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.
If you struggled, review the linked prerequisite first.
| Property | Value |
|---|---|
| Concept | Financial Applications |
| Chapter | 6, Section 5 |
| Difficulty | Intermediate |
| Time | ~15 minutes |
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.
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:
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 $k = r$.
| 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 | $r = \frac{\ln 2}{t_d}$ |
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.
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 = $e^{0.05} - 1 \approx 0.0513 = 5.13\%$.
| 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 $t$ |
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: $$\text{EAR} = e^{0.04} - 1 = 1.0408 - 1 = 0.0408 = 4.08\%$$
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 $e = \lim_{m \to \infty}\left(1 + \frac{1}{m}\right)^m$.
Question 1: If you double the interest rate, does the doubling time get cut in half?
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?
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?
From $3A_0 = A_0 e^{20r}$: $$e^{20r} = 3$$ $$r = \frac{\ln 3}{20} \approx \frac{1.099}{20} = 0.055 = 5.5\%$$
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.
| 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 $e$ |
Looking back:
Looking ahead:
Real-world connections:
| Previous | Up | Next |
|---|---|---|
| Newton's Cooling | Chapter 6 Skills |
Last updated: 2026-01-23