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IB Maths AI · Concept clusters

IB Maths AI concept clusters

Original, past-paper-style synoptic questions. Each one is built around a single setup and needs ideas from several subchapters at once, so you practise connecting topics the way long exam questions expect. 40 questions · 1,028 marks · two per subchapter or merged group · every part with a worked solution · All courses

Topic

Subchapter

Level

Skill

Topic 1 · Number and algebra

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	1.1	1.2	1.3	1.4	1.5	1.6	1.7	Other
A Mars orbiter sending images home: standard form, estimates and doubling storage	●		○	○				2.1
Caffeine after an energy drink: exponential decay and logarithms	○		○	●				2.3
Financing a first electric car: a loan, depreciation and total cost	○	●	○					2.1
Training for a first marathon: arithmetic and geometric build-ups	○	○	●					2.1
An AC test circuit: impedance as a complex number					●	○		2.3 3.1
A grey seal colony: an age-class matrix and its eigenvalues			○			○	●	2.3

1.1 Approximation, percentage error and standard form · 1.4 Exponents and logarithmsmerged group

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A Mars orbiter sending images home: standard form, estimates and doubling storage

1.1 Approximation, percentage error and standard form1.4 Exponents and logarithms1.3 Arithmetic and geometric sequences in context2.1 Linear models and piecewise modelsSL

A spacecraft in orbit around Mars sends its images to Earth by radio. Radio signals travel at \(3.00\times10^{8}\) m s\(^{-1}\). On one particular day the orbiter is \(2.25\times10^{11}\) m from Earth.

In this question 1 megabyte (MB) \(=10^{6}\) bytes and 1 gigabyte (GB) \(=10^{9}\) bytes.

(a)

One image file has a size of 46 800 000 bytes.

Write this number in the form \(a\times10^{k}\), where \(1\le a\lt10\) and \(k\in\mathbb{Z}\).

ROUTINENotes

Show solution

\(4.68\times10^{7}\)

Do not accept \(46.8\times10^{6}\).

Answer: \(4.68\times10^{7}\) bytes

(1)

(b)

Find the time taken for a radio signal to travel from the orbiter to Earth on this day. Give your answer in seconds, in the form \(a\times10^{k}\), where \(1\le a\lt10\) and \(k\in\mathbb{Z}\), and also in minutes.

ROUTINENotes

Show solution

\(\frac{2.25\times10^{11}}{3.00\times10^{8}}\)	M1	Distance divided by speed.
\(7.50\times10^{2}\) s	A1	Accept \(7.5\times10^{2}\) s. 750 s on its own is not in standard form: A0.
12.5 minutes	A1	Allow ECF from their time in seconds.

Answer: \(7.50\times10^{2}\) s \(=12.5\) minutes

(3)

(c)(i)

On this day the orbiter transmits for 9.3 hours at a constant rate of \(4.85\times10^{5}\) bytes per second.

By rounding each of 9.3, 3600 and \(4.85\times10^{5}\) to one significant figure, estimate the total amount of data, in bytes, that the orbiter sends on this day.

ROUTINENotes

Show solution

\(9\times4000\times(5\times10^{5})\)	M1	All three values rounded to 1 s.f. and multiplied.
\(1.8\times10^{10}\) bytes	A1	Accept 18 000 000 000.

Answer: \(1.8\times10^{10}\) bytes

(2)

(c)(ii)

Find the exact amount of data sent on this day. Hence find the percentage error in your estimate from part (c)(i).

MULTINotes

Show solution

\(9.3\times3600\times4.85\times10^{5}=1.62378\times10^{10}\) bytes	B1	Accept \(1.62\times10^{10}\).
\(\varepsilon=\left\|\frac{1.8\times10^{10}-1.62378\times10^{10}}{1.62378\times10^{10}}\right\|\times100\%\)	M1	The exact value must be in the denominator. Allow ECF from (c)(i).
\(10.9\%\)	A1	Accept \(10.85\ldots\%\).

Answer: \(1.62\times10^{10}\) bytes; \(\varepsilon=10.9\%\)

(3)

(d)(i)

On a different day the orbiter uses a faster transmission mode. Figure 1 shows the amount of data \(D\), in megabytes, still stored on the orbiter waiting to be sent, \(t\) minutes after the start of the transmission, for the first 40 minutes.

Find the gradient of the line in Figure 1 and interpret its value in context.

Figure 1: data still stored on the orbiter, D megabytes, t minutes after the start of the transmission

GRAPHNotes

Show solution

\(\frac{900-2100}{40-0}\)	M1	Any two correct points read from the line.
\(-30\)	A1
the stored data decreases by 30 MB every minute (30 MB are sent each minute)	R1	Must refer to a rate per minute in context. Allow ECF from their gradient.

Answer: \(-30\) MB per minute

(3)

(d)(ii)

Show that the transmission rate in this mode is \(5.00\times10^{5}\) bytes per second.

ROUTINENotes

Show solution

\(\frac{30\times10^{6}}{60}\)	M1	Converting MB to bytes and minutes to seconds.
\(=500\,000\) bytes per second	A1	The value 500 000 (or an equivalent value not yet in standard form) must be seen before the given answer.
\(=5.00\times10^{5}\) bytes per second	AG

(2)

(d)(iii)

Assuming the rate stays constant, find the time after the start of the transmission at which all of the stored data has been sent.

MODELNotes

Show solution

\(D=2100-30t\) and \(2100-30t=0\)	M1	Or \(\frac{2100}{30}\). Allow ECF from (d)(i).
70 minutes	A1

Answer: 70 minutes

(2)

(e)

The storage capacity of the orbiter's hardware improves with each new generation. Generation 1 could store 5 GB, and the capacity doubles with each new generation.

Show that generation 6 can store 160 GB.

ROUTINENotes

Show solution

\(u_6=5\times2^{5}\)	M1	Geometric with \(u_1=5\), \(r=2\); or a list 5, 10, 20, 40, 80, 160.
\(=5\times32\)	A1	An intermediate value must be seen. For the list method, all six terms seen scores M1A1.
160 GB	AG

(2)

(f)(i)

Find the greatest number of complete image files of the size given in part (a) that a generation 6 store can hold.

ROUTINENotes

Show solution

\(\frac{160\times10^{9}}{4.68\times10^{7}}=3418.8\ldots\)	M1
3418 files	A1	Must be rounded down to a whole number. 3419 scores A0.

Answer: 3418 files

(2)

(f)(ii)

Find the first generation that can store more than \(4.0\times10^{12}\) bytes.

MULTINotes

Show solution

\(4.0\times10^{12}\) bytes \(=4000\) GB	M1	Or \(u_n\) written in bytes: \(5\times10^{9}\times2^{n-1}\).
\(5\times2^{n-1}\gt4000\Rightarrow n-1\gt9.64\ldots\)	M1	GDC intersection, table, or (HL) \(n-1\gt\frac{\ln800}{\ln2}\). Accept \(u_{10}=2560\) and \(u_{11}=5120\) seen.
generation 11	A1	Must be a whole number; \(n=10.6\) scores A0.

Answer: generation 11

(3)

(g)(i)

Some months later the orbiter is \(3.8\times10^{11}\) m from Earth.

Find the new signal travel time, in minutes.

ROUTINENotes

Show solution

\(\frac{3.8\times10^{11}}{3.00\times10^{8}}=1266.66\ldots\) s	M1
21.1 minutes	A1	Accept \(21.11\ldots\).

Answer: 21.1 minutes

(2)

(g)(ii)

To check quickly, a controller rounds the distance to \(4\times10^{11}\) m before dividing.

Find the percentage error in the travel time this gives. Comment on whether rounding the distance at this stage matters when commands must be timed to the nearest minute.

INTERPRETNotes

Show solution

\(\frac{4\times10^{11}}{3.00\times10^{8}}=1333.3\ldots\) s (22.2 minutes)	M1	Allow the percentage error found directly from the distances, since time is proportional to distance.
\(\varepsilon=\left\|\frac{1333.3-1266.7}{1266.7}\right\|\times100=5.26\%\)	A1
the time is about 67 s (over a minute) too long, so the command would be timed at the wrong minute: rounding early does matter here	R1	Accept 22 minutes against 21 minutes. A comment based on 5.26% alone without reference to the minute does not score.

Answer: \(5.26\%\); it matters (an error of more than one minute)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(4.68\times10^{7}\)	A1	Do not accept \(46.8\times10^{6}\).
Part (b)
\(\frac{2.25\times10^{11}}{3.00\times10^{8}}\)	M1	Distance divided by speed.
\(7.50\times10^{2}\) s	A1	Accept \(7.5\times10^{2}\) s. 750 s on its own is not in standard form: A0.
12.5 minutes	A1	Allow ECF from their time in seconds.
Part (c)(i)
\(9\times4000\times(5\times10^{5})\)	M1	All three values rounded to 1 s.f. and multiplied.
\(1.8\times10^{10}\) bytes	A1	Accept 18 000 000 000.
Part (c)(ii)
\(9.3\times3600\times4.85\times10^{5}=1.62378\times10^{10}\) bytes	B1	Accept \(1.62\times10^{10}\).
\(\varepsilon=\left\|\frac{1.8\times10^{10}-1.62378\times10^{10}}{1.62378\times10^{10}}\right\|\times100\%\)	M1	The exact value must be in the denominator. Allow ECF from (c)(i).
\(10.9\%\)	A1	Accept \(10.85\ldots\%\).
Part (d)(i)
\(\frac{900-2100}{40-0}\)	M1	Any two correct points read from the line.
\(-30\)	A1
the stored data decreases by 30 MB every minute (30 MB are sent each minute)	R1	Must refer to a rate per minute in context. Allow ECF from their gradient.
Part (d)(ii)
\(\frac{30\times10^{6}}{60}\)	M1	Converting MB to bytes and minutes to seconds.
\(=500\,000\) bytes per second	A1	The value 500 000 (or an equivalent value not yet in standard form) must be seen before the given answer.
\(=5.00\times10^{5}\) bytes per second	AG
Part (d)(iii)
\(D=2100-30t\) and \(2100-30t=0\)	M1	Or \(\frac{2100}{30}\). Allow ECF from (d)(i).
70 minutes	A1
Part (e)
\(u_6=5\times2^{5}\)	M1	Geometric with \(u_1=5\), \(r=2\); or a list 5, 10, 20, 40, 80, 160.
\(=5\times32\)	A1	An intermediate value must be seen. For the list method, all six terms seen scores M1A1.
160 GB	AG
Part (f)(i)
\(\frac{160\times10^{9}}{4.68\times10^{7}}=3418.8\ldots\)	M1
3418 files	A1	Must be rounded down to a whole number. 3419 scores A0.
Part (f)(ii)
\(4.0\times10^{12}\) bytes \(=4000\) GB	M1	Or \(u_n\) written in bytes: \(5\times10^{9}\times2^{n-1}\).
\(5\times2^{n-1}\gt4000\Rightarrow n-1\gt9.64\ldots\)	M1	GDC intersection, table, or (HL) \(n-1\gt\frac{\ln800}{\ln2}\). Accept \(u_{10}=2560\) and \(u_{11}=5120\) seen.
generation 11	A1	Must be a whole number; \(n=10.6\) scores A0.
Part (g)(i)
\(\frac{3.8\times10^{11}}{3.00\times10^{8}}=1266.66\ldots\) s	M1
21.1 minutes	A1	Accept \(21.11\ldots\).
Part (g)(ii)
\(\frac{4\times10^{11}}{3.00\times10^{8}}=1333.3\ldots\) s (22.2 minutes)	M1	Allow the percentage error found directly from the distances, since time is proportional to distance.
\(\varepsilon=\left\|\frac{1333.3-1266.7}{1266.7}\right\|\times100=5.26\%\)	A1
the time is about 67 s (over a minute) too long, so the command would be timed at the wrong minute: rounding early does matter here	R1	Accept 22 minutes against 21 minutes. A comment based on 5.26% alone without reference to the minute does not score.

Answers: (a) \(4.68\times10^{7}\) bytes · (b) \(7.50\times10^{2}\) s \(=12.5\) minutes · (c)(i) \(1.8\times10^{10}\) bytes · (c)(ii) \(1.62\times10^{10}\) bytes; \(\varepsilon=10.9\%\) · (d)(i) \(-30\) MB per minute · (d)(iii) 70 minutes · (f)(i) 3418 files · (f)(ii) generation 11 · (g)(i) 21.1 minutes · (g)(ii) \(5.26\%\); it matters (an error of more than one minute)

Common error (c)(ii) Dividing by the estimate instead of the exact value gives 9.79%. In percentage error the exact value is always the denominator.

Exam tip (g)(ii) Keep the full given values on your GDC and round once, at the end. Rounding an input to 1 s.f. passes its whole percentage error on to the answer.

More 1.1 questions in the question bank

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Caffeine after an energy drink: exponential decay and logarithms

1.4 Exponents and logarithms1.1 Approximation, percentage error and standard form2.3 Exponential and sinusoidal models1.3 Arithmetic and geometric sequences in contextSL

A student drinks a can of energy drink containing 160 mg of caffeine. In a simple model, all of the caffeine enters the body at time \(t=0\), and after that the amount of caffeine in the body decreases by 13% each hour.

The amount of caffeine in the body \(t\) hours after the drink is \(A\) mg, where \(A=k\,a^{t}\), \(t\ge0\).

(a)

Write down the value of \(k\) and the value of \(a\).

ROUTINENotes

Show solution

\(k=160\)	A1
\(a=0.87\)	A1	\(a=0.13\) or \(a=1.13\) scores A0.

Answer: \(k=160,\ a=0.87\)

(2)

(b)

Find the amount of caffeine in the body 5 hours after the drink.

ROUTINENotes

Show solution

\(160\times0.87^{5}\)	M1	Allow ECF from (a).
79.7 mg	A1	Accept \(79.747\ldots\).

Answer: 79.7 mg

(2)

(c)

The mass of one caffeine molecule is \(3.22\times10^{-22}\) g.

Find the number of caffeine molecules in the drink. Give your answer in the form \(a\times10^{k}\), where \(1\le a\lt10\) and \(k\in\mathbb{Z}\).

MULTINotes

Show solution

\(160\) mg \(=0.160\) g \(=1.60\times10^{-1}\) g	M1	Correct unit conversion seen or implied.
\(\frac{0.160}{3.22\times10^{-22}}\)	M1	Dividing the total mass by the mass of one molecule, with consistent units.
\(4.97\times10^{20}\)	A1	\(4.97\times10^{23}\) (no conversion) scores M0M1A0.

Answer: \(4.97\times10^{20}\) molecules

(3)

(d)(i)

Figure 1 shows the graph of \(A\) for \(0\le t\le14\).

Write down the equation of the horizontal asymptote of the graph of \(A\), and interpret it in context.

Figure 1: the amount of caffeine in the body, A mg, t hours after the drink

CONCEPTNotes

Show solution

\(A=0\)	A1	Accept \(y=0\) or "the \(t\)-axis".
the amount of caffeine gets closer and closer to 0 mg but the model never predicts it to be completely gone	R1

Answer: \(A=0\)

(2)

(d)(ii)

Use Figure 1 to estimate the time at which 100 mg of caffeine remain in the body.

GRAPHNotes

Show solution

line \(A=100\) meets the curve (seen or implied)	M1
3.4 hours	A1	Accept any value from 3.3 to 3.5.

Answer: about 3.4 hours

(2)

(e)

Show that the time taken for the amount of caffeine in the body to halve is approximately 5 hours.

PROOFNotes

Show solution

\(160\times0.87^{t}=80\) or \(0.87^{t}=0.5\)	M1	Accept a GDC intersection of \(y=160\times0.87^{x}\) and \(y=80\), or (HL) \(t=\frac{\ln0.5}{\ln0.87}\).
\(t=4.977\ldots\)	A1	The unrounded value must be seen. Substituting \(t=5\) to get 79.7 is not enough on its own.
\(\approx5\) hours	AG

(2)

(f)(i)

Show that the amount of caffeine removed from the body during the second hour is 18.096 mg.

