Problem 4 - Advanced Engineering Mathematics

Question

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time \(t\). After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria?

Step-by-Step Solution

Answer

The initial number of bacteria was approximately 224.

Step 1: Identify the Mathematical Model

The given problem involves exponential growth, described by the equation \( P(t) = P_0 e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( k \) is the growth rate, and \( e \) is the base of the natural logarithm.

Step 2: Establish Equations Using Given Data

We know that \( P(3) = 400 \) and \( P(10) = 2000 \). These give us two equations: \( 400 = P_0 e^{3k} \) and \( 2000 = P_0 e^{10k} \).

Step 3: Solve for Growth Rate (k)

Divide the second equation by the first:\[ \frac{2000}{400} = \frac{P_0 e^{10k}}{P_0 e^{3k}} \]This simplifies to:\[ 5 = e^{7k} \]Take the natural logarithm of both sides:\[ \ln(5) = 7k \]\[ k = \frac{\ln(5)}{7} \]

Step 4: Solve for Initial Population (P_0)

Using \( k = \frac{\ln(5)}{7} \), substitute back into the first equation:\[ 400 = P_0 e^{3 \cdot \frac{\ln(5)}{7}} \]Solve for \( P_0 \):\[ P_0 = \frac{400}{e^{3 \cdot \frac{\ln(5)}{7}}} \]

Step 5: Compute the Initial Population

Calculate the expression:\[ P_0 = 400 \times e^{-3 \cdot \frac{\ln(5)}{7}} \]Simplify the exponent:\[ e^{-3 \cdot \frac{\ln(5)}{7}} =\left( e^{\ln(5)}\right)^{\frac{-3}{7}} = 5^{-\frac{3}{7}} \]Therefore, compute:\[ P_0 = 400 \times 5^{-\frac{3}{7}} \]After calculating, \( P_0 \approx 224.19 \).

Key Concepts

Growth Rate Calculation

When dealing with exponential growth, one of the first tasks is often to determine the growth rate. In our exercise, this is represented by the variable \( k \). Exponential growth occurs when the rate of change of a population is proportional to the current size of the population.

To find the growth rate \( k \), we set up two equations using the information provided in the problem: \( P(3) = 400 \) and \( P(10) = 2000 \). Both equations follow the format \( P(t) = P_0 e^{kt} \), leading to:

• \( 400 = P_0 e^{3k} \)
• \( 2000 = P_0 e^{10k} \)

By dividing these equations, the terms involving \( P_0 \) cancel out, simplifying the analysis to just the exponential parts. This gives us the simplified equation \( 5 = e^{7k} \). Taking the natural logarithm of both sides solves for \( k \):

\[ \ln(5) = 7k \]

Which leads to:

\[ k = \frac{\ln(5)}{7} \]

This value of \( k \) characterizes the rapidity of the population increase over time.

Initial Population Estimation

The initial population, typically represented as \( P_0 \), is the size of the population when \( t = 0 \). Finding \( P_0 \) helps us understand how the population size changes relative to a starting point.

Once the growth rate \( k \) is determined, we use the precisely known population after a given time to solve for the initial population. In this case, we use the equation \( 400 = P_0 e^{3 \cdot \frac{\ln(5)}{7}} \). Here, \( 400 \) is the population after 3 hours.

To isolate \( P_0 \), solve for it by rearranging the equation:

\[ P_0 = \frac{400}{e^{3 \cdot \frac{\ln(5)}{7}}} \]

Further simplifying the exponent as:

\[ e^{-3 \cdot \frac{\ln(5)}{7}} = 5^{-\frac{3}{7}} \]

This results in:

\[ P_0 = 400 \times 5^{-\frac{3}{7}} \]

When calculated, the initial population comes out to approximately 224.19 bacteria. This indicates that at the start of the observation (\( t = 0 \)), there were roughly 224 bacteria present.

Exponential Growth Model

The exponential growth model is the foundation of problems involving rapid population changes, such as bacteria growth. It is typically expressed with the formula \( P(t) = P_0 e^{kt} \), where each element represents:

• \( P(t) \): Population at time \( t \)
• \( P_0 \): Initial population
• \( k \): Growth rate
• \( e \): Base of natural logarithms, approximately equal to 2.718

The exponential nature of the formula means that the growth rate acts as an exponent, causing the population to increase at a constantly accelerating rate if \( k \) is positive.

This model assumes that factors such as resources and space do not limit growth, allowing for ongoing and rapid increases. In our exercise, the model accurately portrays the multiplication of bacteria over specific time intervals. By observing the population size at two different times, we can determine both the initial population and growth rate. These knowns allow us to estimate the population at any other time \( t \).

Understanding this model is crucial because exponential growth often appears in biology, finance, and many other fields, indicating how small initial quantities can evolve into much larger amounts over time.