Growth and Decay | Malone Math | Malone Global University

The Central Idea

Suppose a quantity $Q(t)$ changes at a rate proportional to its current value.

The statement

the rate of change of $Q$ is proportional to $Q$

becomes

\[ \boxed{\frac{dQ}{dt}=kQ}, \]

where $k$ is a constant of proportionality.

This simple differential equation produces exponential behavior.

Growth versus Decay

The sign of $k$ determines whether the quantity grows or decays.

\[ k>0, \]

then

\[ \frac{dQ}{dt}>0 \]

whenever $Q>0$, so the quantity grows.

\[ k<0, \]

then

\[ \frac{dQ}{dt}<0 \]

whenever $Q>0$, so the quantity decays.

Solving the Proportional-Rate Equation

Begin with

\[ \frac{dQ}{dt}=kQ. \]

Assuming $Q\neq0$, separate the variables:

\[ \frac{1}{Q}\,dQ=k\,dt. \]

Integrate:

\[ \int\frac{1}{Q}\,dQ = \int k\,dt. \]

Therefore,

\[ \ln|Q|=kt+C. \]

From the Logarithmic Form to the Exponential Form

Exponentiating

\[ \ln|Q|=kt+C \]

gives

\[ |Q|=e^{kt+C}. \]

Since

\[ e^{kt+C}=e^Ce^{kt}, \]

the constant can be absorbed into a single arbitrary constant $A$:

\[ \boxed{Q=Ae^{kt}}. \]

For applications involving positive quantities, $A$ is usually positive.

Using an Initial Condition

Suppose

\[ Q(0)=Q_0. \]

Substitute $t=0$ into

\[ Q=Ae^{kt}. \]

Then

\[ Q_0=Ae^0=A. \]

Therefore,

\[ \boxed{Q(t)=Q_0e^{kt}}. \]

This is the standard continuous growth-and-decay model.

Why the Initial Value Appears as a Multiplier

At time $t=0$,

\[ e^{k(0)}=1. \]

Therefore,

\[ Q(0)=Q_0. \]

The exponential factor

\[ e^{kt} \]

describes how the initial quantity is scaled over time.

Verifying the Solution

Suppose

\[ Q(t)=Q_0e^{kt}. \]

Differentiate:

\[ Q'(t)=kQ_0e^{kt}. \]

Since

\[ Q_0e^{kt}=Q(t), \]

we have

\[ Q'(t)=kQ(t). \]

Therefore,

\[ \boxed{Q'=kQ} \]

is satisfied.

The Relative Growth Rate

From

\[ Q'=kQ, \]

divide by $Q$:

\[ \frac{Q'}{Q}=k. \]

The quantity

\[ \frac{Q'}{Q} \]

is the instantaneous rate of change relative to the current amount.

Thus $k$ is the constant relative growth or decay rate.

Units of the Growth Constant

Suppose $Q$ is measured in kilograms and $t$ in years.

Then

\[ Q' \]

has units

\[ \frac{\text{kilograms}}{\text{year}}. \]

Since

\[ Q'=kQ, \]

$k$ must have units

\[ \boxed{\frac{1}{\text{year}}}. \]

In general, the proportional-rate constant has reciprocal-time units.

Example: Basic Exponential Growth

A population begins at $500$ and grows continuously at a rate proportional to its current size with

\[ k=0.04. \]

The model is

\[ P'=0.04P, \qquad P(0)=500. \]

Therefore,

\[ \boxed{P(t)=500e^{0.04t}}. \]

Interpreting the Growth Model

\[ P(t)=500e^{0.04t}, \]

the value $500$ is the initial population.

The parameter

\[ 0.04 \]

is the continuous relative growth rate.

At any time,

\[ P'=0.04P, \]

so the instantaneous growth rate equals $4\%$ of the current population per unit time.

Example: Basic Exponential Decay

Suppose a chemical substance begins with $80$ grams and decays according to

\[ M'=-0.12M. \]

Since

\[ M(0)=80, \]

the solution is

\[ \boxed{M(t)=80e^{-0.12t}}. \]

The negative sign causes the exponential factor to decrease with time.

