From a Rate to a Quantity
A rate describes how a quantity changes as an input changes. For example, velocity describes how position changes with time, and a flow rate describes how the volume of liquid changes with time. An antiderivative reverses this relationship: it gives a quantity whose derivative is the stated rate.
Units help make this relationship precise. If time is measured in seconds and velocity in metres per second, then position is measured in metres. More generally, the units of a rate multiplied by the units of its input give the units of the quantity being recovered.
This is useful in applications because a rate alone does not specify the quantity at any particular time. For example, knowing a velocity function describes how position changes, but a starting position is also needed to determine position itself.
Why the Units Multiply
A derivative is a limit of change in the output divided by change in the input. If \(Q\) has units of litres and \(t\) has units of hours, then a difference quotient for \(Q'(t)\) has units of litres per hour. Reversing this relationship, a rate measured in litres per hour, multiplied by a time measured in hours, gives a change measured in litres.
For a polynomial rate, each term must have the same units as the rate. If \(t\) is measured in seconds and a velocity is written as \(v(t)=at+b\), then \(b\) has units of metres per second, while \(a\) has units of metres per second squared. The factor \(t\) supplies one second, so \(at\) also has units of metres per second.
After taking an antiderivative, each term has units of position. The integration constant has those same units: it is added to position, not to velocity. This is more than a bookkeeping convention; adding quantities with different units does not make a meaningful model.
Initial Values Select a Particular Antiderivative
Suppose \(F\) is an antiderivative of a rate \(r\), so \(F'(t)=r(t)\). Any quantity \(Q\) with the same rate satisfies \(Q'(t)=r(t)\) as well. Thus \(Q\) and \(F\) are antiderivatives of the same function. On an interval, they differ by a constant. The initial value determines that constant.
If the condition is \(Q(t_0)=Q_0\), substitute \(t_0\) into this expression and solve for \(C\). The units of \(C\) must match the units of \(Q\) and \(Q_0\). Checking the result requires two things: differentiating \(Q\) must recover the rate, and evaluating \(Q\) at \(t_0\) must recover \(Q_0\).
Worked Example: Recovering Position from Velocity
A particle moves along a line. Time \(t\) is measured in seconds, and its velocity is \(v(t)=(2\text{ m/s}^2)t+3\text{ m/s}\). Its position at time zero is \(4\text{ m}\). Find its position function \(s(t)\).
Position satisfies \(s'(t)=v(t)\). Integrate each term with respect to \(t\). Since the coefficient of \(t\) is \(2\text{ m/s}^2\), its antiderivative is \((1\text{ m/s}^2)t^2\). The constant velocity term has antiderivative \((3\text{ m/s})t\). Therefore,
Each displayed term has units of metres, so \(C\) must also have units of metres. Apply the initial condition \(s(0)=4\text{ m}\):
Thus \(C=4\text{ m}\), and the position is
Check the rate by differentiating: \(s'(t)=(2\text{ m/s}^2)t+3\text{ m/s}=v(t)\). Check the initial value: \(s(0)=0+0+4\text{ m}=4\text{ m}\). Both conditions hold.
Rates Can Describe Other Quantities
The same reasoning applies whenever the given function is the rate of a quantity with respect to the input. The units do not have to be metres and seconds. A liquid’s volume might be measured in litres while time is measured in hours; an antiderivative of its flow rate then has units of litres.
It also matters what kind of rate is given. Velocity is a signed rate of change of position. Its antiderivative gives position, not necessarily total distance traveled. If velocity is negative during some interval, position decreases there. Total distance traveled requires different information about the motion than simply evaluating the position antiderivative at the endpoints.
Worked Example: Finding a Volume from a Flow Rate
A tank receives liquid at rate \(r(t)=(4\text{ L/h}^2)t+3\text{ L/h}\), where \(t\) is measured in hours and \(r(t)\) in litres per hour. At \(t=2\text{ h}\), the tank contains \(10\text{ L}\). Find the volume \(V(t)\).
The volume satisfies \(V'(t)=r(t)\). Integrating gives
Both terms shown have units of litres, so \(C\) is measured in litres. Use \(V(2\text{ h})=10\text{ L}\):
Therefore \(C=10\text{ L}-14\text{ L}=-4\text{ L}\), and
Differentiate: \(V'(t)=(4\text{ L/h}^2)t+3\text{ L/h}\), which is the stated flow rate. Evaluate at \(t=2\text{ h}\): \(V(2\text{ h})=8\text{ L}+6\text{ L}-4\text{ L}=10\text{ L}\), as required.
Using a Nonzero Starting Time
An initial condition need not be given at time zero. Substitute the specified time wherever the input appears, keeping its units consistent. When the time is nonzero, the initial value generally determines \(C\) only after all the other terms have been evaluated there.
Worked Example: Recovering Velocity from Acceleration
An object has acceleration \(a(t)=(6\text{ m/s}^3)t-4\text{ m/s}^2\), with \(t\) in seconds. Its velocity at \(t=1\text{ s}\) is \(2\text{ m/s}\). Find \(v(t)\).
Acceleration is the derivative of velocity, so \(v'(t)=a(t)\). Integrate:
Both variable terms have units of metres per second, so \(C\) has units of metres per second. Substitute \(t=1\text{ s}\) and \(v(1\text{ s})=2\text{ m/s}\):
Thus \(2\text{ m/s}=-1\text{ m/s}+C\), so \(C=3\text{ m/s}\). The particular velocity is
Differentiate to check: \(v'(t)=(6\text{ m/s}^3)t-4\text{ m/s}^2=a(t)\). At \(t=1\text{ s}\), \(v(1\text{ s})=3\text{ m/s}-4\text{ m/s}+3\text{ m/s}=2\text{ m/s}\). The rate and initial value are both satisfied.
A Unit Mistake to Avoid
A common error is to add a rate directly to a quantity. Suppose a particle has constant velocity \(3\text{ m/s}\) and initial position \(5\text{ m}\). The expression \(3+5=8\) does not give a position: the first number represents a rate, measured in metres per second, while the second is measured in metres.
To recover position, integrate the velocity with respect to time. The result is \(s(t)=(3\text{ m/s})t+5\text{ m}\), where \(t\) is in seconds. The product \((3\text{ m/s})t\) has units of metres and can be added to the initial position. Differentiating gives \(s'(t)=3\text{ m/s}\), and substituting \(t=0\) gives \(s(0)=5\text{ m}\).
Always check that every term in a quantity model has the same units. Then check the model mathematically: its derivative should be the specified rate, and it should satisfy the given initial value.
Check Your Understanding
For each question, pay attention to both the calculus and the units.
- If a rate is measured in kilograms per minute and the input is time in minutes, what are the units of an antiderivative?
- A position function has derivative \(v(t)\). What physical quantity does the antiderivative represent, and what extra information selects a particular position function?
- A quantity \(Q\) has rate \(Q'(t)=6\text{ L/h}\), with \(t\) measured in hours. What units must the integration constant in \(Q(t)=6t+C\) have?
- Why is it invalid to add a velocity measured in metres per second directly to a position measured in metres?
- What two checks should be made after finding a particular antiderivative from a rate and an initial value?
