Problem 411 - Precalculus

Question

For the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes. Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?

Step-by-Step Solution

Answer

The exponential equation is \( N(t) = 294.627 \, e^{0.0676586t} \). The population doubles in 10 minutes.

Step 1: Understand the Problem

We have two data points for the number of bacteria. At time \( t = 5 \) minutes, there are 360 bacteria, and at time \( t = 20 \) minutes, there are 1000 bacteria. We need to find an exponential equation that models this growth and determine the time it takes for the population to double.

Step 2: Define the Exponential Model

The population of bacteria can be modeled by the exponential equation \( N(t) = N_0 e^{rt} \), where \( N_0 \) is the initial amount, \( r \) is the growth rate, and \( N(t) \) is the population at time \( t \).

Step 3: Set Up the System of Equations

Using the exponential growth model, we have two equations based on the given data:\( 360 = N_0 e^{5r} \)\( 1000 = N_0 e^{20r} \).

Step 4: Solve for the Initial Population \( N_0 \)

Divide the second equation by the first to eliminate \( N_0 \):\( \frac{1000}{360} = \frac{N_0 e^{20r}}{N_0 e^{5r}} = e^{15r} \).This simplifies to \( \frac{1000}{360} = e^{15r} \).

Step 5: Calculate the Growth Rate \( r \)

Take the natural logarithm of both sides to solve for \( r \):\( 15r = ln\left(\frac{1000}{360}\right) \).\( r \approx \frac{ln(2.77778)}{15} \).

Step 6: Solve for \( r \)

Calculate \( r \):\( r \approx \frac{ln(2.77778)}{15} \approx 0.0676586 \).

Step 7: Write the Exponential Equation

With \( r \) known, we can substitute back into the equation to find \( N_0 \):Using \( 360 = N_0 e^{5 \cdot 0.0676586} \), solve for \( N_0 \):\( N_0 \approx \frac{360}{e^{0.338293}} \approx 294.627 \).The exponential model is \( N(t) = 294.627 \, e^{0.0676586t} \).

Step 8: Determine Doubling Time

To find the doubling time, solve for \( t \) when \( N(t) = 2N_0 \). So,\( 2 \cdot 294.627 = 294.627 \, e^{0.0676586t} \).This simplifies to \( 2 = e^{0.0676586t} \), and taking the natural logarithm gives \( t = \frac{ln(2)}{0.0676586} \).

Step 9: Calculate Doubling Time

Calculate the exact time \( t \):\( t \approx \frac{ln(2)}{0.0676586} \approx 10.24 \) minutes.Rounded to the nearest minute, the doubling time is 10 minutes.

Key Concepts

Bacteria Population Modeling

When it comes to understanding how populations of bacteria grow over time, modeling with exponential equations is essential. Bacteria reproduce rapidly under ideal conditions, leading to population sizes that change swiftly. Exponential growth captures this dynamic effectively, always accounting for the continuously compounding nature of reproduction.

An exponential growth model can be represented through an equation of the form:

• \( N(t) = N_0 e^{rt} \)

where:

• \( N(t) \) = the population at time \( t \)
• \( N_0 \) = the initial population size
• \( r \) = the intrinsic growth rate

The crucial detail of the equation is that the population size at any point in time is a function of its initial size, the growth rate, and time. This model highlights how bacteria population follows a rapidly increasing pattern as time progresses.

Population modeling using this equation helps predict how a bacterial culture might behave, which is pivotal in many fields, including ecology, medicine, and biotechnology.

Growth Rate Calculation

The growth rate \( r \) is a fundamental component in understanding how swiftly bacteria populations increase. Calculating it requires using known population figures at specified times.

In the given problem, we're provided with populations at two different times, namely 360 bacteria at 5 minutes and 1000 bacteria at 20 minutes. These data points help create a system of equations, each representing the bacterial population at a specific time using the exponential model equation:

• \( 360 = N_0 e^{5r} \)
• \( 1000 = N_0 e^{20r} \)

By dividing the second equation by the first, we can eliminate \( N_0 \), allowing us to solve for \( r \). The transformation and simplification are handled logarithmically, leading to:

• \( \frac{1000}{360} = e^{15r} \)
• \( ln\left(\frac{1000}{360}\right) = 15r \)

Thus, we solve for \( r \) with precision, rounding to a sensible degree of accuracy. This calculated \( r \) reveals how fast the bacteria are growing in the model and allows users to predict future population sizes.

Understanding the growth rate is vital because it is directly influenced by conditions such as temperature, nutrients, and environmental factors, all of which affect how fast populations expand.

Doubling Time Estimation

Doubling time refers to the period it takes for a population to double in size, which is a key concept in evaluating how efficient bacterial growth is under given conditions. To determine when a population doubles using our growth model, we need to find \( t \) when \( N(t) = 2N_0 \).

Using the growth equation \( N(t) = N_0 e^{rt} \), setting \( N(t) = 2N_0 \) leads to:

• \( 2N_0 = N_0 e^{rt} \)
• \( 2 = e^{rt} \)

The next step involves taking the natural logarithm of both sides, resulting in:

• \( ln(2) = rt \)

Solving for \( t \), we get:

• \( t = \frac{ln(2)}{r} \)

By substituting our previously calculated growth rate \( r \), we derive the exact doubling time. Understanding doubling time provides insights into the bacteria's reproduction rate and is applicable in practical scenarios like controlling infections or managing fermentation processes.Calculating doubling time helps biologists and scientists make critical decisions in research and application, reinforcing its importance in the study of exponential growth.