Function Modelling

A model is a function that describes a real situation well enough to answer questions about it: how low will the tire pressure be by tomorrow, when will the tea be cool enough to drink, how many people will live here in ten years? This lesson pulls together every function type in the course. You’ll choose a type that fits the data, build the equation from a few key points, check whether it’s reasonable, and use it to predict. Trig functions here use radians.

Key ideas

The modelling cycle

1. Look at the data. Plot it, or study the table. Is it increasing, decreasing, or repeating? Does it level off?
2. Choose a type. Use the fingerprints from comparing function types and what you know about the situation.
3. Find the parameters from key points: the starting value, a maximum or minimum, a period, a ratio.
4. Check the fit. Compare the model’s values with the data. The differences (data minus model) are called residuals; for a good model they’re small and don’t follow an obvious pattern.
5. Use and interpret. Predict values, answer the question, and explain what each parameter means in context.
6. Think about limits. Is the model reasonable for the inputs you’re using? Predictions far outside the data (extrapolation) are much riskier than predictions inside it (interpolation).

Choosing a type

The situation or data shows…	Try…
a constant rate of change; constant first differences	linear: $y = mx + b$
constant second differences; one maximum or minimum (like a projectile)	quadratic: $y = a(x - h)^2 + k$
constant ratios; growth or decay by a percentage	exponential: $y = a \cdot b^{x}$
levelling off toward a value other than $0$ (like cooling)	shifted exponential: $y = a \cdot b^{x} + c$
a repeating cycle (tides, seasons, rotation)	sinusoidal: $y = a\cos\big(k(x - d)\big) + c$ with $k = \dfrac{2\pi}{\text{period}}$
constant third (or higher) differences; volumes and other products	polynomial of higher degree

Context matters as much as the numbers. A slow leak loses a percentage of what’s left, so it’s exponential, not linear. A volume made by multiplying three lengths is a cubic.

Technology and regression

Graphing technology can fit a curve of any chosen type to all the data at once (this is called regression), and it can graph residuals. That’s how scientists usually do it. But choosing the type and judging whether the result makes sense is still your job. In the examples below, the models are built by hand from key points, which is often accurate enough and shows exactly where each number comes from.

Worked examples

Example 1: A slow leak

A car tire has a slow leak. Its pressure is measured every $2$ hours.

Time $t$ (h)	$0$	$2$	$4$	$6$	$8$
Pressure $P$ (kPa)	$240$	$216$	$194$	$175$	$157$

• (a) Decide whether a linear or an exponential model fits better, and build it.
• (b) Predict the pressure after $12$ hours.
• (c) The tire is unsafe below $140$ kPa. When does that happen?

Solution.

(a) The differences are $-24, -22, -19, -18$: not constant, and shrinking. The ratios are

$$\frac{216}{240} = 0.9, \quad \frac{194}{216} \approx 0.898, \quad \frac{175}{194} \approx 0.902, \quad \frac{157}{175} \approx 0.897$$

all very close to $0.9$. So the tire loses about $10\%$ of its pressure every $2$ hours, and an exponential model fits. It makes physical sense too: the higher the pressure, the faster air is pushed out.

Start at $240$ and multiply by $0.9$ for every $2$ hours, which is $\tfrac{t}{2}$ two-hour periods:

$$P(t) = 240(0.9)^{t/2}$$

Check the fit: the model gives $216$, $194.4$, $175.0$ and $157.5$ at $t = 2, 4, 6, 8$. Every residual is less than $1$ kPa. ✓

(b) $P(12) = 240(0.9)^6 \approx 127.5$ kPa.

(c) Solve $240(0.9)^{t/2} = 140$ with logarithms:

$$\begin{aligned} (0.9)^{t/2} &= \frac{140}{240} \\ \frac{t}{2}\log 0.9 &= \log\frac{140}{240} \\ t &= \frac{2\log(140/240)}{\log 0.9} \approx 10.23 \end{aligned}$$

The tire becomes unsafe after about $10.2$ hours. (A linear model, losing about $10.4$ kPa per hour on average, would predict the pressure reaching $0$ after about a day, which isn’t how leaks behave.)

Example 2: Cooling tea

A cup of tea is left in a room at $20\,^\circ\text{C}$.

Time $t$ (min)	$0$	$5$	$10$	$15$	$20$
Temperature $T$ (°C)	$90$	$69$	$54.3$	$44.0$	$36.8$

• (a) Build a model for the temperature.
• (b) Predict the temperature after $30$ minutes, and say when the tea reaches $30\,^\circ\text{C}$.

Solution.

