The Central Idea
A derived unit is a unit formed by combining more basic units according to the mathematical definition of a quantity.
For example, speed is defined as distance divided by time:
\[ v=\frac{d}{t}. \]If distance is measured in meters and time in seconds, then speed is measured in
\[ \boxed{\frac{\text{m}}{\text{s}}}. \]The unit follows directly from the formula.
Base Units and Derived Units
A base unit measures a fundamental type of quantity directly. A derived unit is built from one or more base units.
Examples of commonly used base units include:
- meter \((\text{m})\) for length;
- second \((\text{s})\) for time;
- kilogram \((\text{kg})\) for mass;
- kelvin \((\text{K})\) for temperature;
- ampere \((\text{A})\) for electric current.
Derived units include meters per second, kilograms per cubic meter, newtons, joules, watts, and many others.
Derived Units Come from Definitions
The unit of a derived quantity is not arbitrary. It comes from the equation used to define the quantity.
If
\[ Q=\frac{A}{B}, \]then the unit of \(Q\) is obtained by dividing the unit of \(A\) by the unit of \(B\).
If
\[ Q=AB, \]then the unit of \(Q\) is obtained by multiplying the units.
Area as a Derived Quantity
The area of a rectangle is
\[ A=lw. \]If both length and width are measured in meters, then
\[ [A] = \text{m}\cdot\text{m} = \boxed{\text{m}^2}. \]Square meters are therefore a derived unit formed from two factors of length.
Volume as a Derived Quantity
The volume of a rectangular box is
\[ V=lwh. \]If each dimension is measured in meters, then
\[ [V] = \text{m}\cdot\text{m}\cdot\text{m} = \boxed{\text{m}^3}. \]Cubic meters arise because volume depends on three independent length directions.
Speed
Speed is distance divided by time:
\[ v=\frac{d}{t}. \]If distance is measured in meters and time in seconds, then
\[ [v] = \frac{\text{m}}{\text{s}}. \]Other valid speed units include
\[ \frac{\text{km}}{\text{h}}, \qquad \frac{\text{mi}}{\text{h}}, \qquad \frac{\text{ft}}{\text{s}}. \]All represent the same dimensional pattern:
\[ \frac{\text{length}}{\text{time}}. \]Acceleration
Acceleration is the rate of change of velocity with respect to time:
\[ a=\frac{\Delta v}{\Delta t}. \]If velocity is measured in meters per second and time in seconds, then
\[ [a] = \frac{\text{m}/\text{s}}{\text{s}} = \boxed{\frac{\text{m}}{\text{s}^2}}. \]The squared time unit appears because time enters the denominator twice.
Density
Mass density is defined by
\[ \rho=\frac{m}{V}. \]If mass is measured in kilograms and volume in cubic meters, then
\[ [\rho] = \frac{\text{kg}}{\text{m}^3}. \]Thus,
\[ \boxed{\text{density unit}=\frac{\text{kg}}{\text{m}^3}}. \]Flow Rate
A volumetric flow rate measures volume per unit time.
\[ Q=\frac{V}{t}. \]If volume is measured in cubic meters and time in seconds, then
\[ [Q] = \boxed{\frac{\text{m}^3}{\text{s}}}. \]Similar units appear in water systems, ventilation, chemical processing, and many transportation models.
Frequency
Frequency measures repetitions or cycles per unit time.
\[ f=\frac{\text{cycles}}{\text{time}}. \]A cycle count is dimensionless, so the dimensional pattern is
\[ [f]=T^{-1}. \]The SI derived unit is the hertz:
\[ \boxed{1\text{ Hz}=1\text{ s}^{-1}}. \]Force
Force is related to mass and acceleration by
\[ F=ma. \]If mass is measured in kilograms and acceleration in meters per second squared, then
\[ [F] = \text{kg}\cdot\frac{\text{m}}{\text{s}^2}. \]This combination is assigned the special unit name newton:
\[ \boxed{1\text{ N} = 1\frac{\text{kg}\cdot\text{m}}{\text{s}^2}}. \]Special Names for Derived Units
Some derived units are used so often that they receive special names.
