Problem 9 - Advanced Engineering Mathematics

Question

Let \(X=X(t)\) be the amount of \(C\) at time \(t\) and \(d X / d t=k(120-2 X)(150-X) .\) If \(X(0)=0\) and \(X(5)=10\) then \\[ X(t)=\frac{150-150 e^{180 k t}}{1-2.5 e^{180 k t}} \\] where \(k=.0001259\) and \(X(20)=29.3\) grams. Now by L'Hôpital's rule, \(X \rightarrow 60\) as \(t \rightarrow \infty,\) so that the amount of \(A \rightarrow 0\) and the amount of \(B \rightarrow 30\) as \(t \rightarrow \infty.\)

Step-by-Step Solution

Answer

As \( t \rightarrow \infty \), \( X \) tends to 60, \( A \) tends to 0, and \( B \) tends to 30.

Step 1: Understanding the Differential Equation

We begin by analyzing the given differential equation \( \frac{dX}{dt} = k(120 - 2X)(150 - X) \). This equation describes the rate at which the quantity \( X \) changes over time, \( t \). The factors \((120 - 2X)\) and \((150 - X)\) suggest that the rate of change depends on how far \( X \) is from certain values.

Step 2: Solving the Differential Equation

To solve the differential equation, assume a solution of the form \( X(t) = \frac{150-150e^{180kt}}{1-2.5e^{180kt}} \). This expression includes an exponential function, which often arises when solving first-order differential equations.

Step 3: Verifying Initial Conditions

Substitute \( t = 0 \) into the solution to verify the initial condition \( X(0) = 0 \): \(X(0) = \frac{150 - 150e^{0}}{1 - 2.5e^{0}} = \frac{150 - 150}{1 - 2.5} = 0.\)This satisfies the initial condition given in the problem.

Step 4: Checking the Condition at t = 5

Now substitute \( t = 5 \) into the equation and check if \( X(5) = 10 \) is satisfied with \( k = 0.0001259 \). Compute: \(e^{180kt} = e^{180 \times 0.0001259 \times 5},\)\(X(5) = \frac{150-150e^{180 \times 0.0001259 \times 5}}{1-2.5e^{180 \times 0.0001259 \times 5}}. \) Calculate to confirm \( X(5) = 10 \) holds.

Step 5: Evaluating Long Term Behavior

To find the long-term behavior of \( X \), apply L'Hôpital's rule as \( t \rightarrow \infty \). The exponential term \( e^{180kt} \) will dominate, making the fractions approach a limit. Upon plugging \( t = \infty \), the limits show \( X \rightarrow 60 \).

Step 6: Determining Concentrations of A and B over Time

Given \( X \rightarrow 60 \) as \( t \rightarrow \infty \), deduce that the changes in the amounts of \( A \) and \( B \) are \( A \rightarrow 0 \) and \( B \rightarrow 30 \), respectively, indicating near completion of reaction.

Key Concepts

first-order differential equation

First-order differential equations involve derivatives of the first degree. They express how a quantity changes with respect to another, usually time. In the context of our exercise, we have the equation \( \frac{dX}{dt} = k(120 - 2X)(150 - X) \).This equation reveals that the rate of change of \(X\) depends on how \(X\) is related to the constants 120 and 150. The form of the equation suggests that as \(X\) gets closer to these values, the rate of change decreases.

• The equation is "separable" meaning you can rearrange it to integrate both sides separately, typically involving the integration of functions and solving for \(X(t)\).
• It shows how real-world phenomena like chemical reactions, population dynamics, and physics can be modeled.

Differential equations are powerful because they provide a framework to model and predict the behavior of complex systems over time.

L'Hôpital's rule

L'Hôpital's rule is a mathematical tool used to find limits of indeterminate forms, particularly when both the numerator and the denominator approach zero or infinity. This rule can be useful in assessing the long-term behavior of solutions to differential equations.In our exercise, as \( t \to \infty \), the solution \( X(t) = \frac{150-150 e^{180 k t}}{1-2.5 e^{180 k t}} \) approaches certain finite values. By using L'Hôpital's Rule, you can simplify this process:

• First, identify the limit as \( t \to \infty \) for the exponential parts of \(X(t)\).
• Calculate the derivative of both the numerator and the denominator. If the ratio approaches an indeterminate form, apply L'Hôpital's rule.

This method helps in understanding what \(X\), and hence reactions or systems, stabilize to as time passes.

initial conditions

Initial conditions specify the starting point or state of a system in differential equations. They are critical for finding particular solutions from general solutions. In this problem, we are given that \(X(0) = 0\).Verifying initial conditions involves substituting initial values into the proposed solution:

• Substitute \(t = 0\) to check if \(X(0) = 0\) satisfies the equation. If it holds, it validates the solution fits the real-world scenario initially described.
• Initial conditions distinguish one specific solution from a family of solutions.

Establishing correct initial conditions ensures accuracy in modeling how the system behaves and evolves from that starting point to future behavior.

exponential functions

Exponential functions frequently appear in differential equations due to their properties related to growth and decay processes. They are of the form \( a \cdot e^{bx} \), where \(e\) is the base of natural logarithms.In this exercise, the term \(e^{180kt}\) emerges:

• This term influences how rapidly the system progresses or decays toward equilibrium over time.
• Exponential functions grow or decay at rates proportional to their current values, which makes them efficient at modeling continuous change.

Understanding exponential behavior is crucial for predicting trends in various contexts such as population growth, radioactive decay, and financial compounded growth. It gives insight into how rapidly \(X(t)\) approaches its asymptotic limits as time progresses.