Annotated exemplar · AA HL · deliberately mid-band
Where should the 125 ml line go on a wine glass? A mid-band AA HL draft, annotated to show how to improve
Exemplar written by IB Math Revision for teaching — not a real student's IA, not moderated by the IB. Study it — don't reuse it: submitting or copying it is academic misconduct. The data and the student voice are illustrative, and the marks are our judgement. Schools check coursework with similarity software such as Turnitin.
A deliberately mid-band AA HL draft: a volume of revolution about the y-axis and related rates, used to mark 125 ml, 175 ml and 250 ml lines on a restaurant's wine glass. The mathematics is correct; the marks are lost to an unjustified fit, missing rigour, a systematic test error reported as “accurate”, and uneven notation. Every annotation says what lifts that criterion to the next band.
How this exemplar would be marked
| Criterion | Mark | Why |
|---|---|---|
| A · Presentation | 3/4 | Coherent and well organised around a precise aim with a test, and the conclusion gives the three heights. Held at 3: not concise (the same calculation written out three times, GDC keystrokes in the text), and the conclusion does not answer the accuracy part of the aim. |
| B · Mathematical communication | 2/4 | Relevant notation and representations — the integral is displayed and the disc method is written properly — but not consistent: calculator notation for the model and the solver, a table without units, spurious precision in the readings, ml and cm³ mixed, and an unlabelled Figure 2. |
| C · Personal engagement | 2/3 | Significant: a real problem from the student's weekend job, a real person who will use the answer, and the student's own measurements and tests. Held at 2 because after the introduction the work follows the standard path; no decisions of the student's own (a second glass, real pours, a changed method) shape what happens next. |
| D · Reflection | 2/3 | Meaningful: the model is checked against the brim volume and the lines are tested by filling, and the result is interpreted. Held at 2 because the systematic difference in the test is dismissed as “accurate”, and the limitations are listed rather than evaluated or acted on. |
| E · Use of mathematics | 4/6 | Relevant AA HL mathematics, correct throughout and with understanding: the disc method explained, the integral set up about the y-axis, the fundamental theorem of calculus and the chain rule for related rates. Held at 4 because it is not rigorous: the fit is justified only by R², the uniqueness of each height is not argued, no error is carried through, and all integration is left to technology. |
| Total | 13/20 | About 13/20. The mathematics is correct and at HL level; what holds it in the middle is depth — a model choice that is never tested, a systematic error that is not investigated, and calculus that is not connected back to the question. Each annotation shows how to close one gap. |
Marks are our judgement of this teaching exemplar against the current criteria, explained criterion by criterion. They are not IB moderation results.
Excerpts with examiner annotations
Free sections are shown below with comments; the rest is in the full exemplar, available in the protected viewer with the IA package or a Pro plan.
Introduction
At weekends I work in my aunt's small restaurant. Wine is sold by the glass in 125 ml and 175 ml measures, and a large glass is 250 ml (two 125 ml measures), because by law wine sold by the glass has to be sold in these quantities. At the moment the staff use a metal measure and then pour it into the glass, which is slow on busy nights. My aunt asked if we could just mark lines on the glasses instead, and I realised that this is a volumes of revolution question, because a wine glass bowl is a solid of revolution.
Aim: to find the heights at which the 125 ml, 175 ml and 250 ml lines should be marked on the restaurant's wine glass, using a volume of revolution about the \(y\)-axis, and to test how accurate the lines are.
I will also look at how fast the wine level rises when it is poured, because I noticed that the level seems to rise slowly in the middle of the glass.
C Good: a real problem from the student's own life, with a real person who will use the answer. This is a better starting point than most volume-of-revolution IAs. To lift C from 2 to 3, the student's own decisions have to keep driving the work after the introduction — see the later notes.
A Good: a clear aim with a precise answer (three heights) and a test. Sound organisation from here on is why A reaches 3.
A How to improve: “by law” needs a citation at the point of use — the student lists the regulation in the bibliography, but a footnote here is where the reader needs it. Say which country's law applies, since the quantities differ between countries.
Measuring the glass and fitting a model
I photographed the glass next to a ruler and put the photo into GeoGebra. I put the origin at the lowest point inside the bowl, with the \(y\)-axis vertical through the centre of the glass. Let \(y\) be the height above the origin in cm and \(x\) the radius of the inside of the glass at height \(y\), in cm. I read the outside radius every 1 cm from \(y = 0\) to the rim at \(y = 11\) and subtracted the thickness of the glass, which I measured at the rim with my dad's callipers as 0.15 cm.
| y | x |
|---|---|
| 0 | 0.00 |
| 1 | 1.89 |
| 2 | 2.53 |
| 3 | 3.03 |
| 4 | 3.43 |
| 5 | 3.77 |
| 6 | 3.92 |
| 7 | 4.00 |
| 8 | 4.03 |
| 9 | 3.98 |
| 10 | 3.80 |
| 11 | 3.78 |
Data note: the glass and all measurements in this exemplar are illustrative, generated for this annotated draft; a student must use their own.
