From a Rate Schedule to a Final Quantity
A quantity may change according to one rate during one stage, then follow a different rate during the next. A reservoir may fill, drain, and fill again. A stored material may gain mass in one period and lose mass in another. To find the final amount, it is not enough to look at the last rate: we must account for every stage’s contribution.
The key is to choose a time at which the quantity is known, then track each later change from that basepoint. A positive rate increases the quantity; a negative rate decreases it. The final value is the base value plus the signed contributions from all stages between the basepoint and the time of interest.
Rates, Stages, and Signed Accumulation
Suppose \(Q(t)\) is a quantity at time \(t\), and \(r(t)\) is its rate of change. The units of \(r(t)\) are the units of \(Q\) divided by the units of time. For example, if \(Q\) is measured in litres and time in minutes, \(r(t)\) is measured in litres per minute.
Over a time interval on which the rate is constant, the change in the quantity is the rate multiplied by the duration. A positive result represents an increase; a negative result represents a decrease. If the rate varies within a stage, its accumulated change is the definite integral of the rate over that stage. Thus, provided the rate is integrable over the interval, the accumulated change from time \(a\) to time \(b\) is \(\int_a^b r(t)\,dt\).
A piecewise rate is described by different formulas or values on different time intervals. The rate may jump at a stage boundary. For a piecewise-continuous rate, the definite integral exists on each finite stage, and the accumulation can be computed stage by stage. The value assigned to the rate at a single boundary time does not change the integral.
When a rate is constant on each stage, the integral on that stage is just a rate-times-duration calculation. When a rate varies linearly, the signed contribution can be found from the corresponding signed area, such as a triangle or trapezoid. These are two ways to evaluate the same accumulated change.
Why Stage Contributions Add
Suppose the time interval from \(a\) to \(c\) is divided at \(b\), where \(a<b<c\). The accumulated change from \(a\) to \(c\) is the sum of the changes from \(a\) to \(b\) and from \(b\) to \(c\). This additivity is what lets us handle a multistage problem without combining all the rate formulas into one expression.
For known \(Q(a)\), the change through time \(c\) is \(\int_a^c r(t)\,dt\).
The interval from \(a\) to \(c\) consists of the intervals from \(a\) to \(b\) and from \(b\) to \(c\).
Therefore \(\int_a^c r(t)\,dt=\int_a^b r(t)\,dt+\int_b^c r(t)\,dt\), with negative contributions retained as negative.
This identity follows directly from the definition of the definite integral as signed accumulation: splitting an interval separates its accumulated change into the contributions on the two subintervals. The common boundary \(b\) is counted as an endpoint of both intervals, but a single point contributes no area. For finitely many stages, repeated splitting gives the total change as the sum of the stage integrals. Equivalently, if \(a=t_0<t_1<\cdots<t_n=c\), then
Adding the starting quantity to this sum gives the final quantity. The formula works whether a stage raises the quantity, lowers it, or contributes zero.
Worked Examples
Worked Example: A Reservoir with Three Constant Rates
A reservoir contains \(120\text{ L}\) at time \(t=0\). Its net rate of change is \(6\text{ L/min}\) from \(0\) to \(2\) minutes, \(-2\text{ L/min}\) from \(2\) to \(5\) minutes, and \(3\text{ L/min}\) from \(5\) to \(7\) minutes. Find the amount at \(t=7\) minutes.
First, record the base value and calculate each duration. The stage durations are \(2-0=2\text{ min}\), \(5-2=3\text{ min}\), and \(7-5=2\text{ min}\).
The first stage contributes $$ (6\text{ L/min})(2\text{ min})=12\text{ L}. $$ The reservoir contains \(120+12=132\text{ L}\) at \(t=2\) minutes.
The second stage contributes $$ (-2\text{ L/min})(3\text{ min})=-6\text{ L}. $$ At \(t=5\) minutes, the amount is \(132-6=126\text{ L}\).
