IB Mathematics AHL 2.9 modelling AI HL Paper 3- Exam Style Questions- New Syllabus

IBDP Maths AI HL Paper 3 - All Topics

Question

Hem is conducting an investigation into the proliferation of electric vehicles (EVs) across the European Union. His aim is to assess potential infrastructural challenges associated with the projected 2035 phase-out of internal combustion engine vehicles.
To begin his study, Hem develops a projection for the total quantity of electric vehicles expected to be in operation within the EU by the end of 2035.
The historical data for the number of EVs, $N$ (expressed in millions), is presented in the table below. Time $t$ represents the number of years elapsed since the start of 2016, where $t \in \mathbb{R}$.
Year Ending2016201720182019202020212022
$t$1234567
$N$0.20.30.40.61.11.93.1
Hem applies a logistic growth model to this dataset: $N = \frac{310}{1 + Ce^{-kt}}$, where $C, k \in \mathbb{R}^+$.
(a) Within the context of this study, interpret the meaning of the constant 310 in Hem’s model.
(b) (i) By substituting the data point for $t = 1$, verify that $C = 1549e^k$.
(ii) Using the data point for $t = 7$, establish a second mathematical expression for $C$.
(c) Utilising the results from part (b), solve for the values of:
(i) $k$.
(ii) $C$.
Hem subsequently evaluates his model by predicting $N$ for the years 2017 to 2021. The results are displayed below, rounded to one decimal place.
Year Ending20172018201920202021
$t$23456
$N$0.30.40.61.11.9
Predicted $N$0.30.5$a$1.22.0
(d) Determine the value of $a$ to one decimal place.
(e) Find the sum of square residuals ($SS_{res}$) for the predictions made between 2017 and 2021.
Hem calculates an error index $E$ using the formula $E = \sqrt{\frac{SS_{res}}{n}}$, where $n$ represents the number of predictions. He intends to accept the model only if $E < 0.25$.
(f) Determine the value of $E$ and state whether Hem should accept the model.
(g) Project the number of EVs in the EU at the conclusion of 2035 based on this model.
The adequacy of public infrastructure is a significant concern. The EU estimates that $20\%$ of EV owners will require public charging stations, with a target ratio of one station for every 10 such vehicles.
(h) Construct an expression for the required number of public charging points at time $t$.
Historical records show there were 0.22 million public points at the end of 2020, rising to 0.54 million by the end of 2022.
(i) Calculate the average annual growth rate of public charging point installations during this two-year window.
Hem assumes this installation rate will remain constant (linear) through 2035.
(j) Identify the value of $t$ (where $t \geq 5$) when the available charging infrastructure first falls below demand, and state the corresponding year.

Most-appropriate topic codes (IB Mathematics: Applications and Interpretation HL):

• AHL 2.9: Modelling with logistic functions — parts (a)-(c), (g), (h)
• AHL 4.13: Non-linear regression and sum of square residuals — parts (d)-(f) 
• AHL 2.6: Modelling process: choosing, fitting, and using models — parts (a)-(j) 
• SL 2.1: Linear models and rates of change — parts (i)-(j) 
▶️ Answer/Explanation

(a)
In the logistic function \(N = \frac{310}{1 + Ce^{-kt}}\), the value 310 represents the carrying capacity of the model. This is the maximum number of electric vehicles (in millions) that the European Union is predicted to reach in the long term. As \(t \to \infty\), \(e^{-kt} \to 0\), so \(N \to 310\).
Answer: \(\boxed{310 \text{ million is the carrying capacity (maximum number of EVs)}}\)

(b)(i)
Using \(N = 0.2\) when \(t = 1\):
\(0.2 = \frac{310}{1 + Ce^{-k}}\)
\(1 + Ce^{-k} = \frac{310}{0.2} = 1550\)
\(Ce^{-k} = 1549\)
\(C = 1549e^k\)
Shown: \(\boxed{C = 1549e^k}\)

(b)(ii)
Using \(N = 3.1\) when \(t = 7\):
\(3.1 = \frac{310}{1 + Ce^{-7k}}\)
\(1 + Ce^{-7k} = \frac{310}{3.1} = 100\)
\(Ce^{-7k} = 99\)
\(C = 99e^{7k}\)
Answer: \(\boxed{C = 99e^{7k}}\)

(c)(i)
Equating the two expressions for \(C\):
\(1549e^k = 99e^{7k}\)
\(\frac{1549}{99} = e^{6k}\)
\(e^{6k} = \frac{1549}{99} \approx 15.6465\)
\(6k = \ln(15.6465)\)
\(k = \frac{\ln(15.6465)}{6} \approx \frac{2.750}{6} \approx 0.458\)
Answer: \(\boxed{k = 0.458 \ (0.458374…)}\)