Long-Term Behavior

For positive $Q_0$, if

\[ k>0, \]

then

\[ Q(t)=Q_0e^{kt}\to\infty \]

\[ t\to\infty. \]

\[ k<0, \]

then

\[ Q(t)=Q_0e^{kt}\to0. \]

Exponential decay approaches zero asymptotically but never reaches zero in the ideal continuous model.

Doubling Time

For exponential growth, the doubling time is the time required for the quantity to become twice its initial value.

Start with

\[ Q(t)=Q_0e^{kt}. \]

Set

\[ Q(t)=2Q_0. \]

Then

\[ 2Q_0=Q_0e^{kt}. \]

Divide by $Q_0$:

\[ 2=e^{kt}. \]

Doubling-Time Formula

Take the natural logarithm:

\[ \ln2=kt. \]

Therefore,

\[ \boxed{T_d=\frac{\ln2}{k}} \]

for $k>0$.

A larger growth constant produces a shorter doubling time.

Example: Doubling Time

Suppose

\[ P'=0.05P. \]

The doubling time is

\[ T_d=\frac{\ln2}{0.05}. \]

Numerically,

\[ T_d\approx13.86. \]

Therefore, the population doubles approximately every

\[ \boxed{13.86\text{ time units}}. \]

Half-Life

For exponential decay, the half-life is the time required for the quantity to decrease to half its initial value.

Begin with

\[ Q(t)=Q_0e^{kt}, \qquad k<0. \]

Set

\[ Q(t)=\frac12Q_0. \]

Then

\[ \frac12=e^{kt}. \]

Half-Life Formula

Taking natural logarithms gives

\[ \ln\left(\frac12\right)=kt. \]

Since

\[ \ln\left(\frac12\right)=-\ln2, \]

we obtain

\[ \boxed{T_{1/2}=-\frac{\ln2}{k}}. \]

Because $k<0$, the resulting half-life is positive.

Alternative Half-Life Form

If decay is written as

\[ Q'=-\lambda Q, \qquad \lambda>0, \]

then

\[ Q(t)=Q_0e^{-\lambda t}. \]

The half-life becomes

\[ \boxed{T_{1/2}=\frac{\ln2}{\lambda}}. \]

This notation is especially common in radioactive decay.

Example: Half-Life from a Decay Constant

Suppose

\[ M'=-0.08M. \]

Then

\[ \lambda=0.08. \]

The half-life is

\[ T_{1/2} = \frac{\ln2}{0.08}. \]

Thus,

\[ \boxed{T_{1/2}\approx8.66}. \]

Building a Model from Two Measurements

Suppose we know

\[ Q(0)=Q_0 \]

and another value

\[ Q(t_1)=Q_1. \]

Using

\[ Q(t)=Q_0e^{kt}, \]

substitute $t=t_1$:

\[ Q_1=Q_0e^{kt_1}. \]

Solving for the Growth Constant

Divide by $Q_0$:

\[ \frac{Q_1}{Q_0}=e^{kt_1}. \]

Take natural logarithms:

\[ \ln\left(\frac{Q_1}{Q_0}\right)=kt_1. \]

Therefore,

\[ \boxed{ k= \frac{1}{t_1} \ln\left(\frac{Q_1}{Q_0}\right) }. \]

Example: Estimate a Growth Parameter

A population is $1200$ initially and $1500$ after $4$ years.

Then

\[ 1500=1200e^{4k}. \]

Divide:

\[ 1.25=e^{4k}. \]

Take logarithms:

\[ \ln(1.25)=4k. \]

Hence

\[ \boxed{k=\frac{\ln(1.25)}{4}}. \]

Example: Complete the Estimated Model

Using

\[ k=\frac{\ln(1.25)}{4}, \]

the model becomes

\[ \boxed{ P(t)=1200e^{(\ln(1.25)/4)t} }. \]

This can also be written as

\[ \boxed{ P(t)=1200(1.25)^{t/4} }. \]

The two forms are equivalent.

