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IB DP Maths AI · Topic 2 Functions · SL 2.5, 2.6 · AHL 2.9 · ~8 min read

Modelling with functions

The modelling cycle and the families of models IB Maths AI uses: linear and piecewise, quadratic, exponential growth and decay, direct and inverse variation, cubic and sinusoidal models, finding parameters from data, and for HL: logistic, natural logarithm and further piecewise models.

Modelling cycleLinear & piecewise modelsQuadratic modelsExponential growth & decayDirect & inverse variationCubic modelsSinusoidal modelsFitting parameters
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • The modelling cycle: choose a type of function from the shape of the data or the situation; find its parameters; use it to predict; then check whether the answers make sense and where the model breaks down.
  • Recognise the shape: constant rate → linear; symmetric with one turning point → quadratic; constant percentage change → exponential; a repeating cycle → sinusoidal.
  • Exponential models f(x) = kax + c or kerx + c have a horizontal asymptote y = c.
  • Sinusoidal models f(x) = a sin(b(x − c)) + d: amplitude ǀaǀ, period 360°b (or 2πb in radians, HL), principal axis y = d.
  • HL: logistic models L1 + Ce−kx level off at the carrying capacity L; logarithmic models a + b ln x grow ever more slowly.

By the end you can

  • Choose an appropriate model from a description, a graph or data.
  • Find the parameters of linear, quadratic, exponential, direct and inverse variation, cubic and sinusoidal models.
  • Use a model to make predictions, and comment on its validity and limitations.
  • Use logistic, logarithmic and piecewise models. HL

1Which model?

linear: y = mx + cquadratic: ax² + bx + cexponential: ka^x + ccubic: ax³ + bx² + cx + dinverse: y = k/xsinusoidal: a sin(bx) + d
The six model families in the SL course. Learn their shapes so you can spot them in data.
SituationModel
fixed fee + constant ratelinear y = mx + c
projectile, area, profit with one best pricequadratic y = ax2 + bx + c
population, interest, radioactive decay, coolingexponential y = kax + c
y doubles when x doublesdirect variation y = kxn
y halves when x doublesinverse variation y = k/x
tides, daylight, Ferris wheels, temperature over a yearsinusoidal y = a sin(bx) + d

2Quadratic and cubic models

Worked example

A ball's height is h = −4.9t2 + 14t + 1.5 m after t s. Find the maximum height and when the ball lands.

  1. vertex at t = −142(−4.9) = 1.43 s M1
  2. hmax = 11.5 m A1
  3. h = 0 at t = 2.96 s (GDC; the other root is negative, so reject it) A1R1
x20 cm12cut x × x squares from each cornerxV123456100200max V = 263 cm³ at x = 2.43V = x(20 − 2x)(12 − 2x)domain 0 < x < 6
An optimisation model: the GDC's maximum tool finds the best cut.
Worked example

Squares of side x cm are cut from the corners of a 20 cm by 12 cm sheet, and the sides folded up to make an open box. Find the maximum volume.

  1. V = x(20 − 2x)(12 − 2x), 0 < x < 6 M1A1
  2. GDC maximum: x = 2.43 cm A1
  3. Vmax = 263 cm3 A1
Exam tipAlways state the domain of a model from the context. Here 0 < x < 6, because the 12 cm side can't lose more than 12 cm.

3Exponential models

f(x) = kax + c, f(x) = kerx + c
a (or er) is the growth factor; y = c is the asymptote; f(0) = k + c
Worked example

A bacteria colony starts with 500 cells and doubles every 3 hours. Write a model and find the number after 12 hours.

  1. N = 500 × 2t/3 A1
  2. N(12) = 500 × 24 = 8000 cells A1
Worked example

A population is 200 at t = 0 and 450 at t = 4. Model it as P = 200ekt and predict P at t = 10.

  1. 200e4k = 450 ⇒ k = ln 2.254 = 0.203 M1A1
  2. P(10) = 200e2.03 ≈ 1520 A1
  • Cooling follows T = 20 + 70 × 0.9t: the temperature drops towards the room temperature 20 °C, the horizontal asymptote, but never reaches it.
Common mistakePredicting far outside the data. An exponential population model predicts endless growth; real populations run out of food and space. Say this in "comment on the model" questions.

4Direct and inverse variation

  • Direct: y ∝ xn means y = kxn. Inverse: y ∝ 1xn means y = kxn. Find k from one pair of values, then use it.
Worked example

y is directly proportional to x2, and y = 50 when x = 5. Find y when x = 8.