A named unit does not change its underlying structure.
For example,
\[ 1\text{ N} = 1\frac{\text{kg}\cdot\text{m}}{\text{s}^2}. \]The name "newton" is simply a compact way to refer to that derived combination.
Pressure
Pressure is force divided by area:
\[ P=\frac{F}{A}. \]Using newtons and square meters gives
\[ [P] = \frac{\text{N}}{\text{m}^2}. \]This derived unit is called the pascal:
\[ \boxed{1\text{ Pa}=1\frac{\text{N}}{\text{m}^2}}. \]Expanding the newton gives
\[ 1\text{ Pa} = 1\frac{\text{kg}}{\text{m}\cdot\text{s}^2}. \]Work and Energy
In a simple constant-force setting, work is
\[ W=Fd. \]If force is measured in newtons and distance in meters, then
\[ [W] = \text{N}\cdot\text{m}. \]This unit is called the joule:
\[ \boxed{1\text{ J}=1\text{ N}\cdot\text{m}}. \]In base units,
\[ 1\text{ J} = 1\frac{\text{kg}\cdot\text{m}^2}{\text{s}^2}. \]Power
Power measures energy transferred or work performed per unit time:
\[ P=\frac{W}{t}. \]If work is measured in joules and time in seconds, then
\[ [P] = \frac{\text{J}}{\text{s}}. \]This derived unit is called the watt:
\[ \boxed{1\text{ W}=1\frac{\text{J}}{\text{s}}}. \]Momentum
Linear momentum is
\[ p=mv. \]Therefore,
\[ [p] = \text{kg}\cdot\frac{\text{m}}{\text{s}} = \boxed{\frac{\text{kg}\cdot\text{m}}{\text{s}}}. \]Momentum is a useful example of a derived quantity that is often written directly in base-unit form rather than with a separate everyday unit name.
Dynamic Viscosity
In fluid models, dynamic viscosity is often represented by \(\mu\). Its SI unit can be written as
\[ \text{Pa}\cdot\text{s}. \]Expanding the pascal gives
\[ \text{Pa}\cdot\text{s} = \frac{\text{kg}}{\text{m}\cdot\text{s}^2}\cdot\text{s} = \boxed{\frac{\text{kg}}{\text{m}\cdot\text{s}}}. \]This shows how named derived units can be reduced to base-unit combinations.
Electric Charge
Electric charge may be modeled as current multiplied by time:
\[ Q=It. \]With current measured in amperes and time in seconds,
\[ [Q]=\text{A}\cdot\text{s}. \]This unit is called the coulomb:
\[ \boxed{1\text{ C}=1\text{ A}\cdot\text{s}}. \]Rates Are Derived Units
Whenever a quantity is described as "per" another quantity, a quotient of units is usually involved.
Examples include:
- miles per hour;
- dollars per unit;
- people per square kilometer;
- liters per minute;
- kilowatt-hours per household;
- defects per thousand units;
- revenue per employee.
Derived-unit reasoning therefore extends far beyond physics.
Population Density
Population density is commonly modeled as
\[ D=\frac{P}{A}, \]where \(P\) is population and \(A\) is area.
If population is measured in people and area in square kilometers, then
\[ [D] = \boxed{\frac{\text{people}}{\text{km}^2}}. \]This is structurally the same kind of unit construction used for physical density.
Unit Price
A unit price is total cost divided by quantity purchased:
\[ u=\frac{C}{q}. \]If cost is measured in dollars and quantity in kilograms, then
\[ [u] = \boxed{\frac{\$}{\text{kg}}}. \]The same reasoning applies to cost per item, revenue per customer, or dollars per labor hour.
Productivity Rates
Suppose a factory produces \(Q\) units in \(t\) hours.
\[ R=\frac{Q}{t}. \]Then the production rate has units
\[ \boxed{\frac{\text{units}}{\text{hour}}}. \]If labor input is measured instead, a productivity measure might use
\[ \frac{\text{units produced}}{\text{labor hour}}. \]The unit itself tells us how the performance measure is constructed.
