Many populations are facing negative growth rates. If the population of the Ukraine was 60 million in 2008 and will shrink to 59.3 million people in 2009, write the equation which models the Ukraine population over time if the population loss is proportional to population at a given time.
Many populations are facing negative growth rates. If the population of the Ukraine was 60 million in 2008 and will shrink to 59.3 million people in 2009, write the equation which models the Ukraine population over time if the population loss is proportional to population at a given time.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Negative Population Growth Modeling
Many populations are facing negative growth rates. To illustrate this, consider the population of Ukraine. It was 60 million in 2008 and is projected to shrink to 59.3 million people by 2009.
To model Ukraine's population over time, assuming the population loss is proportional to the population at any given time, we will use a differential equation approach.
#### Problem Statement:
If the population of Ukraine was 60 million in 2008 and will shrink to 59.3 million people in 2009, write the equation which models the Ukraine population over time if the population loss is proportional to the population at a given time.
#### Solution Approach:
1. **Identify Parameters:**
- Initial Population \(P(0)\): 60 million
- Population after one year \(P(1)\): 59.3 million
2. **Model Formulation:**
The population decline can be modeled using the differential equation:
\[
\frac{dP}{dt} = -kP
\]
where \(P(t)\) is the population at time \(t\), and \(k\) is the proportionality constant.
3. **Solving Differential Equation:**
By separating variables and integrating, we have:
\[
\int \frac{1}{P} dP = -k \int dt
\]
Solving this yields:
\[
\ln P = -kt + C
\]
where \(C\) is the integration constant. Exponentiating both sides, we get:
\[
P(t) = e^{-kt + C'} = e^{C'} \cdot e^{-kt}
\]
Let \(P(0) = P_0\), thus \(e^{C'} = P_0\). Hence, we have:
\[
P(t) = P_0 e^{-kt}
\]
4. **Applying Initial Conditions:**
- \(P_0 = 60\) million
- \(P(1) = 59.3\) million
Using \(P(1) = 60 e^{-k(1)} = 59.3\), we solve for \(k\):
\[
e^{-k} = \frac{59.3}{60}
\]
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Transcribed Image Text:### Negative Population Growth Modeling
Many populations are facing negative growth rates. To illustrate this, consider the population of Ukraine. It was 60 million in 2008 and is projected to shrink to 59.3 million people by 2009.
To model Ukraine's population over time, assuming the population loss is proportional to the population at any given time, we will use a differential equation approach.
#### Problem Statement:
If the population of Ukraine was 60 million in 2008 and will shrink to 59.3 million people in 2009, write the equation which models the Ukraine population over time if the population loss is proportional to the population at a given time.
#### Solution Approach:
1. **Identify Parameters:**
- Initial Population \(P(0)\): 60 million
- Population after one year \(P(1)\): 59.3 million
2. **Model Formulation:**
The population decline can be modeled using the differential equation:
\[
\frac{dP}{dt} = -kP
\]
where \(P(t)\) is the population at time \(t\), and \(k\) is the proportionality constant.
3. **Solving Differential Equation:**
By separating variables and integrating, we have:
\[
\int \frac{1}{P} dP = -k \int dt
\]
Solving this yields:
\[
\ln P = -kt + C
\]
where \(C\) is the integration constant. Exponentiating both sides, we get:
\[
P(t) = e^{-kt + C'} = e^{C'} \cdot e^{-kt}
\]
Let \(P(0) = P_0\), thus \(e^{C'} = P_0\). Hence, we have:
\[
P(t) = P_0 e^{-kt}
\]
4. **Applying Initial Conditions:**
- \(P_0 = 60\) million
- \(P(1) = 59.3\) million
Using \(P(1) = 60 e^{-k(1)} = 59.3\), we solve for \(k\):
\[
e^{-k} = \frac{59.3}{60}
\]
\[
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