(a) The ratios of the temperatures aren’t constant ($\tfrac{69}{90} \approx 0.767$ but $\tfrac{54.3}{69} \approx 0.787$). But the tea can’t cool below room temperature, so look at how far it is above $20\,^\circ\text{C}$:

$t$ (min)	$0$	$5$	$10$	$15$	$20$
$T - 20$	$70$	$49$	$34.3$	$24.0$	$16.8$

Now the ratios are constant: $\tfrac{49}{70} = \tfrac{34.3}{49} = 0.7$, and $\tfrac{24.0}{34.3} \approx \tfrac{16.8}{24.0} \approx 0.7$. The temperature difference shrinks by $30\%$ every $5$ minutes:

$$T(t) = 20 + 70(0.7)^{t/5}$$

The $20$ is the room temperature (the horizontal asymptote), and the $70$ is the starting difference.

The model levels off at room temperature, as the real tea will.

(b) $T(30) = 20 + 70(0.7)^6 \approx 20 + 8.2 = 28.2\,^\circ\text{C}$.

For $30\,^\circ\text{C}$, solve $20 + 70(0.7)^{t/5} = 30$:

$$(0.7)^{t/5} = \frac{10}{70} \quad\Rightarrow\quad t = \frac{5\log(1/7)}{\log 0.7} \approx 27.3$$

The tea reaches $30\,^\circ\text{C}$ after about $27.3$ minutes. A plain exponential $y = a \cdot b^t$ would wrongly predict the tea cooling toward $0\,^\circ\text{C}$, colder than the room.

Example 3: Most economical speed

A test driver measures a car’s fuel consumption at different speeds.

Speed $v$ (km/h)	$50$	$70$	$90$	$110$	$130$
Consumption $c$ (L/100 km)	$7.0$	$6.2$	$6.6$	$8.2$	$11.0$

Build a model, and find the most economical speed.

Solution. The data falls and then rises, which suggests a quadratic. Check the differences (the speeds go up in equal steps of $20$):

• First differences: $-0.8,\ 0.4,\ 1.6,\ 2.8$
• Second differences: $1.2,\ 1.2,\ 1.2$

Constant second differences confirm a quadratic $c(v) = av^2 + bv + k$. For steps of size $h$, the second difference is $2ah^2$, so $2a(20)^2 = 1.2$, giving $800a = 1.2$ and $a = 0.0015$.

Now use two data points to find $b$ and $k$:

$$\begin{aligned} c(50) &= 0.0015(2500) + 50b + k = 3.75 + 50b + k = 7.0 \\ c(70) &= 0.0015(4900) + 70b + k = 7.35 + 70b + k = 6.2 \end{aligned}$$

Subtracting the first equation from the second: $3.6 + 20b = -0.8$, so $b = -0.22$. Then $k = 7.0 - 3.75 + 11 = 14.25$.

$$c(v) = 0.0015v^2 - 0.22v + 14.25$$

Check: $c(130) = 25.35 - 28.6 + 14.25 = 11.0$. ✓

The minimum is at the vertex:

$$v = -\frac{b}{2a} = \frac{0.22}{0.003} \approx 73.3 \text{ km/h}, \qquad c(73.3) \approx 6.18 \text{ L/100 km}$$

The car is most economical at about $73$ km/h, using about $6.2$ L per $100$ km. The model only makes sense for realistic speeds: it would predict $14.25$ L/100 km at $0$ km/h, which is meaningless for a parked car.

Example 4: Monthly temperatures in radians

The average monthly temperatures for a Canadian city are shown (month $1$ is January).

Month $m$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Temp. (°C)	$-15$	$-13$	$-7$	$2$	$10$	$16$	$18$	$16$	$10$	$2$	$-6$	$-12$

Build a sinusoidal model, check its fit, and use it to estimate the temperature in mid-April ($m = 4.5$).

Solution. Maximum $18$ (July), minimum $-15$ (January), and the cycle repeats every $12$ months.

• Amplitude: $a = \dfrac{18 - (-15)}{2} = 16.5$
• Axis: $c = \dfrac{18 + (-15)}{2} = 1.5$
• $k = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$
• The maximum is at $m = 7$, so use cosine with $d = 7$.

$$T(m) = 16.5\cos\left(\frac{\pi}{6}(m - 7)\right) + 1.5$$

Check the fit for a few months (model values to 1 decimal place):

Month	$1$	$2$	$4$	$6$	$10$	$11$
Data	$-15$	$-13$	$2$	$16$	$2$	$-6$
Model	$-15.0$	$-12.8$	$1.5$	$15.8$	$1.5$	$-6.8$
Residual	$0$	$-0.2$	$0.5$	$0.2$	$0.5$	$0.8$

The residuals are all under $1\,^\circ\text{C}$, so the model fits well.

Mid-April: $T(4.5) = 16.5\cos\left(\dfrac{\pi}{6}(-2.5)\right) + 1.5 = 16.5\cos\left(-\dfrac{5\pi}{12}\right) + 1.5 \approx 16.5(0.2588) + 1.5 \approx 5.8\,^\circ\text{C}$.