Rates of Change Carry Derived Units
A derivative automatically creates a quotient of units.
If \(y\) is measured in dollars and \(t\) in months, then
\[ \frac{dy}{dt} \]has units
\[ \boxed{\frac{\text{dollars}}{\text{month}}}. \]The derivative tells us how many units of the output change per unit of the input.
Second Derivatives Create Higher-Order Units
Differentiating with respect to time twice places time in the denominator twice.
If position \(x\) is measured in meters, then
\[ \frac{dx}{dt} \]has units
\[ \frac{\text{m}}{\text{s}}, \]while
\[ \frac{d^2x}{dt^2} \]has units
\[ \boxed{\frac{\text{m}}{\text{s}^2}}. \]Integrals Also Produce Derived Units
Integration multiplies the units of the integrand by the units of the differential variable.
If a flow rate \(Q(t)\) is measured in cubic meters per second, then
\[ \int Q(t)\,dt \]has units
\[ \frac{\text{m}^3}{\text{s}}\cdot\text{s} = \boxed{\text{m}^3}. \]The integral represents accumulated volume.
Graph Slopes Have Derived Units
If a graph plots quantity \(y\) against quantity \(x\), then its slope has units
\[ \frac{\text{unit of }y}{\text{unit of }x}. \]For example, a cost-versus-output graph may have slope measured in
\[ \frac{\$}{\text{unit produced}}. \]A distance-versus-time graph may have slope measured in
\[ \frac{\text{m}}{\text{s}}. \]Reading the units often reveals the meaning of the slope before any calculation is performed.
Unit Cancellation
Derived-unit calculations often simplify by canceling identical units that appear in a numerator and denominator.
Suppose
\[ v=8\frac{\text{m}}{\text{s}} \]for
\[ t=6\text{ s}. \]Then
\[ vt = 8\frac{\text{m}}{\text{s}} \cdot 6\text{ s} = 48\text{ m}. \]The second unit cancels, leaving the derived result in meters.
Converting Derived Units
Each part of a derived unit must be converted correctly.
Convert
\[ 10\frac{\text{m}}{\text{s}} \]to kilometers per hour.
\[ 10\frac{\text{m}}{\text{s}} \cdot \frac{1\text{ km}}{1000\text{ m}} \cdot \frac{3600\text{ s}}{1\text{ h}}. \]The units reduce to
\[ \frac{\text{km}}{\text{h}}, \]and numerically,
\[ \boxed{10\frac{\text{m}}{\text{s}} = 36\frac{\text{km}}{\text{h}}}. \]Converting Density Units
Suppose a density is given in
\[ \frac{\text{g}}{\text{cm}^3}. \]Converting to kilograms per cubic meter requires converting both the numerator and the cubed denominator.
Since
\[ 1\text{ g}=10^{-3}\text{ kg} \]and
\[ 1\text{ cm}=10^{-2}\text{ m}, \]then
\[ 1\text{ cm}^3=10^{-6}\text{ m}^3. \]Therefore,
\[ \boxed{ 1\frac{\text{g}}{\text{cm}^3} = 1000\frac{\text{kg}}{\text{m}^3} }. \]Derived Units Can Be Expanded
A named derived unit can be expanded into simpler units whenever the underlying structure is useful.
For example,
\[ 1\text{ W} = 1\frac{\text{J}}{\text{s}}. \]Since
\[ 1\text{ J} = 1\frac{\text{kg}\cdot\text{m}^2}{\text{s}^2}, \]we obtain
\[ \boxed{ 1\text{ W} = 1\frac{\text{kg}\cdot\text{m}^2}{\text{s}^3} }. \]Derived Units Can Be Compressed
The reverse process is also useful.
If a calculation produces
\[ \frac{\text{kg}\cdot\text{m}}{\text{s}^2}, \]we can recognize that this is
\[ \boxed{\text{N}}. \]Recognizing named unit structures makes formulas easier to interpret.