The third stage contributes $$ (3\text{ L/min})(2\text{ min})=6\text{ L}. $$ Therefore the amount at \(t=7\) minutes is $$ 120\text{ L}+12\text{ L}-6\text{ L}+6\text{ L}=132\text{ L}. $$ As a check, the stage-by-stage amounts are \(120\text{ L}\), \(132\text{ L}\), \(126\text{ L}\), and \(132\text{ L}\); each change matches its rate and duration. The final amount is \(132\text{ L}\), which is above the starting amount by the net increase of \(12\text{ L}\).
Worked Example: Starting from a Later Basepoint
A storage bin contains \(50\text{ kg}\) of material at time \(t=4\) hours. From \(4\) to \(6\) hours, its net rate of change is \(3\text{ kg/h}\); from \(6\) to \(9\) hours, the rate is \(-4\text{ kg/h}\); and from \(9\) to \(10\) hours, the rate is \(1.5\text{ kg/h}\). Find the amount at \(t=10\) hours.
The known basepoint is \(t=4\) hours, not \(t=0\). Only changes after \(t=4\) are needed. The three durations are \(6-4=2\text{ h}\), \(9-6=3\text{ h}\), and \(10-9=1\text{ h}\).
The first contribution is $$ (3\text{ kg/h})(2\text{ h})=6\text{ kg}, $$ so the amount at \(t=6\) hours is \(50+6=56\text{ kg}\).
The second contribution is $$ (-4\text{ kg/h})(3\text{ h})=-12\text{ kg}, $$ so the amount at \(t=9\) hours is \(56-12=44\text{ kg}\).
The third contribution is $$ (1.5\text{ kg/h})(1\text{ h})=1.5\text{ kg}, $$ so the final amount is \(44+1.5=45.5\text{ kg}\). Combining the changes gives the same result: $$ 50\text{ kg}+6\text{ kg}-12\text{ kg}+1.5\text{ kg}=45.5\text{ kg}. $$ The amount has decreased by \(4.5\text{ kg}\) overall. Checking forward from the basepoint gives \(50\text{ kg}\to56\text{ kg}\to44\text{ kg}\to45.5\text{ kg}\), consistent with each stage’s contribution.
Worked Example: A Rate That Varies Within Stages
A tank’s liquid amount is \(1.2\text{ L}\) at \(t=0\) hours. Its net rate is linear from \(0\text{ L/h}\) at \(t=0\) to \(0.6\text{ L/h}\) at \(t=2\), constant at \(-0.4\text{ L/h}\) from \(t=2\) to \(t=4\), and linear from \(-0.4\text{ L/h}\) at \(t=4\) to \(0.4\text{ L/h}\) at \(t=5\). Find the amount at \(t=5\) hours.
For the first stage, the rate graph forms a triangle above the time axis. Its duration is \(2\text{ h}\), and its average rate is $$ \frac{0+0.6}{2}=0.3\text{ L/h}. $$ Its contribution is $$ (0.3\text{ L/h})(2\text{ h})=0.6\text{ L}. $$ Thus the amount at \(t=2\) is \(1.2+0.6=1.8\text{ L}\).
For the second stage, the constant rate contributes $$ (-0.4\text{ L/h})(4-2)\text{ h} =(-0.4\text{ L/h})(2\text{ h}) =-0.8\text{ L}. $$ The amount at \(t=4\) is \(1.8-0.8=1.0\text{ L}\).
For the last stage, the average of the endpoint rates is $$ \frac{-0.4+0.4}{2}=0\text{ L/h}. $$ Over the duration \(5-4=1\text{ h}\), the signed contribution is \((0\text{ L/h})(1\text{ h})=0\text{ L}\). This stage has a negative part and a positive part of equal area, so they cancel. The amount at \(t=5\) is therefore \(1.0+0=1.0\text{ L}\).
Checking by adding all contributions to the base value gives \(1.2\text{ L}+0.6\text{ L}-0.8\text{ L}+0\text{ L}=1.0\text{ L}\). Every rate-times-duration or area calculation has units of litres, as required for a change in the tank’s liquid amount.
