Continuous Growth and Decay Formula

Most exponential growth and decay problems you meet first involve change happening in discrete jumps: a population that grows by a fixed percentage each year, or an investment that compounds once a month. Continuous growth and decay describe something different: a quantity that changes at every single instant, with no gaps between compounding periods at all. Interest is added, or a substance decays, continuously rather than in steps.

This idea shows up constantly in finance (continuous compound interest), science (radioactive decay, cooling, and population models), and anywhere a rate of change is proportional to the current amount at every moment in time.

The formula for continuous growth and decay is:

\( A = Pe^{rt} \)

where:

• \( A \) is the amount after time \( t \)
• \( P \) is the initial amount (at \( t = 0 \))
• \( e \) is Euler's number, approximately \( 2.71828 \)
• \( r \) is the continuous growth or decay rate, written as a decimal
• \( t \) is time, in the same units used for the rate \( r \)

The number \( e \) is not arbitrary. If you take the ordinary compound growth formula and let the number of compounding periods per year grow without bound, the expression \( \left(1 + \frac{r}{n}\right)^{nt} \) approaches \( e^{rt} \) as \( n \) gets larger and larger. Continuous growth is the limiting case of compounding infinitely often.

The same formula, \( A = Pe^{rt} \), handles both cases:

• If \( r > 0 \), the quantity grows continuously, and \( A \) increases without bound as \( t \) increases.
• If \( r < 0 \), the quantity decays continuously, and \( A \) shrinks toward zero (but never reaches it) as \( t \) increases.

This is the same underlying pattern you see in exponential growth and decay by percentage, except that instead of applying a fixed percentage once per period, the rate \( r \) applies at every instant continuously.

The most common application of this formula is continuous compound interest. If an amount of money \( P \) is invested at an annual interest rate \( r \), compounded continuously, then the balance after \( t \) years is:

\( A = Pe^{rt} \)

This gives the largest possible balance for a given nominal annual rate, since compounding continuously is compounding as often as it is mathematically possible to compound. Compare this to compounding a fixed number of times per year, which is covered in exponential growth and decay by a factor when the growth happens in discrete multiplicative steps.

Suppose \( 500 \) dollars is invested at a continuous annual growth rate of \( 6\% \). Find the balance after \( 8 \) years.

Here \( P = 500 \), \( r = 0.06 \), and \( t = 8 \). Substitute into \( A = Pe^{rt} \):

\( A = 500e^{(0.06)(8)} = 500e^{0.48} \)

Using \( e^{0.48} \approx 1.616 \):

\( A \approx 500 \times 1.616 \approx 808.16 \)

So the investment grows to about \( 808.16 \) dollars after \( 8 \) years.

A radioactive sample starts at \( 200 \) grams and decays continuously at a rate of \( 3\% \) per year. Find the mass remaining after \( 15 \) years.

Here \( P = 200 \), \( r = -0.03 \) (negative because it is decay), and \( t = 15 \):

\( A = 200e^{(-0.03)(15)} = 200e^{-0.45} \)

Using \( e^{-0.45} \approx 0.6376 \):

\( A \approx 200 \times 0.6376 \approx 127.5 \)

So about \( 127.5 \) grams remain after \( 15 \) years. This kind of continuous decay problem is closely related to exponential decay and half-life, where the rate \( r \) is derived from a known half-life instead of being given directly.

Both curves above have the same basic exponential shape you see when you graph exponential functions: growth curves rise faster and faster as \( t \) increases, while decay curves fall quickly at first and then flatten out, approaching the horizontal axis but never crossing it. The starting value \( P \) is always the value of \( A \) at \( t = 0 \), since \( e^{r \cdot 0} = e^0 = 1 \).

• The formula is always \( A = Pe^{rt} \): growth and decay are the same equation, distinguished only by the sign of \( r \).
• \( e \) appears because continuous change is the limit of compounding infinitely often.
• \( A \) never actually reaches zero in continuous decay; it only gets arbitrarily close.
• Always keep the units of \( r \) and \( t \) consistent (for example, an annual rate needs \( t \) measured in years).

Related lessons

• Algebra Help