ROUTINENotes

Show solution

\(A(1)=139.2\) and \(A(2)=121.104\)	M1	Or \(0.13\times139.2\).
\(139.2-121.104=18.096\)	A1	Both values must be seen.
18.096 mg	AG

(2)

(f)(ii)

The amounts of caffeine removed during the 1st, 2nd, 3rd, … hour form a geometric sequence.

Use the sum of this geometric sequence to find the total amount of caffeine removed from the body during the first 8 hours.

MULTINotes

Show solution

\(u_1=20.8\) and \(r=0.87\)	B1	\(r=\frac{18.096}{20.8}\).
\(S_8=\frac{20.8(1-0.87^{8})}{1-0.87}\)	M1	Correct substitution into the sum formula, with their values.
107 mg	A1	Accept \(107.48\ldots\). The direct calculation \(160-160\times0.87^{8}\) without the geometric sequence does not use the method asked for: award 1 mark only for 107 mg.

Answer: 107 mg

(3)

(g)

A blood test 3 hours after the drink suggests that 108 mg of caffeine remain in the body.

Taking this measured value as exact, find the percentage error in the model's prediction for \(t=3\).

ROUTINENotes

Show solution

\(\left\|\frac{105.36-108}{108}\right\|\times100\%\)	M1	Model value \(160\times0.87^{3}=105.36\) must be used as \(v_A\).
\(2.44\%\)	A1

Answer: \(2.44\%\)

(2)

(h)

Suggest one reason why this model may not be appropriate when only a small amount of caffeine remains in the body.

INTERPRETNotes

Show solution

the model never reaches zero, but in reality all of the caffeine is eventually removed from the body

Accept: the body may not remove a constant percentage at very low levels; very small amounts cannot be measured accurately enough to test the model.

(1)

(i)(i)

At \(t=4\) the student drinks a second, smaller can containing 80 mg of caffeine. Assume that the amount from each can decreases by 13% each hour after it is drunk, so that for \(t\ge4\)

\(A=160(0.87)^{t}+80(0.87)^{t-4}\).

Show that, for \(t\ge4\), \(A=c(0.87)^{t}\), where \(c=300\) correct to three significant figures.

ALGEBRANotes

Show solution

\((0.87)^{t-4}=(0.87)^{t}\times(0.87)^{-4}\)	M1	Law of exponents applied.
\(c=160+80\times0.87^{-4}=160+139.64\ldots=299.64\ldots\)	A1	The unrounded value must be seen.
\(c=300\) (3 s.f.)	AG

(2)

(i)(ii)

Find the time at which the amount of caffeine in the body first falls below 50 mg.

MODELNotes

Show solution

\(299.64\ldots\times0.87^{t}=50\)	M1	Or GDC intersection, or \(171.66\times0.87^{t-4}=50\). Using \(c=300\) gives 12.87, which is accepted.
\(t=12.9\) hours	A1	Accept \(12.86\ldots\).

Answer: after 12.9 hours

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(k=160\)	A1
\(a=0.87\)	A1	\(a=0.13\) or \(a=1.13\) scores A0.
Part (b)
\(160\times0.87^{5}\)	M1	Allow ECF from (a).
79.7 mg	A1	Accept \(79.747\ldots\).
Part (c)
\(160\) mg \(=0.160\) g \(=1.60\times10^{-1}\) g	M1	Correct unit conversion seen or implied.
\(\frac{0.160}{3.22\times10^{-22}}\)	M1	Dividing the total mass by the mass of one molecule, with consistent units.
\(4.97\times10^{20}\)	A1	\(4.97\times10^{23}\) (no conversion) scores M0M1A0.
Part (d)(i)
\(A=0\)	A1	Accept \(y=0\) or "the \(t\)-axis".
the amount of caffeine gets closer and closer to 0 mg but the model never predicts it to be completely gone	R1
Part (d)(ii)
line \(A=100\) meets the curve (seen or implied)	M1
3.4 hours	A1	Accept any value from 3.3 to 3.5.
Part (e)
\(160\times0.87^{t}=80\) or \(0.87^{t}=0.5\)	M1	Accept a GDC intersection of \(y=160\times0.87^{x}\) and \(y=80\), or (HL) \(t=\frac{\ln0.5}{\ln0.87}\).
\(t=4.977\ldots\)	A1	The unrounded value must be seen. Substituting \(t=5\) to get 79.7 is not enough on its own.
\(\approx5\) hours	AG
Part (f)(i)
\(A(1)=139.2\) and \(A(2)=121.104\)	M1	Or \(0.13\times139.2\).
\(139.2-121.104=18.096\)	A1	Both values must be seen.
18.096 mg	AG
Part (f)(ii)
\(u_1=20.8\) and \(r=0.87\)	B1	\(r=\frac{18.096}{20.8}\).
\(S_8=\frac{20.8(1-0.87^{8})}{1-0.87}\)	M1	Correct substitution into the sum formula, with their values.
107 mg	A1	Accept \(107.48\ldots\). The direct calculation \(160-160\times0.87^{8}\) without the geometric sequence does not use the method asked for: award 1 mark only for 107 mg.
Part (g)
\(\left\|\frac{105.36-108}{108}\right\|\times100\%\)	M1	Model value \(160\times0.87^{3}=105.36\) must be used as \(v_A\).
\(2.44\%\)	A1
Part (h)
the model never reaches zero, but in reality all of the caffeine is eventually removed from the body	R1	Accept: the body may not remove a constant percentage at very low levels; very small amounts cannot be measured accurately enough to test the model.
Part (i)(i)
\((0.87)^{t-4}=(0.87)^{t}\times(0.87)^{-4}\)	M1	Law of exponents applied.
\(c=160+80\times0.87^{-4}=160+139.64\ldots=299.64\ldots\)	A1	The unrounded value must be seen.
\(c=300\) (3 s.f.)	AG
Part (i)(ii)
\(299.64\ldots\times0.87^{t}=50\)	M1	Or GDC intersection, or \(171.66\times0.87^{t-4}=50\). Using \(c=300\) gives 12.87, which is accepted.
\(t=12.9\) hours	A1	Accept \(12.86\ldots\).

Answers: (a) \(k=160,\ a=0.87\) · (b) 79.7 mg · (c) \(4.97\times10^{20}\) molecules · (d)(i) \(A=0\) · (d)(ii) about 3.4 hours · (f)(ii) 107 mg · (g) \(2.44\%\) · (i)(ii) after 12.9 hours

Exam tip (e) In a "show that" with an approximate answer, write the value to more figures than the one given (here 4.977…), then round it.

More 1.4 questions in the question bank

1.2 Financial mathematics, loans and annuities · 1.3 Arithmetic and geometric sequences in contextmerged group

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Financing a first electric car: a loan, depreciation and total cost

1.2 Financial mathematics, loans and annuities1.3 Arithmetic and geometric sequences in context1.1 Approximation, percentage error and standard form2.1 Linear models and piecewise modelsSL

Lena buys a new electric car for €26 500. She pays a deposit of €4000 and borrows the rest from a bank. The loan is repaid in equal monthly instalments, paid at the end of each month, over 5 years. The nominal annual interest rate is 6.2%, compounded monthly.

(a)(i)

Find the monthly repayment.

ROUTINENotes

Show solution

\(N=60,\ I\%=6.2,\ PV=22\,500,\ FV=0,\ P/Y=12,\ C/Y=12\)	M1	Accept \(PV=-22\,500\) with \(PMT\) positive. \(PV=26\,500\) scores M0.
€437.08	A1

Answer: €437.08

(2)

(a)(ii)

Find the total amount Lena repays to the bank and the total interest she pays. Give your answers correct to the nearest cent.

MULTINotes

Show solution

\(437.0835\ldots\times60\)	M1	Allow ECF from (a)(i).
€26 225.01	A1	Using the rounded payment gives €26 224.80: award A0 for that value.
€3725.01	A1	Accept €3724.80 by ECF from €26 224.80.

Answer: €26 225.01 repaid; €3725.01 interest

(3)

(b)

The value of the car decreases by 14% each year.

Show that the value of the car after 3 years is approximately €16 900.

PROOFNotes

Show solution

\(26\,500\times0.86^{3}\)	M1	Multiplier \(1-0.14\) must be used.
\(=16\,855.48\ldots\)	A1	The unrounded value must be seen.
\(\approx16\,900\)	AG

(2)

(c)

The values of the car at the end of year 1, year 2, year 3, … form a geometric sequence.

Write down the first term and the common ratio of this sequence. Hence find the value of the car at the end of year 8, correct to the nearest euro.

ROUTINENotes

Show solution

\(u_1=22\,790,\ r=0.86\)	A1	Both needed.
\(u_8=22\,790\times0.86^{7}\)	M1	Or \(26\,500\times0.86^{8}\). \(22\,790\times0.86^{8}\) scores M0.
€7929	A1	\(7929.28\ldots\) rounded to the nearest euro.

Answer: €7929

(3)

(d)(i)

The car dealer's guide uses a linear model instead. It values the car at €26 500 when new and at €13 250 after 5 years.

Find the equation of the dealer's model in the form \(L=mt+c\), where \(L\) is the value of the car, in euros, \(t\) years after it was bought. Interpret the value of \(m\) in context.

MODELNotes

Show solution

\(m=\frac{13\,250-26\,500}{5-0}\)	M1
\(L=26\,500-2650t\)	A1
the dealer's model says the car loses €2650 of value each year	R1	Must mention a decrease per year in euros. Allow ECF from their \(m\).

Answer: \(L=26\,500-2650t\)

(3)

(d)(ii)

Figure 1 shows the values given by the two models for \(0\le t\le10\).

Find the value of \(t\), where \(t\gt0\), at which the two models give the same value, and find this value to the nearest euro. State which model gives the lower value of the car before this time.

Figure 1: the value of the car under the two depreciation models

GRAPHNotes

Show solution

\(t=5.88\)	A1	GDC intersection of \(y=26\,500\times0.86^{x}\) and \(y=26\,500-2650x\). Accept 5.9 read from the graph.
€10 915	A1	Accept €10 900 from a graph reading. Allow ECF from their equation in (d)(i).
the percentage model (its curve lies below the line for these times)	R1

Answer: \(t=5.88\) years, value €10 915; the percentage model is lower

(3)

(e)(i)

The bank also offers Lena the same amount over 7 years, at the same rate of 6.2% per year compounded monthly, with payments at the end of each month.

Find the monthly repayment and the total interest for this loan.

MULTINotes

Show solution

\(N=84\) with the other entries as in (a)(i)	M1
€330.85	A1
€5291.74	A1	\(330.854\ldots\times84-22\,500\).

Answer: €330.85 per month; €5291.74 interest

(3)

(e)(ii)

Discuss whether the 7-year loan is better value for Lena than the 5-year loan. Refer to your answers and to the value of the car.

INTERPRETNotes

Show solution

the monthly payment is about €106 lower, so it is easier to afford each month	R1	A valid advantage, with a figure.
but she pays about €1567 more interest, and by the end of year 7 the car is worth only about €9220, so the 5-year loan is better value overall	R1	A valid disadvantage with a figure, and a conclusion. Accept a reasoned conclusion either way.

(2)

(f)

The bank now lowers the rate of the 5-year loan to 4.9% per year, compounded monthly.

Find the new monthly repayment, and the amount Lena saves in total interest compared with the 6.2% loan.

MULTINotes

Show solution

\(N=60,\ I\%=4.9,\ PV=22\,500,\ FV=0\)	M1
€423.57	A1
new interest \(423.5727\ldots\times60-22\,500=2914.36\); saving \(3725.01-2914.36=€810.65\)	A1	Allow ECF from (a)(ii) and from their new repayment.

Answer: €423.57 per month; she saves €810.65

(3)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(N=60,\ I\%=6.2,\ PV=22\,500,\ FV=0,\ P/Y=12,\ C/Y=12\)	M1	Accept \(PV=-22\,500\) with \(PMT\) positive. \(PV=26\,500\) scores M0.
€437.08	A1
Part (a)(ii)
\(437.0835\ldots\times60\)	M1	Allow ECF from (a)(i).
€26 225.01	A1	Using the rounded payment gives €26 224.80: award A0 for that value.
€3725.01	A1	Accept €3724.80 by ECF from €26 224.80.
Part (b)
\(26\,500\times0.86^{3}\)	M1	Multiplier \(1-0.14\) must be used.
\(=16\,855.48\ldots\)	A1	The unrounded value must be seen.
\(\approx16\,900\)	AG
Part (c)
\(u_1=22\,790,\ r=0.86\)	A1	Both needed.
\(u_8=22\,790\times0.86^{7}\)	M1	Or \(26\,500\times0.86^{8}\). \(22\,790\times0.86^{8}\) scores M0.
€7929	A1	\(7929.28\ldots\) rounded to the nearest euro.
Part (d)(i)
\(m=\frac{13\,250-26\,500}{5-0}\)	M1
\(L=26\,500-2650t\)	A1
the dealer's model says the car loses €2650 of value each year	R1	Must mention a decrease per year in euros. Allow ECF from their \(m\).
Part (d)(ii)
\(t=5.88\)	A1	GDC intersection of \(y=26\,500\times0.86^{x}\) and \(y=26\,500-2650x\). Accept 5.9 read from the graph.
€10 915	A1	Accept €10 900 from a graph reading. Allow ECF from their equation in (d)(i).
the percentage model (its curve lies below the line for these times)	R1
Part (e)(i)
\(N=84\) with the other entries as in (a)(i)	M1
€330.85	A1
€5291.74	A1	\(330.854\ldots\times84-22\,500\).
Part (e)(ii)
the monthly payment is about €106 lower, so it is easier to afford each month	R1	A valid advantage, with a figure.
but she pays about €1567 more interest, and by the end of year 7 the car is worth only about €9220, so the 5-year loan is better value overall	R1	A valid disadvantage with a figure, and a conclusion. Accept a reasoned conclusion either way.
Part (f)
\(N=60,\ I\%=4.9,\ PV=22\,500,\ FV=0\)	M1
€423.57	A1
new interest \(423.5727\ldots\times60-22\,500=2914.36\); saving \(3725.01-2914.36=€810.65\)	A1	Allow ECF from (a)(ii) and from their new repayment.

Answers: (a)(i) €437.08 · (a)(ii) €26 225.01 repaid; €3725.01 interest · (c) €7929 · (d)(i) \(L=26\,500-2650t\) · (d)(ii) \(t=5.88\) years, value €10 915; the percentage model is lower · (e)(i) €330.85 per month; €5291.74 interest · (f) €423.57 per month; she saves €810.65

Common error (a)(ii) Multiplying the rounded payment €437.08 by 60 gives €26 224.80, which is 21 cents too low. Use the unrounded value stored in the GDC and round only the final amount.

More 1.2 questions in the question bank

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Training for a first marathon: arithmetic and geometric build-ups

1.3 Arithmetic and geometric sequences in context1.2 Financial mathematics, loans and annuities1.1 Approximation, percentage error and standard form2.1 Linear models and piecewise modelsSL

Nikos is training for his first marathon. Once a week he does a long run.

Plan A: the long run is 8 km in week 1 and increases by 1.5 km every week.
Plan B (suggested by his coach): the long run is 8 km in week 1 and increases by 9% every week.

(a)

Find the length of the long run in week 10 of plan A.

ROUTINENotes

Show solution

\(u_{10}=8+(10-1)\times1.5\)	M1
21.5 km	A1

Answer: 21.5 km

(2)

(b)

Find the total distance of the long runs in the first 16 weeks of plan A.

