Question

The population \(N(t)\) of a region at time \(t\) increases at a rate proportional to the population. If the population doubles every 5 years and is 3 million initially, find \(N(t)\).

Short answer

The population is modeled by \( N(t) = 3e^{\frac{\ln 2}{5}t} \).

Step-by-step solution

Formulate the Problem

The problem states that the rate of population increase is proportional to the current population. This can be modeled with the differential equation: \( \frac{dN}{dt} = kN \), where \( k \) is a constant of proportionality.

Solve the Differential Equation

To solve \( \frac{dN}{dt} = kN \), separate variables to get \( \frac{dN}{N} = k \, dt \). Integrate both sides: \( \int \frac{dN}{N} = \int k \, dt \), which gives \( \ln |N| = kt + C \), where \( C \) is the integration constant.

Exponentiate to Solve for N(t)

Exponentiate both sides to solve for \( N(t) \): \( N = e^{kt+C} = e^C \cdot e^{kt} \). Let \( e^C = N_0 \), the initial population at \( t = 0 \). The solution becomes \( N(t) = N_0 e^{kt} \).

Use Initial Condition

Use the given initial condition \( N(0) = 3 \text{ million} \). Substitute into the equation: \( 3 = 3e^{k \cdot 0} \), confirming \( N_0 = 3 \). Thus, \( N(t) = 3e^{kt} \).

Determine the Growth Constant k

The population doubles in 5 years, so \( N(5) = 2 \times 3 = 6 \). Substituting, \( 6 = 3e^{5k} \), solve for \( k \): \( 2 = e^{5k} \). Take the natural log: \( \ln 2 = 5k \), which gives \( k = \frac{\ln 2}{5} \).

Substitute k back into N(t)

Substitute \( k \) back into the population model: \( N(t) = 3e^{\frac{\ln 2}{5}t} \). This equation represents the population at any time \( t \).

Key concepts

Differential Equations

In a differential equation, we describe how a quantity changes over time. It provides a mathematical way to express relationships involving rates of change. Differential equations can describe a wide range of phenomena, including temperature, velocity, or in this case, population dynamics.
Population growth follows the differential equation \( \frac{dN}{dt} = kN \), where \( N \) represents the population at time \( t \), and \( k \) is the growth rate constant. This equation states that the rate of change in population is proportional to the current population itself. Such equations are commonly used to model exponential growth, where quantities increase continuously over time.
The process of solving these equations involves manipulating them into simpler forms that we can handle with calculus techniques, like integration. This is a vital step to predict future values, such as how a population will evolve over time.

Population Dynamics

Population dynamics is the study of how populations change over time under various influences. These include birth rates, death rates, and migration, among others. In mathematical modeling, it often uses tools like differential equations to forecast future population levels.
In our scenario, the equation \( \frac{dN}{dt} = kN \) perfectly captures unrepressed exponential growth. This means there are no constraints, like limited resources or space, which might slow growth down. Real-world populations often start growing exponentially but may eventually face limitations causing the growth rate to slow.
Understanding these patterns and equations helps researchers or policymakers predict problems before they appear, enabling better preparation and management of population-related issues.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and play a crucial role when solving differential equations related to growth or decay. After separating the variables in the differential equation \( \frac{dN}{dt} = kN \), we reach \( \frac{dN}{N} = k \, dt \), leading to the integral \( \int \frac{dN}{N} = \int k \, dt \).
The result of this integration is \( \ln |N| = kt + C \), where \( C \) is the integration constant. This equation uses the natural logarithm \( \ln \), which is essential for unraveling exponential growth and decay processes. To find \( N(t) \), we exponentiate both sides, transforming the logarithmic form back into an exponential one: \( N = e^{kt + C} \).
This transformation is crucial in turning the raw solution into a form where you can easily apply initial conditions to solve specific problems, such as finding a population size at a given time.

Initial Conditions

Initial conditions are values given at the starting point of an equation or model. They provide the benchmark needed to find specific solutions to differential equations like the one modeling population growth.
In this exercise, the initial condition states that the population is 3 million when \( t = 0 \). This initial condition helps us determine the constant \( N_0 \). By substituting \( N(0) = 3 \) into the equation \( N(t) = N_0 e^{kt} \), we establish that \( N_0 = 3 \).
It allows us to craft a model that accurately reflects the real-world scenario by supporting the calculation of the growth rate constant \( k \). These initial values are critical as they give meaning and context to the mathematical solution, making them practical and relevant.