Introduction
A continuous growth or decay model describes how an amount changes at every instant. A fixed-period percentage describes a different way to express change: it tells how much the amount is multiplied by after a specified interval, such as one year or one month.
The two descriptions are related, but they are not identical. In particular, the continuous rate constant \(k\) is not generally the percentage applied once per period. Confusing these quantities leads to incorrect predictions, even when the numerical value of \(k\) looks like a familiar percentage.
This tutorial derives the exact conversion between them. The interval matters: a continuous rate constant determines a different equivalent percentage for each choice of period length.
Continuous Rate and Fixed-Period Percentage
For a constant continuous rate \(k\), the exponential model is \(y(t)=y_0e^{kt}\), where \(y_0\) is the amount at time \(t=0\). The units of \(k\) are the reciprocal of the time units used for \(t\). For instance, if \(t\) is measured in years, \(k\) is measured in inverse years.
A fixed-period percentage needs a specified period length. Let \(\Delta t>0\) be that length. If the amount is multiplied by \(1+p\) over each such period, then \(p\) is the change written as a decimal: \(p=0.08\) means an increase of \(8\%\), and \(p=-0.08\) means a decrease of \(8\%\). For a positive amount that remains positive, the multiplier \(1+p\) must be positive, so \(p>-1\).
The multiplier applies to the amount at the start of the interval. Thus, if the amount at time \(t\) is \(y(t)\), the model gives \(y(t+\Delta t)=M y(t)\). This equation holds even when \(y(t)=0\). A ratio such as \(y(t+\Delta t)/y(t)\), however, is defined only when \(y(t)\ne 0\). For a positive initial amount \(y_0>0\), the exponential model stays positive, so that ratio is defined at every time.
Deriving the Conversion
The continuous model and the fixed-period multiplier can be compared directly. The model gives the amount after one interval, and the multiplier is the factor relating that amount to the amount at the interval's start.
Starting at time \(t\), the model gives \(y(t+\Delta t)=y_0e^{k(t+\Delta t)}\), while \(y(t)=y_0e^{kt}\).
Using the exponent rule \(e^{a+b}=e^ae^b\), we obtain \(y(t+\Delta t)=y_0e^{kt}e^{k\Delta t}=e^{k\Delta t}y(t)\). This establishes the multiplier equation without dividing by the amount.
The fixed-period multiplier is \(M=e^{k\Delta t}\), so the change as a decimal is \(p=M-1=e^{k\Delta t}-1\).
If \(p\) is known, then \(1+p=e^{k\Delta t}\). Taking natural logarithms gives \(k=\ln(1+p)/\Delta t\), provided \(p>-1\) and \(\Delta t>0\).
This derivation also explains the role of the initial amount. It is not necessary to divide by \(y_0\) to get the multiplier equation. If \(y_0>0\), one can additionally divide by \(y(t)\) and write \(y(t+\Delta t)/y(t)=e^{k\Delta t}\). If \(y_0=0\), the model is identically zero: the multiplier equation still holds, but that ratio is \(0/0\) and is undefined.
Worked Examples
Worked Example: Convert a Continuous Growth Rate to an Annual Percentage
A model has continuous rate constant \(k=0.06\) per year. Find the equivalent percentage increase over one year and the modeled amount after one year if the initial amount is \(500\).
Here \(\Delta t=1\) year. The one-year multiplier is
Therefore, the annual change as a decimal is
As a percentage, this is \(100p\approx 6.18365465\%\), or about \(6.1837\%\) to four decimal places. The modeled amount after one year is
So the amount is about \(530.9183\), to four decimal places. Check: dividing the computed amount by the initial amount gives \(530.91827325/500\approx1.0618365465\), the multiplier from the model. Also, \(\ln(1.0618365465)\approx0.06\), as required for a one-year interval.
Worked Example: Convert a Quarterly Percentage to a Continuous Rate
An amount increases by \(5\%\) every three months. Find the equivalent constant continuous rate in inverse years, then check the multiplier over one quarter and over one year.
A \(5\%\) increase has multiplier \(1+p=1.05\). Three months is \(0.25\) year, so \(\Delta t=0.25\). The conversion formula gives
Thus the equivalent continuous rate is about \(0.1951607\) per year. To check the quarterly multiplier, substitute this rate into the exponential model:
For four quarters, or one year, the multiplier is
This corresponds to a one-year increase of \(1.21550625-1=0.21550625\), or \(21.550625\%\). The check agrees with applying the quarterly multiplier four times: \(1.05^4=1.21550625\).
