Module 15 of 47Lesson 1 of 8 · Models 1 · What is a model? Linear & piecewise linear models
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Mathematical models · Essential understanding 1

What is a model? Linear & piecewise linear models

IB AI SL 2.5with Prof. Kitso
💻 Using the TI-Nspire CX Student Software on a laptop
  • The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
  • menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
  • enter = Enter  ·  tab = Tab (next box)  ·  esc = Esc  ·  del = Backspace.
  • Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
  • Fraction: type /  ·  √: type sqrt(  ·  π: type pi  ·  eˣ: type exp(  ·  store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
  • New page: Ctrl+I (or the Insert menu)  ·  change page: Ctrl+← / Ctrl+→  ·  graph entry line: Ctrl+G  ·  var: Ctrl+L.
  • Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
  • IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
  • If you type nDeriv(, the software shows it as centralDiff( — it is the same command.
The big idea: A mathematical model is an equation or graph that describes a real situation so we can predict and decide. The simplest model is a straight line. When the rate changes part-way, we join several straight lines: a piecewise linear model.
By the end of this lesson you can:
  • explain what a model, a variable and a parameter are
  • write a linear model y = mx + c from a story or from two points
  • interpret the gradient and the intercept in context
  • find where two linear models give the same value
  • build, graph and use a piecewise linear model
🧑🏾 Imagine this…
You jump in a taxi in Gaborone. The meter starts at P12 the moment you sit down, and then adds P6.50 for every kilometre. You do not need to take the trip to know the price of a 10 km ride: you can work it out. The rule “start at 12, add 6.5 for each km” is a model of the taxi fare. It is simple, and it is not perfect (traffic jams may add waiting charges), but it is useful.
Mathematical model
an equation, graph or table that represents a real-life situation. It is always a simplification.
Variable
a quantity that changes: the distance d and the cost C are variables.
Parameter
a fixed number inside the model that we choose or find (for example 12 and 6.5 in C = 12 + 6.5d).
Gradient m
how much y changes when x increases by 1. In context: the rate of change (P per km, litres per minute…).
y-intercept c
the value of y when x = 0. In context: the starting value or fixed charge.
Piecewise function
a function made of different rules for different intervals of x.
Domain
the x-values for which the model makes sense.
★ Linear model
  • A linear model has the form . The graph is a straight line.
  • Gradient from two points: . Then use one point to find c.
  • Use a linear model when equal steps in x give equal steps in y (a constant rate).
  • Two models equal (e.g. two phone plans cost the same): put the right-hand sides equal and solve, or find the intersection of the two graphs on your GDC.
  • Piecewise model: write each rule with its interval, for example y = 2x for 0 ≤ x ≤ 5 and y = x + 5 for 5 < x ≤ 10. At the join the two rules should give the same y (if the real quantity does not jump).
⚠️ Mistakes that lose marks
  • Always give units in your interpretation: “P6.50 per km”, not just “6.5”.
  • In a piecewise model the second rule must start from the value where the first rule ended. Writing 14w for the second rule of a water bill forgets the first 10 kilolitres you already paid for.
  • A model may only make sense for some x-values (a distance cannot be negative). State the domain when asked.
Prof. Kitso says
In an exam, “interpret” means: say what the number means in the story, with its units. “The gradient is 25” earns nothing. “The tank fills at 25 litres per minute” earns the mark.
🧪 Try it yourselfTaxi fare lab
C = b + m·x. Move b (the starting charge) and m (the price per km). Set b = 12 and m = 6.5: the graph passes through (9.4, 73.1) and about (13.54, 100) from Example 1.1. What changes if only b changes? What changes if only m changes?

