Why Units Matter in Differential Equations
A differential equation describes a rate of change. That rate has units, and those units depend on what is changing and on what the change is measured against. Checking units is a practical way to catch errors in a model or in a slope-field label before carrying out further work.
For example, a population measured in people that changes over time measured in hours has a rate measured in people per hour. A number such as \(4\) is not enough to describe that rate unless we know the units of the population and time variables, and the units used for the graph’s axes.
Dimensional checks do not establish that a model accurately describes the real world. They do establish a basic requirement: the units on the two sides of a rate equation must agree, and quantities that are added must have compatible units.
Units of a Derivative
Suppose \(y\) measures a quantity in units \(U_y\), and \(x\) measures a quantity in units \(U_x\). A derivative measures how much \(y\) changes for a small change in \(x\). Its units are therefore \(U_y/U_x\).
This follows from the difference quotient. For a nonzero change \(\Delta x\), the quotient \(\Delta y/\Delta x\) has units \(U_y/U_x\): the numerator is a change in \(y\), and the denominator is a change in \(x\). The derivative is the limit of these quotients as \(\Delta x\) approaches zero, so taking the limit does not change the units.
For instance, if \(V\) is volume in cubic meters and \(t\) is time in seconds, then \(dV/dt\) has units of cubic meters per second. The derivative does not have the same units as \(V\), because it describes change in \(V\) relative to change in time.
Worked Example: Determine the Units of a Rate
A concentration \(C\) is measured in milligrams per liter, and time \(t\) is measured in minutes. Find the units of \(dC/dt\). If \(dC/dt=0.4C\), find the units required for the coefficient \(0.4\).
The dependent variable has units \(\text{mg}/\text{L}\), and the independent variable has units \(\text{min}\). Therefore,
The right side \(0.4C\) must have these same units. Since \(C\) has units \(\text{mg}/\text{L}\), the coefficient must have units \(\text{min}^{-1}\):
For example, if \(C=5\ \text{mg}/\text{L}\), then \(dC/dt=(0.4\ \text{min}^{-1})(5\ \text{mg}/\text{L})=2\ \text{mg}/(\text{L}\cdot\text{min})\). Substitution checks that the rate has the required units.
Checking Both Sides of a Differential Equation
A differential equation is dimensionally consistent only if both sides represent quantities with the same units. In a rate equation, the derivative on the left has the units of the dependent variable divided by the units of the independent variable. Every term on the right must combine to give those same units.
When terms are added or subtracted, they must have the same units. For example, a length can be added to another length, but not to a time. If a constant appears in an expression such as \(y-a\), then \(a\) must have the same units as \(y\), so that the subtraction is meaningful.
Worked Example: Check a Linear Rate Equation
Suppose \(T\) is measured in degrees Celsius, \(t\) in minutes, and a proposed model is \(dT/dt=-0.2(T-20)\). Determine the units the coefficient must have, and check whether the equation is consistent.
The derivative has units \({}^{\circ}\text{C}/\text{min}\). In the difference \(T-20\), the constant \(20\) must be a temperature in degrees Celsius; thus \(T-20\) is a temperature difference measured in degrees Celsius. For the right side to have units \({}^{\circ}\text{C}/\text{min}\), the coefficient must have units \(\text{min}^{-1}\):
So the model is consistent if \(-0.2\) means \(-0.2\ \text{min}^{-1}\). If \(-0.2\) is described as a dimensionless number, the right side has units of degrees Celsius, not degrees Celsius per minute, and the equation is inconsistent.
For a numerical check, take \(T=30\,{}^{\circ}\text{C}\). Then \(T-20=10\,{}^{\circ}\text{C}\), and the right side is \((-0.2\ \text{min}^{-1})(10\,{}^{\circ}\text{C})=-2\,{}^{\circ}\text{C}/\text{min}\). This matches the derivative’s units.
Coefficients and Constants Carry Units
A coefficient in a differential equation is not automatically unitless. Its units are determined by the role it plays in the equation. In an equation such as \(dP/dt=kP\), the units of \(k\) must convert the units of \(P\) into units of \(P\) per time. Thus \(k\) has units of inverse time.
Likewise, an additive constant on the right side of a rate equation must itself have units of rate. In \(dP/dt=aP+b\), the term \(aP\) and the term \(b\) must each have the units of \(dP/dt\). These requirements can reveal missing units or a mistaken term.