I used my GDC to fit regression models to the data with \(x\) as a function of \(y\). The quadratic had \(R^2 = 0.950\), the cubic \(R^2 = 0.982\) and the quartic \(R^2 = 0.990\). I chose the cubic because \(R^2\) is very close to 1 and the quartic only improves it a little but is more complicated:
x = 0.006693y^3 − 0.17236y^2 + 1.41136y + 0.25597
Figure 1 shows that the model fits the points well.
B How to improve: the table has no units and the column headings are just “y” and “x”; the caption should say what was measured and how. The model is written in calculator notation (y^3, no display equation, no number). Write it as a displayed equation, \(x(y) \approx 0.006693y^3 - \dots\), numbered so it can be referred to later, with ≈ because the coefficients are rounded.
B How to improve: radii read from a photograph are given to 0.01 cm, which is more precision than a photo and a ruler can give. State the precision of the readings (about ±0.05 cm) and round to match.
E How to improve: R² alone does not justify a model. Two better reasons are available. First, look at the residuals: the cubic is 0.39 cm too narrow at \(y = 1\). Second, think about the shape: near the bottom a rounded bowl behaves like \(x^2 \approx ky\), so \(x\) has a vertical tangent at the origin, and no polynomial in \(y\) can do that. The cubic has \(x(0) \approx 0.26\) cm, a flat bottom that is not there. Because the volume uses \(x^2\), fitting \(x^2 = py + qy^2 + ry^3\) (through the origin) is the natural HL choice. Here it moves the 125 ml line by only 0.4 mm, since little volume is near the bottom — but saying so, with the number, is what rigour looks like.
D How to improve: the choice of the cubic over the quartic is never tested where it matters. The quartic gives a 125 ml line 0.3 mm lower. Showing that the choice of model changes the answer by less than the measuring precision would turn a guess into a reflection.
D How to improve: the wall thickness is measured only at the rim and assumed to be the same everywhere. Stems join the bowl in a thicker base; this assumption turns out to matter (see the test).
Volume of revolution about the y-axis
In the full exemplar (about 1 page). The disc method for a volume of revolution about the y-axis, and a check of the whole glass against a measuring jug. Open in the protected viewer
Finding the heights of the lines
In the full exemplar (about 1 page). Solving V(h) = 125, 175 and 250 for the heights of the three lines. Open in the protected viewer
Testing the lines
In the full exemplar (about 1 page). Filling the glass with a syringe and comparing measured heights with the predictions. Open in the protected viewer
How fast does the level rise?
In the full exemplar (about 1 page). Related rates: dV/dh from the fundamental theorem of calculus, and the rate at which the level rises at each line. Open in the protected viewer
Conclusion
In conclusion, the lines should be marked at 5.1 cm (125 ml), 6.1 cm (175 ml) and 7.7 cm (250 ml) above the lowest point inside the glass. My test showed that the model is accurate to within 2 mm. I also found that the level rises slowest at the widest part of the glass, which is at 6.7 cm.
Limitations: I only measured the profile every 1 cm, the glass thickness might not be exactly 0.15 cm, and there could be errors reading the photo and the ruler. In the future I could use a more accurate method such as a 3D scan, and I could do the same for other glasses in the restaurant.
A Good: the conclusion answers the aim directly with the three heights. How to improve: it should also answer the second half of the aim — how accurate the lines are — in terms the restaurant cares about (millilitres, not millimetres).
D How to improve: the limitations are listed, not evaluated. The one that matters — the thickness — is here, but without its direction or size. Saying “the lines are probably about 2 mm too low because the wall is thicker near the base, so I would mark them at the measured heights instead” would be critical reflection.
C How to improve: the recommendation to the aunt — the reason for the exploration — is missing. Which heights should she actually use, and would she trust them? Ending with the student's own decision shows engagement to the end.
E How to improve: a 3D scan is a generic extension. The HL extension that fits this exploration is the error analysis: how an error in the radii or in a height reading carries through to the volume, using the derivatives already found.
Bibliography
In the full exemplar (about 0.5 page). Sources and technology used. Open in the protected viewer
What would push it higher?
- Justify the model with residuals and the shape of the glass, not R² alone — at HL, fit x² (what the volume uses) through the origin and compare the heights it gives.
- Add one line of rigour: V′(h) = πx(h)² > 0, so V is strictly increasing and each volume has exactly one height.
- Treat the test as a result: all three measured heights are above the predictions, so find the systematic cause (the wall thickness) and correct the lines.
- Carry errors through: use dV ≈ πx² dh to turn millimetres into millilitres, and say how precise the lines need to be.
- Finish with the student's own decision: the heights to mark and why, tested with real pours.
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All 17 annotated exemplars, including four deliberately mid-band drafts.
Read the whole exploration
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