Estimate totals from unequal-time measurements

Try

A fictional cart moves along a straight marked track. Its velocity is recorded at four times. Positive velocity means motion in the positive track direction; a negative value means motion in the opposite direction. The numbers are synthetic teaching data, and the recorded times are deliberately not equally spaced.

Time t in seconds	0	2	5	9
Velocity v(t) in meters per second	−2	2	4	−2

Predict whether the cart’s final displacement is necessarily zero because its first and last recorded velocities agree. Velocity describes a rate at a time, whereas displacement accumulates velocity over an interval. Equal endpoint rates do not imply equal starting and ending positions. Also predict whether total distance traveled can be smaller than the magnitude of displacement.

Predict

Draw the four measured points on time–velocity axes. Mark widths 2, 3 and 4 seconds between them. A trapezoid estimate joins successive velocity points by straight segments and adds each segment’s average endpoint height times its own width. Predict the contribution of the first interval, whose endpoint velocities have equal magnitude and opposite sign.

State the model before calculating: between the recorded times, velocity is interpolated linearly. This makes the following computation exact for that interpolation. It remains an estimate of an otherwise unknown real velocity history. Merely possessing four measurements does not establish linear behavior between them.

Experiment

Construct the joined graph and this interval ledger. The mean endpoint velocity has units m/s; multiplying by the interval width in seconds produces meters.

Interval in seconds	Width	Mean endpoint velocity	Signed contribution
0 to 2	2	(−2 + 2)/2 = 0	0 m
2 to 5	3	(2 + 4)/2 = 3	9 m
5 to 9	4	(4 − 2)/2 = 1	4 m
Whole interval	9	Not the unweighted mean of table entries	13 m

Cover the final column and recompute it. Then deliberately use width 3 for every interval and identify the contributions that would change. The incorrect total is 12 rather than 13 meters; neither uniform widths nor an unweighted average of the four velocities matches the recorded schedule.

Now shade portions above and below the velocity axis separately. Under the linear model, find each zero crossing before replacing signed areas with positive areas. Do not take absolute values only after summing; that would erase travel that later motion cancels.

Understand

The net displacement estimate is the sum of the signed trapezoids, 13 m. If the cart starts at position 5 m, its modeled final position is 18 m. Position, displacement and distance are different quantities: changing the starting position changes neither the displacement nor the distance traveled.

On the first interval, velocity rises from −2 to 2, crossing zero at t = 1 second. The negative triangle has magnitude 1 m, and the positive triangle also has magnitude 1 m. Net displacement is zero there, but distance is 2 m. On the middle interval, velocity stays positive, so distance and displacement are both 9 m.

On the last interval, velocity changes by −6 m/s over 4 seconds. Its linear rule is v(t) = 4 − (3/2)(t − 5), with numerical time in seconds and velocity in m/s. Solving v(t) = 0 gives t = 23/3 seconds. The positive triangle has base 8/3 seconds and height 4 m/s, so its area is 16/3 m. The negative triangle has base 4/3 seconds and height 2 m/s, so its magnitude is 4/3 m. Their signed difference is 4 m, consistent with the ledger, while their distance contribution is 20/3 m.

Therefore total distance for the interpolated velocity is 2 + 9 + 20/3 = 53/3 m. Its average velocity is 13/9 m/s, while average speed is (53/3)/9 = 53/27 m/s. The magnitude of average velocity cannot exceed average speed because cancellation can reduce a signed total but cannot increase it beyond the sum of the magnitudes.

The composite trapezoid rule expresses the same method for any times t₀, t₁, …, tₙ in increasing order: sum (tᵢ₊₁ − tᵢ)(vᵢ + vᵢ₊₁)/2 over consecutive intervals. The widths need not match. Boundary samples appear once; interior samples contribute to both neighboring trapezoids, with weights determined by both adjacent widths. A formula copied from equally spaced data can therefore assign the wrong weights here.

Apply

Explain what the four samples cannot establish. Between times 2 and 5, define a dimensionless coordinate u = (t − 2 seconds)/(3 seconds). Add εu²(1 − u)² to the interpolated velocity on this interval, where ε is any positive rate in m/s; outside it, leave the interpolation unchanged. This added term is zero at both ends and positive inside, so every recorded velocity remains unchanged while the accumulated displacement increases. Subtracting the term instead decreases the displacement. The samples therefore do not determine a unique trajectory or the sign of the trapezoid error.

Additional shape information can justify one-sided error claims: for a velocity graph known to be convex throughout an interval, the straight chord lies above the graph, so the trapezoid overestimates its signed integral there. A concave graph gives the opposite comparison. Four measurements alone do not establish either condition, nor do they bound unobserved oscillation. Measurement error is another limitation, distinct from the interpolation assumption.

For a new dataset, provide the raw time/rate pairs, widths, signed ledger and a labeled graph or equivalent segment description. State how zero crossings were found when computing distance. If the requested interval ends between samples, explain the extra interpolation needed; if it extends outside the data, identify extrapolation rather than invent a new measured endpoint. A mass-flow or account-rate problem uses the same accumulation mechanism, but its rate units and the meaning of a negative rate must be stated separately.