(c)(ii)
Using \(C = 1549e^k\) with \(k = 0.458374…\):
\(C = 1549 \times e^{0.458374…} \approx 1549 \times 1.5817 \approx 2449.74\)
Or using \(C = 99e^{7k}\):
\(C = 99 \times e^{7 \times 0.458374…} \approx 99 \times e^{3.20862} \approx 99 \times 24.745 \approx 2449.74\)
Answer: \(\boxed{C = 2450 \ (2449.74…)}\)

(d)
Substitute \(t = 4\) into the logistic function:
\(N = \frac{310}{1 + 2449.74e^{-0.458374 \times 4}}\)
\(N = \frac{310}{1 + 2449.74e^{-1.8335}}\)
\(N = \frac{310}{1 + 2449.74 \times 0.1599}\)
\(N = \frac{310}{1 + 391.8} \approx \frac{310}{392.8} \approx 0.7896\)
Rounded to one decimal place: \(a = 0.8\)
Answer: \(\boxed{a = 0.8}\)

(e)
Calculate squared residuals for each year (2017-2021):
• 2017 (\(t=2\)): \((0.3 – 0.3)^2 = 0^2 = 0\)
• 2018 (\(t=3\)): \((0.4 – 0.5)^2 = (-0.1)^2 = 0.01\)
• 2019 (\(t=4\)): \((0.6 – 0.8)^2 = (-0.2)^2 = 0.04\)
• 2020 (\(t=5\)): \((1.1 – 1.2)^2 = (-0.1)^2 = 0.01\)
• 2021 (\(t=6\)): \((1.9 – 2.0)^2 = (-0.1)^2 = 0.01\)
Sum: \(SS_{res} = 0 + 0.01 + 0.04 + 0.01 + 0.01 = 0.07\)
Answer: \(\boxed{SS_{res} = 0.07}\)

(f)
Error function: \(E = \sqrt{\frac{SS_{res}}{n}}\) where \(n = 5\)
\(E = \sqrt{\frac{0.07}{5}} = \sqrt{0.014} \approx 0.1183\)
Since \(0.1183 < 0.25\), the model can be used.
Answer: \(\boxed{E = 0.118 < 0.25 \text{, so the model can be used}}\)

(g)
2035 corresponds to \(t = 2035 – 2015 = 20\) (since \(t=1\) in 2016)
\(N = \frac{310}{1 + 2449.74e^{-0.458374 \times 20}}\)
\(N = \frac{310}{1 + 2449.74e^{-9.16748}}\)
\(N = \frac{310}{1 + 2449.74 \times 0.000106}\)
\(N = \frac{310}{1 + 0.2597} \approx \frac{310}{1.2597} \approx 246.1\)
Rounded to nearest million: 246 million
Answer: \(\boxed{246 \text{ million EVs}}\)

(h)
Number requiring public charging: \(20\%\) of EVs = \(0.2N\)
One charging point for every 10 cars: multiplier = \(0.1\)
Public charging points required = \(0.2N \times 0.1 = 0.02N\)
Using the model: \(N = \frac{310}{1 + 2449.74e^{-0.458374t}}\)
Public charging points = \(\frac{0.02 \times 310}{1 + 2449.74e^{-0.458374t}} = \frac{6.2}{1 + 2449.74e^{-0.458374t}}\) million
Answer: \(\boxed{P(t) = \frac{6.2}{1 + 2449.74e^{-0.458374t}} \text{ million}}\)

(i)
From 2020 (\(t=5\)) to 2022 (\(t=7\)): 2 years
Increase: \(0.54 – 0.22 = 0.32\) million
Average per year: \(\frac{0.32}{2} = 0.16\) million
Answer: \(\boxed{0.16 \text{ million per year}}\)

(j)
Linear model for charging points: Starting at \(t=5\) (2020) with 0.22 million, increasing at 0.16 million/year:
\(P_{\text{actual}}(t) = 0.22 + 0.16(t – 5)\)
Demand: \(P_{\text{demand}}(t) = \frac{6.2}{1 + 2449.74e^{-0.458374t}}\)
Find when demand exceeds supply: \(\frac{6.2}{1 + 2449.74e^{-0.458374t}} > 0.22 + 0.16(t – 5)\)
Solving numerically (using calculator/graphical method):
\(t \approx 15.1492\) (which corresponds to year: \(2015 + 15.1492 \approx 2030.1492\))
At \(t = 15\), supply ≈ 1.82 million, demand ≈ 1.78 million (supply > demand)
At \(t = 16\), supply ≈ 1.98 million, demand ≈ 2.15 million (demand > supply)
First insufficiency occurs between \(t = 15\) and \(t = 16\)
Year: \(2015 + 15 = 2030\) (still sufficient)
Year: \(2015 + 16 = 2031\) (insufficient)
Answer: \(\boxed{t = 15.1, \text{ Year: 2031}}\)

More IBDP MAI Paper 3 More Question
Scroll to Top