MA 16500: Analytic Geometry and Calculus I

7.2. Exponential Models

Goal: Understand exponential growth and decay models and applying it to real-world situations such as population growth and radioactive decay.

1. Introduction

Assume \( y > 0 \). Exponential growth and decay models are obtained by solving the differential equation

\[ \frac{dy}{dt} = ky, \quad \text{initial condition } y(0) = y_0 \]

where \( k > 0 \) corresponds to exponential growth and \( k < 0 \) corresponds to exponential decay.

Solve the Differential Equation: Start with the general differential equation:

\[ \frac{dy}{dt} = ky \]

Separate variables:

\[ \frac{1}{y} \frac{dy}{dt} = k \]

Multiply both sides by \( dt \):

\[ \frac{1}{y} dy = k\,dt \]

Integrate both sides:

\[ \int \frac{1}{y} dy = \int k\,dt \]
\[ \ln|y| = kt + C \]

Since \( y > 0 \),

\[ \ln y = kt + C \]

Exponentiate both sides:

\[ e^{\ln y} = e^{kt + C} \]
\[ y = e^{kt} \cdot e^C \]

Let \( A = e^C \), then:

\[ y(t) = A e^{kt} \]

Finding the Constant \( A \): Use the initial condition \( y(0) \):

\[ y(0) = A e^{k \cdot 0} = A \implies A = y(0) = y_0 \]

Final General Solution:

\[ y(t) = y_0 e^{kt} \]

where \( k > 0 \) gives exponential growth and \( k < 0 \) gives exponential decay.

2. Examples

Exponential Growth

Exponential growth is described by functions of the form:

\[ y(t) = y_0 e^{kt} \]

The initial value is \( y(0) = y_0 \), and the rate constant \( k > 0 \) determines growth. Exponential growth has a constant relative growth rate.

Example 2.1: Population Growth Model

We model the population by an exponential growth model. The population data (in billions) is given below:

\( t \) (the number of years after \( 1999 \)) Year Population (billions)
\(0\) \(1999\) \(6.0\)
\(18\) \(2017\) \(7.4\)

Find the exponential growth model \( P(t) \).

Solution:

\[ P(t) = P_0 e^{kt} \]

From the initial condition \( P(0) = 6.0 \), we get:

\[ P(t) = 6.0 e^{kt} \]

Use the data point \( P(18) = 7.4 \):

\[ 7.4 = 6.0 e^{18k} \implies \frac{7.4}{6.0} = e^{18k} \]

Take natural logarithm:

\[ \ln\left(\frac{7.4}{6.0}\right) = 18k \]

Solve for \( k \):

\[ k = \frac{1}{18}\ln\left(\frac{7.4}{6.0}\right) \]
\[ k \approx 0.01165 \]

Final Model:

\[ P(t) = 6.0 e^{0.01165 t} \]

where \( t \) is the number of years after \( 1999 \) and population is measured in billions.

Exponential Decay Functions

Exponential decay is described by functions of the form:

\[ y(t) = y_0 e^{kt} \]

The initial value is \( y(0) = y_0 \), and the rate constant \( k < 0 \) determines the rate of decay.

Example 2.2: Radioactive Decay (Radium-226)

Radium-226 has a half-life of \( 1590 \) years. This means that the substance decays in such a way that every \( 1590 \) years, half of the remaining material disappears. So if the initial amount is \( m_0 \), then after \( 1590 \) years the amount is \( \frac{1}{2}m_0 \). Find the formula for \( m(t) \), when the initial mass is \( m(0) = 100\) mg.

Solution:

We use the exponential decay model:

\[ m(t) = m_0 e^{kt} \]

Since the initial mass is \( m(0)=100 \), we have \( m_0 = 100 \), so:

\[ m(t) = 100 e^{kt} \]

Use the half-life information: after \( 1590 \) years, half of the original mass remains.

\[ m(1590) = 50 \]

Substitute into the model:

\[ 50 = 100 e^{1590k} \]

Divide both sides by \(100\):

\[ \frac{1}{2} = e^{1590k} \]

Take natural logarithm of both sides:

\[ \ln\left(\frac{1}{2}\right) = 1590k \]

Solve for \( k \):

\[ k = \frac{\ln(1/2)}{1590} = -\frac{\ln 2}{1590} \]

Substitute back into the exponential model:

\[ m(t) = 100 e^{-\frac{\ln 2}{1590} t} =100\left(e^{\ln 2}\right)^{-t/1590} =100\cdot 2^{-t/1590}. \]

Note (Half-life formula: Radioactive Decay):

If the initial amount is \( m_0 \), then after \( h \) years (the half-life), the remaining amount is \( \frac{1}{2} m_0 \). The amount after time \( t \) is given by:

\[ m(t) = m_0 \cdot 2^{-t/h} \]