A numerical workflow for a non-elementary rate model

The next example supplies continuous formulas, rather than four measured rates. A fictional tank initially contains 20 litres. For numerical time t in minutes on [0,4], its inflow is I(t)=6+4exp(−t²/4) litres per minute and its outflow is O(t)=5+t litres per minute. Here exp(u) means e raised to the power u. The constants carry the appropriate units; the exponent is dimensionless when time is expressed in these minutes. The model assumes the stated outflow remains feasible and that neither overflow nor a capacity restriction intervenes.

Define net rate N=I−O=1−t+4exp(−t²/4). Inventory is Q(t)=20+∫₀ᵗN(u)du. These formulas answer three different questions: I measures incoming flow, N measures inventory change, and Q measures the amount already present. The Gaussian exponential does not have an elementary antiderivative, but its definite integral is well defined and can be approximated numerically. Do not force a power-rule primitive that differentiates to the wrong function.

Before using the workbench, predict whether the largest inventory occurs when inflow is largest or when inflow equals outflow. Then execute this reproducible sequence:

1. Write I, O and N, including units and the model interval. Enter the inflow and outflow functions in a numerical tool and choose a window containing 0≤t≤4 and 0≤rate≤11. This window shows both curves without treating its edge as a mathematical endpoint of all possible models.
2. Evaluate N at candidate bracket endpoints. At 2 it is approximately 0.471518; at 3 it is approximately −1.578403. Record the sign change before asking for an intersection.
3. Solve N(t)=0 within that bracket. Record the final bracket as well as the estimate. A plotted crossing alone does not provide a root error bound.
4. Request a definite integral by specifying the expression, variable and two bounds: for example, integrate 6+4exp(-t*t/4) with respect to t from 0 to 4. Exact button names and syntax vary by tool. Integrate O over the same interval, then compare their difference with an integral of N. Add the initial 20 only when reporting inventory.
5. Integrate N from zero to the root estimate to approximate the maximum inventory. Refine the numerical work and retain the method’s error information; changing the time endpoint introduces an additional source of error.

Choose a bracket, calculate, then refine

A synthetic tank starts with 20 liters at t=0. Inflow I(t)=6+4 exp(−t²/4) and outflow O(t)=5+t are in liters per minute, for 0≤t≤4 minutes. Net rate is I−O. Predict when the inventory stops increasing before calculating. These controls are temporary and do not save practice answers.

Numerical graph checkpoints

At t=0: I=10, O=5. At t=2: I=7.47151776468577, O=7, net=0.471517764685769. At t=3: I=6.42159689824746, O=8, net=-1.57840310175254. At t=4: I=6.07326255555494, O=9. The graph uses sampled points; its visual crossing is not an exact root.

1. Bracket the rate intersection

Bracket start (minutes) Bracket end (minutes) Bisection steps48122440

Choose increasing endpoints inside [0,4]. Opposite net-rate signs establish a bracket by continuity. A small residual alone is not a time-error guarantee.

Calculate intersection bracket

2. Integrate the rates over a chosen interval

Interval start (minutes) Interval end (minutes) Even Simpson panels248163264128256

Calculate the selected panel count and twice that count on the same interval. Their difference measures refinement agreement; it is not by itself an error certificate. Inflow and outflow integrals are volumes; their difference is net inventory change.

Calculate and compare refinement

Reset calculator

The workbench uses this fixed model. Its root bracket and integral interval are different inputs: selecting a bracket such as [2,3] must not silently change a requested total over [0,4]. Changing a numerical setting requires a fresh calculation. Keep your prediction and numerical record outside the temporary widget if you want to retain them as written evidence.

On another calculator, select its numerical intersection or zero-finding operation with the recorded bracket rather than accepting an unexplained nearby crossing. Carry the tool’s stored precision into subsequent integration; do not type a heavily rounded root back as though it were exact. Round the final reported amount only after considering the error information. Successfully using this local widget does not establish familiarity with a permitted examination calculator or the separate Bluebook and handwritten-response workflow.

Why the crossing is the global maximum

Continuity and the sign change establish at least one crossing. Differentiation gives N′(t)=−1−2t exp(−t²/4), which is strictly negative for t≥0. Thus N is strictly decreasing throughout the model interval and has exactly one zero, approximately 2.19688298 minutes. Before it, Q′=N is positive; after it, Q′ is negative. This proves that Q reaches its global maximum there. Merely solving I=O would only identify a candidate without the sign argument.

The following are independently calculated numerical reference values, not exact symbolic answers or a promise that every workbench setting prints this many correct digits. Use them to check the setup; the workbench’s current bounds determine what its own run supports.

