Problem 4
Question
Let \(P=P(t)\) be bacteria population at time \(t\) and \(P_{0}\) the initial number. From \(d P / d t=k P\) we obtain \(P=P_{0} e^{k t} .\) Using \(P(3)=400\) and \(P(10)=2000\) we find \(400=P_{0} e^{3 k}\) or \(e^{k}=\left(400 / P_{0}\right)^{1 / 3} .\) From \(P(10)=2000\) we then have \(2000=P_{0} e^{10 k}=P_{0}\left(400 / P_{0}\right)^{10 / 3},\) so $$\frac{2000}{400^{10 / 3}}=P_{0}^{-7 / 3} \quad \text { and } \quad P_{0}=\left(\frac{2000}{400^{10 / 3}}\right)^{-3 / 7} \approx 201$$
Step-by-Step Solution
Verified Answer
The initial population \( P_0 \) is approximately 201.
1Step 1: Understanding the Relationship Between Population and Time
The problem begins with the exponential growth model for bacterial population, given by \( \frac{dP}{dt} = kP \), which leads to a solution of the form \( P = P_0 e^{kt} \). Here, \( P_0 \) is the initial population at time \( t = 0 \), and \( k \) is the growth rate constant.
2Step 2: Using Given Conditions to Establish Equations
We are given two conditions: \( P(3) = 400 \) and \( P(10) = 2000 \). Using these, we can form the equations:1. \( 400 = P_0 e^{3k} \)2. \( 2000 = P_0 e^{10k} \).
3Step 3: Expressing Growth Rate in Terms of Initial Population
By solving the first equation, \( 400 = P_0 e^{3k} \), we can express \( e^k \) in terms of \( P_0 \):\[ e^{k} = \left( \frac{400}{P_0} \right)^{1/3} \]
4Step 4: Solving for Initial Population
Now use the second equation, \( 2000 = P_0 e^{10k} \), and our expression for \( e^k \):\[ 2000 = P_0 \left( \frac{400}{P_0} \right)^{10/3} \]Simplify to find \( P_0 \):\[ \frac{2000}{400^{10/3}} = P_0^{-7/3} \]Solving this gives us:\[ P_0 = \left( \frac{2000}{400^{10/3}} \right)^{-3/7} \approx 201 \]
5Step 5: Conclusion
The initial population \( P_0 \) is approximately 201 based on the provided conditions and the exponential growth model.
Key Concepts
Differential EquationsInitial Value ProblemsMathematical Modeling of Populations
Differential Equations
Differential equations are mathematical expressions that relate a function to its derivatives. In simpler terms, they describe how a quantity changes over time with respect to another changing variable. In the context of our exercise, the differential equation is given by \( \frac{dP}{dt} = kP \). Here, \( \frac{dP}{dt} \) represents the rate of change of the bacterial population \( P \), and \( k \) is a constant that determines the rate at which the population changes. This particular equation is an example of a first-order differential equation.
A first-order differential equation involves only the first derivative of the function. In biological systems, such as the growth of bacteria, differential equations are used to model processes that change continuously over time.
A key point is understanding that solving a differential equation means finding the original function (in this case, \( P(t) \)) that satisfies the relationship given by the equation. This is why we initially find a general solution in the form \( P = P_0 e^{kt} \). The exponential function \( e^{kt} \) suggests growth at a rate proportional to the current population, which is a common behavior in biological systems. Understanding differential equations is fundamental because they help describe how natural processes evolve.
A first-order differential equation involves only the first derivative of the function. In biological systems, such as the growth of bacteria, differential equations are used to model processes that change continuously over time.
A key point is understanding that solving a differential equation means finding the original function (in this case, \( P(t) \)) that satisfies the relationship given by the equation. This is why we initially find a general solution in the form \( P = P_0 e^{kt} \). The exponential function \( e^{kt} \) suggests growth at a rate proportional to the current population, which is a common behavior in biological systems. Understanding differential equations is fundamental because they help describe how natural processes evolve.
Initial Value Problems
Initial value problems are a specific type of problem involving differential equations. They are called this because we use starting conditions (initial values) to find a particular solution to the differential equation. In the exercise, these initial conditions are given by the known populations: \( P(3) = 400 \) and \( P(10) = 2000 \). These conditions allow us to determine specific constants involved in our equations.
Using these values, we derive equations by substituting them back into the general form of our solution, \( P = P_0 e^{kt} \). Solving these equations helps us identify important parameters like \( P_0 \), the initial population, and \( k \), the growth rate.
In the context of a bacterial growth model, an initial value problem tells us where the population starts and helps us predict how it grows or shrinks over time. By solving such problems, we find specific values that precisely describe the situation at hand, rather than just general behavior.
Using these values, we derive equations by substituting them back into the general form of our solution, \( P = P_0 e^{kt} \). Solving these equations helps us identify important parameters like \( P_0 \), the initial population, and \( k \), the growth rate.
In the context of a bacterial growth model, an initial value problem tells us where the population starts and helps us predict how it grows or shrinks over time. By solving such problems, we find specific values that precisely describe the situation at hand, rather than just general behavior.
Mathematical Modeling of Populations
Mathematical modeling in biology uses equations and formulas to represent biological processes. In population dynamics, we utilize mathematical models to estimate or predict how populations change over time. Our exercise uses one of the simplest models – the exponential growth model – to predict bacterial population growth.
The model assumes that the growth rate of the population is constant and proportional to its current size, reflected in the differential equation \( \frac{dP}{dt} = kP \). This leads to the solution \( P = P_0 e^{kt} \), where the exponential function describes uninterupted growth.
While this model is straightforward and useful for showing pure exponential growth, it's important to note that in real-life situations, populations might face limitations, such as resources, which cause the growth rate to decrease. However, for the purpose of understanding basic principles in a controlled environment, this model serves as a valuable tool for learning how populations change in response to various conditions.
The model assumes that the growth rate of the population is constant and proportional to its current size, reflected in the differential equation \( \frac{dP}{dt} = kP \). This leads to the solution \( P = P_0 e^{kt} \), where the exponential function describes uninterupted growth.
While this model is straightforward and useful for showing pure exponential growth, it's important to note that in real-life situations, populations might face limitations, such as resources, which cause the growth rate to decrease. However, for the purpose of understanding basic principles in a controlled environment, this model serves as a valuable tool for learning how populations change in response to various conditions.
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