Units as Structural Information
Units provide information about what a quantity means.
A result measured in
\[ \frac{\text{kg}}{\text{m}^3} \]strongly suggests a density-like quantity.
A result measured in
\[ \frac{\text{J}}{\text{s}} \]suggests power.
A result measured in
\[ \frac{\$}{\text{hour}} \]suggests a rate of cost, revenue, or compensation.
Derived Units Help Interpret Parameters
Suppose
\[ y=mx+b. \]If \(y\) is measured in dollars and \(x\) in units sold, then \(m\) must be measured in
\[ \frac{\$}{\text{unit sold}}. \]The units tell us that \(m\) represents a marginal amount of money per unit of output.
Derived Units Help Check Equations
Suppose
\[ E=Pt. \]If power is measured in watts, then
\[ \text{W}\cdot\text{s} = \frac{\text{J}}{\text{s}}\cdot\text{s} = \text{J}. \]The resulting unit is energy, so the formula is dimensionally plausible.
Common Error: Dropping the Unit Exponent
If a square has side length
\[ 4\text{ m}, \]then its area is
\[ (4\text{ m})^2 = 16\text{ m}^2. \]Writing \(16\text{ m}\) would be incorrect because area requires a squared length unit.
Common Error: Treating a Quotient Unit as Decoration
The unit
\[ \frac{\text{m}}{\text{s}} \]is an actual quotient, not merely a label.
That is why
\[ \frac{\text{m}}{\text{s}} \cdot \text{s} = \text{m}. \]Treating units algebraically makes many modeling relationships easier to understand and verify.
Common Error: Converting Only Part of a Derived Unit
If speed is given in miles per hour and the model requires meters per second, both the distance unit and the time unit must be converted.
Converting miles to meters while leaving hours unchanged produces meters per hour, not meters per second.
Every unit appearing in a derived quantity must be handled explicitly.
Common Error: Ignoring Powers in the Denominator
Acceleration in
\[ \frac{\text{m}}{\text{s}^2} \]contains a squared time unit.
Converting seconds to minutes must therefore account for the square:
\[ \left(\frac{60\text{ s}}{1\text{ min}}\right)^2. \]A power on a unit applies to its conversion factor as well.
A Reliable Procedure for Building a Derived Unit
- Write the defining formula for the quantity.
- Identify the unit of each variable.
- Replace each variable with its unit.
- Multiply, divide, and exponentiate the units exactly as the formula does.
- Cancel compatible units when appropriate.
- Rewrite the result in a standard or named unit if useful.
- Check that the result matches the expected dimensional meaning.
Example: Building a Force Unit from Scratch
Start with
\[ F=ma. \]Suppose
\[ [m]=\text{kg} \]and
\[ [a]=\frac{\text{m}}{\text{s}^2}. \]Then
\[ [F] = \text{kg}\cdot\frac{\text{m}}{\text{s}^2} = \frac{\text{kg}\cdot\text{m}}{\text{s}^2}. \]Recognize the standard name:
\[ \boxed{ \frac{\text{kg}\cdot\text{m}}{\text{s}^2} = \text{N} }. \]Example: Building an Energy Unit from Scratch
Begin with
\[ W=Fd. \]Substitute the force unit:
\[ [W] = \frac{\text{kg}\cdot\text{m}}{\text{s}^2} \cdot \text{m}. \]Therefore,
\[ [W] = \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2}. \]This is the joule:
\[ \boxed{ 1\text{ J} = 1\frac{\text{kg}\cdot\text{m}^2}{\text{s}^2} }. \]Example: Building a Power Unit from Scratch
Power is
\[ P=\frac{W}{t}. \]Since
\[ [W] = \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2}, \]dividing by seconds gives
\[ [P] = \frac{\text{kg}\cdot\text{m}^2}{\text{s}^3}. \]This is the watt:
\[ \boxed{ 1\text{ W} = 1\frac{\text{kg}\cdot\text{m}^2}{\text{s}^3} }. \]Derived Units in Modeling
Derived units help connect formulas to interpretation.