ROUTINENotes

Show solution

\(S_{16}=\frac{16}{2}\left(2\times8+15\times1.5\right)\)	M1	Or \(u_{16}=30.5\) and \(\frac{16}{2}(8+30.5)\).
308 km	A1

Answer: 308 km

(2)

(c)

Show that the total distance of the long runs in the first 12 weeks of plan B is approximately 161 km.

PROOFNotes

Show solution

\(S_{12}=\frac{8(1.09^{12}-1)}{1.09-1}\)	M1	\(r=1.09\) must be used.
\(=161.125\ldots\)	A1	The unrounded value must be seen.
\(\approx161\) km	AG

(2)

(d)

For each plan, find the first week in which the long run is at least 32 km.

MULTINotes

Show solution

\(8+1.5(n-1)\ge32\)	M1	Accept an equation.
week 17 (plan A)	A1
\(8\times1.09^{n-1}\ge32\Rightarrow n-1\ge16.08\ldots\)	M1	GDC intersection or table: \(u_{17}=31.76\), \(u_{18}=34.62\). (HL) \(n-1\ge\frac{\ln4}{\ln1.09}\).
week 18 (plan B)	A1	Must be a whole number. Week 17 for plan B scores A0.

Answer: plan A: week 17; plan B: week 18

(4)

(e)(i)

Figure 1 shows the length of the long run in each of the first 20 weeks for both plans.

The points for plan A lie on a straight line with equation \(y=mx+c\), where \(x\) is the week number and \(y\) is the length of the long run in km. Write down the value of \(m\) and the value of \(c\).

Figure 1: the length of the long run in each of the first 20 weeks for plan A and plan B (in week 1 both plans give 8 km)

ROUTINENotes

Show solution

\(m=1.5\)	A1	The common difference.
\(c=6.5\)	A1	From \(8+1.5(n-1)=1.5n+6.5\).

Answer: \(m=1.5,\ c=6.5\)

(2)

(e)(ii)

Use Figure 1 to identify the first week in which the long run of plan B is longer than the long run of plan A. Verify your answer by calculation.

GRAPHNotes

Show solution

week 17: A \(=32\), B \(=31.76\); week 18: A \(=33.5\), B \(=34.62\)	M1	Both weeks needed for the verification.
week 18	A1

Answer: week 18

(2)

(f)

Nikos estimates the total distance of the long runs in the first 16 weeks of plan B by rounding the common ratio to 1.1.

Find the percentage error in his estimate.

MULTINotes

Show solution

estimate \(\frac{8(1.1^{16}-1)}{0.1}=287.60\ldots\); exact \(\frac{8(1.09^{16}-1)}{0.09}=264.03\ldots\)	M1	Both sums needed.
\(\varepsilon=\left\|\frac{287.60-264.03}{264.03}\right\|\times100\%\)	M1	Exact value in the denominator.
\(8.93\%\)	A1

Answer: \(8.93\%\)

(3)

(g)

Nikos also saves €40 at the end of every month towards a race trip costing €650. His account pays 3% per year, compounded monthly, and he starts with nothing.

Find the number of months he needs to save before he has enough for the trip.

MULTINotes

Show solution

\(I\%=3,\ PV=0,\ PMT=-40,\ FV=650,\ P/Y=12,\ C/Y=12\)	M1	Signs: payments negative, final value positive.
\(N=15.94\ldots\)	A1
16 months	A1	Must be rounded up. After 15 months the balance is €610.61; after 16 it is €652.14.

Answer: 16 months

(3)

(h)

Explain why plan B would not be realistic as a long-term training plan.

INTERPRETNotes

Show solution

a geometric model with \(r\gt1\) grows without limit, faster and faster, e.g. week 30 would be about 97 km	R1	A value from the model, or a clear statement of unbounded accelerating growth.
a runner cannot keep increasing the distance like this (a marathon is about 42 km, and injury or available time limit it)	R1	Must be in context.

(2)

(i)

Nikos decides to follow plan A, but to make every fourth week (weeks 4, 8, 12 and 16) a recovery week, in which the long run is only 60% of the plan A distance for that week.

Find the total distance of his long runs over the 16 weeks.

MULTINotes

Show solution

recovery weeks in plan A: \(12.5,\ 18.5,\ 24.5,\ 30.5\), total 86 km	M1	Arithmetic with \(d=6\).
\(308-0.4\times86\)	M1	Or \((308-86)+0.6\times86\). Allow ECF from (b).
273.6 km	A1

Answer: 273.6 km

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(u_{10}=8+(10-1)\times1.5\)	M1
21.5 km	A1
Part (b)
\(S_{16}=\frac{16}{2}\left(2\times8+15\times1.5\right)\)	M1	Or \(u_{16}=30.5\) and \(\frac{16}{2}(8+30.5)\).
308 km	A1
Part (c)
\(S_{12}=\frac{8(1.09^{12}-1)}{1.09-1}\)	M1	\(r=1.09\) must be used.
\(=161.125\ldots\)	A1	The unrounded value must be seen.
\(\approx161\) km	AG
Part (d)
\(8+1.5(n-1)\ge32\)	M1	Accept an equation.
week 17 (plan A)	A1
\(8\times1.09^{n-1}\ge32\Rightarrow n-1\ge16.08\ldots\)	M1	GDC intersection or table: \(u_{17}=31.76\), \(u_{18}=34.62\). (HL) \(n-1\ge\frac{\ln4}{\ln1.09}\).
week 18 (plan B)	A1	Must be a whole number. Week 17 for plan B scores A0.
Part (e)(i)
\(m=1.5\)	A1	The common difference.
\(c=6.5\)	A1	From \(8+1.5(n-1)=1.5n+6.5\).
Part (e)(ii)
week 17: A \(=32\), B \(=31.76\); week 18: A \(=33.5\), B \(=34.62\)	M1	Both weeks needed for the verification.
week 18	A1
Part (f)
estimate \(\frac{8(1.1^{16}-1)}{0.1}=287.60\ldots\); exact \(\frac{8(1.09^{16}-1)}{0.09}=264.03\ldots\)	M1	Both sums needed.
\(\varepsilon=\left\|\frac{287.60-264.03}{264.03}\right\|\times100\%\)	M1	Exact value in the denominator.
\(8.93\%\)	A1
Part (g)
\(I\%=3,\ PV=0,\ PMT=-40,\ FV=650,\ P/Y=12,\ C/Y=12\)	M1	Signs: payments negative, final value positive.
\(N=15.94\ldots\)	A1
16 months	A1	Must be rounded up. After 15 months the balance is €610.61; after 16 it is €652.14.
Part (h)
a geometric model with \(r\gt1\) grows without limit, faster and faster, e.g. week 30 would be about 97 km	R1	A value from the model, or a clear statement of unbounded accelerating growth.
a runner cannot keep increasing the distance like this (a marathon is about 42 km, and injury or available time limit it)	R1	Must be in context.
Part (i)
recovery weeks in plan A: \(12.5,\ 18.5,\ 24.5,\ 30.5\), total 86 km	M1	Arithmetic with \(d=6\).
\(308-0.4\times86\)	M1	Or \((308-86)+0.6\times86\). Allow ECF from (b).
273.6 km	A1

Answers: (a) 21.5 km · (b) 308 km · (d) plan A: week 17; plan B: week 18 · (e)(i) \(m=1.5,\ c=6.5\) · (e)(ii) week 18 · (f) \(8.93\%\) · (g) 16 months · (i) 273.6 km

Common error (d) For plan B, \(n-1=16.08\) gives \(n=17.08\), and week 17 is still 31.76 km, below 32 km. Round up to the next whole week and check both neighbouring terms.

More 1.3 questions in the question bank

1.5 Complex numbers in Cartesian, polar and exponential form · 1.6 Matrices and matrix algebra · 1.7 Eigenvalues and eigenvectorsmerged group

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An AC test circuit: impedance as a complex number

1.5 Complex numbers in Cartesian, polar and exponential form1.6 Matrices and matrix algebra2.3 Exponential and sinusoidal models3.1 Right-angled trigonometry, bearings and the sine and cosine rulesHL

In AC (alternating current) circuit analysis, each component is given a complex impedance, measured in ohms (Ω). A resistor of resistance \(R\) Ω has impedance \(R\). An inductor has an impedance that is a positive multiple of \(i\), and a capacitor has an impedance that is a negative multiple of \(i\). When components are connected in series, the total impedance is the sum of their impedances.

A test circuit, shown in Figure 1, connects a coil and a protective resistor in series to a 50 Hz supply. The coil is modelled as a resistance of \(R_c\) Ω in series with an inductor of impedance \(24i\) Ω. The protective resistor has resistance \(R\) Ω.

Figure 1: the test circuit (not to scale)

(a)(i)

To find \(R_c\) and \(R\), a technician uses a steady (DC) supply. With a DC supply the inductor has no effect, and the current in amperes is equal to the supply voltage divided by the total resistance in ohms.

Test A: one coil and two protective resistors in series. A 12.0 V supply gives a current of 1.20 A.
Test B: two coils and one protective resistor in series. A 9.9 V supply gives a current of 0.90 A.

Show that \(R_c+2R=10\).

MODELNotes

Show solution

total resistance in Test A is \(R_c+R+R=R_c+2R\)	M1
\(\frac{12.0}{R_c+2R}=1.20\) or \(R_c+2R=\frac{12.0}{1.20}\)	M1
\(R_c+2R=10\)	AG

(2)

(a)(ii)

Write down a second equation in \(R_c\) and \(R\). Hence, by writing your equations as a matrix equation and using an inverse matrix, find \(R_c\) and \(R\).

HL onlyROUTINENotes

Show solution

\(2R_c+R=11\)	B1	From \(\frac{9.9}{0.90}=11\).
\(\begin{pmatrix}1&2\\2&1\end{pmatrix}\begin{pmatrix}R_c\\R\end{pmatrix}=\begin{pmatrix}10\\11\end{pmatrix}\)	M1	Equations may be in either order if consistent.
\(\begin{pmatrix}R_c\\R\end{pmatrix}=-\frac{1}{3}\begin{pmatrix}1&-2\\-2&1\end{pmatrix}\begin{pmatrix}10\\11\end{pmatrix}\)	M1	Correct inverse (seen or implied by GDC) multiplied on the left; \(\det=-3\).
\(R_c=4,\ R=3\)	A1	Both values. A solution with no matrix method seen scores at most B1M0M0A0 (the question requires the inverse-matrix method).

Answer: \(R_c=4\) Ω, \(R=3\) Ω

(4)

(b)

Write down, in the form \(a+bi\), the impedance of the coil, \(Z_c\), and the total impedance of the circuit, \(Z\).

HL onlyROUTINENotes

Show solution

\(Z_c=4+24i\)	A1	Allow ECF from (a)(ii) throughout.
\(Z=7+24i\)	A1

Answer: \(Z_c=4+24i\), \(Z=7+24i\)

(2)

(c)

On the Argand diagram in Figure 2, plot and label the points that represent \(Z_c\), the impedance of the protective resistor, and \(Z\). Describe the geometrical relationship between these three points and the origin.

Figure 2: axes for the Argand diagram

HL onlyGRAPHNotes

Show solution

\(Z_c\) at \((4,\,24)\) and the resistor at \((3,\,0)\)	A1	Points within half a small square; labels needed.
\(Z\) at \((7,\,24)\)	A1	Allow ECF from (b).
the origin, the two component points and \(Z\) form a parallelogram, so \(Z\) is the vector sum of \(Z_c\) and \(3\)	R1	Accept a description of vector addition, e.g. moving \(Z_c\) 3 units to the right gives \(Z\).

(3)

(d)(i)

The angle \(\varphi\) between the line from the origin to \(Z\) and the positive real axis is called the phase angle of the circuit. By considering a right-angled triangle, find \(\varphi\) in degrees.

ROUTINENotes

Show solution

\(\tan\varphi=\frac{24}{7}\)	M1	Or \(\sin\varphi=\frac{24}{25}\) or \(\cos\varphi=\frac{7}{25}\).
\(\varphi=73.7^\circ\)	A1	\(73.739\ldots^\circ\). Allow ECF from (b).

Answer: \(73.7^\circ\)

(2)

(d)(ii)

Find \(|Z|\), and hence write \(Z\) in the form \(re^{i\theta}\), where \(-\pi<\theta\le\pi\).

HL onlyROUTINENotes

Show solution

\(\|Z\|=\sqrt{7^2+24^2}=25\)	A1
\(Z=25e^{1.29i}\)	A1	\(\theta=1.2870\ldots\) radians. Accept \(25e^{0.410\pi i}\). An argument in degrees, e.g. \(25e^{73.7i}\), scores A0.

Answer: \(|Z|=25\); \(Z=25e^{1.29i}\)

(2)

(e)

In this model the supply voltage is represented by the complex number \(V=50\) and the current by the complex number \(I=\dfrac{V}{Z}\). The modulus \(|I|\) is the greatest value of the current, in amperes, and \(\arg I\) gives the phase of the current relative to the voltage.

By first writing \(I\) in the form \(a+bi\), show that the greatest value of the current is 2 A.

HL onlyALGEBRANotes

Show solution

\(I=\frac{50}{7+24i}\times\frac{7-24i}{7-24i}\)	M1	Multiplying numerator and denominator by the conjugate.
\(I=\frac{350-1200i}{625}=0.56-1.92i\)	A1
\(\|I\|=\sqrt{0.56^2+1.92^2}=\sqrt{4}\)	M1	Accept \(\|I\|=\frac{\|V\|}{\|Z\|}=\frac{50}{25}\) as the modulus step only if \(I=0.56-1.92i\) has been found.
\(\|I\|=2\) A	AG

Answer: \(I=0.56-1.92i\)

(3)

(f)(i)

The supply voltage, \(v\) volts, at time \(t\) seconds is \(v=50\sin(100\pi t)\). The current, \(y\) amperes, is \(y=|I|\sin(100\pi t+\arg I)\).

Write \(y\) in the form \(y=a\sin(b(t-c))\), giving the value of \(c\) to 3 significant figures.

HL onlyMULTINotes

Show solution

\(\arg I=-1.287\ldots\) and \(100\pi t-1.287=100\pi\left(t-\frac{1.287}{100\pi}\right)\)	M1	Accept \(\arg I=-\arg Z\). Allow ECF from (d)(ii).
\(y=2\sin(100\pi(t-0.00410))\)	A1	\(c=0.0040966\ldots\); accept \(a=2,\ b=100\pi\ (=314.159\ldots),\ c=0.00410\).

Answer: \(y=2\sin(100\pi(t-0.00410))\)

(2)

(f)(ii)

Find the time at which the current first reaches its greatest value, and interpret the value of \(c\) in context.

INTERPRETNotes

Show solution

\(100\pi(t-0.00410)=\frac{\pi}{2}\), or the voltage peaks first at \(t=\frac{1}{200}=0.005\)	M1
\(t=0.00910\) s (9.10 ms)	A1	Allow ECF from (f)(i).
every maximum of the current occurs 0.00410 s (4.10 ms) after the corresponding maximum of the voltage, so the current lags behind the voltage	R1	Accept about one fifth of a cycle (period 0.02 s). Must mention lag or delay relative to the voltage.

Answer: \(t=0.00910\) s; the current lags the voltage by 4.10 ms

(3)

(g)(i)

The supply frequency is changed to 25 Hz, with the same amplitude of 50 V. At this frequency the impedance of the inductor is \(12i\) Ω. A capacitor of impedance \(-36i\) Ω is added in series, and the resistances are unchanged.

Find the new total impedance, \(Z_2\), and state how \(Z_2\) is related to \(Z\).

HL onlyROUTINENotes

Show solution

\(Z_2=4+3+12i-36i=7-24i\)	A1	Allow ECF from (a)(ii).
\(Z_2=Z^{*}\), the complex conjugate of \(Z\)	A1	Accept the reflection of \(Z\) in the real axis.

