IB Mathematics AHL 5.10 The second derivative-AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
Using quadratic regression on the data (with \(x = 0, 20, 40, \dots\) for years after 1900) gives:
\( f(x) = -0.0989285\ldots x^2 + 45.925\ldots x – 67.7857\ldots \)
Rounded to three significant figures:
\( f(x) = -0.0989x^2 + 45.9x – 67.8 \) (population in thousands).
\( \boxed{f(x) = -0.0989x^2 + 45.9x – 67.8} \)

(b)
The coefficient of determination \(R^2 = 0.98843\) is very close to 1, meaning the quadratic model explains about 98.8% of the variation in the population data. This high value supports Matt’s proposal that a quadratic model is appropriate.
\( \boxed{\text{Yes, because } R^2 \approx 0.988 \text{ is very close to 1, indicating an excellent fit.}} \)

(c)
The parameter \(a = -0.0989\) is negative, which means the quadratic model eventually predicts a decreasing population after reaching a maximum. In the long term, this may not be realistic for a growing city unless there is a known carrying capacity or decline phase.
Alternatively, the constant term \(c = -67.8\) (thousands) suggests a negative population at \(x = 0\) (year 1900), which is impossible.
\( \boxed{\text{The negative } a \text{ suggests the population will eventually decline, which may not be realistic.}} \)

(d)(i)
Differentiate: \( f'(x) = -0.1978x + 45.9 \).
At \(x = 110\): \( f'(110) = -0.1978 \times 110 + 45.9 = 24.142 \approx 24.1\) (thousands per year).
Interpretation: In 2010 (\(x = 110\)), the model predicts an annual growth rate of about 24,100 people per year.
\( \boxed{24.1 \text{ (thousands per year)}} \)

(d)(ii)
From the table, the actual population change from 2000 to 2020 is \(3899 – 3685 = 214\) thousand over 20 years, an average annual growth of \(10.7\) thousand per year. The model’s predicted growth in 2010 (\(24.1\) thousand/year) is more than double this average, suggesting the model overestimates the recent growth rate. This indicates the quadratic model may not accurately reflect the slowing growth seen in the data.
The model predicts much higher growth in 2010 than the recent average, questioning its validity

(e)
A logistic model incorporates a carrying capacity (here 4000 thousand), which reflects the idea that population growth slows as it approaches a maximum sustainable limit, often more realistic for long-term predictions than an unbounded quadratic model.
\( \boxed{\text{The logistic model includes a carrying capacity, which is realistic for populations in a limited environment.}} \ Brooklyn.

(f)(i)
Using the quotient rule or chain rule:
\( g(x) = 4000(1 + 14e^{-0.05x})^{-1} \).
\( g'(x) = 4000 \times (-1)(1 + 14e^{-0.05x})^{-2} \times 14 \times (-0.05) e^{-0.05x} \).
Simplify: \( g'(x) = \frac{2800 e^{-0.05x}}{(1 + 14e^{-0.05x})^2} \).
\( \boxed{g'(x) = \frac{2800 e^{-0.05x}}{(1 + 14e^{-0.05x})^2}} \)

(f)(ii)
The greatest growth rate occurs when \( g”(x) = 0 \) (or by symmetry of the logistic curve, at the point of inflection). For a logistic function \( \frac{L}{1+Ce^{-kx}} \), the maximum growth rate occurs at \( x = \frac{\ln C}{k} \).
Here \( L = 4000, C = 14, k = 0.05 \):
\( x = \frac{\ln 14}{0.05} \approx \frac{2.639057}{0.05} = 52.7811 \).
Year \( = 1900 + 52.7811 \approx 1952.78 \), so during the year 1953 (or rounding down to 1952, as per context).
\( \boxed{\text{The greatest growth rate occurred during 1953 (or 1952).}} \)