Why the Units of the Rate Constant Matter
An exponential model describes a quantity whose rate of change is proportional to its current amount. It is used, for example, to model population growth and the decay of a quantity of substance. The constant of proportionality, \(k\), is not just a number: its units depend on how time is measured.
Keeping track of those units helps distinguish a meaningful model from an equation that only looks plausible. It also gives \(k\) a direct interpretation: it measures the instantaneous rate of change relative to the current amount, per unit of time.
Units in the Proportional-Rate Equation
Suppose \(P(t)\) is an amount that depends on time \(t\), and \(k\) is a constant. The exponential growth-and-decay model is
If \(P\) is measured in units \(U\) and \(t\) in units \(T\), then the derivative \(\frac{dP}{dt}\) has units \(U/T\): it describes a change in the amount per unit of time. The right side \(kP\) must have the same units. Since \(P\) has units \(U\), \(k\) must have units \((U/T)/U=1/T\).
The word relative matters. The derivative \(\frac{dP}{dt}\) is an absolute rate, such as people per year or milligrams per day. Dividing by the current amount removes the units of the amount, leaving a rate per unit of time. The resulting value tells us how rapidly the quantity is changing in proportion to its current size.
Why the Exponent Must Be Dimensionless
The solution to the proportional-rate equation has the form \(P(t)=P(0)e^{kt}\). The exponent of an exponential function must be a dimensionless number. Therefore \(kt\) must have no units. Since \(t\) has units of time, \(k\) must have inverse-time units.
This provides a second way to check the units of \(k\), consistent with checking the differential equation. For example, if time is measured in years, then \(k\) must be in inverse years, often written \(\text{year}^{-1}\) or \(1/\text{year}\). Then \(kt\) is dimensionless when \(t\) is expressed in years.
For example, \(P\) may be measured in people and \(t\) in years.
\(\frac{dP}{dt}\) has units of amount per unit of time.
In \(\frac{dP}{dt}=kP\), this requires \(k\) to have inverse-time units.
In \(e^{kt}\), the product \(kt\) must be dimensionless.
Worked Example: A Population Growing per Year
Worked Example: A Population Growing per Year
A population \(P(t)\) is measured in people, and time \(t\) is measured in years. Suppose it follows \(\frac{dP}{dt}=0.03P\). Find the units and meaning of \(k\), then calculate the instantaneous growth rate when the population is 2,500 people.
The derivative has units of people per year. Since \(P\) is in people, the constant multiplying it must have units of inverse years:
Thus \(k=0.03\ \text{year}^{-1}\). The model says that the instantaneous relative rate is \(0.03\ \text{year}^{-1}\), or 3% per year as a continuous relative rate.
At \(P=2{,}500\) people, substitute into the model:
Check the relative rate by dividing the absolute rate by the current population:
This agrees with \(k\), and the absolute rate has the required units of people per year.
Worked Example: Substance Decaying per Day
Worked Example: Substance Decaying per Day
A sample contains \(M(t)\) milligrams of a substance, with time measured in days. Its model is \(\frac{dM}{dt}=-0.20M\). Interpret \(k\) and find the rate of change when the sample contains 80 milligrams.
Because \(\frac{dM}{dt}\) is measured in milligrams per day and \(M\) in milligrams, \(k\) must be measured in inverse days. Thus \(k=-0.20\ \text{day}^{-1}\). Its negative sign indicates a negative instantaneous relative rate, consistent with decay.
At \(M=80\) milligrams:
Check the relative rate:
The amount is decreasing at 16 milligrams per day at that instant, while its relative rate is \(-0.20\) per day. The units are consistent: multiplying inverse days by milligrams gives milligrams per day.
Changing the Time Unit Changes the Numerical Value of \(k\)
A rate constant must be expressed using the same time unit as the model’s time variable. If time is changed from years to days, the numerical value of \(k\) changes so that the product \(kt\) remains the same. A value in inverse years cannot be inserted directly into a formula whose time variable is measured in days.
