Teacher Guide: Continuous Growth and Decay

Use A = Pe^(rt) and A = Pe^(-kt) to model continuous compounding, bacteria growth and radioactive decay, and compare continuous with annual compounding.

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10 questions on Exponential Functions. Students join with a name, you see everyone's score.

For Teachers

Learning Objectives

Use $A = Pe^{rt}$ to calculate amounts that grow continuously
Use $A = Pe^{-kt}$ to model continuous decay and interpret the sign of the exponent
Compare continuous and annual compounding at the same rate and explain the difference
Find the initial amount $P$ from a known future amount
Evaluate powers of $e$ on a calculator and round as instructed

Prerequisites

• Introduction to Exponential Functions: the form $f(x) = a \cdot b^x$
• Exponential Growth: growth factor $1 + r$ and $A = A_0(1 + r)^t$
• The Natural Base e: $e \approx 2.71828$ as the limit of $\left(1 + \frac{1}{n}\right)^n$
• Converting percents to decimals and the order of operations

Discussion Starters

1. If a bank compounds every second instead of every day, will you get much more money? Why or why not?
2. Why does a negative exponent make an amount smaller? What does $e^{-1}$ mean?
3. Which offer would you choose: 6% compounded annually or 5.85% compounded continuously? How can you decide?
4. Why do you think scientists prefer base $e$ for growth and decay instead of base 10?

Common Misconceptions

Continuous compounding makes money grow without limit, or doubles it much faster

Remediation: Compute $1000(1.05)$ , $1000\left(1 + \frac{0.05}{12}\right)^{12}$ and $1000e^{0.05}$ side by side: 1050, about 1051.16 and about 1051.27. The gain from compounding more often is real but small.

The minus sign in $e^{-kt}$ makes the amount negative

Remediation: Show that $e^{-0.6} \approx 0.5488$ is still positive: a negative exponent gives a fraction between 0 and 1, so the amount shrinks but never goes below zero.

The rate in the exponent can be used as a percent

Remediation: Have students compute $e^{5}$ and $e^{0.05}$ and ask which makes sense as a one-year growth factor for 5% interest.

Differentiation Ideas

For Struggling Students:

• Give the value of $e^{rt}$ in each problem so students focus on setting up $P \times e^{rt}$
• Start with exponents that are whole numbers, like $e^{1}$ and $e^{2}$
• Use a three-column table: $P$ , $e^{rt}$ , $A$

For On-Level Students:

• Compare annual, monthly and continuous compounding for the same deposit
• Solve problems that ask for the initial amount $P$
• Decide between two savings offers by computing one-year growth factors

For Advanced Students:

• Show numerically that $\left(1 + \frac{r}{n}\right)^{n}$ approaches $e^{r}$ as $n$ grows
• Estimate the doubling time of $Pe^{0.07t}$ with a table of values and compare with the rule of 70
• Connect $Pe^{-kt}$ to half-life: find $k$ when the half-life is given, by trying values

Standards Alignment

HSF-LE.A.1c (CCSS.MATH.CONTENT.HSF.LE.A.1.C)
Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another
HSF-LE.B.5 (CCSS.MATH.CONTENT.HSF.LE.B.5)
Interpret the parameters in a linear or exponential function in terms of a context
HSF-IF.C.8b (CCSS.MATH.CONTENT.HSF.IF.C.8.B)
Use the properties of exponents to interpret expressions for exponential functions and classify them as growth or decay
HSA-CED.A.1 (CCSS.MATH.CONTENT.HSA.CED.A.1)
Create equations in one variable and use them to solve problems, including exponential functions

Lesson Resources

activityCompounding Race
Groups compute 1000 dollars at 6% for 10 years compounded yearly, monthly, daily and continuously, then compare
worksheetGrowth or Decay?
Read the sign of the exponent in models like $50e^{-0.3t}$ and compute values with a calculator
visualTwo Curves
Plot $P(1 + r)^t$ and $Pe^{rt}$ on the same axes to see how close they are and which is higher

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

In the lesson on the natural base

e

you saw that compounding more and more often,

\left(1 + \frac{1}{n}\right)^n

, approaches

e \approx 2.71828

. When a quantity grows continuously (every instant, not once a year), we use

e

as the base.