Answer: \(Z_2=7-24i=Z^{*}\)

(2)

(g)(ii)

Without calculating the new current, explain how its greatest value and its phase relative to the voltage compare with your answers to parts (e) and (f).

HL onlyINTERPRETNotes

Show solution

\(\|Z_2\|=\|Z\|=25\), so the greatest value of the current is unchanged at 2 A	R1
\(\arg Z_2=-\arg Z\), so \(\arg I_2=\arg V-\arg Z_2=+1.29\): the current now leads the voltage (by the same phase angle) instead of lagging	R1	Must refer to the change of sign of the argument. Do not accept 'the phase is the same'. In time, the current now leads by \(\frac{1.287}{50\pi}\approx0.00819\) s, not 4.10 ms, because the period has doubled; do not accept 'leads by 4.10 ms'.

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
total resistance in Test A is \(R_c+R+R=R_c+2R\)	M1
\(\frac{12.0}{R_c+2R}=1.20\) or \(R_c+2R=\frac{12.0}{1.20}\)	M1
\(R_c+2R=10\)	AG
Part (a)(ii)
\(2R_c+R=11\)	B1	From \(\frac{9.9}{0.90}=11\).
\(\begin{pmatrix}1&2\\2&1\end{pmatrix}\begin{pmatrix}R_c\\R\end{pmatrix}=\begin{pmatrix}10\\11\end{pmatrix}\)	M1	Equations may be in either order if consistent.
\(\begin{pmatrix}R_c\\R\end{pmatrix}=-\frac{1}{3}\begin{pmatrix}1&-2\\-2&1\end{pmatrix}\begin{pmatrix}10\\11\end{pmatrix}\)	M1	Correct inverse (seen or implied by GDC) multiplied on the left; \(\det=-3\).
\(R_c=4,\ R=3\)	A1	Both values. A solution with no matrix method seen scores at most B1M0M0A0 (the question requires the inverse-matrix method).
Part (b)
\(Z_c=4+24i\)	A1	Allow ECF from (a)(ii) throughout.
\(Z=7+24i\)	A1
Part (c)
\(Z_c\) at \((4,\,24)\) and the resistor at \((3,\,0)\)	A1	Points within half a small square; labels needed.
\(Z\) at \((7,\,24)\)	A1	Allow ECF from (b).
the origin, the two component points and \(Z\) form a parallelogram, so \(Z\) is the vector sum of \(Z_c\) and \(3\)	R1	Accept a description of vector addition, e.g. moving \(Z_c\) 3 units to the right gives \(Z\).
Part (d)(i)
\(\tan\varphi=\frac{24}{7}\)	M1	Or \(\sin\varphi=\frac{24}{25}\) or \(\cos\varphi=\frac{7}{25}\).
\(\varphi=73.7^\circ\)	A1	\(73.739\ldots^\circ\). Allow ECF from (b).
Part (d)(ii)
\(\|Z\|=\sqrt{7^2+24^2}=25\)	A1
\(Z=25e^{1.29i}\)	A1	\(\theta=1.2870\ldots\) radians. Accept \(25e^{0.410\pi i}\). An argument in degrees, e.g. \(25e^{73.7i}\), scores A0.
Part (e)
\(I=\frac{50}{7+24i}\times\frac{7-24i}{7-24i}\)	M1	Multiplying numerator and denominator by the conjugate.
\(I=\frac{350-1200i}{625}=0.56-1.92i\)	A1
\(\|I\|=\sqrt{0.56^2+1.92^2}=\sqrt{4}\)	M1	Accept \(\|I\|=\frac{\|V\|}{\|Z\|}=\frac{50}{25}\) as the modulus step only if \(I=0.56-1.92i\) has been found.
\(\|I\|=2\) A	AG
Part (f)(i)
\(\arg I=-1.287\ldots\) and \(100\pi t-1.287=100\pi\left(t-\frac{1.287}{100\pi}\right)\)	M1	Accept \(\arg I=-\arg Z\). Allow ECF from (d)(ii).
\(y=2\sin(100\pi(t-0.00410))\)	A1	\(c=0.0040966\ldots\); accept \(a=2,\ b=100\pi\ (=314.159\ldots),\ c=0.00410\).
Part (f)(ii)
\(100\pi(t-0.00410)=\frac{\pi}{2}\), or the voltage peaks first at \(t=\frac{1}{200}=0.005\)	M1
\(t=0.00910\) s (9.10 ms)	A1	Allow ECF from (f)(i).
every maximum of the current occurs 0.00410 s (4.10 ms) after the corresponding maximum of the voltage, so the current lags behind the voltage	R1	Accept about one fifth of a cycle (period 0.02 s). Must mention lag or delay relative to the voltage.
Part (g)(i)
\(Z_2=4+3+12i-36i=7-24i\)	A1	Allow ECF from (a)(ii).
\(Z_2=Z^{*}\), the complex conjugate of \(Z\)	A1	Accept the reflection of \(Z\) in the real axis.
Part (g)(ii)
\(\|Z_2\|=\|Z\|=25\), so the greatest value of the current is unchanged at 2 A	R1
\(\arg Z_2=-\arg Z\), so \(\arg I_2=\arg V-\arg Z_2=+1.29\): the current now leads the voltage (by the same phase angle) instead of lagging	R1	Must refer to the change of sign of the argument. Do not accept 'the phase is the same'. In time, the current now leads by \(\frac{1.287}{50\pi}\approx0.00819\) s, not 4.10 ms, because the period has doubled; do not accept 'leads by 4.10 ms'.

Answers: (a)(ii) \(R_c=4\) Ω, \(R=3\) Ω · (b) \(Z_c=4+24i\), \(Z=7+24i\) · (d)(i) \(73.7^\circ\) · (d)(ii) \(|Z|=25\); \(Z=25e^{1.29i}\) · (e) \(I=0.56-1.92i\) · (f)(i) \(y=2\sin(100\pi(t-0.00410))\) · (f)(ii) \(t=0.00910\) s; the current lags the voltage by 4.10 ms · (g)(i) \(Z_2=7-24i=Z^{*}\)

Exam tip (a)(ii) For \(AX=B\) the inverse goes on the left: \(X=A^{-1}B\). Write the full matrix equation before using the GDC, so the method is visible.

Common error (d)(ii) The argument in \(re^{i\theta}\) must be in radians. Convert \(73.7^\circ\) by multiplying by \(\frac{\pi}{180}\), or set the GDC to radian mode before using \(\arg\).

More 1.5 questions in the question bank

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A grey seal colony: an age-class matrix and its eigenvalues

1.7 Eigenvalues and eigenvectors1.6 Matrices and matrix algebra2.3 Exponential and sinusoidal models1.3 Arithmetic and geometric sequences in contextHL

Ecologists count a colony of grey seals once a year. The seals are split into two classes: juveniles (not yet breeding) and adults (breeding). Between one count and the next:

50% of the juveniles survive and remain juveniles, and 25% survive and become adults; the rest die;
90% of the adults survive;
for every adult, an average of 0.33 newborn pups survive to be counted as juveniles at the next count.

These rates are summarised in Figure 1. The numbers of juveniles, \(j_n\), and adults, \(a_n\), \(n\) years after the first count are modelled by \(\mathbf{x}_{n+1}=L\mathbf{x}_n\), where

\(\mathbf{x}_n=\begin{pmatrix}j_n\\a_n\end{pmatrix}\) and \(L=\begin{pmatrix}0.5&0.33\\0.25&0.9\end{pmatrix}\).

At the first count there were 560 juveniles and 700 adults, so \(\mathbf{x}_0=\begin{pmatrix}560\\700\end{pmatrix}\).

Figure 1: yearly changes in the seal colony

(a)

Explain, in context, the meaning of the second row of \(L\).

HL onlyCONCEPTNotes

Show solution

\(a_{n+1}=0.25j_n+0.9a_n\)	A1	Or an equivalent statement in words of the row-by-column product.
next year's adults are the juveniles that mature (25% of this year's juveniles) plus the adults that survive (90% of this year's adults)	R1	Both contributions needed.

(2)

(b)

Find \(\mathbf{x}_1\). Use a power of \(L\) to find the total number of seals predicted 5 years after the first count.

HL onlyROUTINENotes

Show solution

\(\mathbf{x}_1=\begin{pmatrix}511\\770\end{pmatrix}\)	B1
\(\mathbf{x}_5=L^5\mathbf{x}_0=\begin{pmatrix}574.9\ldots\\956.9\ldots\end{pmatrix}\)	M1	Use of \(L^5\mathbf{x}_0\), not \(5L\mathbf{x}_0\) or entries raised to the power 5.
1532 seals	A1	Accept 1530 to 1532 (1531.85…).

Answer: \(\mathbf{x}_1=\begin{pmatrix}511\\770\end{pmatrix}\); about 1532 seals

(3)

(c)(i)

Show that the eigenvalues of \(L\) are 1.05 and 0.35.

HL onlyPROOFNotes

Show solution

\(\det(L-\lambda I)=(0.5-\lambda)(0.9-\lambda)-0.33\times0.25=0\)	M1	Or use of \(\lambda^2-(\operatorname{tr}L)\lambda+\det L=0\) with \(\operatorname{tr}L=1.4\).
\(\lambda^2-1.4\lambda+0.3675=0\)	A1
\(\lambda=\frac{1.4\pm\sqrt{1.96-1.47}}{2}=\frac{1.4\pm0.7}{2}\) or \((\lambda-1.05)(\lambda-0.35)=0\)	A1	A correct solving step must be seen. Substituting both given values to get 0 is acceptable for this mark.
\(\lambda=1.05\) or \(\lambda=0.35\)	AG

(3)

(c)(ii)

Find an eigenvector for each eigenvalue.

HL onlyROUTINENotes

Show solution

\(\lambda=1.05\): \(-0.55x+0.33y=0\)	M1	Use of \((L-\lambda I)\mathbf{v}=\mathbf{0}\) for either eigenvalue.
\(\begin{pmatrix}3\\5\end{pmatrix}\)	A1	Or any non-zero multiple.
\(\lambda=0.35\): \(0.15x+0.33y=0\), \(\begin{pmatrix}11\\-5\end{pmatrix}\)	A1	Or any non-zero multiple.

Answer: \(\lambda=1.05\): \(\begin{pmatrix}3\\5\end{pmatrix}\); \(\lambda=0.35\): \(\begin{pmatrix}11\\-5\end{pmatrix}\)

(3)

(d)

Interpret, in context, the eigenvalue 1.05 and its eigenvector.

HL onlyINTERPRETNotes

Show solution

in the long term the colony grows by 5% per year (it is multiplied by 1.05 each year)	R1	Must refer to the long term.
in the long term the ratio of juveniles to adults approaches 3 : 5 (37.5% juveniles)	R1	Allow ECF from an incorrect eigenvector in (c)(ii).

(2)

(e)(i)

Write down a matrix \(P\) and a diagonal matrix \(D\) such that \(L=PDP^{-1}\).

HL onlyROUTINENotes

Show solution

\(P=\begin{pmatrix}3&11\\5&-5\end{pmatrix}\), \(D=\begin{pmatrix}1.05&0\\0&0.35\end{pmatrix}\)

Columns of \(P\) must be in the same order as the eigenvalues in \(D\). Allow ECF from (c)(ii).

Answer: \(P=\begin{pmatrix}3&11\\5&-5\end{pmatrix}\), \(D=\begin{pmatrix}1.05&0\\0&0.35\end{pmatrix}\)

(1)

(e)(ii)

Hence show that the total number of seals \(n\) years after the first count is \(T_n=1200(1.05)^n+60(0.35)^n\).

HL onlyPROOFNotes

Show solution

\(\mathbf{x}_n=PD^nP^{-1}\mathbf{x}_0\), or \(\mathbf{x}_0=\alpha\begin{pmatrix}3\\5\end{pmatrix}+\beta\begin{pmatrix}11\\-5\end{pmatrix}\)	M1
\(P^{-1}\mathbf{x}_0=\begin{pmatrix}150\\10\end{pmatrix}\), i.e. \(\alpha=150,\ \beta=10\)	A1
\(\mathbf{x}_n=150(1.05)^n\begin{pmatrix}3\\5\end{pmatrix}+10(0.35)^n\begin{pmatrix}11\\-5\end{pmatrix}\), so \(T_n=150(1.05)^n(3+5)+10(0.35)^n(11-5)\)	M1	Adding the two components of each eigenvector term must be seen.
\(T_n=1200(1.05)^n+60(0.35)^n\)	AG

(3)

(f)

Figure 2 shows the values of \(T_n\) for \(0\le n\le20\), together with the curve \(y=1200(1.05)^x\).

With reference to the formula in part (e)(ii), explain why the points lie above the curve, and why the points and the curve are almost indistinguishable for \(n\ge3\).

Figure 2: the total number of seals \(T_n\) (points) and the curve \(y=1200(1.05)^x\) (dashed)

HL onlyGRAPHNotes

Show solution

\(60(0.35)^n>0\) for every \(n\), so \(T_n\) is always greater than \(1200(1.05)^n\)	R1	Accept reference to the gap of 60 at \(n=0\) (1260 against 1200).
\(60(0.35)^n\to0\) quickly (for example \(60(0.35)^3\approx2.6\)), so \(T_n\approx1200(1.05)^n\)	R1	Must refer to the decay of the 0.35 term.

(2)

(g)

For large \(n\), the values of \(T_n\) are approximately a geometric sequence. Write down its common ratio, and use it to find the first year in which the total number of seals is predicted to exceed 2500.

MULTINotes

Show solution

\(r=1.05\)	B1
\(1200(1.05)^n>2500\), giving \(n>15.04\ldots\)	M1	Or a table: \(T_{15}=2494.7\), \(T_{16}=2619.4\).
16 years after the first count	A1	Must be a whole number of years. 15 scores A0.

Answer: 16 years after the first count

(3)

(h)(i)

After an outbreak of disease, the survival rate of adults falls to 80% per year. All other rates are unchanged.

Write down the new matrix \(L_2\), and find its dominant eigenvalue.

HL onlyROUTINENotes

Show solution

\(L_2=\begin{pmatrix}0.5&0.33\\0.25&0.8\end{pmatrix}\)	A1
\(0.974\)	A1	From \(\lambda^2-1.3\lambda+0.3175=0\) or the GDC; \(0.97403\ldots\). The other eigenvalue is \(0.326\).

Answer: \(L_2=\begin{pmatrix}0.5&0.33\\0.25&0.8\end{pmatrix}\); \(\lambda=0.974\)

(2)

(h)(ii)

Discuss whether the colony survives in the long term under each of the two models, and state one limitation of the models.