Worked Example: Converting a Rate Constant to Match Time in Days
A population model has \(k=0.10\ \text{year}^{-1}\), but time is to be entered in days. Use \(365\) days per year to convert \(k\), then check the units and calculate the rate when \(P=36{,}500\) people.
Since \(1\ \text{year}=365\ \text{days}\), the same constant expressed per day is
The converted value has the required units for a model with time measured in days. At \(P=36{,}500\) people, the rate is
Check the same rate using years. First, \(10\ \text{people/day}\) is \(10(365)=3{,}650\ \text{people/year}\). Using the original constant gives
The rates agree after conversion. The exponential exponent also agrees: for 30 days, the converted expression gives \(\left(\frac{1}{3{,}650}\ \text{day}^{-1}\right)(30\ \text{days})=\frac{3}{365}\), a dimensionless number. Expressing 30 days as \(\frac{30}{365}\) years gives \((0.10\ \text{year}^{-1})(\frac{30}{365}\ \text{years})=\frac{3}{365}\), the same result.
A Dimensional-Consistency Check
Dimensional consistency means that the two sides of an equation have matching units. In \(\frac{dP}{dt}=kP\), the left side has units of amount per time. The right side has units of inverse time multiplied by amount, which is also amount per time. The same check applies whether \(P\) represents a population, a mass of substance, or another quantity.
The exponential solution also passes this check. If \(P(0)\) has units of amount, then \(P(0)e^{kt}\) has units of amount because \(e^{kt}\) is dimensionless. The exponential factor changes the numerical size of the initial amount; it does not supply new units.
Common Pitfall: Treating \(k\) as Unitless
A frequent mistake is to write a numerical value for \(k\) without checking which time unit it belongs to. For example, if \(t\) is in days and someone uses \(k=0.10\ \text{year}^{-1}\) directly, then \(kt\) has units of \(\text{day}/\text{year}\), not a dimensionless value. The exponential expression is therefore not dimensionally consistent.
The correction is to convert the rate constant to inverse days before using time in days:
Then multiplying by a time measured in days gives a dimensionless exponent. Another related error is to report a growth or decay rate only as a percentage, omitting “per unit of time.” A relative rate of \(0.03\ \text{year}^{-1}\) is not the same kind of quantity as a bare value of \(0.03\): its time unit is part of its meaning.
Interpreting the Constant Carefully
The equation \(\frac{dP}{dt}=kP\) says that the instantaneous rate of change is proportional to the amount currently present. Dividing by a nonzero \(P\) gives \(\frac{1}{P}\frac{dP}{dt}=k\), so \(k\) is the instantaneous relative rate at every time for which the model applies. Its numerical value is interpreted per unit of the chosen time variable.
For instance, \(k=0.03\ \text{year}^{-1}\) describes an instantaneous relative rate of \(0.03\) per year. It should not be casually treated as a statement that the quantity changes by exactly 3% over every complete year. A fixed-period percentage change and a continuous instantaneous rate are different descriptions; distinguishing them requires additional care.
When checking a model, ask three questions: Are the units of \(k\) inverse time? Does \(kt\) have no units? Do both sides of the differential equation have the same units? These checks do not establish that a model accurately describes reality, but they can reveal an equation that is dimensionally impossible.
Check Your Understanding
For each question, keep the amount units and time units explicit.
- If \(Q\) is measured in liters and \(t\) in minutes in \(\frac{dQ}{dt}=kQ\), what are the units of \(k\)?
- A substance has \(k=-0.08\ \text{hour}^{-1}\). What does the units portion of this value tell you about the time variable?
- If a population is measured in people and \(\frac{dP}{dt}=12\ \text{people/day}\) when \(P=400\) people, what is its instantaneous relative rate at that moment?
- Why must \(kt\) be dimensionless in \(P(t)=P(0)e^{kt}\)?
- A rate constant is given in inverse years, but time is entered in days. What must be done before using the value in the model?