  1. 50 = k × 25 ⇒ k = 2 M1A1
  2. y = 2 × 64 = 128 A1
Worked example

Light intensity I is inversely proportional to the square of the distance d. I = 36 lux at 2 m. Find I at 3 m.

  1. 36 = k4 ⇒ k = 144 M1
  2. I = 1449 = 16 lux A1

5Sinusoidal models

t (min)h (m)24681010203040top 38 m at t = 5bottom 2 m3.446.56h = 20 − 18cos(36t)°axis 20, amplitude 18period 10 min
Read the axis, amplitude and period straight from the graph.
  • From a graph or data: d = max + min2, amplitude a = max − min2, period = time from one peak to the next, b = 360°period (or 2πperiod in radians).
Worked example

A Ferris wheel of diameter 36 m has its lowest point 2 m above the ground and turns once every 10 minutes. A rider starts at the bottom. Find a model for the height h and when the rider first reaches 30 m.

  1. axis 20 m, amplitude 18 m, period 10 min A1
  2. h = 20 − 18cos(36t)° (in radians, HL: h = 20 − 18cos(πt5)) A1
  3. GDC: h = 30 first at t = 3.44 min A1
  • Why −cos? cos starts at its maximum; −cos starts at its minimum, which matches a rider starting at the bottom.
GDC (calculator) methodTo fit a model to data, use the regression tools: linear, quadratic, cubic, exponential, power and sinusoidal regression. Write the model with its parameters to 3 s.f., and give R2 if asked how good the fit is.

6Logistic and logarithmic models HL

f(x) = L1 + Ce−kx, L, C, k > 0
logistic: L is the carrying capacity
tP24681012142505007501000carrying capacity L = 1000P(0) = 100half of L at t = 4.39 (fastest growth)P = 1000/(1 + 9e^(−0.5t)): fast growth at first, then levelling off
Logistic growth: exponential at first, then the curve levels off at the carrying capacity L.
Worked example

A fish population is P = 10001 + 9e−0.5t, t in years. Find the initial population, the long-term population and when it reaches 500.

  1. P(0) = 100010 = 100 A1
  2. as t → ∞, e−0.5t → 0, so P → 1000 A1
  3. 1 + 9e−0.5t = 2 ⇒ t = ln 90.5 = 4.39 years M1A1
  • Logarithmic models f(x) = a + b ln x (x > 0) grow fast at first and then ever more slowly, without an asymptote: for example, perceived loudness against intensity.
  • Piecewise models use different rules on different intervals: a phone plan, a tax system, or a car that accelerates and then cruises. Check that the pieces join where they should.
Linking questionExponential models connect to Sequences and financial mathematics (geometric growth), sinusoidal models need the trig of Trigonometry, 3D geometry and Voronoi diagrams, and HL students can check a model's type by linearising data (Composites, transformations and linearising data). Optimisation with calculus is in Differentiation and optimisation.
Idea for your IA / explorationCollect real data (cooling of a drink, phone battery discharge, daylight hours, a bouncing ball from a video) and compare two or three models. Justify your choice with the context and R², and discuss where the model fails.

Test yourself — cover the answers first

Quick check

  1. Which model fits "the value falls by 12% every year"?
  2. State the asymptote of y = 5 × 0.7x + 3.
  3. y ∝ 1/x and y = 4 when x = 6. Find y when x = 8.
  4. Find the period of y = 4 sin(30x)° + 2.
  5. State the amplitude and axis of y = 7 − 3cos x.
  6. HL What is the carrying capacity of P = 600/(1 + 5e−0.2t)?

Exam-style

  1. The temperature in a city is modelled by T = 8 sin(30(t − 4))° + 18, where t is the month (t = 1 is January). Find the maximum temperature and the month in which it happens.
  2. A model for the value of a car is V = 30 000 × 0.85t + 2000. Find the value after 6 years, interpret the 2000, and comment on using the model for t = 30.
Show answers
  1. exponential, y = k(0.88)ˣ
  2. y = 3
  3. 3
  4. 12
  5. amplitude 3, axis y = 7
  6. 600
  7. max = 18 + 8 = 26 °C
    30(t − 4) = 90 ⇒ t = 7, July
  8. V(6) = 30 000 × 0.856 + 2000 = $13 314.49
    2000 is the lowest value the model allows (scrap value): the horizontal asymptote
    at t = 30 the model gives about $2228; a real 30-year-old car may be worth nothing, or more as a classic, so the prediction is unreliable
Practise this chapter
IB Maths practice papers
Next chapterComposites, transformations and linearising data →
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.