Nothing grows for ever, so model the ceiling

logistic against exponential

167.03logistic at hour 10

2048exponential at hour 10

ceiling 180verdict

Both fitted to the same two points. One of them knows how many students there are.

The figure needs JavaScript. It plots the logistic 180/(1 + 89e^−0.7045t) against 2 × 2ₓ: the logistic gives 2, 4, 7.91, 49.61 and 167.03 at hours 0, 1, 2, 5 and 10 and never passes 180, while the exponential passes 180 at hour 7 and reaches 2048 by hour 10. Later views show a sinusoidal temperature model peaking at 36°C at 2pm and a piecewise model that only joins when a = 0.75. Everything below stands on its own.

The logistic, and what each letter does

P = L / (1 + Ce^−kt), with L, C and k all positive.

L is the carrying capacity: the horizontal asymptote the curve climbs towards. Here the year group, so L = 180.
C comes from the start. At t = 0, P = L/(1 + C) = 2, so 1 + C = 90 and C = 89.
k comes from a second point. At t = 1, P = 4 gives 1 + 89e^−k = 45, so e^−k = 44/89 and k = 0.704.

Then check the point you did not use: at t = 2 it gives 7.91, where the data said 8. Close enough to call the family right.

Early on, a logistic and an exponential are almost the same curve, which is why the exponential looked fine in 2.5. The difference only shows up as the population approaches its limit, and by then it is too late to choose. The limit is a fact about the situation, so put it in the model at the start.

Hour	Logistic	Exponential
0	2	2
1	4	4
2	7.91	8
5	49.61	64
10	167.03	2048
20	179.99	over two million

Half-life

The same exponential idea read as decay. Carbon-14 has a half-life of 5,730 years, so A = A₀(½)^t/5730.

After 5,730 years: half left.
After 11,460 years: a quarter, because (½)² = ¼.
After 17,190: an eighth.

The useful habit is counting half-lives rather than solving an equation: a quarter is two, an eighth is three. If the fraction is not a power of a half, then take logs.

A natural logarithm model

f(x) = a + b ln x grows, but slower and slower. With a = 2 and b = 5:

At x = 1: 2, because ln 1 = 0.
At x = 10: 13.51. At x = 100: 25.03. At x = 1000: 36.54.

Every tenfold increase in x adds the same 11.51. That is the signature: equal multiplications of the input give equal additions to the output, which is the exact mirror of an exponential.

Sinusoidal, now with a phase shift

At Higher Level, radians are assumed unless a degree symbol says otherwise. That is the reverse of Standard Level, where sinusoidal models are in degrees. A machine in the wrong mode here does not give a slightly wrong answer, it gives a period that is out by a factor of about 57.

Bangkok air temperature over a day: T = 6 sin(π(t − 8)/12) + 30, with t the hour.

Amplitude 6°C.
Period 2π ÷ (π/12) = 24 hours, which is the only period a daily model can have.
Principal axis 30°C, the mean.
Phase shift c = 8, so the whole curve is moved 8 hours later. The peak is therefore at t = 8 + 6 = 14, two in the afternoon, and the minimum at 2 in the morning.

So the model runs between 24°C and 36°C, and gives 24.8°C at midnight. The phase shift is what makes it a model of a day rather than of a mathematical curve.

Piecewise, and making it join

A function given in two pieces:

f(x) = ax² for x < 2, and f(x) = 1 + x for x ≥ 2

At x = 2 the second piece gives 1 + 2 = 3.
For the first piece to arrive at the same value, 4a = 3, so a = 0.75.
With a = 1 instead, the left piece arrives at 4 and the right starts at 3: a jump of 1, and the graph is in two disconnected pieces.

You are not asked for the formal definition of continuity. You are asked to make the two pieces meet, which is one equation.

On the GDC: fitting a logistic

L comes from the context, then C and k come from two points, and the machine solves for them. The order matters: get L wrong and nothing else can be right.

When you may use it. Applications. A calculator is allowed in every paper.

TI-Nspire CX II

Set Angle to Radian for this sub-topic: doc → Settings → Document Settings
With L = 180 known, solve for C: nSolve(180/(1+c)=2, c) gives 89
Then for k: nSolve(180/(1+89*e^(-k))=4, k) gives 0.7045 Use the x² key or the right arrow to leave the exponent: typing ^ opens a superscript box and everything after it stays inside.
Define it and predict: p(t):=180/(1+89*e^(-0.7045t)), then p(10) gives 167.03

Casio fx-CG50

MENU → Run-Matrix, then SHIFT MENU SET UP and set Angle to Rad
MENU → Equation → F3 Solver, enter 180÷(1+C)=2 and solve for C: 89
Then 180÷(1+89e^(-K))=4 solved for K: 0.7045
MENU → Table with Y1=180÷(1+89e^(-0.7045X)) reads off every hour at once

The mark people lose. Giving the Solver a starting value that sends it to the wrong root, and not noticing. Both of these equations have one sensible solution, but the solver reports whatever it converges to, so check it: put your C and k back in and confirm P(0) = 2 and P(1) = 4. The other one is degrees mode, which at Higher Level turns a 24-hour period into about 1375 hours, so the model comes out almost flat across a whole day without a single error message.

Your turn

1. For P = 180/(1 + Ce^−kt) with P(0) = 2, what is C?

2. With C = 89 and k = 0.7045, what is P at hour 10, to 2 decimal places?

3. For T = 6 sin(π(t − 8)/12) + 30, at what hour is the temperature highest?

4. f(x) = ax² for x < 2 and 1 + x for x ≥ 2. What value of a makes the pieces join?

5. Why was the exponential in 2.5 a worse model than this logistic, even though it fitted the data just as well?

Question 5. Why was the exponential a worse model than the logistic, even though it fitted the data just as well? Because exponentials are less accurate than logistics. Because the situation has a ceiling of 180 and only the logistic has one, so the exponential had to become impossible. Because the exponential had only two parameters. It was not worse; both are equally good over the range measured.

Where the marks go

1 markIdentifying the model and its limit, L, from the context.

1 markEach parameter found from a condition, with the equation shown.

1 markThe prediction.

1 markAn interpretation of a parameter in the context: what the carrying capacity means, or what the phase shift means.

That last mark is almost always available on an HL modelling question and it is always a sentence. "L = 180 is the size of the year group, which the number of sign-ups cannot exceed" earns it.