Compare local change at two inputs

Try

Two synthetic filling processes both begin with zero liters and reach 24 liters after four minutes. Does that mean they fill at the same rate throughout? Suppose their exact models on 0≤t≤4 are F(t)=t²+2t and G(t)=10t−t², with t in minutes and output in liters. Calculate both outputs at t=2 before looking at the graph. One gives 8 liters and the other 16, despite matching endpoints.

An average rate on an interval is change in output divided by change in input. On the whole interval, both rates are (24−0)/(4−0)=6 liters per minute. That single average does not describe where either process was faster. We will compare short intervals near different times without assuming or calculating an instantaneous derivative.

Predict

For F, predict whether the average rate near minute 3 exceeds the average rate near minute 1. For G, make the same prediction. Then answer a different question: is the amount of water increasing for both processes? A decreasing rate can still be positive, so an amount can continue increasing while its rate becomes smaller.

Use the same interval width for the first comparison: 0.2 minute. Equal widths make changes in output directly comparable, though dividing by width is always necessary to attach the correct units.

Experiment

Model and interval	Endpoint outputs	Change ÷ interval width
F on [1,1.2]	3 and 3.84 liters	0.84/0.2=4.2 L/min
F on [3,3.2]	15 and 16.64 liters	1.64/0.2=8.2 L/min
G on [1,1.2]	9 and 10.56 liters	1.56/0.2=7.8 L/min
G on [3,3.2]	21 and 21.76 liters	0.76/0.2=3.8 L/min

Recompute one row, then halve the interval width to 0.1 while keeping each starting time fixed. For F, the rates become 4.1 and 8.1; for G, they become 7.9 and 3.9. These are rates over new, shorter intervals, not exact instantaneous rates. The comparison persists for these models, but replacing the width changes the statistic being calculated.

Understand

A secant line joins two distinct points on a graph. Its slope equals the interval’s average rate, (output at end−output at start)/(end−start). A positive slope means a net increase across those endpoints. A larger positive slope means a larger increase per unit input. A negative slope describes a net decrease; “larger rate” and “larger magnitude of decrease” are then different statements. For instance, −2 is larger than −5, but −5 has the larger magnitude.

For the stated formulas, algebra explains more than a few samples. If u<v are in [0,4], then F(v)−F(u)=(v−u)(u+v+2)>0 and G(v)−G(u)=(v−u)(10−u−v)>0. Thus both outputs increase over the entire domain. For a fixed positive width h, F’s interval rate is 2u+h+2, increasing with its starting point u. G’s is 10−2u−h, decreasing with u. These identities justify the trend across all eligible intervals of that width, not merely the four table rows.

Sparse observations do not supply those formulas. Suppose all we know is F at integer times: 0,3,8,15,24. One possible complete graph connects successive measurements by straight segments. Another passes additionally through (0.5,10) before returning to (1,3). Both fit all five supplied measurements, but the second decreases on part of the first minute. Joining data smoothly or linearly is an added model assumption, not a discovery forced by the samples.

The shape vocabulary is concave up for F and concave down for G. On an interval, a concave-up graph lies on or below every chord joining two of its points; a concave-down graph lies on or above its chords. Here the increasing secant rates accompany F’s concave-up shape, while decreasing rates accompany G’s concave-down shape. These are different from increasing or decreasing output: both graphs increase throughout this domain.

You can check the chord statement algebraically. At an intermediate fraction s of the way from u to v, with 0≤s≤1, the chord height for F exceeds F((1−s)u+sv) by s(1−s)(v−u)², which is nonnegative. For G the sign reverses, putting its graph above the chord. This verifies the whole-interval shape for these specified quadratics. A few increasing table rates alone would not prove concavity everywhere between the sampled inputs.

Apply

Construct your own piecewise-linear volume graph at times 0,1,2,3,4 for a process that always increases but whose rate decreases on successive one-minute intervals. Begin at zero and finish at ten. Label every vertex and compute all four segment slopes. Then construct another graph with the same endpoints and increasing successive rates.

Compare two complete graph constructions

One first graph uses outputs 0,4,7,9,10, connected by straight segments. Its successive rates are 4,3,2,1 liters per minute, all positive but decreasing. A second uses outputs 0,1,3,6,10 and rates 1,2,3,4. These are full stipulated piecewise-linear models, so the segment slopes apply between vertices. If the outputs were only measurements without the connecting assumption, behavior between measurements would remain undetermined. Both whole-interval averages equal 10/4=2.5 L/min.

For unequal-width intervals, first divide each change by its own width. An increase of three liters over half a minute is faster on average than four liters over one minute. Comparing changes alone would reverse the conclusion.

Continue

In practice, state the interval, units and information source: a complete formula, a stipulated connecting graph, or sparse observations. Distinguish increasing output from increasing average rates. The next lesson derives the quadratic rate identity generally and uses it to explain signs, turning points and restricted domains without calculus.