The Lumenary · Grades 9–12

Mathematical modelling with functions

Name: ______________________

Date: ______________________

You will learn to

Describe and use the modelling cycle (simplify, formulate, solve, interpret, validate)
Formulate linear, quadratic and exponential models from situations and data, and recognise them from differences and ratios in a table
Interpret a model's parameters and make predictions with stated assumptions, including the doubling time $\ln 2 / r$
Fit a line by least squares and explain why the best intercept is set by the means
Validate a model against new data, and explain why a perfect fit or a long extrapolation can mislead

1.
A quantity measured every hour gives $5, 15, 45, 135, 405$ . Which kind of model fits these values exactly?
- A.Linear
- B.Quadratic
- C.Exponential
- D.None of these
2.
A table of a quadratic model shows $2, 5, 10, 17, 26$ at $n = 0, 1, 2, 3, 4$ . What value does the model give at $n = 5$ ?
3.
A $3$ km taxi ride costs $340$ shillings and an $8$ km ride costs $740$ shillings. Assuming a fixed fee plus a constant charge per km, what is the fixed fee, in shillings?
4.
A town's population grows continuously at $3.5\%$ a year, $P(t) = P_0 e^{0.035t}$ . About how long does it take to double?
- A.about 20 years
- B.about 29 years
- C.about 57 years
- D.It depends on the starting population $P_0$ .
5.
In the model $d(v) = v\,t_r + \dfrac{v^2}{2a}$ , the braking part is $\dfrac{v^2}{2a}$ . If a driver doubles the speed (same road, same brakes), what happens to the braking distance?
- A.It doubles.
- B.It is multiplied by 4.
- C.It stays the same, because the brakes are the same.
- D.It increases by 2 metres.
6.
A student fits a polynomial of degree $9$ through all $10$ data points of a sales record, with zero error. What can you conclude about its forecast for next year?
- A.Nothing much: through any 10 points with different x-values there is such a polynomial, so the perfect fit is not evidence that the forecast is good.
- B.It must be excellent, because the model has zero error.
- C.It is exactly right if the data were measured accurately.
- D.It is wrong, because polynomials can never be used for forecasting.
7.
In the proof that the best intercept is $\bar r$ we expand $\sum (r_i - c)^2 = \sum (r_i - \bar r)^2 + 2(\bar r - c)\sum (r_i - \bar r) + n(\bar r - c)^2$ . Why does the middle term vanish?
- A.Because $\sum (r_i - \bar r) = \sum r_i - n\bar r = 0$ : deviations from the mean add up to zero.
- B.Because $\bar r - c = 0$ .
- C.Because squares are never negative.
- D.Because the data lie on a line.
8.
Which Lean statement says “if each value is $b$ times the previous one, the model is exponential”? (Here u : ℕ → ℝ lists the measurements.)
- A.(∀ n, u (n + 1) = b * u n) → ∀ n, u n = u 0 * b ^ n
- B.(∀ n, u (n + 1) - u n = b) → ∀ n, u n = u 0 * b ^ n
- C.(∀ n, u n = u 0 * b ^ n) → ∀ n, u (n + 1) = b * u n
- D.∃ n, u (n + 1) = b * u n
9.
Write fit_line(xs, ys) that returns the least-squares line as a pair (m, c), using $m = S_{xy}/S_{xx}$ and $c = \bar y - m\bar x$ , where $S_{xx} = \sum (x_i - \bar x)^2$ and $S_{xy} = \sum (x_i - \bar x)(y_i - \bar y)$ .
Computational — Practice

Answer key — Mathematical modelling with functions

1.
The ratios $15/5 = 45/15 = \cdots = 3$ are constant, so by the “constant ratios” theorem $u_n = 5 \cdot 3^n$ : an exponential model.
2.
First differences $3, 5, 7, 9$ ; second differences constantly $2$ . The next first difference is $11$ , so the value at $n = 5$ is $26 + 11 = 37$ . Indeed $u_n = n^2 + 2n + 2$ and $25 + 10 + 2 = 37$ .
3.
The slope is $m = \tfrac{740 - 340}{8 - 3} = 80$ shillings per km and the fixed fee is $c = 340 - 80 \cdot 3 = 100$ shillings. Check: $100 + 80 \cdot 8 = 740$ . By the two-point theorem this is the only line through both data points.
4.
$e^{0.035T} = 2$ gives $T = \tfrac{\ln 2}{0.035} \approx 19.8$ years, independent of $P_0$ .
5.
$\tfrac{(2v)^2}{2a} = 4\cdot\tfrac{v^2}{2a}$ : four times as far. This is why speed limits near schools matter so much: at $60$ km/h instead of $30$ km/h the braking distance is four times as long.
6.
Through any $10$ points with distinct $x$ -values passes a polynomial of degree at most $9$ (Lagrange interpolation; perfect_fit_exists in the Lean section). So the zero error was guaranteed in advance, whatever the data, and is no evidence for the forecast. Trust comes from a plausible mechanism, few parameters and successful predictions on data not used in fitting.
7.
$\sum_i (r_i - \bar r) = \sum_i r_i - n\bar r = n\bar r - n\bar r = 0$ . So the middle term is $0$ for every $c$ , and $\sum (r_i - c)^2 = \sum (r_i - \bar r)^2 + n(\bar r - c)^2$ is smallest exactly when $c = \bar r$ .
8.
(∀ n, u (n + 1) = b * u n) → ∀ n, u n = u 0 * b ^ n: the hypothesis is the constant-ratio condition at every step and the conclusion is the exponential formula.

def fit_line(xs, ys):
    n = len(xs)
    mx, my = sum(xs) / n, sum(ys) / n
    sxx = sum((x - mx) ** 2 for x in xs)
    sxy = sum((x - mx) * (y - my) for x, y in zip(xs, ys))
    m = sxy / sxx
    return m, my - m * mx

The last test checks the theorem of the lesson: with the best intercept, the residuals add up to zero, and the line passes through the point of means $(\bar x, \bar y)$ .