IB Mathematics SL 1.6 Approximation decimal places, significant figures AI HL Paper 1- Exam Style Questions- New Syllabus
Question
(i) the displacement of the mass after 1 second.
(ii) the first instant in time when the mass returns to the equilibrium position (\( x = 0 \)).
Most-appropriate topic codes (IB Mathematics: Applications and Interpretation HL ):
• SL 1.6: Percentage error and upper/lower bounds — part (b)
▶️ Answer/Explanation
(a)
First, isolate the second derivative: \( \frac{d^2 x}{dt^2} = 4.9 – 1.2 \frac{dx}{dt} – 4x \).
Applying Euler’s method with \( h=0.1 \):
\( x_{n+1} = x_n + 0.1 \cdot v_n \)
\( v_{n+1} = v_n + 0.1 \cdot (4.9 – 1.2 v_n – 4 x_n) \)
With initial conditions \( t_0=0, x_0=0.5, v_0=9.8 \):
Iterating 10 steps to \( t=1 \) yields a displacement of approximately \( 4.36332 \) m.
(i) \( \boxed{4.36 \text{ m}} \) (to 3 s.f.)
Continuing the iteration, the displacement first becomes negative between \( t=1.9 \) and \( t=2.0 \).
(ii) \( \boxed{2.0 \text{ seconds}} \) (first estimated time reaching \( x=0 \)).
(b)
Actual displacement = 3.85 m. Since this is correct to the nearest cm, the true value \( V_A \) lies in \( 3.845 \leq V_A < 3.855 \).
The estimate \( V_E \) is \( 4.3633 \) m.
To find the maximum percentage error, we use the value in the range that maximizes the absolute difference: \( V_A = 3.845 \).
Absolute Error = \( |4.3633 – 3.845| = 0.5183 \) m.
Percentage Error = \( \frac{0.5183}{3.845} \times 100\% \approx 13.48\% \).
\( \boxed{13.5\%} \) (to 3 s.f.)