They tell us:
- what a computed quantity represents;
- how variables combine;
- how a slope or rate should be interpreted;
- which unit conversions are required;
- whether an equation is dimensionally plausible;
- whether an output has the expected physical or practical meaning.
Practice 1: Area
A rectangle has length \(8\text{ m}\) and width \(3\text{ m}\). Find its area and include the correct derived unit.
Practice 2: Speed
A vehicle travels \(150\text{ km}\) in \(3\text{ h}\). Find its average speed.
Practice 3: Acceleration Unit
If velocity is measured in meters per second and time in seconds, derive the unit of acceleration.
Practice 4: Density
A sample has mass \(12\text{ kg}\) and volume \(0.004\text{ m}^3\). Find its density.
Practice 5: Force
A \(5\text{ kg}\) mass accelerates at \(4\text{ m}/\text{s}^2\). Find the force.
Practice 6: Work
A constant force of \(20\text{ N}\) moves an object \(6\text{ m}\) in the direction of the force. Find the work performed.
Practice 7: Power
A machine performs \(900\text{ J}\) of work in \(3\text{ s}\). Find its average power.
Practice 8: Unit Interpretation
What kind of quantity is suggested by the unit
\[ \frac{\text{people}}{\text{km}^2}? \]Practice 9: Convert Speed
Convert
\[ 5\frac{\text{m}}{\text{s}} \]to kilometers per hour.
Practice 10: Expand a Watt
Express one watt entirely in kilograms, meters, and seconds.
Practice Answers
Practice 1
\[ A=lw=(8\text{ m})(3\text{ m}) =\boxed{24\text{ m}^2}. \]Practice 2
\[ v=\frac{150\text{ km}}{3\text{ h}} =\boxed{50\frac{\text{km}}{\text{h}}}. \]Practice 3
\[ [a] = \frac{\text{m}/\text{s}}{\text{s}} = \boxed{\frac{\text{m}}{\text{s}^2}}. \]Practice 4
\[ \rho = \frac{12\text{ kg}}{0.004\text{ m}^3} = \boxed{3000\frac{\text{kg}}{\text{m}^3}}. \]Practice 5
\[ F = ma = (5\text{ kg}) \left( 4\frac{\text{m}}{\text{s}^2} \right) = \boxed{20\text{ N}}. \]Practice 6
\[ W = Fd = (20\text{ N})(6\text{ m}) = \boxed{120\text{ J}}. \]Practice 7
\[ P = \frac{900\text{ J}}{3\text{ s}} = \boxed{300\text{ W}}. \]Practice 8
The unit represents a population per unit area, so it suggests population density.
Practice 9
\[ 5\frac{\text{m}}{\text{s}} \cdot \frac{1\text{ km}}{1000\text{ m}} \cdot \frac{3600\text{ s}}{1\text{ h}} = \boxed{18\frac{\text{km}}{\text{h}}}. \]Practice 10
\[ 1\text{ W} = 1\frac{\text{J}}{\text{s}} = \boxed{ 1\frac{\text{kg}\cdot\text{m}^2}{\text{s}^3} }. \]What to Remember
Derived units are created by applying algebra to measurement units.
- multiplication of quantities multiplies their units;
- division of quantities divides their units;
- powers apply to units as well as numbers;
- rates naturally produce quotient units;
- derivatives produce output-units per input-unit;
- integrals multiply by the unit of the integration variable;
- named units such as newtons, joules, pascals, and watts can be expanded into base units;
- units help reveal the meaning and plausibility of a model.
The most useful habit is: let the equation tell you what the unit must be.
Where We Go Next
Derived units show how measured quantities combine. The next lesson, Parameters and Coefficients, shifts attention from measured variables to the constants that shape a model, including rates, scaling factors, intercepts, efficiencies, response strengths, and other quantities that control mathematical behavior.