Why Compare Rate Laws in More Than One Way?
A rate law describes how quickly a quantity changes over time. When observations are available, more than one mathematical function may appear to match them. Choosing a model therefore requires more than finding a curve that passes near the observed points.
A useful comparison asks three related questions. Does each rate function track the measured rates across the observation interval? Does its integral predict a reasonable accumulated change? Does it obey the physical restrictions that apply to the system? These tests assess different consequences of a proposed model.
A small pointwise error can be reassuring, but it is not a complete justification. A model may match a few observations and behave implausibly between them, or may predict an unreasonable cumulative total. A physically impossible rate is unsuitable even if its numerical fit looks attractive.
A Framework for Comparing Candidates
The observed and predicted rates must have the same units, and the function must be evaluated at the actual observation times. The score \(S\) does not account for behavior between measurements, physical validity, or whether different observations have different uncertainty. It is a comparison tool, not a proof that a model is correct.
Graphs add information that a single score cannot show. Plot the observed points and each candidate curve on the same axes, with time and rate on labeled axes. Look for whether a curve systematically lies above or below the observations, misses a change in trend, or behaves oddly between or beyond the measured times. In particular, avoid treating a curve’s appearance outside the measured interval as evidence unless extrapolation is part of the question.
Rate Predictions Also Imply Accumulated Predictions
If \(A(t)\) is an accumulated quantity and \(r(t)=A'(t)\) is its rate of change, then the Fundamental Theorem of Calculus gives the accumulated change over \([a,b]\):
This relationship gives a second way to test a candidate rate function. Integrating it predicts the change in the accumulated quantity over an interval. The predicted change can be compared with an observed cumulative difference. When only sampled rates are available, a numerical area estimate, such as the trapezoidal estimate, provides an approximate comparison; it is not the exact integral unless additional conditions make it so.
Physical constraints provide a further test. For example, if a rate represents inflow and outflow is modeled separately, the inflow rate itself may be required to be nonnegative. If a rate represents a net change, however, negative values may be meaningful. The meaning of the modeled quantity determines which restrictions apply.
Worked Example: Pointwise Fit and Integrated Prediction Can Disagree
Worked Example: Pointwise Fit and Integrated Prediction Can Disagree
Suppose rates are observed at \(t=0,1,2,3\) hours, in units per hour. The observations are \(6,4.8,4.2,3.1\). Compare the candidates \(p(t)=6e^{-0.25t}\) and \(q(t)=6-1.5t+0.25t^2\) on \(0\le t\le3\).
At \(t=0\), both predict \(6\). At \(t=1\), \(p(1)=6e^{-0.25}\approx4.6728\), while \(q(1)=6-1.5+0.25=4.75\). At \(t=2\), \(p(2)=6e^{-0.5}\approx3.6392\), while \(q(2)=6-3+1=4\). At \(t=3\), \(p(3)=6e^{-0.75}\approx2.8342\), while \(q(3)=6-4.5+2.25=3.75\).
For \(p\), the residuals (observed minus predicted) are \(0\), \(4.8-4.6728=0.1272\), \(4.2-3.6392=0.5608\), and \(3.1-2.8342=0.2658\). Using unrounded predictions before rounding the final score gives approximately \(S_p=0+0.0162+0.3145+0.0706=0.4013\).
For \(q\), the residuals are \(0\), \(4.8-4.75=0.05\), \(4.2-4=0.2\), and \(3.1-3.75=-0.65\). Thus \(S_q=0+0.0025+0.04+0.4225=0.4650\). By this score, \(p\) fits the listed rate observations better.
Now compare predicted accumulated changes from \(0\) to \(3\). For \(p\), an antiderivative is \(-24e^{-0.25t}\), since its derivative is \(6e^{-0.25t}\). Therefore \(\int_0^3p(t)\,dt=24(1-e^{-0.75})\approx12.6632\). For \(q\), an antiderivative is \(6t-0.75t^2+\frac{1}{12}t^3\); differentiating gives \(6-1.5t+0.25t^2=q(t)\). Hence \(\int_0^3q(t)\,dt=18-6.75+2.25=13.5\).
The trapezoidal estimate from the observed rates is \(\frac{6+4.8}{2}(1)+\frac{4.8+4.2}{2}(1)+\frac{4.2+3.1}{2}(1)=5.4+4.5+3.65=13.55\) units. The estimate is much closer to \(q\)'s integrated prediction than to \(p\)'s. Both candidates are positive on this interval, but the pointwise score favors \(p\) while the cumulative comparison favors \(q\). Neither comparison alone settles the choice; the purpose is to see which evidence matters for the modeling question.
Worked Example: A Good Score Does Not Override a Physical Constraint
Worked Example: A Good Score Does Not Override a Physical Constraint
Suppose a measured inflow rate, in units per hour, is observed at \(t=0,1,2\) hours as \(5,2.3,1.0\). Compare \(p(t)=5e^{-0.87t}\) and \(q(t)=5-3t+0.2t^2\) on \(0\le t\le2\).
Both predict \(5\) at \(t=0\). At \(t=1\), \(p(1)=5e^{-0.87}\approx2.0948\), while \(q(1)=5-3+0.2=2.2\). At \(t=2\), \(p(2)=5e^{-1.74}\approx0.8776\), while \(q(2)=5-6+0.8=-0.2\).
The residuals for \(p\) are \(0\), \(2.3-2.0948=0.2052\), and \(1.0-0.8776=0.1224\). The score is approximately \(S_p=0.2052^2+0.1224^2=0.0421+0.0150=0.0571\). For \(q\), the residuals are \(0\), \(0.1\), and \(1.2\), giving \(S_q=0.1^2+1.2^2=0.01+1.44=1.45\). On pointwise fit, \(p\) is clearly better.
