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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 \(t \geq 0\).

Short Answer

Expert verified
Answer: After 20 days, the bacterial population is approximately 1626 cells, and after 40 days, it's approximately 1685 cells. The general population function is given by \(N(t) = -400e^{-0.25 t} + 1100\).

Step by step solution

01

Integrate the growth rate function

To find the population function, we need to integrate the growth rate, \(N^{\prime}(t)\), with respect to time: \(N(t) = \int N^{\prime}(t) dt = \int 100 e^{-0.25 t} dt\)
02

Calculate the indefinite integral

We can integrate the function \(100 e^{-0.25 t}\) with respect to \(t\) using integration by substitution: Let \(u = -0.25 t\) and, therefore, \(du = -0.25 dt\). We have \(dt = \frac{du}{-0.25}\). Now, substitute the variables and integrate: \(N(t) = \int 100 e^{u} \cdot \frac{du}{-0.25} = -400 \int e^{u} du = -400(e^{u} + C)\) Substitute back \(u = -0.25 t\): \(N(t) = -400(e^{-0.25 t} + C)\)
03

Apply the initial condition

Apply the initial condition that the initial population is 1500 cells when \(t = 0\): \(N(0) = -400(e^{0} + C) = 1500\) Solve for \(C\): \(C = -400 + 1500 = 1100\) Now, we have the population function: \(N(t) = -400e^{-0.25 t} + 1100\)
04

Find the population after 20 and 40 days

Calculate the bacterial population after 20 days: \(N(20) = -400e^{-0.25 \cdot 20} + 1100 \approx 1625.81\) Calculate the bacterial population after 40 days: \(N(40) = -400e^{-0.25 \cdot 40} + 1100 \approx 1685.28\) So, after 20 days, the bacterial population is approximately 1626 cells, and after 40 days, it's approximately 1685 cells.
05

Show the general population function

The population \(N(t)\) at any time \(t\geq 0\) is given by the function: \(N(t) = -400e^{-0.25 t} + 1100\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Growth Rate Functions
When studying the growth of a bacterial population, it's essential to understand how to integrate growth rate functions to predict future populations. In the case of our bacteria in a Petri dish, the growth rate given is N'(t)=100e^{-0.25t}. This formula represents how the number of cells changes over time.

To find the total population N(t) at any time t, we need to integrate the growth rate function with respect to time. This process involves finding the antiderivative, or the original function that, when derived, gives us the growth rate. In this example, integrating the function 100e^{-0.25t} with respect to time gives us the total number of bacterial cells at any given moment.

During the integration process, it might be necessary to use techniques like substitution, which simplifies the integration of certain functions. For example, we set u=-0.25t to make the integration straightforward, eventually finding an expression for N(t) that shows how the population evolves over time.
Initial Value Problems
An initial value problem in calculus is a common practical issue where we have both a differential equation and the value of the function at a specific point, known as the initial condition. This condition allows us to find a particular solution to the differential equation. In context, the bacterial population problem provided mentions that initially, there are 1500 cells, which is our initial condition N(0)=1500.

After integrating the rate of growth, we obtain a general solution that includes an unknown constant C. To find the specific solution that fits our scenario, we apply the initial condition. By setting t=0 into our general solution, we can solve for C. This transforms our population function from a general form with an unknown constant to a precise formula that describes how the population grows from the initial count of 1500 cells at time t=0.
Exponential Growth and Decay
Exponential growth and decay are models that describe how quantities increase or decrease at rates proportional to their current value. Bacterial growth often follows an exponential pattern, at least within certain environmental limits. The rate of growth N'(t) in our problem involves an exponential function, which indicates the population changes exponentially with time.

The key feature of this exponential model is the base of the natural logarithm e raised to a power that includes a negative constant times t, representing time. This negative exponent shows that the rate of growth slows down over time, a common real-world phenomenon due to factors like limited resources or space.

By solving the exponential growth model, we calculate the population at any time t, and we found that after 20 days, the population is approximately 1626 cells, and after 40 days, it is about 1685 cells. The calculation demonstrates that as time increases, the bacterial population approaches an upper limit, determined by the specifics of the growth model and the environmental constraints within the Petri dish.

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