Problem 5 - Precalculus with Calculus Previews

Question

A model for the number of bacteria in a culture after \(t\) hours is given by \(P(t)=P_{0} e^{k t} .\) 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 229.

Step 1: Define the Variables

Identify the given information and variables from the problem. The initial population is represented as \(P_0\). We know that \(P(3) = 400\) and \(P(10) = 2000\), while \(k\) is the growth rate we need to calculate.

Step 2: Express Subsequent Populations

Use the given information to set up equations for the populations after the given times. Given: \(P(3) = P_0 e^{3k} = 400\) and \(P(10) = P_0 e^{10k} = 2000\).

Step 3: Solve for "k"

Divide the second population equation by the first to eliminate \(P_0\): \(\frac{P_0 e^{10k}}{P_0 e^{3k}} = \frac{2000}{400}\). Simplify to find \(e^{7k} = 5\). Take the natural logarithm of both sides: \(7k = \ln(5)\), so \(k = \frac{\ln(5)}{7}\).

Step 4: Solve for "P_0"

Substitute \(k\) back into one of the original equations to solve for \(P_0\). Use \(P(3) = 400\): \(P_0 e^{3(\frac{\ln(5)}{7})} = 400\). Therefore, \(P_0 = \frac{400}{e^{3(\frac{\ln(5)}{7})}}\).

Step 5: Calculate "P_0"

Calculate the numerical value using the expression from Step 4: \(P_0 \approx \frac{400}{(5)^{3/7}}\). Evaluate to find \(P_0 \approx 228.57\). Therefore, the initial number of bacteria \(P_0\) is approximately 229.

Key Concepts

Bacteria population model

When studying the growth of bacteria in a culture, scientists often use an exponential model. This mathematical model helps predict the bacteria count over time. In the population model presented, the bacteria population is expressed as:\[ P(t) = P_0 e^{kt} \]where:

• \(P(t)\): The population at time \(t\).
• \(P_0\): The initial population of the bacteria.
• \(e\): Euler’s number, approximately 2.718, which is the base of natural logarithms.
• \(k\): The growth rate, a specific parameter showing how fast the bacteria are growing.
• \(t\): Time, typically measured in hours or minutes.

By relying on this model, one can make educated guesses about the bacteria count without direct observation over long periods. It's a powerful tool, especially in the fields of microbiology and ecology, where predicting growth patterns is critical.

Initial population calculation

Determining the initial population \(P_0\) in a bacterial growth model is crucial for understanding the starting point of the growth process. To find this value, we need to rearrange and solve the exponential growth equation. Given two observations over time, like in our case:

• After 3 hours, \(P(3) = 400\)
• After 10 hours, \(P(10) = 2000\)

The challenge is to use these observations to first find the growth rate \(k\) and then determine \(P_0\). We start by setting up equations based on these observations, giving: \[ P(3) = P_0 e^{3k} = 400 \]and\[ P(10) = P_0 e^{10k} = 2000 \]With these, we can proceed to find \(k\), which then helps us isolate and solve for \(P_0\). Using such iterative steps, we estimate the initial bacteria quantity to gain insights into the subsequent growth model.

Growth rate determination

The growth rate \(k\) is a vital variable defining how quickly the bacteria population expands over time. In order to determine \(k\), we utilize our two observations:

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

By dividing these equations, \(P_0\) (the initial population) cancels out, allowing us to solve for \(k\):\[ \frac{P_0 e^{10k}}{P_0 e^{3k}} = \frac{2000}{400} \]This simplifies to\[ e^{7k} = 5 \]Then, by taking the natural logarithm of both sides, we have:\[ 7k = \ln(5) \]Solving for \(k\), we find:\[ k = \frac{\ln(5)}{7} \]This growth rate can then be used to make future predictions about the population or to uncover the initial population in the model. It provides a crucial understanding of the dynamics and speed of the bacterial growth process.