Common mistakes

Choosing a type from one or two numbers. Check differences and ratios across the whole table, and think about the situation. A leak or a cooling drink is exponential even if the first few values look nearly linear.

Forgetting the shift in cooling and heating. Objects cool toward room temperature, not toward $0$. Model the difference from room temperature exponentially, then add the room temperature back.

Using the wrong exponent for the time step. If the ratio $0.9$ is per $2$ hours, the exponent is $\tfrac{t}{2}$, not $t$. Check: after $2$ hours, $(0.9)^{2/2} = 0.9$. ✓

Trusting extrapolation. A model that fits data from $50$ to $130$ km/h says nothing reliable about $0$ or $250$ km/h. State the inputs for which the model is reasonable.

Using degrees in a radian model. With $k = \tfrac{2\pi}{12}$, your calculator must be in radian mode. (In degrees you’d use $k = \tfrac{360}{12} = 30$ instead, as in Grade 11.)

Not checking the fit. Always compare the model with at least a couple of data points you didn’t use to build it. Large or patterned residuals mean you should rethink the type.

Practice

1. (Warm-up) Which type of function would you choose to model each situation? Explain briefly.

• (a) The height of the tide over two days.
• (b) The number of bacteria in a culture that doubles every hour.
• (c) A taxi fare with a fixed charge plus a fee per kilometre.
• (d) The volume of a box whose length, width and height all depend on $x$.

Solution

(a) Sinusoidal: tides repeat in a regular cycle.

(b) Exponential: doubling means a constant ratio.

(c) Linear: the fare increases at a constant rate per kilometre.

(d) Polynomial (cubic): volume is a product of three expressions in $x$.

2. (Warm-up) Is each table best modelled by a linear, quadratic, or exponential function?

$x$	$0$	$1$	$2$	$3$	$4$
A	$5$	$8$	$13$	$20$	$29$
B	$5$	$10$	$20$	$40$	$80$
C	$5$	$9$	$13$	$17$	$21$

Solution

A: first differences $3, 5, 7, 9$; second differences $2, 2, 2$. Quadratic.

B: ratios all $2$. Exponential.

C: first differences all $4$. Linear.

3. (Warm-up) A town’s population is modelled by $P(t) = 5000(1.025)^t$, where $t$ is in years. Interpret $5000$ and $1.025$, and predict the population after $10$ years.

Solution

$5000$ is the population at $t = 0$. $1.025$ means the population grows by $2.5\%$ per year.

$P(10) = 5000(1.025)^{10} \approx 6400$ people.

4. (Core) A town’s population is recorded every $5$ years.

Year $t$	$0$	$5$	$10$	$15$	$20$
Population	$12\,000$	$13\,440$	$15\,050$	$16\,860$	$18\,880$

• (a) Show that an exponential model fits, and build it.
• (b) A linear model through the first and last points is $L(t) = 12\,000 + 344t$. Compare both models at $t = 10$ with the data.
• (c) Use both models to predict the population at $t = 30$. Which prediction is more reasonable?

Solution

(a) Ratios: $\tfrac{13\,440}{12\,000} = 1.12$, $\tfrac{15\,050}{13\,440} \approx 1.120$, $\tfrac{16\,860}{15\,050} \approx 1.120$, $\tfrac{18\,880}{16\,860} \approx 1.120$. The population grows by about $12\%$ every $5$ years:

$$E(t) = 12\,000(1.12)^{t/5}$$

(b) At $t = 10$, the data is $15\,050$. $E(10) = 12\,000(1.12)^2 = 15\,052.8$, off by about $3$. $L(10) = 12\,000 + 3440 = 15\,440$, off by about $390$. The exponential model fits much better.

(c) $E(30) = 12\,000(1.12)^6 \approx 23\,686$ and $L(30) = 12\,000 + 10\,320 = 22\,320$. The exponential prediction is more reasonable, because the data shows a constant growth rate (percentage), not a constant number of people per year. Either way, a prediction $10$ years past the data is an extrapolation and should be treated with caution.

5. (Core) At a harbour, high tide is $9$ m at 2:00 a.m., and the next low tide is $3$ m at 8:15 a.m. Build a sinusoidal model for the depth $t$ hours after midnight (in radians), and predict the depth at noon to 2 decimal places.

Solution

$a = \tfrac{9 - 3}{2} = 3$ and $c = \tfrac{9 + 3}{2} = 6$. High to low takes $6.25$ h, which is half a period, so the period is $12.5$ h and $k = \tfrac{2\pi}{12.5}$. The maximum is at $t = 2$:

$$D(t) = 3\cos\left(\frac{2\pi}{12.5}(t - 2)\right) + 6$$

At noon, $t = 12$:

$$D(12) = 3\cos\left(\frac{2\pi}{12.5}(10)\right) + 6 = 3\cos(1.6\pi) + 6 \approx 3(0.3090) + 6 \approx 6.93$$

The depth at noon is about $6.93$ m.

