Fixed distance, force, or work — when a product is locked, two quantities trade off as y = k∕x, on the part of the domain that's real.
The inverse-proportion function turns up everywhere a product is locked. Cover a fixed distance and speed trades against time; press with a fixed force and area trades against pressure; finish a fixed job and the crew size trades against the days. Each of these is y = k∕x in disguise, with k the locked product. The new discipline the real world adds is respecting the domain: speeds, areas, and head-counts can't be negative or zero, so you draw and read only the branch that makes sense. And looking back across this whole strand, you'll meet the one quietly new thing about this curve — its rate of change keeps easing off — which is exactly the door to the bending parabola of the next stage.
To turn a story into a function, hunt for the fixed product. Name the two quantities that vary, write down the constant they always multiply to, and you have y = k∕x on a plate. Take a trip of fixed distance 120 km. Distance, speed, and time are tied by distance = speed × time, so time = distance ∕ speed, which gives
t = 120v (with v·t = 120 always).
The faster you go, the less time it takes — and the product v·t never budges from 120. Read three speeds straight off the rule: at v = 40, t = 120∕40 = 3 h; at v = 60, t = 120∕60 = 2 h; at v = 80, t = 120∕80 = 1.5 h. Here the constant k = 120 is the distance.
A fixed distance makes time an inverse proportion of speed: t = (distance)∕v, with k = the distance. The same template fits any locked product.
Physics is full of locked products. Push down with a fixed force F spread over a contact area S, and the pressure is force per unit area:
p = FS.
Spread the same force over a larger area and the pressure drops — that's why snowshoes keep you on top of soft snow while a stiletto heel sinks in. With F = 600 N the rule is p = 600∕S: at S = 2 m², p = 300 Pa; at S = 6 m², p = 100 Pa. Many laws wear this same shape — Ohm's law I = U∕R at a fixed voltage U, or the volume–pressure trade in a fixed amount of gas. A fixed numerator over a varying denominator is always inverse proportion.
F = 600 N on S = 4 m²: p = 600∕4 = 150 Pa. Double the area to 8 m² and the pressure halves to 75 Pa — because S·p = 600 must hold.
The same trade shows up wherever a fixed total is shared among a varying count. Suppose a job amounts to 36 worker-days of effort. More workers means fewer days each: the days needed are the total divided by the head-count,
d = 36n.
So 6 workers → 6 days, 9 workers → 4 days, 12 workers → 3 days — and in every case n·d = 36, the size of the job. The very same idea splits a fixed budget among items (cost-each = budget ∕ count) or a fixed amount of data across days. A fixed total ÷ a count is an inverse proportion, with k = the total.
A fixed total ÷ a count is inverse proportion: d = (total)∕n, k = the total. Distance, force, budget, data — same skeleton, different letters.
Mathematically y = k∕x has two branches — one in Quadrant Ⅰ, one in Quadrant Ⅲ. But a real model lives on only the sensible part of the domain. A speed is v > 0; a contact area is S > 0; a crew is n a positive whole number. There is no such thing as −40 km/h of distance-covering speed or −5 workers, so the lower-left branch is physically meaningless here.
That means you draw just the first-quadrant branch, and you read answers that fit reality: a head-count rounds to a whole crew, and a negative root is simply thrown away. Before reading any inverse-proportion model, ask the gatekeeper question first.
Always ask "which values of x are physically possible?" before reading the graph. Keep v > 0, S > 0, n a positive whole number — never read off the unreal branch.