HL onlyINTERPRETNotes

Show solution

original model: \(1.05>1\), so the colony grows; after the disease: \(0.974<1\), so the colony declines by about 2.6% per year towards zero if the new rates continue	R1	Both models must be compared. Allow ECF from (h)(i).
a valid limitation, e.g. the rates are assumed constant every year; no limit from food or space (growth cannot continue for ever); no migration; numbers of seals are whole and random	R1

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(a_{n+1}=0.25j_n+0.9a_n\)	A1	Or an equivalent statement in words of the row-by-column product.
next year's adults are the juveniles that mature (25% of this year's juveniles) plus the adults that survive (90% of this year's adults)	R1	Both contributions needed.
Part (b)
\(\mathbf{x}_1=\begin{pmatrix}511\\770\end{pmatrix}\)	B1
\(\mathbf{x}_5=L^5\mathbf{x}_0=\begin{pmatrix}574.9\ldots\\956.9\ldots\end{pmatrix}\)	M1	Use of \(L^5\mathbf{x}_0\), not \(5L\mathbf{x}_0\) or entries raised to the power 5.
1532 seals	A1	Accept 1530 to 1532 (1531.85…).
Part (c)(i)
\(\det(L-\lambda I)=(0.5-\lambda)(0.9-\lambda)-0.33\times0.25=0\)	M1	Or use of \(\lambda^2-(\operatorname{tr}L)\lambda+\det L=0\) with \(\operatorname{tr}L=1.4\).
\(\lambda^2-1.4\lambda+0.3675=0\)	A1
\(\lambda=\frac{1.4\pm\sqrt{1.96-1.47}}{2}=\frac{1.4\pm0.7}{2}\) or \((\lambda-1.05)(\lambda-0.35)=0\)	A1	A correct solving step must be seen. Substituting both given values to get 0 is acceptable for this mark.
\(\lambda=1.05\) or \(\lambda=0.35\)	AG
Part (c)(ii)
\(\lambda=1.05\): \(-0.55x+0.33y=0\)	M1	Use of \((L-\lambda I)\mathbf{v}=\mathbf{0}\) for either eigenvalue.
\(\begin{pmatrix}3\\5\end{pmatrix}\)	A1	Or any non-zero multiple.
\(\lambda=0.35\): \(0.15x+0.33y=0\), \(\begin{pmatrix}11\\-5\end{pmatrix}\)	A1	Or any non-zero multiple.
Part (d)
in the long term the colony grows by 5% per year (it is multiplied by 1.05 each year)	R1	Must refer to the long term.
in the long term the ratio of juveniles to adults approaches 3 : 5 (37.5% juveniles)	R1	Allow ECF from an incorrect eigenvector in (c)(ii).
Part (e)(i)
\(P=\begin{pmatrix}3&11\\5&-5\end{pmatrix}\), \(D=\begin{pmatrix}1.05&0\\0&0.35\end{pmatrix}\)	A1	Columns of \(P\) must be in the same order as the eigenvalues in \(D\). Allow ECF from (c)(ii).
Part (e)(ii)
\(\mathbf{x}_n=PD^nP^{-1}\mathbf{x}_0\), or \(\mathbf{x}_0=\alpha\begin{pmatrix}3\\5\end{pmatrix}+\beta\begin{pmatrix}11\\-5\end{pmatrix}\)	M1
\(P^{-1}\mathbf{x}_0=\begin{pmatrix}150\\10\end{pmatrix}\), i.e. \(\alpha=150,\ \beta=10\)	A1
\(\mathbf{x}_n=150(1.05)^n\begin{pmatrix}3\\5\end{pmatrix}+10(0.35)^n\begin{pmatrix}11\\-5\end{pmatrix}\), so \(T_n=150(1.05)^n(3+5)+10(0.35)^n(11-5)\)	M1	Adding the two components of each eigenvector term must be seen.
\(T_n=1200(1.05)^n+60(0.35)^n\)	AG
Part (f)
\(60(0.35)^n>0\) for every \(n\), so \(T_n\) is always greater than \(1200(1.05)^n\)	R1	Accept reference to the gap of 60 at \(n=0\) (1260 against 1200).
\(60(0.35)^n\to0\) quickly (for example \(60(0.35)^3\approx2.6\)), so \(T_n\approx1200(1.05)^n\)	R1	Must refer to the decay of the 0.35 term.
Part (g)
\(r=1.05\)	B1
\(1200(1.05)^n>2500\), giving \(n>15.04\ldots\)	M1	Or a table: \(T_{15}=2494.7\), \(T_{16}=2619.4\).
16 years after the first count	A1	Must be a whole number of years. 15 scores A0.
Part (h)(i)
\(L_2=\begin{pmatrix}0.5&0.33\\0.25&0.8\end{pmatrix}\)	A1
\(0.974\)	A1	From \(\lambda^2-1.3\lambda+0.3175=0\) or the GDC; \(0.97403\ldots\). The other eigenvalue is \(0.326\).
Part (h)(ii)
original model: \(1.05>1\), so the colony grows; after the disease: \(0.974<1\), so the colony declines by about 2.6% per year towards zero if the new rates continue	R1	Both models must be compared. Allow ECF from (h)(i).
a valid limitation, e.g. the rates are assumed constant every year; no limit from food or space (growth cannot continue for ever); no migration; numbers of seals are whole and random	R1

Answers: (b) \(\mathbf{x}_1=\begin{pmatrix}511\\770\end{pmatrix}\); about 1532 seals · (c)(ii) \(\lambda=1.05\): \(\begin{pmatrix}3\\5\end{pmatrix}\); \(\lambda=0.35\): \(\begin{pmatrix}11\\-5\end{pmatrix}\) · (e)(i) \(P=\begin{pmatrix}3&11\\5&-5\end{pmatrix}\), \(D=\begin{pmatrix}1.05&0\\0&0.35\end{pmatrix}\) · (g) 16 years after the first count · (h)(i) \(L_2=\begin{pmatrix}0.5&0.33\\0.25&0.8\end{pmatrix}\); \(\lambda=0.974\)

More 1.7 questions in the question bank

Topic 2 · Functions

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	2.1	2.2	2.3	2.4	2.5	2.6	Other
Ride-hailing or e-scooter: comparing linear and piecewise fares	●	○		○			1.1
A basketball free throw: building a quadratic model	○	●					5.1 3.1
Daylight hours in Athens and a rooftop solar panel			●	○			5.3 1.1
Electric car registrations: choosing between a linear and an exponential model	○		○	●			4.2
Earthquakes: a logarithmic scale and a linearised frequency model			○		●		1.4 4.2
An invasive plant on a lake: a logistic model and transformations			○	○		●	5.1

2.1 Linear models and piecewise models · 2.2 Quadratic and cubic modelsmerged group

+ Add to test✓ In your testCC-AI-21-22-124 marks~30 min

Ride-hailing or e-scooter: comparing linear and piecewise fares

2.1 Linear models and piecewise models2.2 Quadratic and cubic models1.1 Approximation, percentage error and standard form2.4 Choosing and evaluating modelsSL

In a city, a commuter can travel by a ride-hailing app or by a rental e-scooter.

Ride-hailing app: a fixed charge of €3.00 plus €0.80 per kilometre travelled.
E-scooter: €1.00 to unlock plus €0.28 per minute of riding. The price is capped at €15.00, so any ride of 50 minutes or longer costs €15.00. A single ride may last at most 120 minutes.

(a)(i)

Write down a model for the ride-hailing fare, \(F\) euros, for a trip of \(d\) km.

MODELNotes

Show solution

\(F=3+0.8d\)

Accept \(F(d)=0.8d+3\).

Answer: \(F=3+0.8d\)

(1)

(a)(ii)

Interpret, in context, the gradient and the \(F\)-intercept of your model.

INTERPRETNotes

Show solution

gradient: the fare increases by €0.80 for each extra kilometre	A1	Must refer to the rate per km.
intercept: €3.00 is the fixed charge paid for any trip (the fare before any distance is travelled)	A1

(2)

(b)

Write down a piecewise model for the e-scooter price, \(S\) euros, for a ride of \(t\) minutes, where \(0<t\le120\).

MODELNotes

Show solution

\(S=1+0.28t\) for \(0<t<50\)	A1	Accept \(0<t\le50\), since both pieces give 15 at \(t=50\).
\(S=15\) for \(50\le t\le120\)	A1	Domains must not overlap or leave gaps (other than at the shared value \(t=50\)).

Answer: \(S=\begin{cases}1+0.28t,&0<t<50\\15,&50\le t\le120\end{cases}\)

(2)

(c)

The commuter rides the e-scooter at a constant speed of 15 km/h. Show that a 10 km trip costs €12.20 by e-scooter.

PROOFNotes

Show solution

time \(=\frac{10}{15}\) hours \(=40\) minutes	M1
\(S=1+0.28\times40\)	M1	Correct piece chosen, since \(40<50\).
\(S=12.20\)	AG

(2)

(d)

At 15 km/h, a 50-minute ride covers 12.5 km. Find the distance, less than 12.5 km, for which the e-scooter and the ride-hailing app cost the same.

MULTINotes

Show solution

\(t=4d\), so \(S=1+0.28(4d)=1+1.12d\)	M1
\(1+1.12d=3+0.8d\)	M1	Equating the two models, or reading the intersection from Figure 1 with a supporting calculation.
\(d=6.25\) km	A1	Both fares are €8.00.

Answer: \(6.25\) km

(3)

(e)

Figure 1 shows both prices for trips of up to 20 km at 15 km/h.

Use Figure 1, together with a suitable calculation, to find the set of distances for which the ride-hailing app is cheaper than the e-scooter.

Figure 1: the ride-hailing fare and the e-scooter price for trips of up to 20 km

GRAPHNotes

Show solution

the second intersection is on the capped part: \(3+0.8d=15\)	M1	Accept a correct reading of the graph supported by this equation.
\(d=15\)	A1
\(6.25<d<15\)	A1	Allow ECF from (d). Accept 'between 6.25 km and 15 km'.

Answer: \(6.25<d<15\) km

(3)

(f)

At rush hour the app multiplies the whole fare by a surge factor of 1.4.

Write down the rush-hour fare model, and explain why, at rush hour, the e-scooter at 15 km/h is cheaper for every trip that can be made by e-scooter (up to 30 km).

INTERPRETNotes

Show solution

\(F_{\text{rush}}=1.4(3+0.8d)=4.2+1.12d\)	A1
for \(d\le12.5\) both models have gradient 1.12, so the lines are parallel and the e-scooter is always €3.20 cheaper	R1	Must refer to equal gradients or parallel lines.
for \(d>12.5\) the e-scooter costs €15 while the rush-hour fare is more than \(4.2+1.12\times12.5=\)€18.20	R1

Answer: \(F_{\text{rush}}=4.2+1.12d\)

(3)

(g)(i)

The app company finds that, at €0.80 per km, its customers travel a total of 20 000 km per day. It estimates that each €0.10 increase in the price per km would reduce this daily distance by 2000 km, and each €0.10 decrease would increase it by 2000 km.

Let \(p\) be the price per km, in euros. Show that the daily income from the distance charge is \(R(p)=36\,000p-20\,000p^2\).

PROOFNotes

Show solution

daily distance \(=20\,000-2000\times\frac{p-0.8}{0.1}\)	M1	Or a linear model through \((0.8,\,20\,000)\) with gradient \(-20\,000\).
daily distance \(=36\,000-20\,000p\)	A1
\(R(p)=p(36\,000-20\,000p)=36\,000p-20\,000p^2\)	AG

(2)

(g)(ii)

Find the price per km that gives the greatest daily income, and the greatest daily income.

ROUTINENotes

Show solution

\(p=-\frac{36\,000}{2(-20\,000)}\)	M1	Or a GDC maximum.
€0.90 per km; €16 200	A1	Both values.

Answer: €0.90 per km; €16 200

(2)

(g)(iii)

A manager estimates the daily income at this new price by assuming that customers still travel 20 000 km per day. Taking the value from \(R(p)\) as exact, calculate the percentage error in the manager's estimate.

ROUTINENotes

Show solution

estimate \(=0.90\times20\,000=18\,000\); \(\varepsilon=\left\|\frac{18\,000-16\,200}{16\,200}\right\|\times100\%\)	M1	Allow ECF from (g)(ii).
\(11.1\%\)	A1	\(11.111\ldots\%\)

Answer: \(11.1\%\)

(2)

(h)

Comment on the validity of the model \(R(p)\) for a price of €2.00 per km.

INTERPRETNotes

Show solution

at \(p=2\) the model gives a daily distance of \(-4000\) km and \(R(2)=-8000\), which is impossible	R1	Accept the statement that the distance becomes zero at \(p=1.80\) and negative above it.
the model is not valid there: €2.00 is far outside the prices near €0.80 on which the estimate was based, so this is an unreliable extrapolation	R1

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(F=3+0.8d\)	A1	Accept \(F(d)=0.8d+3\).
Part (a)(ii)
gradient: the fare increases by €0.80 for each extra kilometre	A1	Must refer to the rate per km.
intercept: €3.00 is the fixed charge paid for any trip (the fare before any distance is travelled)	A1
Part (b)
\(S=1+0.28t\) for \(0<t<50\)	A1	Accept \(0<t\le50\), since both pieces give 15 at \(t=50\).
\(S=15\) for \(50\le t\le120\)	A1	Domains must not overlap or leave gaps (other than at the shared value \(t=50\)).
Part (c)
time \(=\frac{10}{15}\) hours \(=40\) minutes	M1
\(S=1+0.28\times40\)	M1	Correct piece chosen, since \(40<50\).
\(S=12.20\)	AG
Part (d)
\(t=4d\), so \(S=1+0.28(4d)=1+1.12d\)	M1
\(1+1.12d=3+0.8d\)	M1	Equating the two models, or reading the intersection from Figure 1 with a supporting calculation.
\(d=6.25\) km	A1	Both fares are €8.00.
Part (e)
the second intersection is on the capped part: \(3+0.8d=15\)	M1	Accept a correct reading of the graph supported by this equation.
\(d=15\)	A1
\(6.25<d<15\)	A1	Allow ECF from (d). Accept 'between 6.25 km and 15 km'.
Part (f)
\(F_{\text{rush}}=1.4(3+0.8d)=4.2+1.12d\)	A1
for \(d\le12.5\) both models have gradient 1.12, so the lines are parallel and the e-scooter is always €3.20 cheaper	R1	Must refer to equal gradients or parallel lines.
for \(d>12.5\) the e-scooter costs €15 while the rush-hour fare is more than \(4.2+1.12\times12.5=\)€18.20	R1
Part (g)(i)
daily distance \(=20\,000-2000\times\frac{p-0.8}{0.1}\)	M1	Or a linear model through \((0.8,\,20\,000)\) with gradient \(-20\,000\).
daily distance \(=36\,000-20\,000p\)	A1
\(R(p)=p(36\,000-20\,000p)=36\,000p-20\,000p^2\)	AG
Part (g)(ii)
\(p=-\frac{36\,000}{2(-20\,000)}\)	M1	Or a GDC maximum.
€0.90 per km; €16 200	A1	Both values.
Part (g)(iii)
estimate \(=0.90\times20\,000=18\,000\); \(\varepsilon=\left\|\frac{18\,000-16\,200}{16\,200}\right\|\times100\%\)	M1	Allow ECF from (g)(ii).
\(11.1\%\)	A1	\(11.111\ldots\%\)
Part (h)
at \(p=2\) the model gives a daily distance of \(-4000\) km and \(R(2)=-8000\), which is impossible	R1	Accept the statement that the distance becomes zero at \(p=1.80\) and negative above it.
the model is not valid there: €2.00 is far outside the prices near €0.80 on which the estimate was based, so this is an unreliable extrapolation	R1

Answers: (a)(i) \(F=3+0.8d\) · (b) \(S=\begin{cases}1+0.28t,&0<t<50\\15,&50\le t\le120\end{cases}\) · (d) \(6.25\) km · (e) \(6.25<d<15\) km · (f) \(F_{\text{rush}}=4.2+1.12d\) · (g)(ii) €0.90 per km; €16 200 · (g)(iii) \(11.1\%\)

Common error (g)(iii) Percentage error divides by the exact value (16 200), not by the estimate. Dividing by 18 000 gives 10%, which is wrong.

More 2.1 questions in the question bank

+ Add to test✓ In your testCC-AI-21-22-226 marks~30 min

A basketball free throw: building a quadratic model

2.2 Quadratic and cubic models2.1 Linear models and piecewise models5.1 Gradients, tangents and rates of change3.1 Right-angled trigonometry, bearings and the sine and cosine rulesSL

A basketball coach uses motion-tracking software to study a player's free throw. The path of the centre of the ball is modelled by

\(h(x)=ax^2+bx+c\), \(\quad 0\le x\le3.8\),

where \(h\) is the height of the centre of the ball above the floor and \(x\) is the horizontal distance from the point of release, both in metres. The software records three positions of the centre of the ball.

