Back to Lesson

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.

Present LessonShare to ClassroomView Student Page

Use this lesson with your class

Free, no student accounts needed.

Share with students

Students open the lesson and practise with instant feedback.

Printable worksheet

All practice problems on paper, with a separate answer key.

Worksheet Answer key

Class quiz

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

For Teachers

Learning Objectives
  • Use A=PertA = Pe^{rt} to calculate amounts that grow continuously
  • Use A=Pe−ktA = 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 PP from a known future amount
  • Evaluate powers of ee on a calculator and round as instructed
Prerequisites
  • • Introduction to Exponential Functions: the form f(x)=a⋅bxf(x) = a \cdot b^x
  • • Exponential Growth: growth factor 1+r1 + r and A=A0(1+r)tA = A_0(1 + r)^t
  • • The Natural Base e: e≈2.71828e \approx 2.71828 as the limit of (1+1n)n\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−1e^{-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 ee 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(1.05), 1000(1+0.0512)121000\left(1 + \frac{0.05}{12}\right)^{12} and 1000e0.051000e^{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−kte^{-kt} makes the amount negative

Remediation: Show that e−0.6≈0.5488e^{-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 e5e^{5} and e0.05e^{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 erte^{rt} in each problem so students focus on setting up P×ertP \times e^{rt}
  • • Start with exponents that are whole numbers, like e1e^{1} and e2e^{2}
  • • Use a three-column table: PP, erte^{rt}, AA

For On-Level Students:

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

For Advanced Students:

  • • Show numerically that (1+rn)n\left(1 + \frac{r}{n}\right)^{n} approaches ere^{r} as nn grows
  • • Estimate the doubling time of Pe0.07tPe^{0.07t} with a table of values and compare with the rule of 70
  • • Connect Pe−ktPe^{-kt} to half-life: find kk 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−0.3t50e^{-0.3t} and compute values with a calculator

  • visualTwo Curves

    Plot P(1+r)tP(1 + r)^t and PertPe^{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 ee you saw that compounding more and more often, (1+1n)n\left(1 + \frac{1}{n}\right)^n, approaches e≈2.71828e \approx 2.71828. When a quantity grows continuously (every instant, not once a year), we use ee as the base.
Continuous growth:
A=PertA = P e^{rt}
Continuous decay:
A=Pe−ktA = P e^{-kt}
where:
  • PP is the initial amount (at t=0t = 0)
  • rr is the continuous growth rate and kk the continuous decay rate, both written as decimals (5%=0.055\% = 0.05)
  • tt is the time, in the same unit as the rate (years, hours, days)
  • AA is the amount after time tt
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=PertA = Pe^{rt} with a negative rate, for example r=−0.03r = -0.03.
In the calculator you can type exe^{x} directly as e^(x), for example e^0.5 gives ≈1.6487\approx 1.6487. On a basic calculator without an ee key, type 2.71828x2.71828^{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.
1

Identify the values

P=1000P = 1000, r=0.05r = 0.05, t=10t = 10 → Values identified
2

Find the exponent rtrt

rt=0.05×10=0.5rt = 0.05 \times 10 = 0.5 → rt=0.5rt = 0.5
3

Find e0.5e^{0.5} with the calculator

2.718280.5≈1.6487212.71828^{0.5} \approx 1.648721 → e0.5≈1.648721e^{0.5} \approx 1.648721
4

Multiply by PP

A=1000×1.648721≈1648.72A = 1000 \times 1.648721 \approx 1648.72 → A≈1648.72A \approx 1648.72

Common Mistakes

Writing the rate as a percent in the exponent: 1000e5×101000 e^{5 \times 10}

Why it's wrong: The formula needs the rate as a decimal. e50e^{50} is a gigantic number (about 5×10215 \times 10^{21}), which makes no sense for a savings account.

Correct: Convert first: 5%=0.055\% = 0.05, so A=1000e0.05×10=1000e0.5A = 1000 e^{0.05 \times 10} = 1000 e^{0.5}.

Forgetting the minus sign for decay: 80e0.03×2080 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=80e−0.03×20=80e−0.6≈43.90A = 80 e^{-0.03 \times 20} = 80 e^{-0.6} \approx 43.90.

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

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

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

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

Why it's wrong: (1+r)t(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 PertPe^{rt}. Once a year means P(1+r)tP(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 PertPe^{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 ee
  • Physics and medicine: radioactive atoms decay at random moments, and a drug leaves the bloodstream gradually, so both follow Pe−ktPe^{-kt}
  • Science in general: almost every growth or decay formula in chemistry, physics and economics is written with base ee
Once you can read PertPe^{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 e0.05≈1.0513e^{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×e0.045×44000 \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 1000e0.6≈18221000 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×e0.4×52000 \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≈0.918e^{-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×e−0.086×10100 \times e^{-0.086 \times 10}:

Key Takeaways

  • 1Continuous growth: A=PertA = Pe^{rt}; continuous decay: A=Pe−ktA = Pe^{-kt}, with rates written as decimals
  • 2A positive exponent means growth, a negative exponent means decay
  • 3In the calculator, type exe^{x} as e^(x) (or 2.71828x2.71828^{x} without an ee 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=AertP = \frac{A}{e^{rt}}
  • 6Solving for the time tt needs logarithms, which come in later lessons

Frequently Asked Questions

Why do we use ee for continuous growth?

Compounding nn times a year at rate rr gives (1+rn)nt\left(1 + \frac{r}{n}\right)^{nt}. As nn gets larger and larger, this approaches erte^{rt}. So ee 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 1.051.05 and continuous compounding by e0.05≈1.0513e^{0.05} \approx 1.0513. Over long periods and large amounts the difference becomes noticeable.

Is A=Pe−ktA = Pe^{-kt} the same as A=PertA = Pe^{rt} with a negative rr?

Yes. Some books write decay with a positive decay rate kk and a minus sign, others use one formula PertPe^{rt} and let rr be negative, for example r=−0.03r = -0.03. Both give the same result.

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

That means solving A=PertA = Pe^{rt} for tt, which needs the natural logarithm ln⁡\ln. You will learn it in the logarithm lessons. For now you can estimate tt by trying values in a table.

Glossary

Continuous growth
Growth that happens at every instant, modeled by A=PertA = Pe^{rt} with r>0r > 0
Continuous decay
Decrease that happens at every instant, modeled by A=Pe−ktA = Pe^{-kt} with k>0k > 0
Continuous compounding
Interest added at every instant; the limit of compounding more and more often
Continuous rate
The rate rr or kk in the exponent, written as a decimal (4%=0.044\% = 0.04)
Euler's number ee
The constant e≈2.71828e \approx 2.71828, the natural base for continuous growth and decay
Initial amount
The starting value PP at time t=0t = 0

Formula Card

Continuous growth

A=PertA = P e^{rt}

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

Continuous decay

A=Pe−ktA = P e^{-kt}

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

Annual compounding (for comparison)

A=P(1+r)tA = P(1 + r)^t

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

Initial amount

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

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

More in This Topic

Introduction to Exponential Functions

Advanced25 min

Exponential Growth

Advanced25 min

Graphing Exponential Functions

Advanced30 min

Exponential Decay

Advanced25 min