Exponential Form versus Growth-Factor Form

Continuous growth is often written

\[ Q(t)=Q_0e^{kt}. \]

If a quantity changes by a factor $b$ over a time interval $T$, the same model can be written

\[ \boxed{ Q(t)=Q_0b^{t/T} }. \]

Since

\[ b=e^{kT}, \]

these are simply two representations of the same exponential relationship.

Radioactive Decay

Radioactive decay is modeled by

\[ \frac{dN}{dt}=-\lambda N, \]

where $N(t)$ is the amount of radioactive material and $\lambda>0$ is the decay constant.

The solution is

\[ \boxed{N(t)=N_0e^{-\lambda t}}. \]

The model assumes each unit of material has the same constant probability rate of decay over time.

Example: Radioactive Decay from Half-Life

Suppose a substance has half-life $10$ years.

Then

\[ \lambda=\frac{\ln2}{10}. \]

Therefore,

\[ N(t) = N_0e^{-(\ln2/10)t}. \]

Equivalently,

\[ \boxed{ N(t)=N_0\left(\frac12\right)^{t/10} }. \]

Repeated Half-Lives

After one half-life,

\[ N=\frac12N_0. \]

After two half-lives,

\[ N=\frac14N_0. \]

After three half-lives,

\[ N=\frac18N_0. \]

In general, after $n$ half-lives,

\[ \boxed{ N=N_0\left(\frac12\right)^n }. \]

Continuous Compounding

If money grows continuously at a constant relative rate $r$, the balance $A(t)$ satisfies

\[ A'=rA. \]

With

\[ A(0)=P, \]

the solution is

\[ \boxed{A(t)=Pe^{rt}}. \]

Here $P$ is the principal and $r$ is the continuously compounded annual rate.

Example: Continuous Compounding

Suppose $\$10{,}000$ is invested at a continuously compounded rate of $6\%$ per year.

Then

\[ A'=0.06A, \qquad A(0)=10000. \]

Therefore,

\[ \boxed{ A(t)=10000e^{0.06t} }. \]

Continuous versus Discrete Growth

A discrete compounding model may have the form

\[ A_n=A_0(1+r)^n. \]

A continuous growth model has the form

\[ A(t)=A_0e^{kt}. \]

The distinction is important:

discrete models update at separated time steps;
continuous models describe change at every instant.

Differential equations naturally describe the continuous model.

Population Growth

If births and deaths combine to produce a constant net per-capita growth rate, then population may be modeled by

\[ P'=kP. \]

The solution

\[ P=P_0e^{kt} \]

represents unrestricted exponential growth when $k>0$.

This model is often useful over limited ranges of time but may become unrealistic when resources or space become limiting.

Why Exponential Growth Eventually Becomes Unrealistic

In the model

\[ P'=kP, \]

the proportional growth rate never changes.

There is no built-in limitation from:

food;
space;
capital;
competition;
resource depletion;
market saturation;
environmental capacity.

The next lesson on logistic models introduces a limiting mechanism.

Decay Does Not Mean Constant Subtraction

Exponential decay does not remove the same amount during every time interval.

Instead, the amount lost over a short interval is proportional to the amount currently present.

Thus a larger quantity initially loses more per unit time than a smaller quantity under the same decay constant.

Constant Change versus Proportional Change

Compare

\[ Q'=c \]

with

\[ Q'=kQ. \]

The first equation produces linear change:

\[ Q=ct+C. \]

The second produces exponential change:

\[ Q=Ae^{kt}. \]

Constant absolute change and constant relative change are fundamentally different models.

Percentage Rates and Continuous Rates

\[ Q'=0.03Q, \]

the instantaneous relative rate is

\[ 0.03 \]

per unit time.

This is commonly described as a continuous rate of $3\%$ per unit time.

It should not automatically be interpreted as exactly the same as increasing by $3\%$ once at the end of each discrete period.