Position	\(x\) (m)	\(h\) (m)
release point	0	2.10
mid-flight	2.0	4.04
centre of the hoop	3.8	3.05

Figure 1: side view of the free throw (not to scale)

(a)(i)

Write down the value of \(c\), and show that \(4a+2b=1.94\).

ALGEBRANotes

Show solution

\(c=2.1\)	B1
\(h(2)=4a+2b+2.1=4.04\)	M1	Substitution of \((2.0,\,4.04)\) must be seen.
\(4a+2b=1.94\)	AG

Answer: \(c=2.1\)

(2)

(a)(ii)

Write down a second equation in \(a\) and \(b\), and hence find the values of \(a\) and \(b\).

ROUTINENotes

Show solution

\(14.44a+3.8b+2.1=3.05\), i.e. \(14.44a+3.8b=0.95\)	B1
solving the two equations simultaneously	(M1)	Implied by correct answers from the GDC.
\(a=-0.4,\ b=1.77\)	A1	Both values.

Answer: \(a=-0.4\), \(b=1.77\)

(3)

(b)

Find the maximum height of the centre of the ball, and the horizontal distance from the point of release at which it occurs.

ROUTINENotes

Show solution

\(x=-\frac{1.77}{2(-0.4)}=2.2125\)	M1	Or a GDC maximum. Allow ECF from (a)(ii).
maximum height \(4.06\) m at \(x=2.21\) m	A1	\(4.0580\ldots\) m. Both values needed.

Answer: \(4.06\) m at \(x=2.21\) m

(2)

(c)

The equation \(h(x)=0\) has one positive solution. Find this solution, and explain why it has no meaning for this shot.

INTERPRETNotes

Show solution

\(x=5.40\)	A1	\(5.3976\ldots\). Allow ECF from (a)(ii).
it is the distance at which the ball would reach floor level if the curve continued, but \(5.40>3.8\): the ball reaches the hoop first, so it lies outside the domain of the model	R1	Must refer to the domain or to the hoop/backboard stopping the path.

Answer: \(x=5.40\) m

(2)

(d)

On the axes in Figure 2, sketch the graph of \(h\) for \(0\le x\le3.8\). Label the point of release, the maximum point and the centre of the hoop with their coordinates.

Figure 2: axes for the sketch of the path of the ball

GRAPHNotes

Show solution

concave-down parabola drawn only for \(0\le x\le3.8\)	A1	Curve must not continue beyond the end points.
end points \((0,\,2.1)\) and \((3.8,\,3.05)\) labelled	A1
maximum point \((2.21,\,4.06)\) labelled, with the curve symmetrical about it	A1	Allow ECF from (b).

(3)

(e)(i)

The coach compares the path of the ball with the straight line from the point of release to the centre of the hoop. Find the equation of this line.

ROUTINENotes

Show solution

\(m=\frac{3.05-2.1}{3.8-0}=0.25\)	M1
\(y=0.25x+2.1\)	A1

Answer: \(y=0.25x+2.1\)

(2)

(e)(ii)

Find the greatest vertical distance between the path of the ball and this line.

MULTINotes

Show solution

\(h(x)-(0.25x+2.1)=-0.4x^2+1.52x\)	M1	Or a GDC maximum of the difference.
\(1.44\) m	A1	\(1.444\) m, at \(x=1.9\).

Answer: \(1.44\) m

(2)

(f)(i)

Find \(h'(x)\), and the gradient of the path at the centre of the hoop.

ROUTINENotes

Show solution

\(h'(x)=-0.8x+1.77\)	A1	Allow ECF from (a)(ii).
\(h'(3.8)=-1.27\)	A1

Answer: \(h'(x)=-0.8x+1.77\); \(-1.27\)

(2)

(f)(ii)

The angle of entry is the acute angle between the path of the ball and the horizontal at the centre of the hoop. Find the angle of entry.

ROUTINENotes

Show solution

\(\tan\theta=1.27\)	M1	Use of the size of the gradient from (f)(i).
\(\theta=51.8^\circ\)	A1	\(51.783\ldots^\circ\). Allow ECF.

Answer: \(51.8^\circ\)

(2)

(g)

A taller player releases the ball 2.40 m above the floor, from the same horizontal distance to the hoop. The software shows that the centre of the ball passes through the point \((2.0,\,3.95)\) and through the centre of the hoop.

Find the angle of entry for this shot.

MULTINotes

Show solution

\(c=2.4\), \(4a+2b=1.55\), \(14.44a+3.8b=0.65\)	M1
\(a=-0.336,\ b=1.45\)	A1	\(a=-0.33552\ldots,\ b=1.44605\ldots\)
\(h'(3.8)=2a(3.8)+b=-1.10\)	M1
\(\theta=47.8^\circ\)	A1	\(47.83\ldots^\circ\). Using rounded coefficients may give \(47.9^\circ\); accept.

Answer: \(47.8^\circ\)

(4)

(h)

The hoop has an inner diameter of 0.45 m and the ball has a diameter of 0.24 m. Seen along the direction in which the ball is travelling, the hoop appears as an opening of width \(0.45\sin\theta\) metres, where \(\theta\) is the angle of entry.

Use this to explain whether a steeper angle of entry helps the ball to go through the hoop, and compare the two shots.

INTERPRETNotes

Show solution

\(0.45\sin\theta\) increases as \(\theta\) increases (for acute angles), so a steeper entry gives a wider opening and more room for error	R1
first shot: \(0.45\sin51.8^\circ=0.354\) m; second shot: \(0.45\sin47.8^\circ=0.334\) m; both exceed 0.24 m, but the first shot leaves the larger margin	R1	Allow ECF from (f)(ii) and (g).

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(c=2.1\)	B1
\(h(2)=4a+2b+2.1=4.04\)	M1	Substitution of \((2.0,\,4.04)\) must be seen.
\(4a+2b=1.94\)	AG
Part (a)(ii)
\(14.44a+3.8b+2.1=3.05\), i.e. \(14.44a+3.8b=0.95\)	B1
solving the two equations simultaneously	(M1)	Implied by correct answers from the GDC.
\(a=-0.4,\ b=1.77\)	A1	Both values.
Part (b)
\(x=-\frac{1.77}{2(-0.4)}=2.2125\)	M1	Or a GDC maximum. Allow ECF from (a)(ii).
maximum height \(4.06\) m at \(x=2.21\) m	A1	\(4.0580\ldots\) m. Both values needed.
Part (c)
\(x=5.40\)	A1	\(5.3976\ldots\). Allow ECF from (a)(ii).
it is the distance at which the ball would reach floor level if the curve continued, but \(5.40>3.8\): the ball reaches the hoop first, so it lies outside the domain of the model	R1	Must refer to the domain or to the hoop/backboard stopping the path.
Part (d)
concave-down parabola drawn only for \(0\le x\le3.8\)	A1	Curve must not continue beyond the end points.
end points \((0,\,2.1)\) and \((3.8,\,3.05)\) labelled	A1
maximum point \((2.21,\,4.06)\) labelled, with the curve symmetrical about it	A1	Allow ECF from (b).
Part (e)(i)
\(m=\frac{3.05-2.1}{3.8-0}=0.25\)	M1
\(y=0.25x+2.1\)	A1
Part (e)(ii)
\(h(x)-(0.25x+2.1)=-0.4x^2+1.52x\)	M1	Or a GDC maximum of the difference.
\(1.44\) m	A1	\(1.444\) m, at \(x=1.9\).
Part (f)(i)
\(h'(x)=-0.8x+1.77\)	A1	Allow ECF from (a)(ii).
\(h'(3.8)=-1.27\)	A1
Part (f)(ii)
\(\tan\theta=1.27\)	M1	Use of the size of the gradient from (f)(i).
\(\theta=51.8^\circ\)	A1	\(51.783\ldots^\circ\). Allow ECF.
Part (g)
\(c=2.4\), \(4a+2b=1.55\), \(14.44a+3.8b=0.65\)	M1
\(a=-0.336,\ b=1.45\)	A1	\(a=-0.33552\ldots,\ b=1.44605\ldots\)
\(h'(3.8)=2a(3.8)+b=-1.10\)	M1
\(\theta=47.8^\circ\)	A1	\(47.83\ldots^\circ\). Using rounded coefficients may give \(47.9^\circ\); accept.
Part (h)
\(0.45\sin\theta\) increases as \(\theta\) increases (for acute angles), so a steeper entry gives a wider opening and more room for error	R1
first shot: \(0.45\sin51.8^\circ=0.354\) m; second shot: \(0.45\sin47.8^\circ=0.334\) m; both exceed 0.24 m, but the first shot leaves the larger margin	R1	Allow ECF from (f)(ii) and (g).

Answers: (a)(i) \(c=2.1\) · (a)(ii) \(a=-0.4\), \(b=1.77\) · (b) \(4.06\) m at \(x=2.21\) m · (c) \(x=5.40\) m · (e)(i) \(y=0.25x+2.1\) · (e)(ii) \(1.44\) m · (f)(i) \(h'(x)=-0.8x+1.77\); \(-1.27\) · (f)(ii) \(51.8^\circ\) · (g) \(47.8^\circ\)

Exam tip (f)(ii) The gradient of a path is \(\tan\) of its angle with the horizontal. Use the size of the gradient, \(1.27\), so that the angle of entry comes out acute.

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2.3 Exponential and sinusoidal models · 2.4 Choosing and evaluating modelsmerged group

+ Add to test✓ In your testCC-AI-23-24-124 marks~30 min

Daylight hours in Athens and a rooftop solar panel

2.3 Exponential and sinusoidal models2.4 Choosing and evaluating models5.3 Integration and the trapezoidal rule1.1 Approximation, percentage error and standard formSL

The number of hours of daylight in Athens, \(D\), is modelled by

\(D(t)=a\sin\big(b(t-c)\big)+d\),

where \(t\) is the number of days after 1 January (so \(t=0\) on 1 January) and \(a\), \(b\), \(c\) and \(d\) are positive constants. The argument of the sine function is in radians.

The longest day of the year is 21 June (\(t=171\)), with 14.7 hours of daylight. The shortest day is half a year later, with 9.6 hours of daylight. The pattern repeats every 365 days.

(a)(i)

Write down the value of \(d\) and the value of \(a\).

ROUTINENotes

Show solution

\(d=\frac{14.7+9.6}{2}=12.15\)	A1
\(a=\frac{14.7-9.6}{2}=2.55\)	A1

Answer: \(d=12.15\), \(a=2.55\)

(2)

(a)(ii)

Interpret the value of \(d\) in this context.

CONCEPTNotes

Show solution

12.15 hours is the mean daylight in Athens over the year (halfway between the longest and the shortest day)

Accept "the daylight on the days when it is halfway between its greatest and least values". A bare "the principal axis" scores R0.

(1)

(b)

Show that \(b\approx0.0172\).

ROUTINENotes

Show solution

period \(=365\) days, so \(\frac{2\pi}{b}=365\)	M1
\(b=\frac{2\pi}{365}=0.017214\ldots\)	A1	At least 4 s.f. must be seen.
\(b\approx0.0172\)	AG

(2)

(c)

Find the value of \(c\).

ALGEBRANotes

Show solution

maximum when \(b(171-c)=\frac{\pi}{2}\) or the curve crosses the principal axis rising a quarter of a period (91.25 days) before the maximum	M1
\(c=171-91.25=79.75\)	A1	Accept 79.7 or 79.8 (from \(b=0.0172\): 79.67). Accept any other correct value \(79.75+365n\) only if a value with \(0\lt c\lt365\) is also given.

Answer: \(c=79.75\)

(2)

(d)

Figure 1 shows the graph of \(D\) for \(0\le t\le365\).

Use the model to find the number of days in the year on which Athens has more than 13 hours of daylight.

Figure 1: the modelled daylight in Athens, \(D\) hours, against \(t\) days after 1 January

GRAPHNotes

Show solution

\(2.55\sin\big(0.017214(t-79.75)\big)+12.15=13\) or intersection of the graph of \(D\) with the line \(D=13\)	M1	Line \(D=13\) drawn on Figure 1 is enough for M1.
\(t=99.5\) and \(t=242.5\)	A1	Both values needed. Allow ECF from (c).
143 days	A1	Days \(t=100\) to \(t=242\) inclusive. Accept 143.0 (from \(242.5-99.5\)).

Answer: 143 days

(3)

(e)(i)

The table shows the measured daylight in Athens on six days, the value given by the model, and the residual, where residual \(=\) measured value \(-\) model value.

\(t\) (days)	20	79	140	201	263	324
Measured \(D\) (hours)	10.0	12.1	14.3	14.5	12.2	9.9
Model \(D(t)\) (hours)	9.97	12.12	14.35	?	12.12	9.92
Residual (hours)	0.03	−0.02	−0.05	?	0.08	−0.02

Find the model value and the residual for \(t=201\).

ROUTINENotes

Show solution

\(D(201)=2.55\sin\big(0.017214(201-79.75)\big)+12.15=14.37\)	A1	Accept 14.4.
residual \(=14.5-14.37=0.133\)	A1	Accept 0.13 or 0.1. Allow ECF from their \(D(201)\) if the sign convention is correct.

Answer: \(D(201)=14.37\) hours; residual \(0.13\) hours

(2)

(e)(ii)

Comment on how well the model fits the measured data.

INTERPRETNotes

Show solution

all residuals are small: at most about 0.13 hours (8 minutes), around 1% of the daylight	R1	Must refer to the size of the residuals.
residuals are both positive and negative with no clear pattern, so the model fits the data well	R1	A conclusion supported by the sign or pattern of the residuals.

(2)

(f)(i)

A house in Athens has solar panels on its roof. On every 60th day the owner records the energy produced by the panels on that day, \(E\) kWh per day.

\(t\) (days)	0	60	120	180	240	300	360
\(E\) (kWh per day)	10.8	12.9	16.2	17.9	16.0	12.4	11.0

Use the trapezoidal rule with 6 strips to estimate the total energy produced from \(t=0\) to \(t=360\).

ROUTINENotes

Show solution

\(h=60\)	(A1)
\(\frac{60}{2}\big[10.8+11.0+2(12.9+16.2+17.9+16.0+12.4)\big]\)	M1	Correct substitution into the trapezoidal rule.
5178 kWh	A1	Accept 5180.

Answer: 5178 kWh

(3)

(f)(ii)

The electricity meter shows that the panels produced 5240 kWh from \(t=0\) to \(t=360\).

Find the percentage error in the estimate from part (f)(i).

ROUTINENotes

Show solution

\(\left\|\frac{5178-5240}{5240}\right\|\times100\%\)	M1	The meter reading must be the denominator. Allow ECF from (f)(i).
1.18%	A1	Accept 1.2%.

Answer: 1.18%

(2)

(g)

In a city further north, the longest day is also 21 June, with 18.8 hours of daylight, and the shortest day has 6.0 hours of daylight. Its daylight is modelled by \(D_{N}(t)=p\sin\big(b(t-c)\big)+q\), with the same values of \(b\) and \(c\) as the Athens model.

Find the number of days in the year on which this city has more daylight than Athens.

MULTINotes

Show solution

\(p=6.4\) and \(q=12.4\)	A1
\(6.4\sin\big(b(t-79.75)\big)+12.4=2.55\sin\big(b(t-79.75)\big)+12.15\) or intersection of the graphs, giving \(t=76.0\) and \(t=266.0\)	M1
190 days	A1	Accept 191 (days \(t=76\) to \(t=266\) inclusive). Allow ECF from (b) and (c).

Answer: about 190 days

(3)

(h)

Close to the North Pole there are several weeks each summer when the sun never sets (24 hours of daylight) and several weeks each winter when it never rises.

Explain why a model of the form \(D(t)=a\sin\big(b(t-c)\big)+d\) would not be suitable for a city there.

