Edexcel A level Maths (9MA0) · Pure mathematics › Numerical methods
Practise Numerical methods in context. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
Water flows into a large tank, which is empty at time \(t = 0\). The rate of flow, \(r\) litres per minute, is measured every 8 minutes for 48 minutes, where \(t\) is the time in minutes. The results are shown in the table.
[object Object]
(a) Use the trapezium rule with all the values in the table to estimate the volume of water in the tank after 48 minutes. Give your answer to 3 significant figures.[3]
(b) The tank holds 1800 litres. After 48 minutes, water continues to flow into the tank at a constant rate of 22.7 litres per minute. Use your answer to part (a) to estimate how many more minutes it takes for the tank to become full. Give your answer to the nearest minute.[2]
(c) Suggest how the estimate in part (a) could be made more accurate.[1]
Show the answer and mark scheme
(a)Answer: \(1510\) litres
B1 for the strip width \(h = 8\) (or \(\frac{1}{2} \times 8\) seen)
M1 for the correct bracket structure \(\left[y_0 + y_{6} + 2(y_1 + \cdots + y_{5})\right]\) with the values from the table
A1 for awrt \(1510\) litres
Worked solution: The volume is \(\int_{0}^{48} r \, \mathrm{d}t\), the area under the graph of \(r\) against \(t\). \(\frac{1}{2} \times 8 \times \left[50.0 + 22.7 + 2(40.1 + 33.5 + 29.0 + 26.1 + 24.1)\right] = 1513.2\), so the volume is approximately \(1510\) litres (3 s.f.).
(b)Answer: 13 minutes
M1 for \(\frac{1800 - \text{their (a)}}{22.7}\)
A1 for 13 minutes
Worked solution: The volume still needed is about \(1800 - 1510 = 290\) litres, so the time is \(\frac{290}{22.7} = 12.775330\ldots\) minutes, i.e. about 13 more minutes.
(c)Answer: Measure the rate of flow at shorter time intervals, so that more, narrower strips are used
B1 for measuring the rate more frequently (more strips of smaller width)
Worked solution: Measure the rate of flow at shorter time intervals, so that the trapezium rule can use more, narrower strips.
Question 2Medium7 marks
The power output, \(P\) kilowatts, of a set of solar panels is recorded on a sunny day, every 3 hours from 6 am to 6 pm. The time \(t\) is measured in hours after 6 am. The results are shown in the table. The energy generated, in kilowatt-hours (kWh), is given by \(\int_{0}^{12} P \, \mathrm{d}t\).
[object Object]
(a) Use the trapezium rule with all the values in the table to estimate the energy generated between 6 am and 6 pm. Give your answer to 3 significant figures.[3]
(b) Each kWh generated saves the household 30p. Use your answer to part (a) to estimate the amount saved on this day, giving your answer to the nearest penny.[2]
(c) State whether your answer to part (a) is likely to be an overestimate or an underestimate of the energy generated.[1]
(d) Give a reason for your answer to part (c).[1]
Show the answer and mark scheme
(a)Answer: \(20.3\) kWh
B1 for the strip width \(h = 3\) (or \(\frac{1}{2} \times 3\) seen)
M1 for the correct bracket structure \(\left[y_0 + y_{4} + 2(y_1 + \cdots + y_{3})\right]\) with the values from the table
A1 for awrt \(20.3\) kWh
Worked solution: \(\frac{1}{2} \times 3 \times \left[0.00 + 0.00 + 2(1.98 + 2.80 + 1.98)\right] = 20.28\), so the energy generated is approximately \(20.3\) kWh (3 s.f.).
(b)Answer: £6.08
M1 for their energy \(\times 30\) (pence) or \(\times 0.3\) (pounds)
A1 for £6.08 (or £6.09 from the rounded answer to (a))
Worked solution: \(20.28 \times 30 = 608.4\) pence, so the saving is about £6.08.
(c)Answer: Underestimate
B1 for underestimate
Worked solution: Underestimate.