Finding Time from a Target Value

Suppose

\[ Q(t)=Q_0e^{kt} \]

and we want to know when the quantity reaches $Q_T$.

Set

\[ Q_T=Q_0e^{kt}. \]

Divide:

\[ \frac{Q_T}{Q_0}=e^{kt}. \]

Take logarithms:

\[ \ln\left(\frac{Q_T}{Q_0}\right)=kt. \]

Therefore,

\[ \boxed{ t= \frac{1}{k} \ln\left(\frac{Q_T}{Q_0}\right) }. \]

Example: Time to Reach a Target

Suppose

\[ P(t)=800e^{0.06t}. \]

Find when $P=1600$.

Set

\[ 1600=800e^{0.06t}. \]

Then

\[ 2=e^{0.06t}. \]

Thus,

\[ t=\frac{\ln2}{0.06}. \]

Therefore,

\[ \boxed{t\approx11.55}. \]

Example: Time to Decay to a Percentage

Suppose

\[ M(t)=M_0e^{-0.1t}. \]

Find when $20\%$ of the original amount remains.

Set

\[ 0.2M_0=M_0e^{-0.1t}. \]

Then

\[ 0.2=e^{-0.1t}. \]

Taking logarithms:

\[ \ln(0.2)=-0.1t. \]

Therefore,

\[ \boxed{ t=-\frac{\ln(0.2)}{0.1} }. \]

Estimating the Initial Quantity

If $k$, $t$, and a later value $Q(t)$ are known, solve

\[ Q(t)=Q_0e^{kt} \]

for $Q_0$:

\[ \boxed{ Q_0=Q(t)e^{-kt} }. \]

This is useful when reconstructing an earlier quantity from a known exponential model.

Linearizing Exponential Data

\[ Q(t)=Q_0e^{kt}, \]

take natural logarithms:

\[ \ln Q = \ln Q_0+kt. \]

Thus a plot of $\ln Q$ against $t$ should be approximately linear when exponential growth or decay is a good model.

The slope is $k$, and the vertical intercept is $\ln Q_0$.

Why Linearization Is Useful

The transformed relationship

\[ \ln Q=\ln Q_0+kt \]

allows exponential behavior to be studied using linear methods.

This can help with:

estimating parameters;
checking whether exponential behavior is plausible;
comparing growth rates;
interpreting data over several orders of magnitude.

Sensitivity to the Growth Constant

Because

\[ Q(t)=Q_0e^{kt}, \]

small changes in $k$ can produce large differences over long periods.

The ratio of two models with constants $k_1$ and $k_2$ is

\[ \frac{Q_0e^{k_1t}}{Q_0e^{k_2t}} = e^{(k_1-k_2)t}. \]

Even a modest difference in continuous growth rate compounds exponentially over time.

Model Assumptions Matter

The equation

\[ Q'=kQ \]

assumes that $k$ remains constant.

In real systems, the rate parameter may change because of:

environmental changes;
resource limits;
policy changes;
market conditions;
technology;
temperature;
intervention;
measurement error.

A mathematically correct solution can still be a poor real-world model if its assumptions are inappropriate.

Common Error: Using a Positive Constant for Decay

If a positive quantity is decreasing according to

\[ Q'=kQ, \]

then $k$ must be negative.

Alternatively, write

\[ Q'=-\lambda Q, \qquad \lambda>0. \]

Both conventions describe exponential decay.

Common Error: Forgetting the Initial Value

The general solution is

\[ Q=Ae^{kt}. \]

\[ Q(0)=Q_0, \]

then

\[ A=Q_0. \]

Therefore,

\[ \boxed{Q=Q_0e^{kt}}. \]

Do not leave the arbitrary constant undetermined when an initial value is supplied.

Common Error: Confusing Half-Life with the Decay Constant

The half-life is a time.

The decay constant has reciprocal-time units.

They are related by

\[ \boxed{ T_{1/2}=\frac{\ln2}{\lambda} }. \]

They are not numerically interchangeable.

Common Error: Treating Exponential Decay as Linear

If a quantity loses half its value every $10$ years, it does not lose a fixed absolute amount every $10$ years.

