Why Check the Units in a Formula?
A function formula can look algebraically neat and still fail to represent the quantity it is supposed to describe. One useful check is to track the units of its inputs, coefficients, and output. This is called a dimensional check.
For example, if \(C(t)\) represents a cost in dollars and \(t\) is measured in hours, then every term added to another term in the formula for \(C(t)\) must represent dollars. A term measured in dollars cannot be added to one measured in dollars per hour. The numerical calculation might be possible, but its result would not have a coherent meaning.
A dimensional check does not establish that a formula accurately describes a real situation. It can, however, rule out formulas whose terms cannot be combined consistently, and it can help locate a likely error.
Units in a Function Formula
Suppose an input \(x\) has units \(X\), and the output \(y=f(x)\) has units \(Y\). A coefficient in the formula may itself have units. Those units are what allow a term involving \(x\) to have the same units as the output.
In multiplication, units multiply along with numerical values. In division, units divide. For instance, multiplying a speed in miles per hour by a time in hours gives miles:
When a quantity is raised to a power, its units are raised to the same power. If \(t\) is measured in seconds, then \(t^2\) has units of seconds squared. A coefficient may carry units that cancel or adjust those powers.
For an output measured in units \(Y\), an input measured in units \(X\), and a term \(a x^k\), the coefficient \(a\) must have units \(Y/X^k\). This makes the whole term have output units:
Here, the brackets indicate the units of a quantity. The coefficient is not automatically unitless: its units depend on the input and the output.
Why Added Terms Must Have the Same Units
Addition combines quantities of the same kind. For example, adding two lengths is meaningful: \(1\) meter plus \(30\) centimeters can be computed after expressing both lengths in a common unit. Adding a length to a time is not meaningful as a physical quantity. Converting both to numerical values does not make them quantities of the same kind.
The same requirement applies term by term in a function formula. If \(f(x)\) is a distance, each term in a sum defining \(f(x)\) must be a distance. If a formula includes terms measured in different units, adding them does not produce the intended output quantity.
Worked Example: Checking a Cost Formula
Worked Example: Checking a Cost Formula
A repair shop charges a fixed fee of \(12\) dollars plus \(8\) dollars per hour. Let \(t\) be the repair time in hours, and let \(C(t)\) be the total cost in dollars. Check the formula and find the cost for a \(3\)-hour repair.
The hourly rate has units dollars per hour. The fixed fee has units dollars, so the proposed formula is
Since \(t\) is measured in hours, the first term has units
The second term is also measured in dollars. Thus the terms can be added, and the output has the stated units. For \(t=3\) hours,
Check: the calculated charge consists of \(24\) dollars for three hours of work and the \(12\)-dollar fixed fee. Both components are dollars, and their total is \(36\) dollars, as required.
Worked Example: Finding an Inconsistent Term
Worked Example: Finding an Inconsistent Term
Suppose \(D(t)\) is intended to be a distance in meters, and \(t\) is time in seconds. Consider the expression
Check the units of each term. The first term has units
The last term has units of meters. But the middle term has units
That is not a distance, so the three terms cannot be added as written. A likely repair is that the coefficient \(3\) should have units meters per second, rather than meters. With that change, the formula becomes
Now the middle term has units
so all terms have units of distance. For \(t=2\) seconds, substitute with units shown:
Check: each term in the corrected calculation is a distance in meters, and the result is in meters. The dimensional check identifies a plausible correction, although only the context or source of the formula can determine whether that correction is actually intended.
A General Rule for Coefficient Units
The previous example illustrates a reusable calculation. Suppose the output \(f(x)\) has units \(Y\), the input \(x\) has units \(X\), and a term has the form \(a x^k\). The units of \(x^k\) are \(X^k\). To make the product have units \(Y\), the coefficient must have units \(Y/X^k\).
Write down the units the entire formula is supposed to produce.
For a term involving \(x^k\), raise the input units to the \(k\)th power.
Choose units for the coefficient that make the product match the output units.
Verify that every term being added or subtracted has the same units.
For example, if \(f(x)\) is a cost in dollars and \(x\) is time in hours, a term \(a x^2\) requires \(a\) to have units dollars per hour squared. Then
The coefficient's units are determined by the term, not by the coefficient's numerical value. Writing only a number beside a variable can hide this important information in an applied formula.
Worked Example: Checking a Distance Formula
Worked Example: Checking a Distance Formula
A model for distance is given by \(s(t)=v t\), where \(v=60\) miles per hour and \(t\) is measured in hours. Check the units and find the distance after \(1.5\) hours.
The product has units
Thus the formula has the right output units for a distance. Substituting \(t=1.5\) hours gives
Check: the hours cancel, leaving miles, and \(60\times1.5=90\). The result is \(90\) miles, a distance in the required unit.
Unit Conversions Matter
Two rates can describe the same kind of quantity while using different time units. Before substituting a time into a rate, make sure those units match. For example, a rate in dollars per minute must be paired with time in minutes, or converted before it is used with time in hours.
A common mistake is to use a numerical rate without checking its unit. Suppose a service costs \(0.20\) dollars per minute, and the service lasts \(2\) hours. The calculation \(0.20\times2=0.40\) dollars is wrong because it treats \(2\) hours as \(2\) minutes. Convert first:
Now the cost is
The units cancel to leave dollars. This correction shows why the unit label must travel with the number throughout the calculation.
What a Dimensional Check Can and Cannot Show
If terms in a formula have incompatible units, the formula cannot represent the stated output quantity in the stated way. This makes dimensional analysis a useful error-detection tool. It can reveal, for instance, a missing time denominator in a rate or a missing power on a coefficient's units.
However, passing the check does not prove a formula is correct. Two different formulas can have consistent units while describing different relationships. A dimensional check cannot establish that the numerical coefficients are appropriate, that the formula fits observations, or that its assumptions match the situation. It is a necessary check on the structure of an applied formula, not a complete validation of the model.
Check Your Understanding
For each question, track the units as well as the numerical values.
- If \(t\) is in hours and \(C(t)\) is in dollars, what units must a coefficient \(a\) have in the term \(at\)?
- A distance formula includes a term \((4\text{ meters})t\), where \(t\) is in seconds. What units does this term have, and is it a distance?
- If a rate is \(72\) kilometers per hour, what time unit should be used directly with it to calculate a distance?
- Why is a formula with dimensionally consistent terms not automatically a correct model?
- A cost formula adds a fixed fee in dollars to a term measured in dollars per hour. What should you check or change before adding the terms?