6. (Core) An open box is made from a $20$ cm by $30$ cm sheet of cardboard by cutting a square of side $x$ cm from each corner and folding up the sides.

• (a) Write a model for the volume $V(x)$, and state a reasonable domain.
• (b) Make a table for $x = 1, 2, 3, 4, 5, 6$, and confirm that the third differences are constant.
• (c) Estimate the value of $x$ that gives the largest volume.

Solution

(a) The base is $(30 - 2x)$ by $(20 - 2x)$ and the height is $x$:

$$V(x) = x(20 - 2x)(30 - 2x)$$

All lengths must be positive, so $0 \lt x \lt 10$.

(b)

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$V$ (cm³)	$504$	$832$	$1008$	$1056$	$1000$	$864$

First differences: $328, 176, 48, -56, -136$. Second: $-152, -128, -104, -80$. Third: $24, 24, 24$. Constant third differences confirm a cubic (and $24 = 4 \cdot 3!$ matches the leading coefficient $4$ of $V(x) = 4x^3 - 100x^2 + 600x$).

(c) The volume is largest near $x = 4$ cm (about $1056$ cm³). Graphing technology puts the maximum at $x \approx 3.92$ cm, with $V \approx 1056.3$ cm³.

7. (Core) The concentration of a medication in a patient’s blood is modelled by $C(t) = 50(0.8)^t$ mg/L, where $t$ is in hours. Interpret the model, and find when the concentration drops below $10$ mg/L.

Solution

The starting concentration is $50$ mg/L, and $20\%$ of the medication is removed each hour (since $1 - 0.8 = 0.2$).

Solve $50(0.8)^t = 10$:

$$(0.8)^t = 0.2 \quad\Rightarrow\quad t = \frac{\log 0.2}{\log 0.8} \approx 7.21$$

The concentration drops below $10$ mg/L after about $7.2$ hours.

8. (Challenge) A leaking tire has a pressure of $250$ kPa at $t = 0$ and $200$ kPa after $5$ hours.

• (a) Build a linear model and an exponential model through these two points.
• (b) Use both to predict the pressure after $20$ hours.
• (c) Which model is more reasonable, and why?

Solution

(a) Linear: the slope is $\tfrac{200 - 250}{5} = -10$ kPa/h, so $L(t) = 250 - 10t$.

Exponential: the ratio over $5$ hours is $\tfrac{200}{250} = 0.8$, so $E(t) = 250(0.8)^{t/5}$.

(b) $L(20) = 250 - 200 = 50$ kPa. $E(20) = 250(0.8)^4 = 102.4$ kPa.

(c) The exponential model. Air leaks faster when the pressure is higher and slower as it drops, so the tire loses a percentage of its pressure, not a fixed amount. The linear model also predicts a pressure of $0$ at $t = 25$ h and negative pressures after that, which is impossible.

9. (Challenge) A ball is thrown upward. Its height is $1.2$ m at $t = 0$, $15.3$ m at $t = 1$ s, and $19.6$ m at $t = 2$ s.

• (a) Find a quadratic model $h(t) = at^2 + bt + c$.
• (b) Find the maximum height and when the ball lands, to 2 decimal places.

Solution

(a) $h(0) = c = 1.2$. Then

$$\begin{aligned} h(1) &= a + b + 1.2 = 15.3 &&\Rightarrow\quad a + b = 14.1 \\ h(2) &= 4a + 2b + 1.2 = 19.6 &&\Rightarrow\quad 2a + b = 9.2 \end{aligned}$$

Subtracting: $a = 9.2 - 14.1 = -4.9$, and then $b = 14.1 + 4.9 = 19$.

$$h(t) = -4.9t^2 + 19t + 1.2$$

(The $-4.9$ matches half the acceleration due to gravity, $9.8$ m/s², a good sign the model is reasonable.)

(b) The vertex is at $t = \dfrac{19}{9.8} \approx 1.94$ s, and

$$h\left(\frac{19}{9.8}\right) = 1.2 + \frac{19^2}{4(4.9)} = 1.2 + \frac{361}{19.6} \approx 19.62 \text{ m}$$

The ball lands when $h(t) = 0$. By the quadratic formula, taking the positive root:

$$t = \frac{-19 - \sqrt{19^2 - 4(-4.9)(1.2)}}{2(-4.9)} = \frac{19 + \sqrt{384.52}}{9.8} \approx 3.94 \text{ s}$$

The maximum height is about $19.62$ m, at about $1.94$ s, and the ball lands after about $3.94$ s.