt} = P(a - bP) \)
• The goal is to separate the variables \(P\) and \(t\).

By expressing it as \( \frac{dP}{P(a-bP)} = dt \), we have effectively separated the variables, enabling us to integrate each side independently. This makes it possible to transform the problem from differential form into integrable functions. Integration leads us to a more manageable expression that describes the relationship between population \(P\) and time \(t\). This technique is particularly useful for nonlinear equations, often found in real-world modeling scenarios.

Exponential Growth Model

Exponential growth models are widely used in population modeling due to their ability to depict rapid increase over time. In this exercise, the resulting equation from separation of variables leads us to an exponential form. The population \( P(t) \) is given by:

• \( P(t) = \frac{c}{1 + (c/3.929 - 1)e^{-at}} \)
• This formulation includes a parameter \( c = a/b \).

The term \( e^{-at} \) signifies the exponential decay factor, which competes with linear growth to stabilize the population at higher values. The model's structure ensures that as population size \(P\) increases, the growth rate diminishes, preventing indefinite exponential growth. This mirrors the logistic model often used to represent population capped by carrying capacity, a limit dictated by available resources.

Population Prediction

Predicting future populations relies on both mathematical models and previously obtained data values. By solving the differential equation, we derived the function:

• \( P(t) = \frac{197.274}{1 + 49.21e^{-0.0313t}} \)

This function can estimate the population at any given time \(t\). In our task, the model predicts for specific years, such as 1960 and 1970. Comparing predicted values to actual census data, there were errors, such as a 12.8% discrepancy in 1960. This highlights a key aspect of modeling: validating predictions against real-world data is crucial. Models are simplifications, and acknowledging their limitations is vital to improve future predictions.

Initial Conditions in Mathematical Models

Initial conditions provide the starting point for solving differential equations and are vital to accurate modeling. In this exercise:

• The initial population at time \( t = 0 \) is \( 3.929 \) million.

This condition is integral for finding the specific parameters \(c\) and \(a\) critical to the model. Without initial conditions, mathematical solutions would be too general, lacking precision for real-world scenarios. Initial conditions ensure that the model aligns well with historical data, allowing for meaningful predictions. They are fundamental in determining both short-term dynamics and long-term behavior in a population model, reflecting the unique initial state of the system.