Continuously Compounded Interest
When Interest Earns Interest Every Instant
Banks compound interest annually, monthly, or daily, but what happens if interest compounds continuously, every instant? This isn’t just a theoretical curiosity. Continuous compounding simplifies calculations and reveals a deep connection: money grows at a rate proportional to how much you have, which is exactly the exponential growth model.
The formula $A(t) = A_0 e^{rt}$ is used throughout finance, from pricing bonds to valuing annuities.
Before You Start
Quick Self-Check
What is the solution to $\frac{dA}{dt} = 0.05A$ with $A(0) = 1000$? → Exponential Growth ODE
Solve $e^{5r} = 2$ for $r$. → Natural Logarithm
If you struggled, review the linked prerequisite first.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Financial Applications |
| Chapter | 6, Section 5 |
| Difficulty | Intermediate |
| Time | ~15 minutes |
Key Concepts
From Discrete to Continuous Compounding
If you invest $A_0$ dollars at annual interest rate $r$, compounded $n$ times per year, after $t$ years you have:
$$A = A_0 \left(1 + \frac{r}{n}\right)^{nt}$$
| Compounding | $n$ | Formula for $A_0 = \$5000$, $r = 0.02$, $t = 3$ |
|---|---|---|
| Annual | 1 | $5000(1.02)^3 = \$5306.04$ |
| Quarterly | 4 | $5000(1.005)^{12} = \$5308.39$ |
| Monthly | 12 | $5000(1 + 0.02/12)^{36} = \$5308.92$ |
| Daily | 365 | $5000(1 + 0.02/365)^{1095} = \$5309.17$ |
| Continuous | $\infty$ | $5000e^{0.06} = \$5309.18$ |
Notice: As $n$ increases, the values converge to the continuous limit.
The Continuous Compounding Formula
As $n \to \infty$:
$$A(t) = \lim_{n \to \infty} A_0 \left(1 + \frac{r}{n}\right)^{nt} = A_0 e^{rt}$$
This uses the fundamental limit $\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.
$$\boxed{A(t) = A_0 e^{rt}}$$
where:
- $A_0$ = initial principal (amount invested)
- $r$ = annual interest rate (as a decimal)
- $t$ = time in years
- $A(t)$ = amount after $t$ years
Why This Is Exponential Growth
Differentiating $A(t) = A_0 e^{rt}$:
$$\frac{dA}{dt} = rA_0 e^{rt} = rA(t)$$
So: the rate of increase of an investment is proportional to its current size. This is exactly $\frac{dy}{dt} = ky$ with $k = r$.
Key Formulas
| To Find | Formula |
|---|---|
| Future Value | $A(t) = A_0 e^{rt}$ |
| Time to reach target | $t = \frac{\ln(A/A_0)}{r}$ |
| Doubling time | $t_d = \frac{\ln 2}{r} \approx \frac{0.693}{r}$ |
| Rate from doubling time | $r = \frac{\ln 2}{t_d}$ |
The Rule of 72
For quick mental math: If interest rate is $r\%$ per year, the doubling time is approximately $\frac{72}{r}$ years.
Example: At 6% interest, money doubles in about $72/6 = 12$ years.
Effective Annual Rate
The effective annual rate (EAR) tells you the equivalent annual rate for continuous compounding:
$$\text{EAR} = e^r - 1$$
Example: A 5% continuously compounded rate has EAR = $e^{0.05} - 1 \approx 0.0513 = 5.13\%$.
Common Pitfalls
| Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Using $A = A_0(1 + r)^t$ for continuous | That’s annual compounding | Use $A = A_0 e^{rt}$ for continuous |
| Confusing $r$ as a percentage | If rate is 5%, use $r = 0.05$ | Convert percentage to decimal |
| Mixing continuous and discrete formulas | They give slightly different answers | Pick one and be consistent |
| Forgetting $r$ is annual | If given monthly rate, multiply by 12 | Ensure $r$ and $t$ are in matching units |
| Expecting big differences | Continuous vs daily compounding differs by pennies | The difference matters more for large $A_0$ or long $t$ |
Worked Example
Problem: You invest \$10,000 at 4% annual interest, compounded continuously.
(a) How much will you have after 5 years? (b) How long until the investment doubles? (c) What is the effective annual rate?
