IBDP Maths SL 5.8 Local maximum and minimum values AA HL Paper 2- Exam Style Questions- New Syllabus

(a) Numerical Approximation using Euler’s Method

We use the iterative formula: \( x_{n+1} = x_n + h \) and \( y_{n+1} = y_n + h \cdot f(x_n, y_n) \), where \( h = 0.25 \) and \( f(x,y) = \frac{2x}{x^2 + y} \).

Step 1: Starting at \( (x_0, y_0) = (1, 0) \)
\( f(1, 0) = \frac{2(1)}{1^2 + 0} = 2 \)
\( y_1 = 0 + 0.25(2) = 0.5 \); \( x_1 = 1.25 \)

Step 2: At \( (1.25, 0.5) \)
\( f(1.25, 0.5) = \frac{2(1.25)}{(1.25)^2 + 0.5} = \frac{2.5}{1.5625 + 0.5} = \frac{2.5}{2.0625} \approx 1.21212 \dots \)
\( y_2 = 0.5 + 0.25(1.21212) \approx 0.80303 \); \( x_2 = 1.5 \)

Step 3: At \( (1.5, 0.80303) \)
\( f(1.5, 0.80303) = \frac{2(1.5)}{(1.5)^2 + 0.80303} = \frac{3}{2.25 + 0.80303} = \frac{3}{3.05303} \approx 0.98256 \dots \)
\( y_3 = 0.80303 + 0.25(0.98256) \approx 1.04867 \); \( x_3 = 1.75 \)

Step 4: At \( (1.75, 1.04867) \)
\( f(1.75, 1.04867) = \frac{2(1.75)}{(1.75)^2 + 1.04867} = \frac{3.5}{3.0625 + 1.04867} = \frac{3.5}{4.11117} \approx 0.85132 \dots \)
\( y_4 = 1.04867 + 0.25(0.85132) \approx 1.26150 \dots \); \( x_4 = 2.0 \)

Final Answer (3 s.f.): \( \boxed{y \approx 1.26} \)

(b) Justification of Results

(i) Analysis of Error:
From the diagram, the solution curve is concave down (\( f”(x) < 0 \)) over the interval \( [1, 2] \). In Euler’s method, the approximation is built using tangent line segments. Since tangents to a concave-down curve always lie above the curve, the numerical result is an overestimate.

(ii) Behaviour of the Method:
At the initial point \( (1, 0) \), the gradient is \( \frac{dy}{dx} = 2 \), which is positive. Euler’s method calculates the next point by following this positive slope in the direction of increasing \( x \). As long as \( x > 1 \), the calculated gradients remain positive, leading to positive \( y \)-values. The method is “trapped” in the upper branch and cannot jump to the negative branch of the relation without a change in the sign of the step or a singularity.