But an inflow rate cannot be negative. Since \(q(2)=-0.2\), \(q\) violates that constraint on the observation interval and cannot represent this inflow law there, regardless of its score. The exponential candidate is positive because \(5e^{-0.87t}>0\) for every real \(t\). For a further check, the integrated predictions are \(\int_0^2p(t)\,dt=\frac{5}{0.87}(1-e^{-1.74})\approx4.7384\) and \(\int_0^2q(t)\,dt=[5t-1.5t^2+\frac{0.2}{3}t^3]_0^2=10-6+\frac{0.8}{3}\approx4.2667\). The trapezoidal estimate from observations is \(\frac{5+2.3}{2}+\frac{2.3+1.0}{2}=3.65+1.65=5.30\). These comparisons do not remove \(q\)'s physical violation; they show why fit, integrated evidence, and constraints should be considered separately.
Worked Example: Two Plausible Decreasing Laws
Worked Example: Two Plausible Decreasing Laws
Suppose the observed rates at \(t=0,1,2\) hours are \(4,2.5,1.8\) units per hour. Compare \(u(t)=\frac{4}{1+t}\) and \(v(t)=4e^{-0.6t}\) on \(0\le t\le2\).
The predictions for \(u\) are \(4\), \(2\), and \(\frac{4}{3}\approx1.3333\). Its residuals are \(0\), \(0.5\), and \(1.8-\frac{4}{3}=\frac{7}{15}\approx0.4667\). Thus \(S_u=0.5^2+(\frac{7}{15})^2=0.25+\frac{49}{225}\approx0.4678\).
The predictions for \(v\) are \(4\), \(4e^{-0.6}\approx2.1952\), and \(4e^{-1.2}\approx1.2048\). Its residuals are \(0\), \(2.5-2.1952=0.3048\), and \(1.8-1.2048=0.5952\). Using the unrounded exponential values, \(S_v\approx0.0929+0.3543=0.4472\). The exponential candidate has the slightly lower squared-residual score.
The integrated predictions are \(\int_0^2u(t)\,dt=4\ln(1+t)\big|_0^2=4\ln3\approx4.3944\) and \(\int_0^2v(t)\,dt=\frac{4}{0.6}(1-e^{-1.2})\approx4.6587\). The antiderivatives check: \(\frac{d}{dt}[4\ln(1+t)]=4/(1+t)\), and \(\frac{d}{dt}[-\frac{4}{0.6}e^{-0.6t}]=4e^{-0.6t}\). The trapezoidal estimate from observed rates is \(\frac{4+2.5}{2}+\frac{2.5+1.8}{2}=3.25+2.15=5.40\). Here both candidates are positive and decreasing on the interval, and \(v\) has both the lower pointwise score and the integrated prediction closer to the sampled-rate estimate. This is useful agreement, but it still does not prove either law is correct between observations.
A Common Mistake: Treating a Score as a Verdict
A common error is to conclude that the candidate with the smallest residual score must be the correct model. For example, in the inflow comparison above, \(q\)'s score is \(1.45\), so the incorrect reasoning might be: “A score is only a numerical ranking; a larger score is still acceptable if the graph looks reasonable.” Yet the calculation \(q(2)=5-6+0.8=-0.2\) directly contradicts the requirement that inflow be nonnegative. The correction is to reject \(q\) as an inflow law on the stated interval, then compare any remaining candidates using fit and accumulation evidence. Constraints are requirements, not optional scoring preferences.
Another error is to compare accumulated changes with an observed cumulative total without using the same time interval and starting value. A rate integral gives a change, \(A(b)-A(a)\), not necessarily the total \(A(b)\). If the initial stock \(A(a)\) is known, the predicted endpoint is \(A(a)+\int_a^b r(t)\,dt\). Keeping this distinction prevents a unit-consistent calculation from being applied to the wrong quantity.
A Practical Comparison Sequence
Identify the observation interval and ensure every candidate rate is expressed in the same units as the measurements.
Evaluate each model at the observed times, calculate residuals, and inspect the curves against the data over the modeled interval.
Integrate each candidate over the relevant interval and compare the result with cumulative observations or a clearly identified estimate.
Check required signs, bounds, and any balance restrictions that follow from the meaning of the rates and stocks.
Explain which evidence favors each candidate and avoid claiming that finite observations establish a unique rate law.
Why This Comparison Matters
In model construction, a rate function is not merely a curve drawn through observations. It makes predictions at unobserved times and determines changes in cumulative quantities. Two laws can be close at measured times but imply different totals; a law can also fit observed values yet violate a physical restriction. Comparing pointwise behavior, integral consequences, and constraints makes those differences explicit.
The result is a reasoned model assessment rather than an automatic winner. A model is more credible when it tracks the observations, gives plausible accumulated predictions, and respects the physical meaning of the system. If the evidence conflicts, report the conflict and the assumptions involved instead of hiding it behind a single score.
Check Your Understanding
For each question, explain the reasoning as well as any calculation.
- What does a smaller sum of squared residuals establish, and what does it not establish?
- If \(r(t)=A'(t)\), what quantity is predicted by \(\int_a^b r(t)\,dt\)?
- Why must a rate constraint be checked across the modeled interval rather than only at some observation times?
- In the first worked example, which candidate has the smaller rate-residual score, and which integrated prediction is closer to the trapezoidal estimate?
- If a candidate rate represents an inflow, how would you assess a model that fits the data well but becomes negative within the interval?