Solution:
(a) Using $A(t) = A_0 e^{rt}$: $$A(5) = 10000 \cdot e^{0.04 \times 5} = 10000 \cdot e^{0.2} \approx 10000 \times 1.2214 = \$12,214.03$$
(b) For doubling, solve $20000 = 10000 e^{0.04t}$: $$e^{0.04t} = 2$$ $$0.04t = \ln 2$$ $$t = \frac{\ln 2}{0.04} = \frac{0.693}{0.04} = 17.33 \text{ years}$$
Or use the Rule of 72: $72/4 = 18$ years (close approximation).
(c) Effective annual rate: $$\text{EAR} = e^{0.04} - 1 = 1.0408 - 1 = 0.0408 = 4.08\%$$
Practice Problems
You invest \$2,000 at 3% annual interest, compounded continuously. How much do you have after 10 years?
How long will it take for $5,000 to grow to $8,000 at 5% annual interest, compounded continuously?
An investment of $6,000 grows to $7,500 in 4 years with continuous compounding. What is the annual interest rate?
You have \$20,000 to invest for 10 years. Bank A offers 4.8% compounded monthly. Bank B offers 4.7% compounded continuously.
(a) How much would you have at each bank after 10 years? (b) Which is the better deal? (c) What continuously compounded rate would be equivalent to Bank A’s offer?
Prove that $\lim_{n \to \infty} A_0\left(1 + \frac{r}{n}\right)^{nt} = A_0 e^{rt}$.
Hint: Use the substitution $m = n/r$ and the limit definition $e = \lim_{m \to \infty}\left(1 + \frac{1}{m}\right)^m$.
CCI-Style Conceptual Questions
Question 1: If you double the interest rate, does the doubling time get cut in half?
Answer
Yes. Doubling time is $t_d = \frac{\ln 2}{r}$. If you double $r$, the new doubling time is $\frac{\ln 2}{2r} = \frac{t_d}{2}$.
Question 2: Is continuous compounding always better than daily compounding at the same stated rate?
Answer
Yes, but barely. The difference is tiny. For $\$10,000$ at 5% for 1 year: daily gives $\$10,512.67$, continuous gives $\$10,512.71$, a difference of 4 cents.
Question 3: An investment triples in 20 years with continuous compounding. What is the annual rate?
Answer
From $3A_0 = A_0 e^{20r}$: $$e^{20r} = 3$$ $$r = \frac{\ln 3}{20} \approx \frac{1.099}{20} = 0.055 = 5.5\%$$
Common Misconceptions
the continuously compounded growth rate $r$ equals the actual percentage increase per year.
This is the rate-as-fixed-number error. The rate $r$ in $A(t) = A_0 e^{rt}$ is a proportionality constant satisfying $\frac{dA}{dt} = rA$, meaning $r$ is the instantaneous relative rate, not the amount added each year. After one year at rate $r = 0.05$, the account grows by a factor of $e^{0.05} \approx 1.0513$, a 5.13% increase, not exactly 5%. This is the effective annual rate, and it exceeds $r$ because interest accrues continuously throughout the year.
Mastery Checklist
Mental Model
The “Infinite Installments” Analogy:
Imagine getting paid your annual interest in smaller and smaller installments: monthly, daily, hourly, every second. Each payment immediately starts earning its own interest. As installments become infinitely frequent, you approach continuous compounding. The magic number $e$ emerges naturally from this infinite subdivision process.
Still Confused?
| If you’re struggling with... | Review this |
|---|---|
| The exponential model $A' = rA$ | Exponential Growth ODE |
| Solving $e^{rt} = c$ for $t$ | Natural Logarithm |
| The limit definition of $e$ | Section 6.4: Definition of $e$ |
Connections
Looking back:
- Exponential Growth ODE provides the mathematical foundation
- The limit $(1 + 1/n)^n \to e$ connects to the definition of $e$
Looking ahead:
- Present value calculations reverse this process
- Annuities involve summing many exponential terms
- Differential equations generalize this to variable rates
Real-world connections:
- Bond pricing and yield calculations
- Mortgage and loan analysis
- Investment portfolio growth projections
- Inflation-adjusted returns
| Previous | Up | Next |
|---|---|---|
| Newton’s Cooling | Chapter 6 Skills |
Last updated: 2026-01-23