Units and a Common Mistake
Units provide a useful check on the calculation. A rate has units of quantity per time, and a duration has units of time. Their product has units of quantity. For example, \((\text{L/min})(\text{min})=\text{L}\). Each stage contribution must therefore have the same units as the quantity being tracked, so it can be added to the base value.
A common mistake is to calculate the numerical duration but omit its units when writing a rate-times-duration product. For the reservoir’s second stage, the endpoints are \(2\) minutes and \(5\) minutes; their difference is a duration of \(3\) minutes. If the rate is \(-2\text{ L/min}\), writing \((-2\text{ L/min})(5-2)=-6\text{ L}\) omits the duration’s units. The corrected calculation is $$ (-2\text{ L/min})((5-2)\text{ min}) =(-2\text{ L/min})(3\text{ min}) =-6\text{ L}. $$ The endpoints are both times, and their difference is a duration; the missing units belong on that duration in the product.
Another frequent error is to treat a negative rate as a positive contribution because the quantity itself is nonnegative. A tank may hold a positive volume while its net rate is negative. The rate’s sign describes the direction of change, not whether the amount is below zero. Keep each stage contribution signed, and interpret the final amount only after all contributions have been added to the base value.
A Reliable Method for Contextual Problems
Organize a multistage accumulation problem before doing arithmetic. The following steps make the model and its interpretation clear:
For example, \(Q\) might be a volume measured in litres or a mass measured in kilograms.
Record the time and quantity known there. Do not assume the basepoint is time zero if the problem gives a value at another time.
For each stage, identify its endpoints, rate information, and signed contribution.
Use a rate times duration for a constant rate, or signed area/integration for a varying rate. Keep increases positive and decreases negative.
Confirm that every contribution has quantity units, add them to the base value, and state what the final number means in context.
It can also help to keep a running total after each stage, as in the examples. The running total is not a new rate calculation; it is a bookkeeping check that makes it easier to detect an omitted stage or an incorrect sign. If the final amount has a physical constraint, such as a reservoir volume that cannot be negative, compare the calculated amounts with that constraint and decide whether the model or its assumptions need further attention.
Why the Basepoint and Signs Matter
The basepoint anchors the accumulation to an actual quantity. A rate schedule alone describes changes, not the absolute amount: adding the same net change to two different starting amounts produces two different final amounts. Conversely, once a base value is known, the stage contributions determine how it changes.
Signed accumulation also distinguishes net change from total movement. A stage with a negative rate reverses some earlier increase; it does not erase the fact that the increase occurred. In Example 1, the reservoir gained \(12\text{ L}\), then lost \(6\text{ L}\), then gained \(6\text{ L}\). The net change is \(12\text{ L}\), even though the total amount moved into or out of the reservoir over the stages is \(24\text{ L}\). The latter counts the magnitudes of the changes, not their signed sum.
For a contextual answer, state both the numerical value and what it represents. “\(132\text{ L}\) at \(t=7\) minutes” is more informative than “132,” because the units and time identify the physical quantity being reported.
Check Your Understanding
Use signed stage contributions and include units in your reasoning.
- A container holds \(30\text{ L}\) at \(t=1\) minute. Its rate is \(4\text{ L/min}\) from \(1\) to \(3\) minutes and \(-1\text{ L/min}\) from \(3\) to \(5\) minutes. What amount does the model give at \(t=5\) minutes?
- A quantity is known to be \(18\text{ kg}\) at \(t=6\) hours. Why can its value at \(t=9\) be found without knowing its value at \(t=0\), provided the rate is known from \(t=6\) to \(t=9\)?
- What are the units of the contribution from a rate measured in \(\text{m}^3/\text{min}\) acting over \(8\text{ min}\)?
- A rate is negative during one stage. What does its sign tell you, and what does it not tell you about the quantity itself?
- Explain why the accumulated change across three consecutive stages is the sum of the three signed stage contributions.
