Higher Level only, and it answers the question 2.5 deliberately left open.
Start by reminding them what 2.5 concluded: the data could not choose, so the context did. Then point out that the context handed over a number, 180, and ask where that number goes in a model. It goes in as L, before anything is fitted.
Fit C and k with them from the same two points the exponential used, then check hour 2: 7.91 against a measured 8. The logistic was available all along and fits just as well. The exponential was never the only model that fitted; it was the only one anybody tried.
Then step the figure and let the exponential run off the top of the window while the logistic flattens. The sentence to land: early on the two are indistinguishable, and by the time they differ it is too late to choose, so the limit has to go in at the start.
Views three and four are the piecewise pair. Show a = 1 first, with the visible jump, and ask what value would close it.
| Question | Answer |
|---|---|
| 1. C from P(0) = 2 | 180/(1 + C) = 2, so C = 89. |
| 2. P at hour 10 | 167.03. |
| 3. Hottest hour | 8 + 24/4 = 14, so 2pm. |
| 4. a for continuity | 4a = 3, so 0.75. |
| 5. Why the exponential was worse | B. The situation has a ceiling and only one model does. |
1 markL, identified from the context rather than the data.
1 markEach parameter from a condition, with the equation shown.
1 markThe prediction.
1 markInterpreting a parameter in context.
The interpretation mark is on almost every HL modelling question and is always a sentence. Give them the form: "L = 180 is the carrying capacity, the size of the year group, which the number of sign-ups cannot exceed."
| They wrote | What happened |
|---|---|
| 90 on question 1 | Gave 1 + C. |
| 178 | Subtracted instead of dividing. The relationship is a fraction. |
| 180 on question 2 | Gave the ceiling. Worth asking whether the curve ever reaches it: at hour 10 it is still 13 short. |
| 180.08 | Brackets, and the dangerous one: 180/1 + 89e−7.045 looks like the ceiling, so it passes a sanity check it should fail. |
| 2048 | Used the exponential from 2.5. |
| 8 on question 3 | Gave the phase shift, which is where the curve crosses its mean going up, not the peak. |
| 2 | Gave the minimum, twelve hours away. |
| 20 | Gave where it crosses the mean coming back down. |
| 3 on question 4 | Gave the value the pieces must meet at, not a. |
| 1.5 | Divided by 2 instead of 2². The left piece at x = 2 is 4a. |
"Where does L come from?" Never from the data. From the situation: a year group, a petri dish, a market, a stadium. If a question gives you a maximum anywhere in the stem, that is L, and a student hunting for it in the numbers has misread the model.
"Why is C so big?" Because the start is far below the ceiling. C = L/P₀ − 1, so starting at 2 out of 180 gives 89. It is worth deriving that once: it turns C from an arbitrary letter into the ratio of the ceiling to the start.
"Radians or degrees?" Radians, at HL, unless a degree symbol appears. Say it on this page and say it again in 3.7. In degrees mode the daily temperature model has a period of about 1375 hours, so the curve is almost flat across a day and no error message appears.
| Stage | What to do |
|---|---|
| Demonstrate | Set radians first, visibly. Then solve for C and for k in two solver calls, substitute both back to confirm P(0) = 2 and P(1) = 4, and only then predict. Doing the check in front of them is what makes them do it. |
| Where they stick | Brackets round the whole denominator. 180/(1 + 89e^(-kt)) typed without them gives about 180, which looks like the carrying capacity and therefore survives a glance. Make them read the expression back before pressing enter. |
| The check | Substitute the fitted parameters back into the conditions you fitted from. If P(0) is not 2, the arithmetic is wrong, and the whole question is downstream of it. |
A table is better than a graph for a logistic. The flattening is more convincing as a column of numbers creeping towards 180 than as a curve that looks flat for other reasons.
| Step | What |
|---|---|
| 1 | Recall 2.5. The context gave a number: 180. Ask where it goes. |
| 2 | Fit C, then k, from the same two points the exponential used. |
| 3 | Check hour 2. 7.91 against 8. The logistic fitted all along. |
| 4 | Step the figure. Early agreement, late divergence, and why that order matters. |
| 5 | Half-life, by counting halvings rather than solving. |
| 6 | The natural log model, and equal multiplications giving equal additions. |
| 7 | Radians, loudly. Then the temperature model and its phase shift. |
| 8 | Piecewise with a = 1, the visible jump, then solve for 0.75. |
Do not say the logistic "is" exponential at the start. It is close to one, and the difference is exactly what makes it the better model. Saying they are the same early on invites students to use whichever is easier, which is how 2.5 went wrong.
Do not let a parameter be found without a condition written down. C = 89 with no equation beside it is unmarkable, and it is also how students lose track of which letter came from where.