Quantity on the stated interval	Numerical reference	Interpretation
∫₀⁴I(t)dt	31.05665113 L	Total incoming amount
∫₀⁴O(t)dt	28 L exactly	Total outgoing amount
∫₀⁴N(t)dt	3.05665113 L	Net inventory change
Q(4)	23.05665113 L	Final inventory
Maximum Q	26.02050270 L	Amount at the unique crossing
(1/4)∫₀⁴I(t)dt	7.76416278 L/min	Average inflow rate
(1/4)∫₀⁴N(t)dt	0.76416278 L/min	Average net rate

At the crossing, I=O≈7.19688298 L/min; neither flow is zero. The maximum inventory therefore need not occur at maximum inflow. Both endpoint inventories are positive and the only interior extremum is a maximum, so this model never empties over [0,4]. An actual finite tank would additionally need enough capacity for the predicted maximum.

Precision, accuracy and a defensible record

With bisection on [2,3], k sign-preserving halvings leave bracket width 2⁻ᵏ. Its midpoint is within half that width of the unique root, in exact arithmetic. Printing ten decimal places after only a few halvings does not shrink this uncertainty. A small function residual and a small time error are also different quantities; a bracket supplies a direct time bound.

A quadrature estimate needs its own analysis. This workbench uses composite Simpson integration with an even number n of equal panels. For an interval of length L, the error is at most M L⁵/(180n⁴) if the fourth derivative has absolute value at most M throughout that interval. The model’s fourth derivative is I⁽⁴⁾(t)=(t⁴/4−3t²+3)exp(−t²/4), while O⁽⁴⁾=0. Bounding each Gaussian-weighted term separately gives |I⁽⁴⁾|≤3+12/e+16/e²<10, so M=10 is a valid conservative choice for both I and N. For [0,4] and n=32 the resulting bound is 1/18432 litres, approximately 0.0000543 L. Doubling n divides this bound by 16. It concerns integration of the stated smooth functions, not measurement or modeling error.

The earlier trapezoid method also remains valid, with a different estimate and bound. Here I″=(t²−2)exp(−t²/4) has absolute value at most 2, giving trapezoid error at most L³/(6n²). Do not label a Simpson run with a trapezoid estimate or assume that the two methods share the same convergence rate.

The maximum calculation also uses an approximate upper limit. If that time is within ε of the root, changing the endpoint changes the exact net integral by at most 5ε litres, because |N|≤5 on [0,4]. Add an applicable quadrature bound to this endpoint bound when claiming an enclosure. Floating-point roundoff is distinct from these exact-arithmetic truncation arguments. Agreement between two refinements is useful evidence but does not, by itself, replace a bound. Report only a precision your evidence supports, with the mathematical integral beside the decimal result.

Fresh manual transfer

Keep the same inflow but change the outflow to 6+t L/min, the initial amount to 12 L, and the interval to [0,3]. This is a separate case; the fixed workbench above does not edit its outflow to represent it. Use a suitable numerical tool or a documented manual quadrature procedure. Supply the net-rate formula, a sign-changing bracket, a uniqueness/maximality argument, total inflow/outflow, final amount, maximum amount and average net rate. Record numerical settings and explain which uncertainty your reported digits reflect.

Check the transfer setup and reference values

The new net rate is M(t)=4exp(−t²/4)−t. Its derivative is again −1−2t exp(−t²/4), so it is strictly decreasing. M(1)≈2.115203 and M(2)≈−0.528482 give a unique crossing in (1,2), approximately 1.79210044 minutes. The sign change from positive to negative establishes the inventory maximum.

Over [0,3], incoming amount is approximately 24.84950715 L, outgoing amount is exactly 22.5 L, and net change is approximately 2.34950715 L. Final inventory is approximately 14.34950715 L, while its maximum is approximately 16.03001338 L. Average net rate is approximately 0.78316905 L/min. These are independent reference approximations, not exact values. The same fourth-derivative bound of 10 applies for Simpson integration; a width-one root bracket again supplies a bisection time bound, and |M|≤4 on this shorter interval supplies an endpoint-integral bound of 4ε.

Review the submitted record against these observable requirements:

• Define the changed inflow, outflow, net rate and units on [0,3].
• Show endpoint signs and strict decrease to justify a unique crossing and the inventory maximum.
• Record the initial and final root brackets, numerical estimate and time-error interpretation.
• Give every definite-integral expression with its bounds, numerical method and applicable error information.
• Include the initial 12 L when reporting amounts, and divide by three minutes when reporting average net rate.
• Carry precision through intermediate work and state the limits of numerical, physical-model and device-specific evidence.

Copying the worked tank’s maximum or treating total inflow as final inventory fails the changed-case reasoning. This public activity adds practice inside the existing lesson; it is not a new completion requirement or a certified assessment of calculator proficiency.

Continue

Keep the modeled result and the evidence limitation together: the trapezoid sum is exactly the accumulation of the joined linear graph, while the unknown process may differ between observations. The written practice asks you to defend that distinction, not merely supply a sum. Use the derivative bounds above when claiming a guaranteed discretization error for the supplied smooth model. Those bounds do not transfer automatically to the earlier measured data.