From Uncertain Rates to a Range of Accumulations
A rate tells us how quickly a quantity changes. If the rate is measured with error, its accumulated effect is uncertain too. The useful conclusion is usually not one exact total, but a lower and an upper estimate that the true total must lie between.
The key is to express the measurement uncertainty as bounds on the true rate throughout the time interval of interest. Once those bounds are available, they can be accumulated. A rate uncertainty matters more over a longer interval, so the duration of each interval must be included in the calculation.
Describe the Measurement Uncertainty
Let \(r(t)\) be the true rate of change of a quantity, and let \(m(t)\) be the measured rate. Suppose the measurement error is known to be no larger than \(\varepsilon(t)\), where \(\varepsilon(t)\ge 0\). This means that at each relevant time, the true rate differs from the measured rate by at most \(\varepsilon(t)\).
Here, “pointwise” means that the stated bounds hold at every time in the interval (or except at isolated jump points, which do not change the accumulated value). The bounds must describe the rate throughout the interval, not just at a few measurement times.
For a rate \(r(t)\) on \([a,b]\), the accumulated change is $$ \Delta Q=\int_a^b r(t)\,dt. $$ If the quantity has initial value \(Q(a)\), its later value is \(Q(b)=Q(a)+\Delta Q\). The rate may be positive or negative: a negative rate contributes a decrease, so its accumulated contribution is negative.
The Accumulation Bounds
Because the true rate lies between \(L(t)\) and \(U(t)\), its accumulated change lies between the accumulations of those rate bounds.
The reason is that subtracting the lower rate bound from the true rate gives a nonnegative function, as does subtracting the true rate from the upper bound. Accumulating either nonnegative difference cannot produce a negative result.
From \(L(t)\le r(t)\le U(t)\), we have \(r(t)-L(t)\ge 0\) and \(U(t)-r(t)\ge 0\) throughout the interval.
Since the integral of a nonnegative integrable function over an interval is nonnegative, $$ \int_a^b (r(t)-L(t))\,dt\ge 0,\qquad \int_a^b (U(t)-r(t))\,dt\ge 0. $$
By linearity of accumulation, the first inequality gives \(\int_a^b L(t)\,dt\le\int_a^b r(t)\,dt\), and the second gives \(\int_a^b r(t)\,dt\le\int_a^b U(t)\,dt\). Together they give the claimed bounds.
This argument explains why the method works; it does not require the rate to be positive. The inequalities compare signed accumulated changes, so negative contributions are handled automatically.
Worked Examples
Worked Example: Several Constant Rate Measurements
A tank’s volume changes over three consecutive intervals. The measured rates and their error bounds are \(5\pm0.4\) litres per minute for 2 minutes, \(2\pm0.2\) litres per minute for 3 minutes, and \(4\pm0.5\) litres per minute for 1 minute. The initial volume is 50 litres. Find a guaranteed range for the final volume.
First calculate the lower and upper rate on each interval:
For the first interval, the rate lies between \(5-0.4=4.6\) and \(5+0.4=5.4\) litres per minute. For the second, it lies between \(2-0.2=1.8\) and \(2+0.2=2.2\). For the third, it lies between \(4-0.5=3.5\) and \(4+0.5=4.5\).
Multiply each rate bound by its interval duration and add the contributions. The lower bound on the volume increase is $$ (4.6)(2)+(1.8)(3)+(3.5)(1) =9.2+5.4+3.5 =18.1\text{ litres}. $$
The upper bound on the increase is $$ (5.4)(2)+(2.2)(3)+(4.5)(1) =10.8+6.6+4.5 =21.9\text{ litres}. $$
Therefore the final volume is between $$ 50+18.1=68.1\text{ litres} \quad\text{and}\quad 50+21.9=71.9\text{ litres}. $$
Check: Using the measured rates themselves gives an estimated increase of \((5)(2)+(2)(3)+(4)(1)=10+6+4=20\) litres, between 18.1 and 21.9. The total uncertainty is \((0.4)(2)+(0.2)(3)+(0.5)(1)=0.8+0.6+0.5=1.9\) litres, so the increase range is \(20\pm1.9\), exactly the range calculated above.
Worked Example: Signed Rates and a Changing Quantity
A substance is removed from a container during the first 2 minutes and added during the next 4 minutes. The measured rate of change is \(-2\pm0.5\) grams per minute during the first interval and \(1\pm0.25\) grams per minute during the second. The initial amount is 12 grams. Bound the final amount.
