Introduction

This topic explores exponential models that arise from differential equations, focusing on situations where the rate of change of a quantity is proportional to the size of the quantity itself. The essential knowledge tells us that specific applications of finding general and particular solutions to differential equations include motion along a line and exponential growth and decay. Additionally, the model for exponential growth and decay that arises from the statement "The rate of change of a quantity is proportional to the size of the quantity" is $\frac{dy}{dt}=ky$, and the exponential growth and decay model $\frac{dy}{dt}=ky$ with initial condition $y=y_0$ when $t=0$ has solutions of the form $y=y_0{e}^{kt}$.

Understanding Exponential Growth and Decay Models

The Fundamental Differential Equation

The cornerstone of exponential modeling is the differential equation:

$\frac{dy}{dt}=ky$

This equation states that the rate of change of quantity $y$ with respect to time $t$ is directly proportional to the current amount of $y$. The constant $k$ is called the growth constant (when $k>0$) or decay constant (when $k<0$).

Interpreting the Variables and Constants

Variable $y$: Represents the quantity being modeled (population, radioactive material, temperature difference, etc.)

Variable $t$: Represents time

Constant $k$: The proportionality constant that determines the rate of growth or decay

Example 1:

A bacterial culture grows at a rate proportional to its current size. If $P(t)$ represents the population at time $t$, then:

$\frac{dP}{dt}=kP$

where $k>0$ represents the growth rate constant.

Solving the Exponential Model

General Solution Derivation

To solve $\frac{dy}{dt}=ky$, we use separation of variables:

Step 1: Separate the variables $\frac{dy}{y}=kdt$

Step 2: Integrate both sides $\int \frac{dy}{y}=\int kdt$

$\ln |y|=kt+C$

Step 3: Solve for $y$ $|y|=e^{kt+C}=e^C\cdot e^{kt}$

$y=\pm e^C\cdot e^{kt}$

Let $A=\pm e^C$, then the general solution is:

$y=A{e}^{kt}$

Particular Solution with Initial Conditions

When we have an initial condition $y(0)=y_0$, we can find the particular solution:

$y_0=A{e}^{k\cdot 0}=A{e}^0=A$

Therefore, $A=y_0$, and the particular solution is:

$y=y_0{e}^{kt}$

Example 2:

Solve $\frac{dy}{dt}=3y$ with $y(0)=5$.

Using the formula: $y=y_0{e}^{kt}$

With $y_0=5$ and $k=3$:

$y=5e^{3t}$

Applications of Exponential Growth Models

Population Growth

Population growth often follows an exponential model when resources are unlimited.

Example 3:

A city's population grows at a rate proportional to its current population. In 2020, the population was 50,000, and the growth rate is 2% per year. Find the population model.

The differential equation is $\frac{dP}{dt}=kP$ where $k=0.02$ (2% = 0.02).

With initial condition $P(0)=50000$:

$P(t)=50000e^{0.02t}$

where $t$ is measured in years since 2020.

Investment Growth with Continuous Compounding

Money invested with continuous compounding follows exponential growth.

Example 4:

An investment grows continuously at an annual rate of 5.5%. If $1000 is initially invested, find the value after $t$ years.

The model is $\frac{dA}{dt}=0.055A$ with $A(0)=1000$.

Solution: $A(t)=1000e^{0.055t}$

Applications of Exponential Decay Models

Radioactive Decay

Radioactive substances decay at a rate proportional to the amount present.

Example 5:

Carbon-14 has a half-life of 5,730 years. If a sample initially contains 100 grams, find the decay model.

The model is $\frac{dC}{dt}=kC$ with $C(0)=100$.

Since half-life is 5,730 years: $C(5730)=50$

$50=100e^{5730k}$

$0.5=e^{5730k}$

$\ln (0.5)=5730k$

$k=\frac{-\ln (2)}{5730}$

Therefore: $C(t)=100e^{-\frac{\ln (2)}{5730}t}$

Newton's Law of Cooling

The temperature difference between an object and its surroundings decreases exponentially.

Example 6:

A cup of coffee at 90°C is placed in a room at 20°C. The temperature difference decreases at a rate proportional to the current temperature difference.

Let $T(t)$ be the temperature of the coffee at time $t$, and let $D(t)=T(t)-20$ be the temperature difference.

The model is $\frac{dD}{dt}=kD$ where $k<0$.

With $D(0)=90-20=70$:

$D(t)=70e^{kt}$

Therefore: $T(t)=20+70e^{kt}$

Determining Growth and Decay Constants

Using Doubling Time and Half-Life

For exponential growth, if the doubling time is $T_d$, then:

$k=\frac{\ln (2)}{T_d}$

For exponential decay, if the half-life is $T_h$, then:

$k=\frac{-\ln (2)}{T_h}$

Example 7:

If a population doubles every 8 years, find the growth constant.

$k=\frac{\ln (2)}{8}\approx 0.0866$ per year

Working with Partial Information

Example 8:

A radioactive sample decreases from 80 grams to 60 grams in 5 hours. Find the half-life.

Set up: $N(t)=N_0{e}^{kt}$ with $N_0=80$

Given: $N(5)=60$

$60=80e^{5k}$

$0.75=e^{5k}$

$k=\frac{\ln (0.75)}{5}$

For half-life $T_h$: $40=80e^{k{T}_h}$

$0.5=e^{k{T}_h}$

$T_h=\frac{\ln (0.5)}{k}=\frac{5\ln (0.5)}{\ln (0.75)}\approx 12.06$ hours

Interpreting Solutions in Context

Understanding the Behavior

The solution $y=y_0{e}^{kt}$ exhibits different behaviors based on the sign of $k$:

When $\text{k > 0}$ (Growth):

When $\text{k < 0}$ (Decay):

When $\text{k = 0}$:

Verification of Solutions

Example 9:

Verify that $y=3e^{-2t}$ solves $\frac{dy}{dt}=-2y$ with $y(0)=3$.

Check differential equation:

$\frac{dy}{dt}=\frac{d}{dt}(3e^{-2t})=3(-2)e^{-2t}=-6e^{-2t}$

$-2y=-2(3e^{-2t})=-6e^{-2t}$ ✓

Check initial condition: $y(0)=3e^{-2(0)}=3e^0=3$ ✓

Problem-Solving Strategy

Step-by-Step Approach

Practice Section