Continuous Compounding
Continuous compounding is a Grade 7 advanced math topic from Yoshiwara Intermediate Algebra using the formula A = Pe^(rt) to model interest that compounds infinitely often. The number e ≈ 2.718 arises naturally as the limit of (1 + 1/n)^n as n approaches infinity.
Key Concepts
Property When interest is compounded continuously, the amount $A(t)$ in an account after $t$ years is given by the function $$A(t) = Pe^{rt}$$ where $P$ is the principal invested and $r$ is the annual interest rate expressed as a decimal.
Examples If you invest 2000 dollars at 4% interest compounded continuously, the value of your account after $t$ years is $A(t) = 2000e^{0.04t}$. To find how long it takes for a 1000 dollar investment to double to 2000 dollars at 6% interest compounded continuously, you solve the equation $2000 = 1000e^{0.06t}$. The value of a 5000 dollar investment after 8 years at 3.5% interest compounded continuously is $A(8) = 5000e^{0.035(8)} \approx 6615.65$ dollars.
Explanation Continuous compounding is the theoretical limit of earning interest. It's as if interest is being calculated and added to your balance at every single moment. This formula gives the maximum amount of money an investment can earn at a given rate.
Common Questions
What is the continuous compounding formula?
A = Pe^(rt), where P is principal, e ≈ 2.718 is Eulers number, r is the annual interest rate, and t is time in years.
How does continuous compounding differ from regular compounding?
Continuous compounding uses infinitely many compounding periods per year, resulting in slightly more interest than any finite compounding schedule.
How does $500 grow at 4% continuously for 5 years?
A = 500·e^(0.04 × 5) = 500·e^0.2 ≈ 500 × 1.2214 ≈ $610.70.
Why does e appear in continuous compounding?
As you increase compounding frequency n without limit, (1 + r/n)^(nt) approaches Pe^(rt). The number e emerges as this limiting value.