It loses half of whatever amount remains.

Thus

\[ Q_0,\quad \frac12Q_0,\quad \frac14Q_0,\quad \frac18Q_0,\ldots \]

forms a multiplicative pattern, not an additive one.

Common Error: Mixing Discrete and Continuous Percentage Rates

A discrete annual growth factor

\[ 1+r \]

and a continuous rate parameter $k$ are related but not identical.

If the one-period growth factor is $1+r$, the equivalent continuous rate satisfies

\[ e^k=1+r. \]

Therefore,

\[ \boxed{k=\ln(1+r)}. \]

Common Error: Ignoring Units

If $t$ is measured in years, then $k$ must be expressed per year.

A rate given per month cannot be inserted directly into a model whose time variable is measured in years without conversion.

Dimensional consistency is a basic model check.

A Reliable Growth-and-Decay Workflow

Identify the changing quantity.
Identify the time variable and its units.
Translate proportional change into $Q'=kQ$.
Determine whether $k$ should be positive or negative.
Write the general solution $Q=Ae^{kt}$.
Use the initial condition to determine $A$.
Use additional measurements to determine $k$ when necessary.
Use logarithms to solve for unknown times.
Check units and signs.
Differentiate to verify the differential equation.
Interpret the result in the context of the model.
Assess whether constant proportional change is a reasonable assumption.

Example: Full Growth Workflow

A culture contains $2000$ cells initially and $2600$ cells after $3$ hours. Assume exponential growth.

Begin with

\[ P(t)=2000e^{kt}. \]

Use

\[ P(3)=2600: \] \[ 2600=2000e^{3k}. \]

Divide:

\[ 1.3=e^{3k}. \]

Therefore,

\[ k=\frac{\ln1.3}{3}. \]

Complete the Full Growth Workflow

The model is

\[ \boxed{ P(t)=2000e^{(\ln1.3/3)t} }. \]

The doubling time is

\[ T_d = \frac{\ln2}{k}. \]

Substitute $k$:

\[ T_d = \frac{3\ln2}{\ln1.3}. \]

Thus the same fitted model can answer both prediction and timing questions.

Example: Full Decay Workflow

A sample begins with $120$ grams and has a half-life of $6$ hours.

Write

\[ M(t)=120e^{-\lambda t}. \]

Since

\[ \lambda=\frac{\ln2}{6}, \]

the model is

\[ \boxed{ M(t)=120e^{-(\ln2/6)t} }. \]

Equivalently,

\[ \boxed{ M(t)=120\left(\frac12\right)^{t/6} }. \]

Practice 1: Solve a Growth Equation

Solve

\[ P'=0.03P, \qquad P(0)=400. \]

Practice 2: Solve a Decay Equation

Solve

\[ M'=-0.2M, \qquad M(0)=50. \]

Practice 3: Doubling Time

Find the doubling time for

\[ Q'=0.08Q. \]

Practice 4: Half-Life

Find the half-life for

\[ Q'=-0.05Q. \]

Practice 5: Find the Growth Constant

A quantity increases from $100$ to $140$ in $5$ years.

Assuming continuous exponential growth, find $k$.

Practice 6: Build the Model

A population is $3000$ initially and $3600$ after $4$ years.

Construct an exponential growth model.

Practice 7: Radioactive Decay

A radioactive sample has half-life $12$ years and initially contains $80$ grams.

Construct a decay model.

Practice 8: Continuous Compounding

Write the balance function for $\$5000$ invested at a continuously compounded annual rate of $4.5\%$.

Practice 9: Time to Reach a Target

\[ P(t)=1000e^{0.07t}, \]

find the time required for $P$ to reach $2000$.

Practice 10: Remaining Percentage

A substance satisfies

\[ M(t)=M_0e^{-0.04t}. \]

Find the time required for $25\%$ of the original quantity to remain.

Practice 11: Units

A population model is

\[ P'=kP, \]

where $t$ is measured in months.

What units must $k$ have?