Worked examples

Example 1.1A taxi fare model
A taxi in Gaborone charges a fixed P12 plus P6.50 per kilometre. The cost for a journey of d km is C pula.
  1. (a) Write a model for C in terms of d.
  2. (b) Find the cost of a 9.4 km journey.
  3. (c) Neo has P100. What is the longest journey she can pay for?
🧠 Think it through
  1. Fixed charge is the y-intercept (12). Price per km is the gradient (6.5).
  2. Substitute d = 9.4.
  3. Put C = 100 and solve for d. This is a linear equation, so you can do it by hand or with nSolve.
✍️ By hand
🔢 On your TI-Nspire CX II
  1. Open a Calculator page and type 6.5*9.4+12 enter ↵💻 Enter.
  2. Use menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve or type nSolve(6.5*d+12=100,d) enter ↵💻 Enter.
1.1*DocDEG6.5·9.4+1273.1nSolve(6.5·d+12=100,d)13.5385
Journey cost, and the distance Neo can afford
✅ Answer (a) C = 6.5d + 12. (b) P73.10. (c) 13.5 km (3 s.f.).
She can pay for 13.5 km, not 14 km: round down here, because 14 km would cost P103.
Example 1.2A model from two measurements
Kagiso fills a water tank at a steady rate. After 2 minutes it holds 60 litres and after 5 minutes it holds 135 litres. Let V be the volume in litres after t minutes.
  1. (a) Find the gradient and say what it means.
  2. (b) Find the equation of the model.
  3. (c) How long until the tank holds 500 litres?
🧠 Think it through
  1. Gradient .
  2. Use the point (2, 60) in V = mt + c to find c. Interpret c as the volume at t = 0.
  3. Set V = 500 and solve.
✍️ By hand
, so
🔢 On your TI-Nspire CX II
  1. On a Calculator page type (135-60)/(5-2) enter ↵💻 Enter to get m.
  2. Type nSolve(60=25*2+c,c) enter ↵💻 Enter to get c.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
  3. Type nSolve(25*t+10=500,t) enter ↵💻 Enter.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
1.1*DocDEG135−605−225nSolve(60=25·2+c,c)10nSolve(25·t+10=500,t)19.6
✅ Answer (a) m = 25: the tank fills at 25 litres per minute. (b) V = 25t + 10 (the tank already held 10 litres at t = 0). (c) 19.6 minutes.
Check: V(5) = 25 × 5 + 10 = 135 ✓. Always test your equation on the other point.
Example 1.3Which phone plan is cheaper?
Plan A costs P50 per month plus P0.80 per minute of calls. Plan B has no monthly fee but costs P1.30 per minute. Let x be the number of minutes used in a month.
  1. (a) Write the monthly cost of each plan.
  2. (b) For how many minutes do the plans cost the same? What is that cost?
  3. (c) Mpho uses about 150 minutes a month. Which plan should she choose, and how much does she save?
🧠 Think it through
  1. Plan A: C = 0.8x + 50. Plan B: C = 1.3x.
  2. Equal cost: put 0.8x + 50 = 1.3x. On the GDC, graph both and find the intersection.
  3. Compare the two costs at x = 150.
✍️ By hand
Cost at 100 minutes:
At 150 minutes: A ; B
🔢 On your TI-Nspire CX II
  1. Open a Graphs page (ctrl💻 Ctrl doc ▾💻 File menu then 2). Type 0.8x+50 in f1 and 1.3x in f2, then press enter ↵💻 Enter.
    1.1*DocDEG02002800xy5050f1=0.8x+50f2=1.3x
  2. Set the window so both lines are visible: menu💻 click menu on keypad ▸ 4: Window / Zoom ▸ 1: Window Settings (x from 0 to 200, y from 0 to 280).
  3. Find the crossing point: menu💻 click menu on keypad▸6: Analyze Graph▸4: Intersection. Click to the left of the point, then to the right, and press enter ↵💻 Enter.
    1.1*DocDEG02002800xy5050f1=0.8x+50f2=1.3x(100, 130)
✅ Answer (a) A: C = 0.8x + 50; B: C = 1.3x. (b) 100 minutes, costing P130 each. (c) Plan A (P170 against P195), saving P25.
Below 100 minutes plan B is cheaper; above 100 minutes plan A is cheaper. The line with the larger gradient (B) is the one that grows faster.
Example 1.4A piecewise water bill
In Francistown the water bill is P8 per kilolitre (kL) for the first 10 kL used in a month. Every kL above 10 kL costs P14. Let B be the bill in pula for w kL.
  1. (a) Write B as a piecewise function of w.
  2. (b) Find the bill for 7 kL and for 16 kL.
  3. (c) Thato's bill is P290. How much water did she use?
🧠 Think it through
  1. For 0 ≤ w ≤ 10 the rule is B = 8w.
  2. At w = 10 the bill is P80. Above 10 kL you pay P80 plus P14 for each extra kL: B = 80 + 14(w − 10).
  3. P290 is more than P80, so Thato is on the second rule.
✍️ By hand
;
🔢 On your TI-Nspire CX II
  1. Check the join: the first rule at w = 10 gives 80 and the second rule at w = 10 gives 80 + 0 = 80, so the graph has no jump.
  2. To graph on the GDC, use the piecewise template from the template palette in a Graphs entry line, or enter each piece with the “such that” bar and its interval (for example 8x | 0 ≤ x ≤ 10).
  3. Type 80+14*(16-10) enter ↵💻 Enter, then nSolve(80+14*(w-10)=290,w) enter ↵💻 Enter.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
1.1*DocDEG80+14·(16−10)164nSolve(80+14·(w−10)=290,w)25
1.1*DocDEG0303500xy550B(w)(10, 80)(25, 290)
✅ Answer (a) B = 8w for 0 ≤ w ≤ 10; B = 80 + 14(w − 10) for w > 10. (b) P56 and P164. (c) 25 kL.
The gradient of the graph jumps from 8 to 14 at w = 10, but the graph itself is joined. The kink is where the price changes.
Example 1.5A cycling journey (distance–time graph)
Kabo cycles from Kasane. For the first 1.5 hours she rides at 20 km/h. She then rests for half an hour. After the rest she rides at 15 km/h. Let d be her distance in km from Kasane after t hours.
  1. (a) Write d as a piecewise function of t for 0 ≤ t ≤ 4.
  2. (b) Sketch the graph and explain the flat part.
  3. (c) At what time has she travelled 60 km?
🧠 Think it through
  1. Piece 1 (0 ≤ t ≤ 1.5): d = 20t. At t = 1.5 she has gone 30 km.
  2. Piece 2 (1.5 < t ≤ 2): she is resting, so d stays at 30 (gradient 0).
  3. Piece 3 (t > 2): she starts again from 30 km: d = 30 + 15(t − 2).
✍️ By hand
🔢 On your TI-Nspire CX II
  1. Graph the three pieces and check that the corners are at (1.5, 30) and (2, 30).
  2. Type nSolve(30+15*(t-2)=60,t) enter ↵💻 Enter.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
1.1*DocDEG04.5700xy0.510d(t)(2, 30)(4, 60)
1.1*DocDEGnSolve(30+15·(t−2)=60,t)4
✅ Answer (a) as above. (b) A horizontal segment means no distance is gained: Kabo is resting. (c) After 4 hours.
The steepness of each piece is the speed: 20 km/h, then 0 km/h, then 15 km/h.