During the first interval, the true rate is between \(-2.5\) and \(-1.5\) grams per minute. Its accumulated change is therefore between $$ (-2.5)(2)=-5\text{ grams} \quad\text{and}\quad (-1.5)(2)=-3\text{ grams}. $$
During the second interval, the rate is between \(0.75\) and \(1.25\) grams per minute. The accumulated change there is between $$ (0.75)(4)=3\text{ grams} \quad\text{and}\quad (1.25)(4)=5\text{ grams}. $$
Add the lower contributions to get the lower bound for the net change, and add the upper contributions to get the upper bound: $$ -5+3=-2\text{ grams}, \qquad -3+5=2\text{ grams}. $$ Thus the final amount lies between \(12-2=10\) grams and \(12+2=14\) grams.
Check: The measured rates give a first-interval change of \((-2)(2)=-4\) grams and a second-interval change of \((1)(4)=4\) grams, for a net change of 0 grams. This lies within \([-2,2]\), so the measured-rate estimate of 12 grams is inside the final range \([10,14]\). The first interval’s negative contributions have not been treated as positive areas.
Worked Example: A Varying Measured Rate and Error Bound
Over 4 minutes, a measured rate increases linearly from 2 to 6 millilitres per minute. The error bound is 0.2 millilitres per minute during the first 2 minutes and 0.6 millilitres per minute during the next 2 minutes. Find bounds for the accumulated volume change.
Use the trapezoid areas for the measured rate. On the first 2-minute interval, the measured rate goes from 2 to 4, so its accumulated value is $$ \frac{2+4}{2}(2)=6\text{ millilitres}. $$ On the next 2-minute interval, it goes from 4 to 6, giving $$ \frac{4+6}{2}(2)=10\text{ millilitres}. $$ Thus the measured curve accumulates \(6+10=16\) millilitres.
The uncertainty contributes at most \((0.2)(2)=0.4\) millilitres on the first interval and \((0.6)(2)=1.2\) millilitres on the second. The total uncertainty is \(0.4+1.2=1.6\) millilitres. Therefore the accumulated change is bounded by $$ 16-1.6=14.4\text{ millilitres} \quad\text{and}\quad 16+1.6=17.6\text{ millilitres}. $$
Check: The lower rate curve is 0.2 below the measured curve for 2 minutes and 0.6 below it for 2 minutes, so its accumulated value is \(16-0.4-1.2=14.4\) millilitres. The upper curve gives \(16+0.4+1.2=17.6\) millilitres. Both agree with the stated bounds.
What the Range Does—and Does Not—Say
The interval \([Q_{\min},Q_{\max}]\) is a guaranteed range based on the stated rate bounds. It is not automatically a probability interval, and it does not say that every value in the range is equally likely. The uncertainty information is deterministic: the true rate is constrained to remain between the lower and upper bounds.
If the only restrictions on the true rate are the pointwise bounds, the lower and upper accumulated values are reached by taking the rate equal to the lower or upper bound throughout. Additional information, such as a requirement that the rate be nonnegative, may narrow the possible range. Apply any such extra restrictions before interpreting the result.
There is also an important limitation: isolated readings do not, by themselves, control what happens between readings. A set of measurements at individual times gives an accumulation bound only if there is also enough information to bound the rate between those times—for example, a stated rate curve, a justified interpolation model, or a bound on how quickly the rate can vary. Without that, the rate could change substantially between recorded measurements.
A Common Error: Forgetting the Duration
A rate uncertainty has units of quantity per time, not units of quantity. It must be multiplied by the time over which it applies before it can contribute to an accumulation uncertainty.
In the first example, adding the error bounds \(0.4+0.2+0.5=1.1\) litres would be incorrect: those numbers are in litres per minute, and the intervals have different durations. The correct calculation is $$ (0.4)(2)+(0.2)(3)+(0.5)(1)=1.9\text{ litres}. $$ The units confirm the correction: \((\text{litres per minute})(\text{minutes})=\text{litres}\).
A Reliable Calculation
For piecewise-constant measurements, a practical procedure is to calculate the lower and upper rate on each interval, multiply each by that interval’s duration, and then add the lower contributions and the upper contributions separately. For rates that vary over time, accumulate the lower and upper rate curves over the full interval instead. In either case, check the signs, units, and initial value before reporting the range.
Check Your Understanding
Try each question by writing the rate bounds before calculating the accumulated bounds.
- A rate is measured as \(7\pm0.3\) metres per second for 5 seconds. What range does this give for the distance accumulated during that interval?
- A measured rate is \(-1\pm0.2\) kilograms per minute for 3 minutes. What are the lower and upper bounds on the accumulated change?
- Why must a rate error bound be multiplied by the duration of its interval?
- Measurements are recorded only at the start and end of an interval. What additional information is needed to guarantee an accumulation bound between those measurements?
- If the rate is bounded above and below at every time, why do the same inequalities hold for its accumulated change?
