Problem 43 - Calculus Early Transcendentals

Question

A culture of bacteria in a Petri dish has an initial population of 1500 cells and grows at a rate (in cells/day) of \(N^{\prime}(t)=100 e^{-0.25 t} .\) Assume \(t\) is measured in days. a. What is the population after 20 days? After 40 days? b. Find the population \(N(t)\) at any time \(t \geq 0\)

Step-by-Step Solution

Answer

Answer: The population after 20 days is approximately 1600.97 cells; after 40 days it is approximately 1537.47 cells. The population function is given by: \(N(t) = -400 e^{-0.25 t} + 1900\) for any time \(t \geq 0\).

Step 1: Integrate the rate function

To find the population function \(N(t)\), we need to integrate the rate function \(N^{\prime}(t)\): \(\int N^{\prime}(t) dt = \int 100 e^{-0.25 t} dt\)

Step 2: Finding the antiderivative

We need to find the antiderivative of the integrand in the given equation: \(\int 100 e^{-0.25 t} dt = -400 e^{-0.25 t} + C\) Here, \(C\) is the constant of integration.

Step 3: Use the initial condition

Now we use the given initial condition that the population is 1500 cells at \(t = 0\): \(N(0) = -400 e^{-0.25 \times 0} + C = -400 + C = 1500\) Solving for \(C\): \(C = 1900\)

Step 4: Write the population function

Now that we have the constant of integration \(C\), we can write down the population function \(N(t)\): \(N(t) = -400 e^{-0.25 t} + 1900\)

Step 5: Find the population after 20 days and 40 days

Now, we need to find the population after 20 days and 40 days using the population function \(N(t)\): - Population after 20 days: \( N(20) = -400 e^{-0.25 \times 20} + 1900 \approx 1600.97\) - Population after 40 days: \( N(40) = -400 e^{-0.25 \times 40} + 1900 \approx 1537.47\) So, after 20 days, the population is approximately 1600.97 cells, and after 40 days, it is approximately 1537.47 cells. a. The population after 20 days is approximately 1600.97 cells; after 40 days it is approximately 1537.47 cells. b. The population function is given by: \(N(t) = -400 e^{-0.25 t} + 1900\) for any time \(t \geq 0\).

Key Concepts

Population Growth

Population growth is a fascinating area of study in mathematics, particularly when analyzing how populations change over time. These changes can manifest in various forms, such as a bacterial culture in a lab or human demographics. In the context of calculus, population growth is often modeled using differential equations. This mathematical representation allows us to understand and predict changes based on initial conditions and growth rates.
When discussing population growth in bacteria, we start with an initial population, which refers to the number of individuals initially present. In our exercise, we have a starting point of 1,500 bacterial cells in a Petri dish. Then, the bacteria grow at a certain rate, defined by the differential equation provided. The growth is presented in terms of cells per day, indicating how many new cells are expected to appear each day.
It's essential to note that different factors can influence this growth rate, including environmental conditions and resources. In calculus, we assume our growth conditions are fixed for simplicity, allowing us to solve the problem mathematically. Such modeling can help predict future population sizes, manage resources effectively, and understand broader biological trends.

Differential Equations

Differential equations are a powerful tool in calculus used to describe how things change continuously. In essence, they relate the rate of change of a quantity to the quantity itself. In the context of population growth, the differential equation tells us how the population size changes with time.
In our exercise, the rate of change of the bacterial population is given by the function \(N^{\prime}(t) = 100 e^{-0.25 t}\).
Here, \(N^{\prime}(t)\) represents the change in population concerning time, \(t\). This expression involves an exponential function, which often appears in natural growth and decay processes. Understanding how to work with these equations can not only help predict future population sizes but also reveal insights into the nature of growth processes.
To solve such differential equations, integration techniques are employed. They help us find the original function \(N(t)\), representing the population size at any given time \(t\). These solutions allow us to evaluate the population under different scenarios, such as after 20 or 40 days, as illustrated in our problem.

Integration Techniques

Integration is a fundamental technique in calculus that helps find quantities from their rates of change. It is essentially the reverse process of differentiation. When we have a differential equation, like the one in our exercise, integration helps us determine the original function from its derivative.
In the bacterial growth scenario, the integration process involves finding an antiderivative. Given the rate \(N^{\prime}(t) = 100 e^{-0.25 t}\), we need to find \(N(t)\). The integration step provides us with the expression \[\int 100 e^{-0.25 t} \, dt = -400 e^{-0.25 t} + C\]
where \(C\) is the constant of integration. This constant is determined by applying initial conditions such as \(N(0) = 1500\), leading us to the full population function \(N(t)\).
This function, \(N(t) = -400 e^{-0.25 t} + 1900\), allows us to calculate the population at any time \(t\), and it is a crucial step for understanding changes over time. Through integration, we transform information about rates into meaningful data about quantities, illustrating how populations evolve.