From a Rate Range to an Accumulation Range
A rate describes how quickly a quantity changes. If a flow rate stays between two known values during a time interval, the total amount that flows must also stay within limits. The interval’s width converts the rate limits into accumulation limits.
The same idea applies to signed rates. For example, velocity can be negative or positive, and the resulting net displacement can also be negative or positive. The bounds still work, but it is important to distinguish a signed net change from a nonnegative total amount.
The Basic Bounds
Suppose the rate has a minimum value \(m\) and a maximum value \(M\) throughout the interval. This means that at every time \(t\) in \([a,b]\), the rate satisfies \(m\le r(t)\le M\). If the rate is continuous on the closed interval, it attains a minimum and maximum there. More generally, it is enough to know constants \(m\) and \(M\) that bound every rate value, even if those values are not attained.
The width \(b-a\) is nonnegative. Thus, when each rate value is multiplied by the width of a time slice, the order of the inequalities is preserved. In particular, a negative rate does not reverse the bounds: it is the positive interval width being multiplied, not a negative number.
When the rate is nonnegative, the minimum and maximum are also nonnegative, so the lower and upper accumulation bounds are nonnegative. If rates can have either sign, the same theorem bounds the net change. It does not, by itself, give bounds on the total amount of movement or total amount flowing in both directions.
Why the Bounds Hold
The reasoning begins with a partition of the interval. A partition divides it into smaller intervals, each with a positive width. On each small interval, a rectangle sum uses one rate value as the rectangle’s height and the subinterval’s width as its base. Since every rate value lies between \(m\) and \(M\), every such rectangle has its signed value between the corresponding lower and upper rate times its width.
If \(a=b\), there is no time interval over which the quantity can change. The accumulated change is \(A=0\), and both bounds equal zero: \(m(b-a)=0=M(b-a)\). Thus the claimed inequalities hold with equality.
Now suppose \(a<b\). Choose subintervals with positive widths \(\Delta t_1,\ldots,\Delta t_n\). Their widths add to the full width: \(\sum_{i=1}^{n}\Delta t_i=b-a\).
For any sample time \(\xi_i\) in subinterval \(i\), the rate satisfies \(m\le r(\xi_i)\le M\). Because \(\Delta t_i>0\), multiplication gives \(m\Delta t_i\le r(\xi_i)\Delta t_i\le M\Delta t_i\).
Adding all the subinterval inequalities preserves their order and gives \(m(b-a)\le\sum_{i=1}^{n}r(\xi_i)\Delta t_i\le M(b-a)\).
By assumption, the rectangle sums approach \(A\) as the partitions are refined. Since every such sum lies between the same two bounds, its limit does too: \(m(b-a)\le A\le M(b-a)\).
This argument explains why the bounds require the rate restriction to hold throughout the interval, not merely at its endpoints or at one selected time. A rate value outside the claimed range could make a rectangle fall outside the corresponding bounds.
Worked Examples
Worked Example: Bounding a Nonnegative Flow
A pump’s flow rate stays between \(2.4\) and \(3.1\) litres per hour for \(3\) hours. Find bounds on the amount of water pumped.
Here the minimum rate is \(m=2.4\) litres per hour, the maximum rate is \(M=3.1\) litres per hour, and the interval width is \(3\) hours. The rate is nonnegative, so the accumulated amount is bounded by multiplying each rate bound by the duration:
The lower calculation is \(2.4\times3=7.2\) litres, and the upper calculation is \(3.1\times3=9.3\) litres. Therefore \(7.2\le A\le9.3\) litres.
Check: The hours cancel, leaving litres. A constant flow rate of \(2.8\) litres per hour, which lies between the stated limits, would pump \(2.8\times3=8.4\) litres. This value lies between \(7.2\) and \(9.3\) litres, as required.
Worked Example: Bounding Signed Displacement
A vehicle’s velocity is between \(-1.5\) and \(2\) metres per second for \(4\) seconds. Find bounds on its net displacement.
The rate bounds are \(m=-1.5\) metres per second and \(M=2\) metres per second. The interval width is \(4\) seconds, which is positive. The theorem gives:
Multiplying gives \(-6\) metres on the left and \(8\) metres on the right, so \(-6\le\Delta x\le8\) metres.
Check: The seconds cancel, leaving metres. If the velocity were constantly \(-1\) metre per second, the displacement would be \((-1)(4)=-4\) metres, within the bounds. These are bounds on net displacement, not a claim that the vehicle’s total distance travelled is between \(-6\) and \(8\) metres.
Worked Example: Adding Bounds on Adjacent Intervals
A nonnegative flow rate has bounds of \(1.2\) to \(2.0\) litres per minute from minute \(0\) to minute \(2\). From minute \(2\) to minute \(5\), its bounds are \(0.5\) to \(1.4\) litres per minute. Find bounds on the total amount that flows during the five minutes.
The first interval has width \(2-0=2\) minutes. Its accumulation is between \(1.2\times2=2.4\) litres and \(2.0\times2=4.0\) litres. The second interval has width \(5-2=3\) minutes. Its accumulation is between \(0.5\times3=1.5\) litres and \(1.4\times3=4.2\) litres.
Add the two lower bounds and the two upper bounds separately:
Check: The amounts from the two intervals have the same unit, so they can be added. For example, constant rates of \(1.6\) litres per minute on the first interval and \(1.0\) litre per minute on the second satisfy the stated rate limits. They give \(1.6\times2+1.0\times3=3.2+3.0=6.2\) litres, which lies between \(3.9\) and \(8.2\) litres.
Using the Tightest Available Bounds
Sometimes a single pair of rate bounds applies over the whole interval; sometimes different bounds are known on different parts. A valid pair of bounds over a larger interval can be obtained from the smallest lower bound and the largest upper bound across all its parts. However, this may be less informative than calculating separate bounds on each part and then adding them, as in the third example.
For instance, the third example has an overall lower rate bound of \(0.5\) litres per minute and an overall upper rate bound of \(2.0\) litres per minute over \(5\) minutes. Using only those rates gives \(2.5\le A\le10\) litres. These bounds are valid, but the separate-interval calculation gives the tighter range \(3.9\le A\le8.2\) litres because it uses more information about when the flow rate changes.
Common Mistakes
A common error is to reverse the order of the bounds just because both rates are negative. Suppose a velocity is between \(-3\) and \(-1\) metres per second for \(5\) seconds. The incorrect calculation \(-5\le\Delta x\le-15\) reverses the endpoints. The correct multiplication is by the positive duration \(5\): \((-3)(5)=-15\) and \((-1)(5)=-5\). Therefore \(-15\le\Delta x\le-5\) metres.
Another error is to treat the upper-rate bound as an exact answer. If a rate is known only to lie between \(m\) and \(M\), the accumulation is not determined exactly; the information supports an interval of possible values. A further error is to use a rate bound that is not valid throughout the interval. Check both the rate range and the interval width before multiplying.
Check Your Understanding
Use the rate bounds and interval widths to answer each question. Keep the units in your calculations.
- A nonnegative rate is between \(4\) and \(6\) kilograms per minute for \(3\) minutes. What bounds do you get for the accumulated amount?
- A velocity is between \(-2\) and \(1\) metres per second for \(5\) seconds. What bounds do you get for net displacement?
- Why does multiplying each rate bound by an interval width preserve the order of the inequalities?
- A rate has different bounds on two adjacent intervals. What calculation combines the separate lower and upper accumulation bounds?
- If \(a=b\), what is the accumulated change, and what do the rate-based bounds become?