Continuous growth:

A = P e^{rt}

Continuous decay:

A = P e^{-kt}

where:

$P$ is the initial amount (at $t = 0$ )
$r$ is the continuous growth rate and $k$ the continuous decay rate, both written as decimals ( $5\% = 0.05$ )
$t$ is the time, in the same unit as the rate (years, hours, days)
$A$ is the amount after time $t$

The sign of the exponent tells you the direction: a positive exponent means growth, a negative exponent means decay. You can also write decay as

A = Pe^{rt}

with a negative rate, for example

r = -0.03

In the calculator you can type

e^{x}

directly as e^(x), for example e^0.5 gives

\approx 1.6487

. On a basic calculator without an

e

key, type

2.71828^{x}

instead.

Worked Examples

You deposit 1000 dollars in an account that pays 5% interest compounded continuously. How much is in the account after 10 years? Round to the nearest cent.

Identify the values

$P = 1000$ , $r = 0.05$ , $t = 10$ → Values identified

Find the exponent $rt$

$rt = 0.05 \times 10 = 0.5$ → $rt = 0.5$

Find $e^{0.5}$ with the calculator

$2.71828^{0.5} \approx 1.648721$ → $e^{0.5} \approx 1.648721$

Multiply by $P$

$A = 1000 \times 1.648721 \approx 1648.72$ → $A \approx 1648.72$

Answer: After 10 years the account holds about 1648.72 dollars.

Common Mistakes

Writing the rate as a percent in the exponent: $1000 e^{5 \times 10}$

Why it's wrong: The formula needs the rate as a decimal. $e^{50}$ is a gigantic number (about $5 \times 10^{21}$ ), which makes no sense for a savings account.

Correct: Convert first: $5\% = 0.05$ , so $A = 1000 e^{0.05 \times 10} = 1000 e^{0.5}$ .

Forgetting the minus sign for decay: $80 e^{0.03 \times 20}$

Why it's wrong: Without the minus sign the exponent is positive and the amount grows to about 145.77 grams, more than the 80 grams you started with.

Correct: Decay needs a negative exponent: $A = 80 e^{-0.03 \times 20} = 80 e^{-0.6} \approx 43.90$ .

Multiplying before raising to the power: $(1000 \times 2.71828)^{0.5}$

Why it's wrong: The exponent belongs only to $e$ , not to $P$ . Order of operations says powers come before multiplication.

Correct: First compute $e^{rt}$ , then multiply by $P$ . On the calculator: $1000 \times e^{(0.05 \times 10)}$ , with brackets around the exponent.

Using $(1 + r)^t$ when the problem says compounded continuously

Why it's wrong: $(1 + r)^t$ is compounding once per period. Continuous compounding always gives a little more at the same rate.

Correct: Look for the words continuously or continuous rate: they mean $Pe^{rt}$ . Once a year means $P(1 + r)^t$ .

Why It Matters

Many processes do not wait for the end of the year to change: they change at every moment.

Finance: some savings accounts and loans compound interest continuously, and $Pe^{rt}$ is the upper limit of what any compounding schedule can give at the same rate
Biology: bacteria in a lab divide all the time, not on a timetable, so their growth is modeled with $e$
Physics and medicine: radioactive atoms decay at random moments, and a drug leaves the bloodstream gradually, so both follow $Pe^{-kt}$
Science in general: almost every growth or decay formula in chemistry, physics and economics is written with base $e$

Once you can read

Pe^{rt}

, you can compare bank offers, predict lab results and understand how scientists date old objects.

Real World Applications

Banking: Continuously Compounded Savings

Banks compare offers using the most generous possible compounding, which is continuous. It shows the ceiling of what a given rate can earn.

Example:

At 5% compounded continuously, 1 dollar grows to $e^{0.05} \approx 1.0513$ dollars in a year, so the effective yearly rate is about 5.13%, a bit more than 5%.

1Try It Yourself

You put 4000 dollars into an account paying 4.5% compounded continuously and leave it for 4 years.

How much is in the account after 4 years? Round to the nearest dollar.

Step 1: Write the mathematical expression

Calculate $4000 \times e^{0.045 \times 4}$ :

Biology: Bacteria in the Lab

Microbiologists model a bacteria culture that has plenty of food with $N(t) = N_0 e^{rt}$, where $r$ is the continuous growth rate per hour.

Example:

A culture of 1000 bacteria growing at a continuous rate of 30% per hour has $1000 e^{0.6} \approx 1822$ bacteria after 2 hours.

