Problem 30 - Thomas Calculus

Question

Growth of bacteria A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are \(10,000\) bacteria. At the end of 5 hours there are \(40,000 .\) How many bacteria were present initially?

Step-by-Step Solution

Answer

The initial population of bacteria was 1250.

Step 1: Understand the Exponential Growth Model

The population of bacteria grows exponentially, which can be modeled using the equation \( P(t) = P_0 \times e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( k \) is the growth rate, and \( t \) is time.

Step 2: Set Up Equations Using Given Conditions

We are given two conditions: \( P(3) = 10000 \) and \( P(5) = 40000 \). Substituting these into the exponential model gives us two equations: \( 10000 = P_0 \times e^{3k} \) and \( 40000 = P_0 \times e^{5k} \).

Step 3: Solve for the Growth Rate \( k \)

Divide the second equation by the first to eliminate \( P_0 \): \( \frac{40000}{10000} = \frac{P_0 \times e^{5k}}{P_0 \times e^{3k}} \). Simplify to get \( 4 = e^{2k} \). Taking the natural logarithm on both sides, \( \ln(4) = 2k \), so \( k = \frac{\ln(4)}{2} \).

Step 4: Find the Initial Population \( P_0 \)

Now that we have \( k \), substitute it back into either equation to solve for \( P_0 \). Using the first equation: \( 10000 = P_0 \times e^{3\left(\frac{\ln(4)}{2}\right)} \). Compute \( e^{3\left(\frac{\ln(4)}{2}\right)} = \left(e^{\ln(4)}\right)^{\frac{3}{2}} = 4^{1.5} \). Hence, \( 10000 = P_0 \times 4^{1.5} \). Simplify to find \( P_0 = \frac{10000}{4^{1.5}} \). Compute \( 4^{1.5} = 8 \), so \( P_0 = \frac{10000}{8} = 1250 \).

Key Concepts

Growth Rate

In the context of exponential growth, the growth rate, denoted as \( k \), is a crucial parameter. It measures how quickly the population of bacteria increases over time. A higher growth rate means that the population is increasing more rapidly.

• Mathematically, the growth rate \( k \) is found using the formula \( k = \frac{\ln(4)}{2} \), derived from the natural logarithm of the ratio of population changes over specific time intervals.
• This formula highlights the importance of logarithms in decomposing exponential equations into linear ones to solve for \( k \).

Understanding growth rates can help predict how populations will evolve over time, which is invaluable in fields like biology and ecology.

Initial Population

Initial population, denoted \( P_0 \), is the starting number of bacteria before any growth has occurred. It is a foundational component of the exponential growth model, allowing us to calculate and predict future population sizes.

• In an exponential model \( P(t) = P_0 \times e^{kt} \), \( P_0 \) sets the baseline from which population growth begins.
• In our exercise, solving for \( P_0 \) involved substituting known values into the growth equation and isolating \( P_0 \).
• We found \( P_0 = \frac{10000}{8} = 1250 \) after calculating \( 4^{1.5} = 8 \).

Understanding the initial population not only provides insights into starting conditions but also aids in predicting the trajectory of growth.

Bacteria Population Model

The bacteria population model is an exponential function used to describe how populations grow under ideal conditions. The general model is \( P(t) = P_0 \times e^{kt} \), which reflects continuous growth.

• It's characterized by the current population size \( P(t) \), initial population \( P_0 \), growth rate \( k \), and time \( t \).
• This formula indicates that as time \( t \) increases, the exponential factor \( e^{kt} \) leads to significant growth, assuming \( k \) is positive.
• The exponential nature means that small changes in \( k \) or \( t \) can lead to large changes in \( P(t) \).

This model is particularly helpful in predicting bacterial growth in a stable environment, allowing scientists to prepare for various outcomes.

Natural Logarithm

The natural logarithm, represented as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. It is particularly useful in exponential growth equations for its natural properties.

• In solving the exponential growth equation, \( \ln(x) \) helps simplify equations involving exponential terms.
• For example, finding the growth rate involved using \( \ln(4) \) to convert the equation \( 4 = e^{2k} \) to a form solvable for \( k \).
• The property \( \ln(e^x) = x \) is frequently used to simplify exponential equations.

Natural logarithms are essential tools in calculus and advanced mathematics for solving problems involving exponential and logarithmic relationships.