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 \(N^{\prime}(t)=100 e^{-0.25 t}\). 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

Answer: The population of bacteria after 20 days is approximately 1668.63 cells, and after 40 days, it is approximately 1629.99 cells. The function for the population at any given time t ≥ 0 is N(t) = -400e^{-0.25t} + 1900.

Step-by-step solution

Find the integral of the growth rate function

To determine the population at any time t, we need to integrate the growth rate function, N'(t). N'(t)= 100e^{-0.25t} Integrate N'(t) with respect to t:

Add the constant of integration

After integrating, we obtain the following function: N(t) = -400e^{-0.25t} + C Next, we need to find the constant of integration, C. We know the initial population is 1500.

Find the constant of integration using the initial population

Plug in the initial values: when t = 0, N(t) = 1500: 1500 = -400e^{-0.25(0)} + C Solve for C: C = 1900 Now we have the population function, N(t): N(t) = -400e^{-0.25t} + 1900

Determine the population after 20 and 40 days

First find the population after 20 days: t = 20; N(20) = -400e^{-0.25(20)} + 1900 ≈ 1668.63 Next, find the population after 40 days: t = 40; N(40) = -400e^{-0.25(40)} + 1900 ≈ 1629.99 The population after 20 days is approximately 1668.63 cells, and after 40 days, it is approximately 1629.99 cells. #b. Population N(t) at any time t ≥ 0#

State the population function

From the previous steps, we found the population function to be: N(t) = -400e^{-0.25t} + 1900 The population N(t) at any time t ≥ 0 can be found using this function.

Key concepts

Exponential Growth

Exponential growth describes how quantities increase at a constant proportional rate. It's a common pattern in nature, especially for populations of living organisms. In our exercise, the bacteria population in a Petri dish is an example of exponential growth. The rate of change is expressed using the differential equation \( N'(t) = 100 e^{-0.25t} \). This equation shows that the growth rate decreases over time, due to the negative exponent. This pattern ensures that while the bacteria population increases, it does so at a diminishing rate.
A key insight into exponential growth is that it often involves rapidly changing conditions that eventually slow down or plateau. In practical terms, this could mean that the population quickly uses up resources, like nutrients in the dish.

• Exponential growth: rapid increase initially
• Continuous growth rate adjustment
• Eventually slows due to limitations

Understanding exponential growth is crucial for predicting long-term outcomes of population dynamics.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used to model various phenomena, including growth, where rates of change play a pivotal role. In the context of our exercise, the differential equation \( N'(t) = 100 e^{-0.25t} \) models how the bacteria population changes over time.
To find the population function \( N(t) \) from this differential equation, we integrate \( N'(t) \). Integration enables us to move from capturing changes (derivative) to finding the actual population (function).
The solution to a differential equation can describe a dynamic process.

• Shows relationship between current state and rate of change
• Foundation for solving growth problems
• Transforms rates into actual values

Differential equations help us model complex systems and predict future behaviors. They require finding constants post-integration, often using initial conditions, which are known values at specific points.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation with a specified value at a starting point, essential for determining unique solutions. In the given exercise, we have an IVP:
- Given initial population: 1500 bacteria when \( t = 0 \).
This initial value is used to find the constant of integration after integrating the growth rate. For our equation, \( N(t) = -400e^{-0.25t} + C \), the initial population helps us determine that \( C = 1900 \).
Initial conditions ensure our solution is not only mathematically valid but also practically applicable. It effectively 'anchors' the solution to a known point. By substituting the initial value into the integrated equation, we can solve for \( C \), giving a complete function that predicts future populations.

• Initial value provides a starting reference point
• Ensures unique, applicable solutions
• Essential for determining constants in integration

Understanding initial value problems is key to solving real-life scenarios modeled by differential equations.