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AI SL · 12.7

Mathematical Models

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12.7.1

Linear, quadratic, cubic, variation and exponential models

12.7.1

Linear, quadratic, cubic, variation and exponential models

Target: I can build a linear model from two data points, and say what its gradient and intercept mean. (+5 more)

By the end of this section

Key points

Choosing the model

  • A constant rate of change gives a linear model \(f(x)=mx+c\): m is the change in y for each 1 in x, c is the value when x = 0.
  • A rise then a fall (or a fall then a rise) gives a quadratic model \(ax^2+bx+c\). The turning point is the maximum or minimum. a < 0 gives a maximum.
  • A volume built from three lengths gives a cubic model.
  • A quantity that settles at a fixed level gives an exponential model with an asymptote: \(f(x)=c+ka^x\) with \(0<a<1\), or \(f(x)=c+ke^{rx}\) with r < 0. The asymptote is \(y=c\).

Variation

  • \(f(x)=ax^n\). If n > 0, y grows with x (direct variation). If n < 0, y falls as x grows (inverse variation).
  • Find a from one pair of values: put them in and solve. Then use the model for any other value.
  • For n < 0 the y-axis is a vertical asymptote: y gets very large as x gets close to 0, and the model has no value at x = 0.

Using a model well

  • Domain: only the inputs that make sense. Time cannot be negative, a length must be positive, a height must not be negative.
  • From the calculator: graph the model, then use Maximum, Zero or Intersection. Or write the model equal to the target and solve for x.
  • A prediction is reasonable only inside the domain, and when it does not break the situation (a negative temperature below the room, a negative height).
  • State the assumption: a linear model assumes a constant rate, an exponential model with an asymptote assumes the quantity settles at c.

On the calculator

Graphing a model and finding a maximum or a zero

  • Enter the model. Set a window that shows the whole domain.
  • Use the maximum or the zero on the graph, and read both coordinates.
  • Check the value makes sense in the context: inside the domain, and not negative if it is a length or a height.

Screenshot and video to come

Solving a model for the input

  • Write the model equal to the target value.
  • Solve for the unknown. For a quadratic or a cubic there may be more than one solution.
  • Keep the solutions that are inside the domain.

Screenshot and video to come

Finding the constants from two points

  • Put each point into the model to write two equations.
  • Solve the pair for the two unknown constants.
  • Check by putting one of the points back into the model.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A call-out fee and an hourly rate

A plumber charges a fixed call-out fee plus a fee for each hour of work. A 2-hour job costs £135 and a 5-hour job costs £270. The cost is C = mt + c, where t is in hours. (a) Find m and c. (b) Say what m and c mean. (c) Find the cost of a 3.5-hour job.

Working
  1. Find m

    \(m=\dfrac{270-135}{5-2}=45\)

    Change in cost ÷ change in time.

  2. Find c

    \(135=45\times2+c\)

    So c = 45.

  3. Meaning

    m = 45: £45 for each hour. c = 45: the call-out fee, the cost at t = 0.

    Say it with the units and the context.

  4. Part (c)

    \(C=45\times3.5+45=202.50\)

    A 3.5-hour job costs £202.50.

Video walkthrough coming
A kicked ball

A ball is kicked from a height of 0.5 m. Its height h metres after t seconds is \(h=-4.9t^2+11.2t+0.5\), until it lands. (a) Find the greatest height and the time it is reached. (b) Find when the ball lands. (c) Write down a reasonable domain for the model.

0.511.522.512345678thh(1.14, 6.9)
Working
  1. The graph

    Graph the model. The graph is a downward parabola, so the vertex is the maximum.

    a < 0 means a maximum.

  2. Maximum

    From the calculator: maximum at t = 1.14, h = 6.9

    The ball reaches 6.9 m after 1.14 s.

  3. Landing

    Solve for t: h = 0 gives t = 2.33 (the other solution is negative)

    Time cannot be negative, so keep the positive solution.

  4. Domain

    \(0\le t\le2.33\)

    From the kick until it lands.

Video walkthrough coming
An open box

An open box is made from a rectangular card 30 cm by 20 cm. A square of side x cm is cut from each corner and the sides are folded up. The volume of the box is \(V=x(30-2x)(20-2x)\) cm³. (a) Write down a reasonable domain for x. (b) Find the volume when x = 3. (c) Find the greatest volume and the value of x that gives it.

246810200400600800100012001400xVV(3.92, 1060)
Working
  1. Domain

    \(0<x<10\)

    The sides 30 − 2x and 20 − 2x must both be positive, and the shorter side gives x < 10.

  2. Part (b)

    \(V=3\times24\times14=1008\)

    30 − 6 = 24 and 20 − 6 = 14.

  3. Part (c)

    From the calculator: maximum at x = 3.92, V = 1060

    Graph V on the domain and use Maximum.

  4. Answer

    The greatest volume is 1060 cm³, when x = 3.92 cm.

    State both with units.

Video walkthrough coming
Light and distance

The brightness I (lux) of a lamp at a distance d metres is modelled by \(I=kd^{-2}\). At 2 m the brightness is 45 lux. (a) Find k. (b) Find the brightness at 5 m. (c) Find the distance at which the brightness is 20 lux. (d) What happens to the model as d gets close to 0?

Working
  1. Find k

    \(45=k\times2^{-2}\)

    k = 45 × 4 = 180.

  2. Part (b)

    \(I=180\times5^{-2}=7.2\)

    7.2 lux at 5 m.

  3. Part (c)

    \(180d^{-2}=20\)

    Solve for d: d = 3 m.

  4. Part (d)

    I gets very large. The y-axis is a vertical asymptote.

    The model has no value at d = 0: a lamp is not infinitely bright.

Video walkthrough coming
A bottle warming up

A bottle of water is taken from a fridge at 4 °C into a room at 22 °C. Its temperature after t minutes is modelled by \(T=c+ka^t\), with 0 < a < 1. After 10 minutes it is 11 °C. (a) Write down c and k. (b) Find a. (c) Write down the equation of the horizontal asymptote and say what it means. (d) Find when the temperature is 20 °C.

Working
  1. The constants

    c = 22 (the room), and T(0) = 4 gives c + k = 4, so k = −18.

    The temperature settles at the room temperature.

  2. Find a

    \(11=22-18a^{10}\)

    \(a^{10}=11/18\) so a = 0.9519.

  3. Asymptote

    T = 22: the temperature of the room. The bottle never quite reaches it.

    c is the asymptote.

  4. Part (d)

    \(22-18(0.9519)^t=20\)

    Solve for t: t = 44.6 minutes.

Video walkthrough coming
How far can a model be trusted?

The height h cm of a bean plant after t days is modelled by \(h=-2.5t^2+30t+10\). (a) Find the time at which the model has its greatest height, and the height. (b) The model is used for \(t\ge0\). Explain why it cannot be used for t = 14. (c) Write down a reasonable domain.

Working
  1. Maximum

    From the calculator: maximum at t = 6, h = 100

    A parabola turns at t = 6.

  2. Part (b)

    h(14) = −60 cm

    A negative height is impossible, and the plant would have shrunk since day 6.

  3. Part (c)

    \(0\le t\le6\)

    A plant does not get shorter. The model stops being reasonable after its maximum.

Video walkthrough coming