Challenge questions

⭐ Challenge 1.6A burning candlePaper 1 style
A candle is 20 cm tall when lit. After 3 hours it is 14 cm tall. Assume it burns at a constant rate. Let h cm be its height after t hours.
  1. (a) Find the equation of the linear model h = mt + c.
  2. (b) Interpret the gradient.
  3. (c) State a suitable domain for t.
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Two points: (0, 20) and (3, 14).
  2. Negative gradient means the height is falling.
  3. The candle cannot have negative height: solve h ≥ 0.
🔢 On your TI-Nspire CX II
  1. Type (14-20)/(3-0) enter ↵💻 Enter for the gradient. Then c = 20 because t = 0 gives h = 20.
1.1*DocDEG14−203−0-2nSolve(20−2·t=0,t)10
✅ Answer (a) h = 20 − 2t. (b) The candle burns down by 2 cm every hour. (c) 0 ≤ t ≤ 10 (it is used up after 10 hours).
⭐ Challenge 1.7Delivery chargesPaper 2 style
A courier company in Maun charges for a parcel of mass w kg as follows: a flat P25 for parcels up to and including 2 kg; for parcels over 2 kg and up to 10 kg, P25 plus P8 for each kilogram above 2 kg; for parcels over 10 kg, a flat P89.
  1. (a) Write the charge C as a piecewise function of w.
  2. (b) Find the charge for a 6.375 kg parcel.
  3. (c) Show that the model has no jumps at w = 2 and at w = 10.
  4. (d) A parcel costs P73 to send. Find its mass.
  5. (e) Give one reason why the company may use a flat charge above 10 kg.
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Piece 1: C = 25. Piece 2: C = 25 + 8(w − 2). At w = 10 piece 2 gives 25 + 64 = 89. Piece 3: C = 89.
  2. Substitute w = 6.375 into piece 2.
  3. Evaluate both neighbouring rules at the join: they must agree.
  4. P73 is between 25 and 89, so use piece 2.
✍️ By hand
At : and ✓. At : and ✓.
🔢 On your TI-Nspire CX II
  1. Type 25+8*(6.375-2) enter ↵💻 Enter and 25+8*(10-2) enter ↵💻 Enter.
  2. Solve part (d) with nSolve(25+8*(w-2)=73,w) enter ↵💻 Enter.
    📍 Where to find it on the TI-Nspire
    nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve
1.1*DocDEG25+8·(6.375−2)6025+8·(10−2)89nSolve(25+8·(w−2)=73,w)8
1.1*DocDEG0141000xy220C(w)(2, 25)(10, 89)
✅ Answer (a) as above. (b) P60. (c) Both rules give 25 at w = 2 and 89 at w = 10, so there is no jump. (d) 8 kg. (e) For example: very heavy parcels are put on a pallet and sent at a fixed rate, or the company wants a simple price for bulk freight.
“Show that” means you must display the working. Writing only “it joins” earns no marks.
Next: Lesson 2 · Quadratic models
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