# The turn **Teaches:** `the-turn-to-coordinates` · **Assumes:** everything in this arc · **~45 min** --- ## What you have, before anything is drawn Take stock. Without ever plotting a point, you have established all of the following about $x(500-x)$ on $[0,500]$: 1. It is zero at $x=0$ and $x=500$, and positive in between. 2. Its largest value is 62500, attained only at $x=250$. 3. That maximum is *proved*, not found — the quantity is $62500 - (x-250)^2$, and a square is never negative. 4. It takes every value between 0 and 62500, each of them at exactly two cuts, one either side of 250, equally far from it. 5. Except 62500 itself, which it takes at one cut. 6. It never takes any value above 62500, and asking for one produces a negative square. 7. Its square, $x^2(500-x)^2$, is zero at the two ends to second order and is never negative anywhere. That is a complete description. There is no fact about this quantity that a graph is going to add. Bear that in mind for the next hour, because you are about to be shown a picture that will feel like an explanation, and it is not one. It is a **transcription**. ## Two things that had to happen first ### The curve existed, and nobody knew it was about this Around 200 BC, Apollonius of Perga wrote eight books on the sections of a cone, and named three of them. The names were not new words. They were borrowed, deliberately, from the Greek vocabulary for the application-of-areas problems in Euclid Book VI — the same problems you met in `quad-05-completing-the-square`: | Greek term | in area problems it meant | conic named | |---|---|---| | *elleipsis* | the applied area **falls short** | ellipse | | *parabolē* | the applied area **fits exactly** | parabola | | *hyperbolē* | the applied area **exceeds** | hyperbola | So the parabola was named after the exact-fit case of the very construction that solves quadratics. And then, for roughly eighteen hundred years, essentially nothing came of it. The curve was a slice of a cone, studied by geometers. The area problem was a matter of cutting lines and completing squares. Nobody joined them, because there was no way to write down what "joining them" would even mean. ### Somebody had to draw a quantity against a quantity In the 1350s Nicole Oresme, in Paris, worked out how to represent a quality that varies — a speed that changes along a stretch of time, a heat that changes along a rod. He drew the stretch as a horizontal line and, at each point, erected a vertical segment whose length represented the intensity there. The top of those segments traced a shape, and the *area* of that shape was the total. That is a graph. He got real results with it, including the mean-speed theorem. What he did not have was an *equation* to graph — no unknown, no algebra, nothing to solve. He was drawing qualities, not functions. ## 1637 Two things appear within about a year of each other. Fermat circulates a manuscript on plane and solid loci; Descartes publishes *La Géométrie* as an appendix to the *Discourse on Method*. Both do a version of the same thing: take an equation in two unknown lengths, and identify it with a curve. But the move that actually breaks the dam is smaller and easier to miss. It is this: > **Descartes picks a segment and calls it one.** Recall from `quad-01-species` why $x^2 + x$ was unwriteable: $x$ is a length, $x^2$ is an area, and you cannot add a length to an area any more than you can add an hour to a second. The two live in different species and there is no conversion, because the conversion factor would be $x$ itself, which is the thing you do not know. A unit length breaks the circle. Once one segment is called 1, you can construct, from lengths $1$ and $x$, a fourth proportional: $1 : x = x : ?$, and the $?$ is a **length** which behaves in every way like $x^2$. Do it again for $x^3$. Every power of the unknown is now a length, marked on the same line as every other, measured against the same unit. The species collapse. There is one kind of quantity. $x^2 + x$ is a length you can point at, $x^4 - 1000x^3 + 250000x^2$ is a length you can point at, and — critically — a *pair* of such quantities can be laid out at right angles and marked as a point. **That is the Cartesian revolution.** Not the axes. The unit. ## Now turn the ledger Take the table from `quad-02b-the-table`, or the one you filled in by hand. Each row has a cut and a product. Lay the cuts along a horizontal line and, above each, mark a height equal to the product. Do about ten of them. Then join them. Here is what appears, and here is the discipline of this lesson — **for each feature, name the numbered fact from the top of this page that you already proved.