A line changes at a constant rate: equal steps in x always add the same amount to y (that's exactly what slope means, back in Stage 21). A hyperbola does not — we saw the two graphs share a plane back in 22.4, but they move very differently. Walk along t = 120∕v in equal steps of 20 km/h and watch the drop in time shrink:
| v goes from … to … | t goes from … to … | change in t |
|---|---|---|
| 40 → 60 km/h | 3 h → 2 h | −1 h |
| 60 → 80 km/h | 2 h → 1.5 h | −0.5 h |
| 80 → 100 km/h | 1.5 h → 1.2 h | −0.3 h |
Same 20 km/h step each time, yet the time saved keeps easing off: a full hour, then half an hour, then under twenty minutes. Because the rate of change is not constant, the graph cannot be straight — it must bend. That bending, changing rate is the signature of the curves still to come: the next stage meets the quadratic function, whose parabola bends the other way and whose rate of change marches in equal steps of its own. The steady stride of the line, the easing curve of the hyperbola, the accelerating sweep of the parabola — three different ways for a function to move.
An inverse-proportion model keeps its product constant but its rate of change is not — so its graph bends. That is the door from the straight line of Stage 21 to the parabola of Stage 23.
· A locked product makes one quantity an inverse proportion of the other: y = k∕x, with k the constant product.
· Fixed distance → t = 120∕v (k = distance); fixed force → p = 600∕S (k = force); fixed job → d = 36∕n (k = the total).
· A real model keeps only the sensible domain — v > 0, S > 0, n a positive whole number — so you draw and read just the first-quadrant branch.
· A line changes at a constant rate; a hyperbola's rate keeps easing off, so its graph bends — the door to the parabola ahead.
A trip is 240 km long. Write the time as a function of speed, then find the time at 80 km/h.
Time = distance ∕ speed, so t = 240∕v (k = 240). At v = 80: t = 240∕80 = 3 h.
A force of 600 N presses on a contact area of 4 m². Find the pressure.
p = F∕S = 600∕4 = 150 Pa. (Double the area and the pressure halves, since S·p = 600.)
A job is 48 worker-days. How many days will it take 8 workers?
d = (total)∕n = 48∕8 = 6 days. The model is d = 48∕n, with n·d = 48 always.
For t = 120∕v, why do we draw only the v > 0 branch?
Speed can't be zero or negative — there is no real v ≤ 0 for a trip — so only positive speeds make sense, and we keep just the first-quadrant branch.
On t = 120∕v, how much does t change from v = 30 to v = 60, and from v = 60 to v = 90? Which drop is bigger?
30 → 60: t goes 4 h → 2 h, a drop of 2 h. 60 → 90: t goes 2 h → 1⅓ h, a drop of only ⅔ h. The first drop is bigger — the rate of change keeps shrinking.
Explain why a line's graph is straight but a hyperbola's bends.
A line changes at a constant rate — equal steps in x add equal amounts to y. A hyperbola's rate of change keeps shrinking (the drops ease off), so equal steps no longer give equal changes, and the graph must curve.
Six questions to lock it in. Tap the answer you think is right.
The big idea of this closing lesson is modeling: a great many everyday relationships hide a locked product, and spotting it lets a student write y = k∕x on sight — fixed distance gives t = (distance)∕v, fixed force gives p = F∕S, a fixed job gives d = (total)∕n. The constant k is never arbitrary; it is the distance, the force, the size of the job. Encourage learners to say the product out loud ("v·t = 120, always") before touching the graph.
The specific misconception to watch is twofold. First, students reach for a linear model out of habit ("twice the speed, twice the something") when the situation is really inverse — anchor them on the product test: does x·y stay fixed? Second, they draw or read the negative branch where it has no physical meaning; insist on the gatekeeper question "which x are possible?" so they keep v > 0, S > 0, n a positive whole number. A third, subtler point: a curve does not have a constant rate of change, so resist averaging it like a line.
Common Core: F-LE.A.1 / F-LE.B.5 (recognize non-linear relationships and interpret a model's parameters), A-CED.A.2 / N-Q.A.2 (model with y = k∕x; choose a sensible domain and units), and F-IF.B.6 (average rate of change is not constant for a curve). This lesson also quietly sets up Stage 23's quadratic functions, where a different — but again non-constant — rate of change produces the parabola.