Worked Example: Check Units in a Sum of Rates
Let \(P\) be measured in people and \(t\) in hours. Consider \(dP/dt=aP+b\). Find the required units of \(a\) and \(b\).
The derivative has units \(\text{people}/\text{hour}\). Since \(aP\) must have those units and \(P\) has units people, \(a\) must have units \(\text{hour}^{-1}\):
The term \(b\) is added to \(aP\), so it must also have units \(\text{people}/\text{hour}\). Therefore,
As a check, if \(P=100\) people, \(a=0.03\ \text{hour}^{-1}\), and \(b=2\ \text{people}/\text{hour}\), then \(aP=3\ \text{people}/\text{hour}\), and \(aP+b=5\ \text{people}/\text{hour}\). Both terms can be added, and the result has the units required for \(dP/dt\).
Units and Slope-Field Labels
A slope-field label represents the value of a derivative at a point. It therefore has the units of the dependent variable per unit of the independent variable. A label may be written as a plain number in a diagram, but its meaning depends on the units used for the axes.
This matters especially when an axis uses scaled units. If a graph’s vertical coordinate measures population in thousands of people, its derivative with respect to time in hours is measured in thousands of people per hour. A slope of \(0.5\) on that graph corresponds to a rate of \(0.5\) thousand people per hour, or \(500\) people per hour.
Worked Example: Interpret a Slope-Field Label with Scaled Units
A slope-field graph uses \(Q\), measured in thousands of people, on the vertical axis and \(t\), measured in hours, on the horizontal axis. At one point, the differential equation gives \(dQ/dt=0.5\). Interpret the label in people per hour.
The derivative’s units are thousands of people per hour. Thus the label means
Since \(1\) thousand people is \(1000\) people,
Therefore, the slope-field label \(0.5\) corresponds to an increase of \(500\) people per hour. The conversion is consistent: the numerical vertical coordinate counts thousands, so a change of \(0.5\) in \(Q\) is a change of \(500\) people.
A Quick Dimensional-Check Method
When checking a rate equation or a slope-field label, use the units systematically rather than relying on the appearance of the numbers. The following sequence makes the check explicit.
State the units of the dependent variable and the independent variable.
Divide the dependent-variable units by the independent-variable units.
Each term on the right side must have the derivative’s units, and terms being added or subtracted must match.
Include the axis units and convert scaled units if the context requires it.
Why Dimensional Consistency Is Necessary but Not Sufficient
A dimensional check can reject an equation that adds incompatible quantities or gives a rate the wrong units. Passing the check is useful, but it does not prove that the equation is mathematically or scientifically correct. Different equations can have the same units while describing different behavior.
For example, two different formulas for a population’s rate might both produce people per hour. Units alone cannot decide which formula describes the population accurately. They check consistency of measurement, not the evidence behind a model or the truth of its predictions.
Common Mistake: Treating a Slope as Unitless
A common error is to identify the derivative’s units with the units of the dependent variable, forgetting to divide by the independent-variable units. For instance, if \(V\) is measured in cubic meters and \(t\) in seconds, the incorrect calculation is to say that \(dV/dt\) has units \(\text{m}^3\). The derivative describes volume change per time change, so the correct units are \(\text{m}^3/\text{s}\).
This mistake can also distort a slope-field interpretation. A label of \(3\) on axes measured in meters and seconds means \(3\ \text{m}/\text{s}\), not simply \(3\) meters. If the same graph uses minutes instead of seconds, the numerical rate for the same physical change must be expressed per minute. Always include both axis units before interpreting a label.
Check Your Understanding
For each question, identify the relevant units before deciding whether the equation or rate is consistent.
- If \(A\) is measured in square centimeters and \(t\) in minutes, what are the units of \(dA/dt\)?
- If \(dP/dt=kP\), where \(P\) is measured in people and \(t\) in days, what units must \(k\) have?
- In \(dy/dt=c+y\), can \(c\) and \(y\) be added if \(y\) is measured in meters and \(t\) in seconds? Explain what units \(c\) must have for the equation to be consistent.
- A slope-field graph measures \(y\) in thousands of units and \(t\) in hours. What physical rate does the label \(0.2\) represent in units per hour?
- Why does an equation passing a dimensional check not prove that it is an accurate model?
