Problem 14 - Advanced Engineering Mathematics

Question

A mathematical model for the area \(A\) (in \(\mathrm{cm}^{2}\) ) that a colony of bacteria \((B .\) dendroides \()\) occupies is given by $$ \frac{d A}{d t}=A(2.128-0.0432 A) $$ Suppose that the initial area is \(0.24 \mathrm{~cm}^{2}\). (a) Use the \(\mathrm{RK} 4\) method with \(h=0.5\) to complete the following table. $$ \begin{array}{|l|lllll|} \hline t \text { (days) } & 1 & 2 & 3 & 4 & 5 \\ \hline A \text { (observed) } & 2.78 & 13.53 & 36.30 & 47.50 & 49.40 \\ \hline A \text { (approximated) } & & & & & \\ \hline \end{array} $$ (b) Use a numerical solver to graph the solution of the initialvalue problem. Estimate the values \(A(1), A(2), A(3), A(4)\), and \(A(5)\) from the graph. (c) Use separation of variables to solve the initial-value problem and compute the values \(A(1), A(2), A(3), A(4)\), and \(A(5)\)

Step-by-Step Solution

Answer

Use RK4, solve analytically, and graph to find values. RK4 provides approximations to match the observed data.

Step 1: Understanding the RK4 Method

The Runge-Kutta method (RK4) is an iterative technique for solving differential equations. It involves calculating four intermediate values (slopes) and using them to approximate the next step in the solution. It combines these slopes weighted to account for their relative reliability to calculate the function value at the next step. The formula for RK4 is:\[A_{n+1} = A_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4)\]where:- \(k_1 = f(t_n, A_n)\)- \(k_2 = f(t_n + \frac{h}{2}, A_n + \frac{h}{2}k_1)\)- \(k_3 = f(t_n + \frac{h}{2}, A_n + \frac{h}{2}k_2)\)- \(k_4 = f(t_n + h, A_n + hk_3)\)for the given differential equation.

Step 2: Applying RK4 for t = 1 day

Starting with the initial value \(A_0 = 0.24\) at \(t = 0\), apply the RK4 method with a step size \(h = 0.5\) to find \(A_{0.5}\) and then \(A_1\).1. Calculate \(k_1 = A_0(2.128 - 0.0432A_0)\).2. Calculate \(k_2 = (A_0 + \frac{h}{2}k_1)(2.128 - 0.0432(A_0 + \frac{h}{2}k_1))\).3. Calculate \(k_3 = (A_0 + \frac{h}{2}k_2)(2.128 - 0.0432(A_0 + \frac{h}{2}k_2))\).4. Calculate \(k_4 = (A_0 + hk_3)(2.128 - 0.0432(A_0 + hk_3))\).5. Compute \(A_{0.5}\) and \(A_1\) using the RK4 formula.

Step 3: Iterating RK4 for t = 2, 3, 4, 5 days

Repeat the RK4 steps for each subsequent half-day interval using the most recent computed value of A as the initial condition for the next iteration to find \(A_{1.5},\) \(A_2\), and so on. Continue this process to fill the table for points at \(t = 2, 3, 4, 5\).

Step 4: Solving the Differential Equation Analytically

Use separation of variables to solve the given differential equation:\[\frac{dA}{A(2.128 - 0.0432A)} = dt\]Integrate both sides and apply the initial condition \(A(0) = 0.24\) to find an explicit expression for \(A(t)\). Then, evaluate \(A(t)\) at \(t = 1, 2, 3, 4, 5\).

Step 5: Graphical Method and Solver

Use a numerical solver or graphing software to solve the differential equation and plot the solution curve over the desired time range. Approximate the values \(A(1), A(2), A(3), A(4),\) and \(A(5)\) directly from the graph, ensuring these values are consistent with the analytical and RK4 results.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In simple terms, they describe how a particular quantity changes over time or space. They are essential in modeling real-world phenomena where the rate of change is often more important than the change itself.
For example, in this exercise, the equation \( \frac{dA}{dt} = A(2.128 - 0.0432A) \) describes how the area \( A \) of a bacterial colony changes with time \( t \). The left side of the equation is the derivative of \( A \) with respect to \( t \), representing the rate of change of the area.
On the right side, the expression \( A(2.128 - 0.0432A) \) suggests a logistic growth equation, a common model for population dynamics. Here, the growth rate is affected by both the current size of the colony and a logistic term \( 2.128 - 0.0432A \), which accounts for factors such as limited resources or space.

• Independent variable: Often represents time; in this case, \( t \).
• Dependent variable: The quantity that changes over time; here, \( A \) for area.
• Form: Can be ordinary (ODE) or partial (PDE).

Understanding differential equations involves solving them, which can be complex and requires methods like the Runge-Kutta technique or separation of variables.

Numerical Methods

Numerical methods are approaches used to find approximate solutions to mathematical problems, especially when an exact solution is too difficult or impossible to obtain. The RK4, or fourth-order Runge-Kutta method, is a popular numerical method for solving ordinary differential equations (ODEs).
This technique is especially useful for finding solutions to complex or non-linear differential equations where analytic solutions are not feasible. The RK4 method involves calculating four intermediate slopes at each step and combining them to provide a more accurate solution estimate.
When using the RK4, you calculate the following:

• \(k_1\): The slope at the beginning of the interval.
• \(k_2\): The slope at the midpoint of the interval, calculated with \(k_1\).
• \(k_3\): Another slope at the midpoint, calculated with \(k_2\).
• \(k_4\): The slope at the end of the interval, calculated with \(k_3\).

The final approximation for the next step is a weighted average of these slopes. This makes RK4 a robust method for many applications, offering good accuracy with relatively fewer calculations compared to other lower-order methods.
Using RK4 in the given exercise involves starting with an initial area \( A_0 = 0.24 \) and calculating the area over time using the step size \( h = 0.5 \). You compute successive values for \( A(t) \) at each time step using the RK4 formula.

Separation of Variables

Separation of variables is a classical technique used to solve certain types of differential equations analytically. The method involves rearranging an equation so that all terms containing one variable are on one side and terms containing another are on the opposite side. This allows for the integration of both sides separately.
In the context of the exercise, you start with the differential equation:
\[ \frac{dA}{A(2.128 - 0.0432A)} = dt \]
The primary goal in separation of variables is to split the equation into two parts. You integrate the left side with respect to \( A \) and the right side with respect to \( t \). After integration, both sides are equated to find the function that solves the initial value problem.
For example, solving this differential equation using separation of variables and integrating both sides will provide a function \( A(t) \) that can be evaluated at specific time points \( t = 1, 2, 3, 4, 5 \).
Steps in Separation of Variables:

• Rearrange the equation to allow integration of both sides separately.
• Integrate the left side with respect to \( A \) and the right side with respect to \( t \).
• Apply initial conditions to solve for constants of integration and find the explicit solution.

This method is efficient for separable equations and provides an exact solution, which can then be used to verify the results obtained from numerical methods like RK4.