** | what you see | what it is | proved where | |---|---|---| | the curve meets the base line at 0 and 500 | the degenerate fields | fact 1 | | a single highest point, above $x=250$ | the ceiling and where it is reached | facts 2, 3 | | the curve is a mirror image about the vertical through 250 | the $\pm d$ pairing | fact 4 | | a horizontal line at height 40000 cuts it **twice** | two cuts give that product | fact 4 | | a horizontal line at 62500 **touches** it once | the two cuts have collided | fact 5 | | a horizontal line at 70000 **misses** it entirely | the impossible demand | fact 6 | | the horizontal line at 62500 lies above the curve everywhere else | the assessor's rule | `quad-03` | | the vertical gap between that line and the curve | the overcharge, $(x-250)^2$ | `quad-03` | Every row of that table is a thing you knew. The picture has told you nothing. What it has done — and it is not nothing — is make all eight of them **visible at once**, so that a question like "how many cuts give a product of 55000?" is answered by looking rather than by computing. That is a genuine gain in speed and in the ability to see a whole situation at a glance. It is not a gain in knowledge. > **Do this.** Draw the horizontal line at height 62500 and shade the region between it and > the curve. You are shading $(x-250)^2$. Now, without measuring anything, say where the > shaded region is thickest and why. Then draw, on separate axes, the height of that shaded > gap against $x$. What curve is it, and which way up? ## The quartic, and why tangency is not a new idea Now plot $x^2(500-x)^2$ — the squared area, from `quad-04-do-not-expand`. It touches the base line at $x=0$ and $x=500$ and does not cross, and it stays above the line everywhere, including outside $[0,500]$. You knew both. It is a square, so it is never negative — that was fact 7, and it took one sentence. Nothing about the drawing established it; the drawing is only obeying it. The tangency is the visual form of the doubled factor. Compare: $x(500-x)$ *crosses* the line at 0 and 500, while $x^2(500-x)^2$ *touches* and turns back. The difference between crossing and touching is the difference between a factor appearing once and appearing twice, which you established in `quad-04` by evaluating signs on either side, with no picture at all. This is worth insisting on because the usual order of teaching is backwards. Students are shown that repeated roots "bounce" and are asked to remember it. The bounce is a *consequence*. The fact is about the sign of a product near a zero, and it is provable in two lines. ## An honest note about what Descartes actually did The phrase "Cartesian coordinates" does a lot of retrospective work. Descartes has no $y$-axis. His figures show one reference line and lengths erected from it at a fixed angle, not necessarily a right angle. He does not plot negative values. He is not, in our sense, graphing a function — the idea of a function as an object with a graph is a century and a half later, and belongs largely to Euler. What Descartes does is associate an equation in two unknown lengths with the locus of points satisfying it, in order to solve *geometric construction problems*, which is what he cared about. So the tidy classroom story — Descartes invented the coordinate plane and thereby graphs — is a compression of about 150 years of work by many people. What is genuinely his, and genuinely revolutionary, is the unit length and the resulting freedom to write and manipulate expressions of mixed degree. Everything else followed from having somewhere to put them. There is a reason to tell students this rather than the tidy version. The tidy version makes the graph look like the point of the whole enterprise, which invites exactly the habit this arc was built to prevent: reaching for a picture before reading a form. --- ## Exit 1. From the graph alone, and without computing: for how many $x$ is the true area 55000? Then confirm your answer by the method of `quad-02-parts-of-500`, and say which was faster and which was more certain. 2. Sketch $x(500-x)$ and $x^2(500-x)^2$ on the same axes, scaled so both fit. Where do they cross? Explain the crossings *algebraically* before you check them. 3. A student says "the maximum is at 250 because that's the vertex of the parabola." What is wrong with this as a justification? Give the justification that is not circular. 4. Everything in this arc was proved before the picture. Name one thing the picture makes *easier* and one thing it makes *harder* to see clearly. 5. Descartes' unit length dissolved the species. Write $x^2 + x$ in the old grammar and explain, in two sentences, what the unit segment supplies that makes the sum legal. 6. *Open.* You have now seen the maximum of $x(500-x)$ established four separate ways: by table, by the deviation identity, by Euclid VI.27, and by the vertex of a parabola. Rank them by how much you would trust each if you had never seen the other three, and defend your ranking. Answers: [`ANSWERS.md`](ANSWERS.md). ## Related - The picture itself, drawn from a ledger you fill in: `quad-06b-the-ledger-turned`. - The refusal that the unit length lifts: `quad-01-species`. - Every fact the picture transcribes: `quad-02-parts-of-500`, `quad-03-assessors-rule`, `quad-04-do-not-expand`.