(d)Answer: The graph of \(P\) against \(t\) is concave, so the tops of the trapezia lie below the curve
B1 for a reason: the graph of \(P\) against \(t\) is concave (it rises at a decreasing rate and falls at an increasing rate), so the tops of the trapezia lie below the curve
Worked solution: The values rise more and more slowly and then fall more and more quickly, so the graph of \(P\) against \(t\) bends downwards (it is concave). The straight tops of the trapezia lie below the curve, so the trapezium rule gives an underestimate.
Question 3Hard11 marks
A circular pond has centre \(O\) and radius \(r\) metres. A straight fence \(AB\) is built across the pond, where \(A\) and \(B\) are points on the edge of the pond and angle \(AOB = \theta\) radians, \(0 \lt \theta \lt \pi\). The smaller region cut off by the fence (the minor segment) is to have an area equal to one quarter of the area of the pond.
(a) Show that \(\theta - \sin\theta = \dfrac{\pi}{2}\).[3]
(b) Show that the equation \(\theta - \sin\theta = \frac{\pi}{2}\) has a root between 2.2 and 2.4.[2]
(c) Using the iteration formula \(\theta_{n + 1} = \dfrac{\pi}{2} + \sin\theta_n\) with \(\theta_1 = 2\), find \(\theta_2\) and \(\theta_3\), giving your answers to 4 decimal places.[2]
(d) The iteration converges to the root \(\alpha\). Find \(\alpha\) correct to 3 decimal places, justifying the accuracy of your answer.[2]
(e) Find the length of the fence \(AB\) as a multiple of \(r\), giving your answer to 3 significant figures.[2]
M1 for segment area = sector area − triangle area: \(\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\)
M1 for equating to \(\frac{1}{4}\pi r^2\)
A1* for cancelling \(\frac{1}{2}r^2\) to obtain the given equation
Worked solution: Area of segment = area of sector \(AOB\) − area of triangle \(AOB\) \(= \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\). Setting this equal to \(\frac{1}{4}\pi r^2\) and dividing by \(\frac{1}{2}r^2\): \(\theta - \sin\theta = \frac{\pi}{2}\).
(b)Answer: With \(\mathrm{f}(\theta) = \theta - \sin\theta - \frac{\pi}{2}\): \(\mathrm{f}(2.2) = -0.179\), \(\mathrm{f}(2.4) = 0.154\); sign change and \(\mathrm{f}\) continuous.
M1 for evaluating \(\mathrm{f}(2.2)\) and \(\mathrm{f}(2.4)\) for a suitable \(\mathrm{f}\) (awrt \(-0.18\) and \(0.15\))
A1 for a change of sign, \(\mathrm{f}\) continuous, so a root lies between 2.2 and 2.4
Worked solution: Let \(\mathrm{f}(\theta) = \theta - \sin\theta - \frac{\pi}{2}\). \(\mathrm{f}(2.2) = -0.179\), \(\mathrm{f}(2.4) = 0.154\). \(\mathrm{f}\) is continuous and changes sign, so there is a root between 2.2 and 2.4.
M1 for iterating further (or another method) and choosing the interval \([2.3095, 2.3105]\), evaluating \(\mathrm{f}(2.3095) = -0.0006\) and \(\mathrm{f}(2.3105) = 0.0010\)
A1 for a sign change, so \(\alpha = 2.310\) (3 d.p.)
Worked solution: Continuing the iteration gives values closing in on 2.3099. Check: \(\mathrm{f}(2.3095) = -0.00064 \lt 0\) and \(\mathrm{f}(2.3105) = 0.00104 \gt 0\). There is a change of sign, so \(2.3095 \lt \alpha \lt 2.3105\) and \(\alpha = 2.310\) to 3 d.p.
(e)Answer: \(AB = 1.83r\)
M1 for \(AB = 2r\sin\frac{\alpha}{2}\) (or the cosine rule \(AB^2 = 2r^2 - 2r^2\cos\alpha\))
A1 for \(1.83r\)
Worked solution: \(AB = 2r\sin\dfrac{\alpha}{2} = 2r\sin(1.1549\ldots) = 1.83r\).