Worked Example: Find the Equivalent Two-Day Decay Percentage
A quantity has a half-life of \(10\) days. Find the percentage decrease over a two-day interval under the corresponding continuous decay model.
After \(10\) days the remaining fraction is \(1/2\), so the model's rate constant satisfies \(e^{10k}=1/2\). Taking natural logarithms gives
Over two days, the multiplier is
The change as a decimal is \(p=M-1\approx0.8705505633-1=-0.1294494367\). Thus the quantity decreases by about \(12.94494367\%\), or \(12.9449\%\) to four decimal places. Check: the remaining fraction over five two-day intervals is \(M^5=(2^{-1/5})^5=1/2\), exactly the stated half-life.
Why \(k\) Is Not the Periodic Percentage
The most common error is to treat a numerical value of \(k\) as though it were already the change over one chosen period. For a one-year interval, the exact percentage as a decimal is \(e^k-1\), not \(k\). For a shorter interval, the exact change is \(e^{k\Delta t}-1\), not \(k\Delta t\).
Worked Example: Correcting the Direct-Percentage Error
Suppose \(k=0.12\) per year and the initial amount is \(100\). A mistaken calculation says that \(k=0.12\) means a \(12\%\) increase in one year, giving \(100(1.12)=112\).
The model instead gives
The exact one-year multiplier is \(e^{0.12}\approx1.1274968516\), so the equivalent annual percentage increase is \((1.1274968516-1)100\%\approx12.74968516\%\). The mistaken answer \(112\) corresponds to a multiplier of \(1.12\), not \(e^{0.12}\). Checking the correction, \(\ln(1.1274968516)\approx0.12\), which recovers the continuous rate for a one-year period.
The direct identification can be a useful approximation when \(|k\Delta t|\) is small, because \(e^x\) is close to \(1+x\) for \(x\) near zero. But an approximation is not the exact conversion. When a problem asks for the exact equivalent percentage or multiplier, use the exponential formula.
Reading the Units and Interpreting the Result
The product \(k\Delta t\) must be dimensionless: the time units in \(k\) and \(\Delta t\) must match. If \(k\) is in inverse years, then \(\Delta t\) must be expressed in years before using \(e^{k\Delta t}\). This is not a new rate conversion rule; it is a check that the exponent is formed consistently.
The equivalent periodic percentage depends on the length of the period. For a fixed \(k\), a one-month percentage and a one-year percentage are different quantities. If \(\Delta t\) is one month expressed in years, the monthly multiplier is \(e^{k/12}\), and the monthly percentage change as a decimal is \(e^{k/12}-1\). Repeating that multiplier for twelve months gives \(e^k\), the same multiplier as the continuous model over one year.
For decay, the signed rate constant is negative, so \(e^{k\Delta t}<1\) for \(\Delta t>0\). Consequently, \(p=e^{k\Delta t}-1\) is negative. That negative value records a decrease; the positive percentage decrease is its opposite, \(1-e^{k\Delta t}\). Keeping the signed change and the positive size of a decrease distinct helps preserve the meaning of the conversion.
Summary
A continuous rate constant describes instantaneous relative change per unit time; it is not itself a fixed-period percentage. Over an interval of length \(\Delta t\), the continuous model multiplies the amount by \(e^{k\Delta t}\). The corresponding periodic change is \(p=e^{k\Delta t}-1\), and the reverse conversion is \(k=\ln(1+p)/\Delta t\).
To avoid the common error, identify the period, match the time units, calculate the multiplier, and only then express the change as a percentage. The multiplier equation \(y(t+\Delta t)=e^{k\Delta t}y(t)\) works even for a zero amount; a ratio of amounts is justified only when the denominator is nonzero.
Check Your Understanding
For each question, state the period and units you are using, and keep the multiplier separate from the percentage change.
- A model has \(k=0.04\) per year. What is its exact multiplier over one year, and what is the equivalent annual percentage change?
- An amount increases by \(3\%\) every two months. Write the formula for its equivalent continuous rate constant, using years as the time unit.
- Why is it invalid to calculate \(y(t+\Delta t)/y(t)\) when the initial amount is zero, even though the multiplier equation still holds?
- A decay model has \(k<0\). Is its fixed-period multiplier above or below \(1\), and what sign does \(p=e^{k\Delta t}-1\) have for \(\Delta t>0\)?
- If a rate constant is given per day, what must be checked before using it with an interval measured in weeks?
