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.

Advanced30 min18 problems
Practice Problems

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.

Try it now

A substance decays continuously at a rate of 4% per year. Which formula models the amount AA after tt years, starting from PP?

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.

Interactive Visual

Function graph

05101520050010001500200025003000tyy = 0

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Asymptote

Table of values

t05101520
A = P·e^(rt)10001284.031648.7221172718.28
A = P(1 + r)^t10001276.281628.892078.932653.3

Interest Calculator

Formula
Compound Interest:A = Pe^(rt)A = 1000 · e^(0.05 · 10) = $1,648.72

Compound Interest

Interest Earned:$648.72
Total Amount:$1,648.72

Same money, different compounding

CompoundingnTotal AmountInterest Earned
Annually1$1,628.89$628.89
Monthly12$1,647.01$647.01
Daily365$1,648.66$648.66
Continuously∞$1,648.72$648.72

Growth Over Time

yearsAmount

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Practice Problems

18 problems
Problem 1 of 18
Easy
A substance decays continuously at a rate of 4% per year. Which formula models the amount AA after tt years, starting from PP?

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

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.
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.
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.
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.
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 $e^{rt}$.

Today's challengeOne problem, three tries

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