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The population of a town grows at a rate proportional to the population present at time \(t .\) The initial population of 500 increases by \(15 \%\) in 10 years. What will be the population in 30 years? How fast is the population growing at \(t=30 ?\)

Short Answer

Expert verified
Population in 30 years will be approximately 765, growing at a rate of about 10.73 per year at that time.

Step by step solution

01

Understand the Problem

We have a population that grows at a rate proportional to its current size. This is a classic example of exponential growth, where the rate of change of the population is proportional to the current population size. We know the population after 10 years and need to find it after 30 years, as well as the growth rate at that time.
02

Set Up the Differential Equation

Since the growth is proportional to the population, we express it as a differential equation: \( \frac{dP}{dt} = kP \), where \( k \) is the constant of proportionality. \( P(t) \) represents the population at time \( t \).
03

Solve the Differential Equation

To solve the differential equation \( \frac{dP}{dt} = kP \), separate variables and integrate: \( \int \frac{1}{P} dP = \int k dt \). This integration gives us \( \ln |P| = kt + C \). Exponentiating both sides, we find \( P(t) = Ce^{kt} \).
04

Find Constant of Integration \(C\)

We know that at \( t=0 \), \( P(0) = 500 \). Substituting into the equation \( P(t) = Ce^{kt} \), we get \( 500 = Ce^0 \), thus \( C = 500 \).
05

Determine \(k\) Using Given Information

We also know that after 10 years, the population has increased by 15%, giving us \( P(10) = 500 \times 1.15 = 575 \). Substituting into the equation \( 575 = 500e^{10k} \), we solve for \( k \): \( e^{10k} = \frac{575}{500} \). Taking the natural logarithm, \( 10k = \ln(1.15) \), so \( k = \frac{\ln(1.15)}{10} \).
06

Compute Population at \(t=30\)

Substitute \( k \) back into the population equation to find \( P(30) \): \( P(30) = 500e^{30k} = 500e^{30 \times \frac{\ln(1.15)}{10}} \). Simplifying, \( P(30) = 500e^{3 \ln(1.15)} = 500(1.15)^3 \approx 764.56 \).
07

Calculate Growth Rate at \(t=30\)

The growth rate at any time \( t \) is given by the derivative \( \frac{dP}{dt} = kP(t) \). So at \( t=30 \), \( \frac{dP}{dt} = \frac{\ln(1.15)}{10} \times 764.56 \approx 10.73 \).

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. In the context of population growth, they are utilized to model situations where a quantity changes at a rate proportional to its current size.

This type of equation allows us to understand and predict the behavior of dynamically changing populations by expressing the growth rate as a differential equation. For example, in our problem, the rate of population increase \( \frac{dP}{dt} = kP \) represents a simple differential equation. Here, \( P(t) \) is the population at time \(t\), and \( k \) is a constant representing the growth rate.

These equations are vital because they provide a framework to solve for \(P(t)\), which estimates the population at any future time given the initial population conditions.
Population Growth Modeling
Population growth modeling involves using mathematical concepts to forecast how populations change over time. In exponential growth models, populations increase without any restrictions at a constant rate proportional to their current size.

This type of growth is described mathematically by the equation \( P(t) = Ce^{kt} \), where \( C \) is the initial population, \( k \) is the growth constant, and \( t \) is the time elapsed.

Understanding population dynamics through such models helps to predict future population sizes and assess growth trends over short or long periods.
Separable Equations
Separable equations are a class of differential equations that can be rearranged to allow separation of variables on each side of the equation. This method aims to isolate \(dP\) on one side and \(dt\) on the other, enabling integration.

In the population growth example, \( \frac{dP}{dt} = kP \) can be separated into \( \int \frac{1}{P} \,dP = \int k \,dt \).

Integration of these separately yields \( \ln |P| = kt + C \), leading to the solution \( P(t) = Ce^{kt} \). This method is efficient for solving simple differential equations and finding explicit solutions.
Initial Value Problem
An initial value problem provides specific starting conditions to solve a differential equation uniquely. This means that instead of having indefinite solutions with arbitrary constants, we identify a particular solution.

In population growth, the initial condition is often the known population size at \( t=0 \). For instance, if \( P(0) = 500 \), this allows us to substitute and solve for \( C \) in the equation \( P(t) = Ce^{kt} \).

Such problems lead to solutions that accurately reflect a real-world scenario from which predictions are made, such as calculating future population sizes.

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Most popular questions from this chapter

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by \(3 \%\). If the rate of decay is proportional to the amount of the substance present at time \(t,\) find the amount remaining after 24 hours.

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