Practice 12: Compare Models

Which model represents constant absolute change?

\[ Q'=5 \]

\[ Q'=0.05Q? \]

Which represents constant relative change?

Practice Answers

Practice 1

Using

\[ P(t)=P_0e^{kt}, \]

we obtain

\[ \boxed{ P(t)=400e^{0.03t} }. \]

Practice 2

The solution is

\[ \boxed{ M(t)=50e^{-0.2t} }. \]

Practice 3

Use

\[ T_d=\frac{\ln2}{k}. \]

Therefore,

\[ \boxed{ T_d=\frac{\ln2}{0.08}\approx8.66 }. \]

Practice 4

Since $k=-0.05$,

\[ T_{1/2} = -\frac{\ln2}{-0.05}. \]

Thus,

\[ \boxed{ T_{1/2}=\frac{\ln2}{0.05}\approx13.86 }. \]

Practice 5

Use

\[ 140=100e^{5k}. \]

Then

\[ 1.4=e^{5k}. \]

Therefore,

\[ \boxed{ k=\frac{\ln1.4}{5} }. \]

Practice 6

Start with

\[ P(t)=3000e^{kt}. \]

Since

\[ 3600=3000e^{4k}, \]

we get

\[ 1.2=e^{4k}. \]

Thus,

\[ k=\frac{\ln1.2}{4}. \]

The model is

\[ \boxed{ P(t)=3000e^{(\ln1.2/4)t} }. \]

Practice 7

With half-life $12$,

\[ \lambda=\frac{\ln2}{12}. \]

Therefore,

\[ \boxed{ M(t)=80e^{-(\ln2/12)t} }. \]

Equivalently,

\[ \boxed{ M(t)=80\left(\frac12\right)^{t/12} }. \]

Practice 8

Continuous compounding gives

\[ \boxed{ A(t)=5000e^{0.045t} }. \]

Practice 9

Set

\[ 2000=1000e^{0.07t}. \]

Then

\[ 2=e^{0.07t}. \]

Therefore,

\[ \boxed{ t=\frac{\ln2}{0.07}\approx9.90 }. \]

Practice 10

Set

\[ 0.25M_0=M_0e^{-0.04t}. \]

Therefore,

\[ 0.25=e^{-0.04t}. \]

Taking logarithms:

\[ \ln0.25=-0.04t. \]

Thus,

\[ \boxed{ t=-\frac{\ln0.25}{0.04} }. \]

Practice 11

Since $t$ is measured in months,

\[ \boxed{k\text{ has units of }1/\text{month}}. \]

Practice 12

The equation

\[ \boxed{Q'=5} \]

represents constant absolute change.

The equation

\[ \boxed{Q'=0.05Q} \]

represents constant relative change.

What to Remember

Continuous exponential growth and decay begin with the proportional-rate equation

\[ \boxed{Q'=kQ}. \]

With initial value $Q(0)=Q_0$, the solution is

\[ \boxed{Q(t)=Q_0e^{kt}}. \]

$k>0$ produces exponential growth.
$k<0$ produces exponential decay.
$k$ is the constant relative rate of change.
The doubling time is $T_d=\ln2/k$ for $k>0$.
The half-life is $T_{1/2}=-\ln2/k$ for $k<0$.
Additional measurements can be used to estimate $k$.
Logarithms are used to solve for unknown times.
Continuous compounding is a direct application of the same model.
Exponential models assume the proportional-rate parameter remains constant.

The central relationship is

\[ \boxed{ \frac{Q'}{Q}=k }. \]

Exponential behavior occurs because the rate of change remains proportional to the quantity currently present.

Where We Go Next

Exponential growth assumes that the relative growth rate remains constant regardless of how large the system becomes. Many real systems cannot grow without limit.

The next lesson, Logistic Models, introduces a first-order nonlinear model in which growth slows as the state approaches a limiting capacity:

\[ \frac{dP}{dt} = rP\left(1-\frac{P}{K}\right). \]

This model combines early exponential-like growth with equilibrium behavior and a natural upper scale.