INTERPRETNotes

Show solution

a sine curve reaches its maximum and minimum only at single instants, so it cannot stay at 24 hours (or at 0 hours) for several weeks: the real graph has flat sections	R1
to reach 24 and 0 hours the model would need \(a=12\) and \(d=12\), and any larger amplitude gives values above 24 or below 0 hours, which are impossible; a model built for Athens cannot simply be extrapolated to other latitudes	R1	Accept any valid second reason about impossible values or the domain of validity of the model.

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(d=\frac{14.7+9.6}{2}=12.15\)	A1
\(a=\frac{14.7-9.6}{2}=2.55\)	A1
Part (a)(ii)
12.15 hours is the mean daylight in Athens over the year (halfway between the longest and the shortest day)	R1	Accept "the daylight on the days when it is halfway between its greatest and least values". A bare "the principal axis" scores R0.
Part (b)
period \(=365\) days, so \(\frac{2\pi}{b}=365\)	M1
\(b=\frac{2\pi}{365}=0.017214\ldots\)	A1	At least 4 s.f. must be seen.
\(b\approx0.0172\)	AG
Part (c)
maximum when \(b(171-c)=\frac{\pi}{2}\) or the curve crosses the principal axis rising a quarter of a period (91.25 days) before the maximum	M1
\(c=171-91.25=79.75\)	A1	Accept 79.7 or 79.8 (from \(b=0.0172\): 79.67). Accept any other correct value \(79.75+365n\) only if a value with \(0\lt c\lt365\) is also given.
Part (d)
\(2.55\sin\big(0.017214(t-79.75)\big)+12.15=13\) or intersection of the graph of \(D\) with the line \(D=13\)	M1	Line \(D=13\) drawn on Figure 1 is enough for M1.
\(t=99.5\) and \(t=242.5\)	A1	Both values needed. Allow ECF from (c).
143 days	A1	Days \(t=100\) to \(t=242\) inclusive. Accept 143.0 (from \(242.5-99.5\)).
Part (e)(i)
\(D(201)=2.55\sin\big(0.017214(201-79.75)\big)+12.15=14.37\)	A1	Accept 14.4.
residual \(=14.5-14.37=0.133\)	A1	Accept 0.13 or 0.1. Allow ECF from their \(D(201)\) if the sign convention is correct.
Part (e)(ii)
all residuals are small: at most about 0.13 hours (8 minutes), around 1% of the daylight	R1	Must refer to the size of the residuals.
residuals are both positive and negative with no clear pattern, so the model fits the data well	R1	A conclusion supported by the sign or pattern of the residuals.
Part (f)(i)
\(h=60\)	(A1)
\(\frac{60}{2}\big[10.8+11.0+2(12.9+16.2+17.9+16.0+12.4)\big]\)	M1	Correct substitution into the trapezoidal rule.
5178 kWh	A1	Accept 5180.
Part (f)(ii)
\(\left\|\frac{5178-5240}{5240}\right\|\times100\%\)	M1	The meter reading must be the denominator. Allow ECF from (f)(i).
1.18%	A1	Accept 1.2%.
Part (g)
\(p=6.4\) and \(q=12.4\)	A1
\(6.4\sin\big(b(t-79.75)\big)+12.4=2.55\sin\big(b(t-79.75)\big)+12.15\) or intersection of the graphs, giving \(t=76.0\) and \(t=266.0\)	M1
190 days	A1	Accept 191 (days \(t=76\) to \(t=266\) inclusive). Allow ECF from (b) and (c).
Part (h)
a sine curve reaches its maximum and minimum only at single instants, so it cannot stay at 24 hours (or at 0 hours) for several weeks: the real graph has flat sections	R1
to reach 24 and 0 hours the model would need \(a=12\) and \(d=12\), and any larger amplitude gives values above 24 or below 0 hours, which are impossible; a model built for Athens cannot simply be extrapolated to other latitudes	R1	Accept any valid second reason about impossible values or the domain of validity of the model.

Answers: (a)(i) \(d=12.15\), \(a=2.55\) · (c) \(c=79.75\) · (d) 143 days · (e)(i) \(D(201)=14.37\) hours; residual \(0.13\) hours · (f)(i) 5178 kWh · (f)(ii) 1.18% · (g) about 190 days

Common error (b) Using a period of 12 or 360. The period is the length of one cycle in the units of \(t\), here 365 days.

Exam tip (d) A sine equation has two solutions in each cycle. Use the graph to see that the daylight is above 13 hours between the two crossings, then subtract.

Common error (f)(ii) Dividing by the estimate instead of the exact value. The meter reading is the exact value, so it goes in the denominator.

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Electric car registrations: choosing between a linear and an exponential model

2.4 Choosing and evaluating models2.3 Exponential and sinusoidal models2.1 Linear models and piecewise models4.2 Correlation and regressionSL

The table shows the number of new electric cars, \(y\) thousand, registered in a country in each year from 2017 to 2024, where \(x\) is the number of years after 2017.

Year	2017	2018	2019	2020	2021	2022	2023	2024
\(x\)	0	1	2	3	4	5	6	7
\(y\) (thousands)	2.4	3.5	4.8	6.7	9.5	13.4	18.6	26.2

(a)

By considering the first differences and the ratios of consecutive values of \(y\), explain why an exponential model is more suitable than a linear model for these data.

CONCEPTNotes

Show solution

first differences 1.1, 1.3, 1.9, 2.8, 3.9, 5.2, 7.6: increasing, not constant	M1	At least three differences seen.
ratios 1.46, 1.37, 1.40, 1.42, 1.41, 1.39, 1.41	M1	At least three ratios seen.
the ratios are approximately constant (about 1.4), so the data are close to exponential, not linear	R1	Must refer to both differences and ratios.

(3)

(b)(i)

A linear model \(y=mx+c\) is fitted to the data.

Find the value of \(m\), the value of \(c\) and the value of Pearson's product-moment correlation coefficient, \(r\).

ROUTINENotes

Show solution

\(m=3.22\)	A1	3.22262…
\(c=-0.642\)	A1	−0.641666…
\(r=0.950\)	A1	0.950398…

Answer: \(m=3.22\), \(c=-0.642\), \(r=0.950\)

(3)

(b)(ii)

Interpret the value of \(m\) in this context.

CONCEPTNotes

Show solution

the linear model predicts that registrations increase by about 3220 cars each year

Must include "per year" and the quantity (thousands of cars or 3220 cars).

(1)

(c)

An exponential model \(y=ka^{x}\) is fitted to the data.

Find the value of \(k\) and the value of \(a\).

ROUTINENotes

Show solution

\(k=2.44\)	A1	2.43818…
\(a=1.40\)	A1	1.40416…

Answer: \(y=2.44(1.40)^{x}\)

(2)

(d)

Show that the exponential model corresponds to an increase in registrations of approximately 40% per year.

PROOFNotes

Show solution

percentage change \(=(a-1)\times100\%=(1.404-1)\times100\%\)	M1	Or: each year the number is multiplied by 1.404.
\(=40.4\%\)	A1
\(\approx40\%\) per year	AG

(2)

(e)

Figure 1 shows the data together with the two models.

Using Figure 1, explain why the linear model is not appropriate, even though the value of \(r\) is close to 1.

Figure 1: the registration data with the linear and exponential models

GRAPHNotes

Show solution

the data lie above the line at both ends and below it in the middle: the residuals form a pattern, so the data are curved	R1
\(r\) only measures how close the points are to some straight line; a high \(r\) does not show that a straight line is the right shape	R1

(2)

(f)

Use each model to predict the number of new electric cars registered in 2025.

ROUTINENotes

Show solution

linear: \(3.2226(8)-0.6417=25.1\) thousand	A1	Accept 25 100 cars.
exponential: \(2.4382(1.4042)^{8}=36.8\) thousand	A1	Accept 36.0 to 36.9 (from rounded parameters). Allow ECF from (b)(i) and (c).

Answer: linear: 25.1 thousand; exponential: 36.8 thousand

(2)

(g)(i)

A government subsidy for electric cars ended at the start of 2025. The number registered in 2025 was 29.5 thousand.

Find the residual of each model at \(x=8\).

ROUTINENotes

Show solution

linear: \(29.5-25.14=4.36\) thousand	A1	Accept 4.4.
exponential: \(29.5-36.85=-7.35\) thousand	A1	Accept −7.3 or −7.4. Allow ECF from (f).

Answer: linear: 4.36 thousand; exponential: −7.35 thousand

(2)

(g)(ii)

The linear model gave the closer prediction for 2025. Comment on this, with reference to the reliability of both predictions.

INTERPRETNotes

Show solution

both predictions are extrapolations beyond the data (2017–2024), and the situation changed when the subsidy ended, so growth slowed	R1
the linear model was closer only by chance: it fits the 2017–2024 data badly and was already underestimating recent years; neither model allows for the change	R1	Accept any valid reason referring to the residual pattern or to the change in conditions.

(2)

(h)

About 110 thousand new cars of all types are registered in this country each year.

Find the first year in which the exponential model predicts more than 110 thousand electric car registrations. Hence explain why the exponential model cannot continue to hold.

MODELNotes

Show solution

\(2.4382(1.4042)^{x}=110\) gives \(x=11.2\)	M1	Or table: \(x=11\) gives 102 thousand and \(x=12\) gives 143 thousand.
2029	A1	Allow ECF from (c). 2028 (reading \(x=11.2\) as a year during 2028) scores A0: the yearly total first exceeds 110 thousand at \(x=12\).
electric car registrations cannot exceed the total number of new cars, so growth must slow down as the market fills; an exponential model grows without limit	R1

Answer: 2029

(3)

Show full solution

Working	Mark	Notes
Part (a)
first differences 1.1, 1.3, 1.9, 2.8, 3.9, 5.2, 7.6: increasing, not constant	M1	At least three differences seen.
ratios 1.46, 1.37, 1.40, 1.42, 1.41, 1.39, 1.41	M1	At least three ratios seen.
the ratios are approximately constant (about 1.4), so the data are close to exponential, not linear	R1	Must refer to both differences and ratios.
Part (b)(i)
\(m=3.22\)	A1	3.22262…
\(c=-0.642\)	A1	−0.641666…
\(r=0.950\)	A1	0.950398…
Part (b)(ii)
the linear model predicts that registrations increase by about 3220 cars each year	R1	Must include "per year" and the quantity (thousands of cars or 3220 cars).
Part (c)
\(k=2.44\)	A1	2.43818…
\(a=1.40\)	A1	1.40416…
Part (d)
percentage change \(=(a-1)\times100\%=(1.404-1)\times100\%\)	M1	Or: each year the number is multiplied by 1.404.
\(=40.4\%\)	A1
\(\approx40\%\) per year	AG
Part (e)
the data lie above the line at both ends and below it in the middle: the residuals form a pattern, so the data are curved	R1
\(r\) only measures how close the points are to some straight line; a high \(r\) does not show that a straight line is the right shape	R1
Part (f)
linear: \(3.2226(8)-0.6417=25.1\) thousand	A1	Accept 25 100 cars.
exponential: \(2.4382(1.4042)^{8}=36.8\) thousand	A1	Accept 36.0 to 36.9 (from rounded parameters). Allow ECF from (b)(i) and (c).
Part (g)(i)
linear: \(29.5-25.14=4.36\) thousand	A1	Accept 4.4.
exponential: \(29.5-36.85=-7.35\) thousand	A1	Accept −7.3 or −7.4. Allow ECF from (f).
Part (g)(ii)
both predictions are extrapolations beyond the data (2017–2024), and the situation changed when the subsidy ended, so growth slowed	R1
the linear model was closer only by chance: it fits the 2017–2024 data badly and was already underestimating recent years; neither model allows for the change	R1	Accept any valid reason referring to the residual pattern or to the change in conditions.
Part (h)
\(2.4382(1.4042)^{x}=110\) gives \(x=11.2\)	M1	Or table: \(x=11\) gives 102 thousand and \(x=12\) gives 143 thousand.
2029	A1	Allow ECF from (c). 2028 (reading \(x=11.2\) as a year during 2028) scores A0: the yearly total first exceeds 110 thousand at \(x=12\).
electric car registrations cannot exceed the total number of new cars, so growth must slow down as the market fills; an exponential model grows without limit	R1

Answers: (b)(i) \(m=3.22\), \(c=-0.642\), \(r=0.950\) · (c) \(y=2.44(1.40)^{x}\) · (f) linear: 25.1 thousand; exponential: 36.8 thousand · (g)(i) linear: 4.36 thousand; exponential: −7.35 thousand · (h) 2029

Exam tip (c) Store the regression equation in your GDC and use the stored function in later parts. Re-typing rounded values makes later predictions drift.

Exam tip (g)(ii) When commenting on a prediction, say whether it is interpolation or extrapolation and give one reason from the context why the pattern may not continue.

More 2.4 questions in the question bank

2.5 Logarithmic scales and linearising data · 2.6 Transformations of graphs and logistic modelsmerged group

+ Add to test✓ In your testCC-AI-25-26-125 marks~30 min

Earthquakes: a logarithmic scale and a linearised frequency model

2.5 Logarithmic scales and linearising data1.4 Exponents and logarithms2.3 Exponential and sinusoidal models4.2 Correlation and regressionHL

The magnitude \(M\) of an earthquake is defined by \(M=\log_{10}A\), where \(A\) is the largest ground movement, in micrometres, recorded by a standard seismometer at a fixed distance from the earthquake.

(a)

Two earthquakes have magnitudes 6.8 and 5.1.

Find how many times larger the ground movement is for the earthquake of magnitude 6.8.

ROUTINENotes

Show solution

\(\frac{10^{6.8}}{10^{5.1}}=10^{1.7}\)	M1
50.1 times	A1	Accept 50.

Answer: 50.1 times

(2)

(b)

The energy \(E\) joules released by an earthquake of magnitude \(M\) satisfies \(\log_{10}E=4.8+1.5M\).

Show that the earthquake of magnitude 6.8 releases approximately 355 times as much energy as the earthquake of magnitude 5.1.

PROOFNotes

Show solution

\(\log_{10}E_1-\log_{10}E_2=(4.8+1.5\times6.8)-(4.8+1.5\times5.1)=2.55\)	M1	Alternative: \(E_1=10^{15.0}=1.00\times10^{15}\) and \(E_2=10^{12.45}=2.82\times10^{12}\) (M1).
\(\log_{10}\left(\frac{E_1}{E_2}\right)=2.55\)	M1	Use of the quotient law, or \(\frac{E_1}{E_2}=\frac{10^{15.0}}{10^{12.45}}\) in the alternative method (M1).
\(\frac{E_1}{E_2}=10^{2.55}=354.8\ldots\)	A1	At least 4 s.f. (354.8) must be seen; the given value 355 alone scores A0.
\(\approx355\)	AG

(3)

(c)

For one region, the table shows the mean number of earthquakes per year, \(N\), with magnitude at least \(M\), calculated from a 40-year record.

\(M\)	3.0	3.5	4.0	4.5	5.0	5.5	6.0
\(N\)	1520	510	148	52	15.5	4.8	1.6
\(\log_{10}N\)	3.182	2.708	?	1.716	1.190	?	0.204

Write down the two missing values of \(\log_{10}N\), correct to 3 decimal places.

ROUTINENotes

Show solution

\(\log_{10}148=2.170\)	A1
\(\log_{10}4.8=0.681\)	A1

Answer: 2.170 and 0.681

(2)

(d)

Figure 1 shows the points \((M,\ \log_{10}N)\) from the table, except those for \(M=4.0\) and \(M=5.5\).

Plot these two points on Figure 1. By drawing a line of best fit on Figure 1, estimate the magnitude \(M\) for which \(N=10\).

