June 2024 Paper 1 Q16

AQACurrent spec5 marksModelling Numerical Methods

16 Figure 2 below shows a 1.5 metre length of pipe.

Figure 2: a length of pipe with a U-shaped (parabolic) open cross-section, length labelled 1.5 m — Figure 2

The symmetrical cross-section of the pipe is shown below, in Figure 3, where \(x\) and \(y\) are measured in centimetres.

Figure 3: U-shaped curve below the x-axis meeting it at x = −2 and x = 2, with minimum point at (0, −3) — Figure 3

Use the trapezium rule, with the values shown in the table below, to find the best estimate for the volume of the pipe.

\(x\)	0	0.4	0.8	1.2	1.6	2
\(y\)	\(-3\)	\(-2.943\)	\(-2.752\)	\(-2.353\)	\(-1.572\)	0

[5 marks]

Mark scheme

Scheme	Marks	AO
Uses the symmetry of the curve. Evidenced by doubling area from \(x\) = 0 to \(x\) = 2 Or considering the whole region from \(x = -2\) to \(x\) = 2	M1	3.1a
States or uses \(h\) = 0.4 OE Accept 0.2 as the multiplier. PI by 4.448 or 8.896 Accept use of \(h\)=0.8 or multiplier of 0.4 provided their answer is not then doubled.	B1	2.2a
Substitutes given y values or absolute y values to achieve \(3 + 0 + 2[2.943 + 2.752 + 2.353 + 1.572]\) or \(-3 + 0 + 2[-2.943 - 2.752 - 2.353 - 1.572]\) or \(0 + 0 + 2\begin{bmatrix} 2.943 + 2.752 + 2.353 + 1.572 \\ +3 + 2.943 + 2.752 + 2.353 + 1.572 \end{bmatrix}\) or \(0 + 0 + 2\begin{bmatrix} -2.943 - 2.752 - 2.353 - 1.572 \\ -3 - 2.943 - 2.752 - 2.353 - 1.572 \end{bmatrix}\) Condone missing or misplaced zeros PI by \(\pm 22.24\) or \(\pm 44.48\)	M1	1.1a
Obtains \(\pm 8.896\) or \(\pm 4.448\) Do not award this mark if they go on to obtain \(\pm 17.792\)	A1	1.1b
Obtains AWRT 1300cm³ Or AWRT 0.0013m³ Must include units	A1	3.2a
	(5 marks)

Typical solution

\[\text{Area} \approx \frac{0.4}{2}\left(3 + 0 + 2[2.943 + 2.752 + 2.353 + 1.572]\right)\]\[= 4.448\]\[\text{Total area} = 8.896\]\[\text{Volume} = 8.896 \times 150\]\[= 1334.4\ \text{cm}^3\]\[\approx 1300\ \text{cm}^3\]