IB Mathematics SL 5.8 Approximating areas using the trapezoidal rule AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
(a)
Trapezoidal rule with \( h = 1.9 \):
\[ \text{Area} \approx \frac{1.9}{2} \left[ 0 + 0 + 2(1.68 + 2.81 + 2.32) \right] \]
\[ = 0.95 \times 2 \times 6.81 = 12.939 \]
\( \boxed{12.9\ \text{m}^2} \) (3 sf)
(b)
Volume per second = Area × Velocity.
\( 12.939 \times 0.3 = 3.8817 \).
\( \boxed{3.88\ \text{m}^3\ \text{s}^{-1}} \)
(c)
Lowest part of boat is 1.6 m below water.
River depth at centre (\( x = 3.8 \)) is 2.81 m.
Clearance = \( 2.81 – 1.6 = 1.21 \) m.
\( \boxed{1.21\ \text{m}} \)
(d)
\( y = -0.15x^2 + 1.14x + 0.9 \).
Maximum at \( x = -\frac{b}{2a} = \frac{1.14}{0.3} = 3.8 \).
\( y(3.8) = -0.15(3.8)^2 + 1.14(3.8) + 0.9 = 3.066 \).
\( \boxed{3.07\ \text{m}} \)
(e)
Boat width = 5 m, centred at \( x = 3.8 \).
Boat edges are at \( x = 3.8 \pm 2.5 \), so \( x = 1.3 \) and \( x = 6.3 \).
Bridge height at \( x = 1.3 \): \( y = -0.15(1.3)^2 + 1.14(1.3) + 0.9 = 2.1285 \) m.
Boat height above water = 2.35 m.
Since \( 2.1285 < 2.35 \), the boat will hit the arch at its corners.
\( \boxed{\text{No}} \)