Problem 6 - Advanced Engineering Mathematics

Question

Let \(A=A(t)\) be the amount present at time \(t .\) From \(d A / d t=k A\) and \(A(0)=100\) we obtain \(A=100 e^{k t} .\) Using \(A(6)=97\) we find \(k=\frac{1}{6} \ln 0.97 .\) Then \(A(24)=100 e^{(1 / 6)(\ln 0.97) 24}=100(0.97)^{4} \approx 88.5 \mathrm{mg}\).

Step-by-Step Solution

Answer

After 24 units of time, the amount is approximately 88.5 mg.

Step 1: Understand the Given Differential Equation

We start with the differential equation \(\frac{dA}{dt} = kA\), which represents exponential growth or decay. Here, \(A\) is the amount present at time \(t\), and \(k\) is a constant that we need to find.

Step 2: Solve the Differential Equation

To solve this differential equation, we assume a solution of the form \(A(t) = A_0 e^{kt}\), where \(A_0\) is the initial amount. Given \(A(0) = 100\), we substitute and confirm that \(A(t) = 100 e^{kt}\).

Step 3: Use Given Condition to Find \(k\)

We are given that \(A(6) = 97\). Using the solution \(A(t) = 100 e^{kt}\), we substitute \(t = 6\) and \(A(6) = 97\) to find: \[ 97 = 100 e^{6k} \]Solving for \(k\), we have:\[ e^{6k} = \frac{97}{100} \]Taking the natural logarithm of both sides gives: \[ 6k = \ln 0.97 \]Thus, \(k = \frac{1}{6} \ln 0.97\).

Step 4: Calculate \(A(24)\) Using \(k\)

Now that we have \(k = \frac{1}{6} \ln 0.97\), we calculate \(A(24) = 100 e^{24k}\). Substituting \(k\), we get: \[ A(24) = 100 e^{\left( \frac{1}{6} \ln 0.97 \right) 24} \]Simplifying further, \(24 \times \frac{1}{6} = 4\), so we have:\[ A(24) = 100 e^{4 \ln 0.97} = 100 (0.97)^4 \]

Step 5: Compute the Final Amount

Calculate \( (0.97)^4 \) using a calculator:\[ (0.97)^4 = 0.88529281 \]Therefore, \( A(24) \approx 100 \times 0.8853 = 88.5 \).

Step 6: Conclusion

The amount present at time \( t = 24 \) is approximately \( 88.5 \) mg.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In our example, \( \frac{dA}{dt} = kA \) is a differential equation that describes exponential growth or decay.This specific form means that the rate of change of the amount \( A \) is proportional to the amount \( A \) itself.

Key points to note about this differential equation include:

• \( A \) represents the quantity that changes over time.
• \( t \) is the time variable.
• \( k \) is a constant that determines the rate of growth or decay.

Solving differential equations often involves finding a specific function \( A(t) \) that satisfies the equation for all \( t \). In our solution process, we assume \( A(t) = A_0 e^{kt} \) as a solution to capture either exponential growth or decay. This is a common step when you encounter a differential equation involving exponential change.

Initial Condition

An initial condition is a value that specifies the state of a system at the beginning of an analysis. In our problem, we start with \( A(0) = 100 \), indicating that at time \( t = 0 \), the amount present is 100.

Initial conditions are crucial because they allow us to find the unique solution to a differential equation by helping identify unknown constants like \( A_0 \) or \( k \). For this particular equation, it tells us:

• The specific starting amount of the substance, which in this case is 100 units.
• \( A(0) = 100 \) ensures the accuracy of the exponential model we derive.

Using the initial conditions, we plug into our assumed solution \( A(t) = A_0 e^{kt} \) to deduce that \( A_0 = 100 \).This simplifies our model to \( A(t) = 100 e^{kt} \), laying the groundwork for further analysis of the system's behavior over time.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is approximately 2.718. It is a fundamental concept in calculus, especially when working with exponential equations, such as solving for \( k \) in our problem.

Why is the natural logarithm used here?

• When we have an equation of the form \( e^{6k} = \frac{97}{100} \), taking the natural logarithm on both sides allows usto solve for the exponent \( 6k \).
• The property \( \ln(e^x) = x \) is applied, because the natural logarithm is the inverse function of the exponential function.
• It simplifies multiplication of exponents in expressions like \( e^{(1/6) \ln 0.97} \).

In this problem, by applying the natural logarithm, we isolated \( k \):\[ 6k = \ln 0.97 \]Solving it gives:\[ k = \frac{1}{6}\ln 0.97 \]This operation is essential for accurately describing exponential decay or growth processes.

Exponential Growth Solution

An exponential growth solution captures how quantities grow or decline over time. The general form is \( A(t) = A_0 e^{kt} \), where \( A_0 \) is the initial value, and \( k \) is the growth (or decay) constant.In our exercise, it describes either increasing or decreasing amounts, depending on whether \( k \) is positive or negative.

In exponential decay, like our example:

• \( k \) is negative, reflecting that the quantity decreases as time progresses.
• The amount at time \( t = 24 \) was found using \( A(t) = 100 e^{24k} \), which simplifies to \(100(0.97)^4\).
• Calculating \( (0.97)^4 \) shows the rate of decay, leading to approximately 88.5 mg remaining.

An exponential growth solution is pivotal in modeling real-world systems, as it provides insightful predictions about declining populations, radiation decay, or other natural and economic phenomena, essentially illustrating how a system evolves over time.