2Try It Yourself

A lab culture starts with 2000 bacteria and grows continuously at 40% per hour.

How many bacteria are there after 5 hours? Round to the nearest whole number.

Step 1: Write the mathematical expression

Calculate $2000 \times e^{0.4 \times 5}$ :

Medicine: Radioactive Tracers

Hospitals use radioactive substances such as iodine-131 for diagnosis and therapy. The amount left in a sample follows $A = Pe^{-kt}$, so doctors know how much remains on any day.

Example:

Iodine-131 decays continuously at about 8.6% per day. After 1 day, $e^{-0.086} \approx 0.918$ , so about 91.8% of the sample is left.

3Try It Yourself

A hospital receives 100 mg of iodine-131, which decays continuously at a rate of 8.6% per day.

How much is left after 10 days? Round to one decimal place.

Step 1: Write the mathematical expression

Calculate $100 \times e^{-0.086 \times 10}$ :

Key Takeaways

1Continuous growth: $A = Pe^{rt}$ ; continuous decay: $A = Pe^{-kt}$ , with rates written as decimals
2A positive exponent means growth, a negative exponent means decay
3In the calculator, type $e^{x}$ as e^(x) (or $2.71828^{x}$ without an $e$ key) and put brackets around the exponent
4At the same rate and time, continuous compounding always gives a little more than annual compounding
5To find the initial amount, divide: $P = \frac{A}{e^{rt}}$
6Solving for the time $t$ needs logarithms, which come in later lessons

Frequently Asked Questions

Why do we use $e$ for continuous growth?

Compounding

n

times a year at rate

r

gives

\left(1 + \frac{r}{n}\right)^{nt}

. As

n

gets larger and larger, this approaches

e^{rt}

. So

e

is what you get when growth happens at every instant.

How much more does continuous compounding really give?

Not much more, but always more. At 5% for one year, annual compounding multiplies by

1.05

and continuous compounding by

e^{0.05} \approx 1.0513

. Over long periods and large amounts the difference becomes noticeable.

Is $A = Pe^{-kt}$ the same as $A = Pe^{rt}$ with a negative $r$ ?

Yes. Some books write decay with a positive decay rate

k

and a minus sign, others use one formula

Pe^{rt}

and let

r

be negative, for example

r = -0.03

. Both give the same result.

How do I find how long it takes to reach a certain amount?

That means solving

A = Pe^{rt}

for

t

, which needs the natural logarithm

\ln

. You will learn it in the logarithm lessons. For now you can estimate

t

by trying values in a table.

Glossary

Continuous growth: Growth that happens at every instant, modeled by $A = Pe^{rt}$ with $r > 0$
Continuous decay: Decrease that happens at every instant, modeled by $A = Pe^{-kt}$ with $k > 0$
Continuous compounding: Interest added at every instant; the limit of compounding more and more often
Continuous rate: The rate $r$ or $k$ in the exponent, written as a decimal ( $4\% = 0.04$ )
Euler's number $e$: The constant $e \approx 2.71828$ , the natural base for continuous growth and decay
Initial amount: The starting value $P$ at time $t = 0$

Formula Card

Continuous growth

A = P e^{rt}

$P$ = initial amount, $r$ = continuous growth rate (decimal), $t$ = time, $A$ = amount after time $t$.

Continuous decay

A = P e^{-kt}

$k$ = continuous decay rate (decimal, positive). The minus sign makes the amount shrink.

Annual compounding (for comparison)

A = P(1 + r)^t

Interest added once per year. Gives slightly less than $Pe^{rt}$ at the same rate.

Initial amount

P = \frac{A}{e^{rt}}

Divide the final amount by the growth factor $e^{rt}$.

Teacher Guide: Continuous Growth and Decay

Use this lesson with your class

For Teachers

For Struggling Students:

For On-Level Students:

For Advanced Students:

Lesson Content

Definition

Worked Examples

Common Mistakes

Why It Matters

Real World Applications

Banking: Continuously Compounded Savings

Biology: Bacteria in the Lab

Medicine: Radioactive Tracers

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Continuous growth

Continuous decay

Annual compounding (for comparison)

Initial amount

More in This Topic

Introduction to Exponential Functions

Exponential Growth

Graphing Exponential Functions

Exponential Decay