Figure 1: \(\log_{10}N\) against \(M\) (the points for \(M=4.0\) and \(M=5.5\) are not shown)

GRAPHNotes

Show solution

points \((4.0,\ 2.170)\) and \((5.5,\ 0.681)\) plotted correctly	A1	Both points needed, within half a small square. Allow ECF from (c).
\(N=10\) gives \(\log_{10}N=1\); line of best fit read at a height of 1	M1
\(M\approx5.2\)	A1	Accept 5.1 to 5.3. (The regression line in (e) gives 5.19.)

Answer: \(M\approx5.2\)

(3)

(e)

Find the equation of the regression line of \(\log_{10}N\) on \(M\), in the form \(\log_{10}N=p+qM\), and the value of Pearson's product-moment correlation coefficient, \(r\).

ROUTINENotes

Show solution

\(p=6.18\)	A1	6.18204…
\(q=-0.998\)	A1	−0.997554…
\(r=-0.9999\)	A1	−0.999884…; accept −1.00 (3 s.f.). Using the unrounded values of \(\log_{10}N\) gives the same results to 3 s.f.

Answer: \(\log_{10}N=6.18-0.998M\), \(r=-0.9999\)

(3)

(f)(i)

Show that the regression line gives a model of the form \(N=k\times c^{M}\), where \(c\approx0.1\), and find the value of \(k\).

PROOFNotes

Show solution

\(N=10^{6.182-0.9976M}\)	M1	Undoing the logarithm with a power of 10.
\(N=10^{6.182}\times\left(10^{-0.9976}\right)^{M}\), so \(k=10^{6.182}=1.52\times10^{6}\)	A1	Laws of exponents must be seen. Accept \(1.52\times10^{6}\) to \(1.53\times10^{6}\).
\(c=10^{-0.9976}=0.1006\)	A1
\(\approx0.1\)	AG

Answer: \(N=1.52\times10^{6}\times(0.101)^{M}\)

(3)

(f)(ii)

Interpret the value of \(c\) in this context.

CONCEPTNotes

Show solution

each increase of 1 in magnitude multiplies the yearly number of earthquakes by about 0.1: there are about one tenth as many

Accept "decreases by about 90% for each unit of magnitude".

(1)

(g)

Use the model to estimate the mean number of earthquakes per year with magnitude at least 7.0 in this region. Hence estimate the mean time between such earthquakes, and comment on the reliability of your estimate.

MODELNotes

Show solution

\(\log_{10}N=6.182-0.9976(7)\), so \(N=0.158\) per year	A1	Accept 0.16. Allow ECF from (e).
mean time \(=\frac{1}{0.158}=6.32\) years	A1	Accept 6.3 years.
\(M=7.0\) is outside the data range (3.0 to 6.0), so this is extrapolation and may be unreliable	R1

Answer: 0.158 per year; about one every 6.3 years

(3)

(h)

Explain why the model was fitted using the graph of \(\log_{10}N\) against \(M\), rather than the graph of \(N\) against \(M\).

INTERPRETNotes

Show solution

\(N\) ranges from 1.6 to 1520 (about three powers of 10), so on a graph of \(N\) against \(M\) the points for large \(M\) would be squashed near the axis	R1
if \(N\) is exponential in \(M\), the graph of \(\log_{10}N\) against \(M\) is a straight line, which is easy to judge and can be fitted by linear regression	R1

(2)

(i)(i)

A student suggests a power model \(N=hM^{n}\) instead, and plots \(\log_{10}N\) against \(\log_{10}M\).

Find the value of \(r\) for \(\log_{10}N\) against \(\log_{10}M\).

ROUTINENotes

Show solution

\(r=-0.995\)

−0.994798…

Answer: \(r=-0.995\)

(1)

(i)(ii)

State, with a reason, whether the exponential model or the power model is more appropriate for these data.

CONCEPTNotes

Show solution

\(\|r\|\) is closer to 1 for \(\log_{10}N\) against \(M\) (0.9999) than against \(\log_{10}M\) (0.995), so the semi-log points are closer to a straight line	R1	Accept: the log–log points curve (residuals negative at both ends and positive in the middle).
the exponential model is more appropriate	A1	Do not award R0A1.

Answer: the exponential model

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\frac{10^{6.8}}{10^{5.1}}=10^{1.7}\)	M1
50.1 times	A1	Accept 50.
Part (b)
\(\log_{10}E_1-\log_{10}E_2=(4.8+1.5\times6.8)-(4.8+1.5\times5.1)=2.55\)	M1	Alternative: \(E_1=10^{15.0}=1.00\times10^{15}\) and \(E_2=10^{12.45}=2.82\times10^{12}\) (M1).
\(\log_{10}\left(\frac{E_1}{E_2}\right)=2.55\)	M1	Use of the quotient law, or \(\frac{E_1}{E_2}=\frac{10^{15.0}}{10^{12.45}}\) in the alternative method (M1).
\(\frac{E_1}{E_2}=10^{2.55}=354.8\ldots\)	A1	At least 4 s.f. (354.8) must be seen; the given value 355 alone scores A0.
\(\approx355\)	AG
Part (c)
\(\log_{10}148=2.170\)	A1
\(\log_{10}4.8=0.681\)	A1
Part (d)
points \((4.0,\ 2.170)\) and \((5.5,\ 0.681)\) plotted correctly	A1	Both points needed, within half a small square. Allow ECF from (c).
\(N=10\) gives \(\log_{10}N=1\); line of best fit read at a height of 1	M1
\(M\approx5.2\)	A1	Accept 5.1 to 5.3. (The regression line in (e) gives 5.19.)
Part (e)
\(p=6.18\)	A1	6.18204…
\(q=-0.998\)	A1	−0.997554…
\(r=-0.9999\)	A1	−0.999884…; accept −1.00 (3 s.f.). Using the unrounded values of \(\log_{10}N\) gives the same results to 3 s.f.
Part (f)(i)
\(N=10^{6.182-0.9976M}\)	M1	Undoing the logarithm with a power of 10.
\(N=10^{6.182}\times\left(10^{-0.9976}\right)^{M}\), so \(k=10^{6.182}=1.52\times10^{6}\)	A1	Laws of exponents must be seen. Accept \(1.52\times10^{6}\) to \(1.53\times10^{6}\).
\(c=10^{-0.9976}=0.1006\)	A1
\(\approx0.1\)	AG
Part (f)(ii)
each increase of 1 in magnitude multiplies the yearly number of earthquakes by about 0.1: there are about one tenth as many	R1	Accept "decreases by about 90% for each unit of magnitude".
Part (g)
\(\log_{10}N=6.182-0.9976(7)\), so \(N=0.158\) per year	A1	Accept 0.16. Allow ECF from (e).
mean time \(=\frac{1}{0.158}=6.32\) years	A1	Accept 6.3 years.
\(M=7.0\) is outside the data range (3.0 to 6.0), so this is extrapolation and may be unreliable	R1
Part (h)
\(N\) ranges from 1.6 to 1520 (about three powers of 10), so on a graph of \(N\) against \(M\) the points for large \(M\) would be squashed near the axis	R1
if \(N\) is exponential in \(M\), the graph of \(\log_{10}N\) against \(M\) is a straight line, which is easy to judge and can be fitted by linear regression	R1
Part (i)(i)
\(r=-0.995\)	A1	−0.994798…
Part (i)(ii)
\(\|r\|\) is closer to 1 for \(\log_{10}N\) against \(M\) (0.9999) than against \(\log_{10}M\) (0.995), so the semi-log points are closer to a straight line	R1	Accept: the log–log points curve (residuals negative at both ends and positive in the middle).
the exponential model is more appropriate	A1	Do not award R0A1.

Answers: (a) 50.1 times · (c) 2.170 and 0.681 · (d) \(M\approx5.2\) · (e) \(\log_{10}N=6.18-0.998M\), \(r=-0.9999\) · (f)(i) \(N=1.52\times10^{6}\times(0.101)^{M}\) · (g) 0.158 per year; about one every 6.3 years · (i)(i) \(r=-0.995\) · (i)(ii) the exponential model

Common error (a) Reading a difference of 1.7 in magnitude as "1.7 times larger". On a base-10 logarithmic scale, a difference of 1.7 means a factor of \(10^{1.7}\).

Exam tip (f)(i) Keep the base the same throughout: the line was found with \(\log_{10}\), so convert back with powers of 10, not with \(e\).

More 2.5 questions in the question bank

+ Add to test✓ In your testCC-AI-25-26-228 marks~35 min

An invasive plant on a lake: a logistic model and transformations

2.6 Transformations of graphs and logistic models2.3 Exponential and sinusoidal models5.1 Gradients, tangents and rates of change2.4 Choosing and evaluating modelsHL

A lake has a surface area of 240 hectares (ha). An invasive floating plant is found covering 4 ha of the lake. The area covered by the plant, \(A\) ha, \(t\) weeks later is modelled by

\(A(t)=\dfrac{180}{1+Ce^{-0.35t}}\), \(t\ge0\),

where \(C\) is a constant.

(a)

Find the value of \(C\).

ROUTINENotes

Show solution

\(\frac{180}{1+C}=4\)	M1
\(C=44\)	A1

Answer: \(C=44\)

(2)

(b)

Write down the equation of the horizontal asymptote of the graph of \(A\), and interpret it in this context.

CONCEPTNotes

Show solution

\(A=180\)	A1	Accept \(y=180\).
the plant cannot cover more than 180 ha: the area approaches 180 ha in the long term (the carrying capacity of the lake for this plant)	R1

Answer: \(A=180\)

(2)

(c)

Find the area covered after 10 weeks, and the time taken for the plant to cover half of the lake.

ROUTINENotes

Show solution

\(A(10)=77.3\) ha	A1
\(\frac{180}{1+44e^{-0.35t}}=120\)	M1	Half of the lake is 120 ha. Using 90 scores M0.
\(t=12.8\) weeks	A1	12.7923…

Answer: 77.3 ha; 12.8 weeks

(3)

(d)

In the first weeks, an ecologist used the exponential model \(E(t)=4e^{0.35t}\).

Find \(E(4)\) and \(A(4)\). Explain why the two models give similar values when \(t\) is small, but not when \(t\) is large.

MULTINotes

Show solution

\(E(4)=16.2\) ha	A1
\(A(4)=15.2\) ha	A1
while \(A\) is small compared with 180, logistic growth is almost exponential; later the logistic growth slows as \(A\) approaches 180, but \(E\) keeps growing without limit (it would exceed the whole lake)	R1

Answer: \(E(4)=16.2\) ha, \(A(4)=15.2\) ha

(3)

(e)(i)

The area grows fastest when it is half of its limiting value.

Show that the area grows fastest when \(t=\dfrac{\ln44}{0.35}\).

PROOFNotes

Show solution

\(\frac{180}{1+44e^{-0.35t}}=90\), so \(1+44e^{-0.35t}=2\)	M1
\(e^{-0.35t}=\frac{1}{44}\), so \(-0.35t=\ln\frac{1}{44}=-\ln44\)	A1	Correct use of logarithms to reach \(-0.35t=-\ln44\).
\(t=\frac{\ln44}{0.35}\)	AG

(2)

(e)(ii)

Hence write down the time at which the area grows fastest.

ROUTINENotes

Show solution

\(t=10.8\) weeks

10.8119…

Answer: 10.8 weeks

(1)

(f)(i)

On the axes in Figure 1, sketch the graph of \(A\) for \(0\le t\le30\). Show the horizontal asymptote, the value of \(A\) at \(t=0\), and the point of fastest growth with its coordinates.

Figure 1: axes for the sketch of \(A\)

GRAPHNotes

Show solution

increasing S-shaped curve, levelling off below the asymptote and close to it at \(t=30\)	A1	\(A(30)=179.8\).
intercept \((0,\ 4)\) and asymptote \(A=180\) shown	A1
point of fastest growth marked at \((10.8,\ 90)\), where the curve is steepest and changes from bending up to bending down	A1	Allow ECF from (e)(ii).

(3)

(f)(ii)

Use your GDC to find the gradient of the graph of \(A\) at the point of fastest growth. Interpret this value in context.

ROUTINENotes

Show solution

15.75	A1	Accept 15.7 or 15.8.
at 10.8 weeks the area covered is increasing at about 15.8 ha per week, the greatest rate of growth	R1	Units (ha per week) required.

Answer: 15.75 ha per week

(2)

(g)(i)

An environmental agency reports the same data in a different way. It measures time as \(s\) days after a date that is 14 days before \(t=0\), so that \(s=7t+14\), and it gives the area, \(B\), in km², where 1 km² \(=100\) ha.

Describe a sequence of transformations that maps the graph of \(A\) against \(t\) onto the graph of \(B\) against \(s\).

CONCEPTNotes

Show solution

horizontal stretch, scale factor 7	A1
followed by a horizontal translation of 14 units to the right	A1	Accept a translation of 2 units right followed by a stretch of scale factor 7. The order must be consistent with the size of the translation.
vertical stretch, scale factor 0.01	A1	Accept \(\frac{1}{100}\).

(3)

(g)(ii)

Hence write down an equation for \(B\) in terms of \(s\). State the domain of this function and the equation of its horizontal asymptote.

ALGEBRANotes

Show solution

\(B(s)=\frac{1.8}{1+44e^{-0.05(s-14)}}\)	A1	Accept equivalent forms, e.g. \(0.01A\left(\frac{s-14}{7}\right)\) written out in full, or \(\frac{1.8}{1+44e^{-0.35(s-14)/7}}\). Allow ECF from (a).
domain \(s\ge14\)	A1
asymptote \(B=1.8\)	A1

Answer: \(B(s)=\dfrac{1.8}{1+44e^{-0.05(s-14)}}\), \(s\ge14\), asymptote \(B=1.8\)

(3)

(h)

Suppose instead that a logistic model had been fitted with the GDC using only the areas measured in the first 5 weeks.

Evaluate how reliable such a model would be for predicting the area after 20 weeks.

INTERPRETNotes

Show solution

in the first 5 weeks the growth is almost exponential and shows no sign of levelling off, so the limiting value (180) cannot be estimated reliably from these data	R1
\(t=20\) is far outside the data range, so the prediction is an extrapolation and could be badly wrong; the model should be refitted as more data arrive	R1

(2)

(i)

A removal programme starts at \(t=0\). With the programme in place, the area is modelled by \(A_{R}(t)=\dfrac{120}{1+29e^{-0.35t}}\).

Explain why, under this model, the plant never covers half of the lake, and find the time at which the area grows fastest.

MULTINotes

Show solution

the limiting value is 120 ha, half of the lake; \(A_{R}(t)\lt120\) for all \(t\) (the asymptote is never reached)	R1	Must refer to the limiting value 120 ha being equal to (not more than) half of the 240 ha lake.
\(t=\frac{\ln29}{0.35}=9.62\) weeks	A1	Or solving \(A_{R}(t)=60\) on the GDC. Accept 9.6 weeks.

Answer: 9.62 weeks

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\frac{180}{1+C}=4\)	M1
\(C=44\)	A1
Part (b)
\(A=180\)	A1	Accept \(y=180\).
the plant cannot cover more than 180 ha: the area approaches 180 ha in the long term (the carrying capacity of the lake for this plant)	R1
Part (c)
\(A(10)=77.3\) ha	A1
\(\frac{180}{1+44e^{-0.35t}}=120\)	M1	Half of the lake is 120 ha. Using 90 scores M0.
\(t=12.8\) weeks	A1	12.7923…
Part (d)
\(E(4)=16.2\) ha	A1
\(A(4)=15.2\) ha	A1
while \(A\) is small compared with 180, logistic growth is almost exponential; later the logistic growth slows as \(A\) approaches 180, but \(E\) keeps growing without limit (it would exceed the whole lake)	R1
Part (e)(i)
\(\frac{180}{1+44e^{-0.35t}}=90\), so \(1+44e^{-0.35t}=2\)	M1
\(e^{-0.35t}=\frac{1}{44}\), so \(-0.35t=\ln\frac{1}{44}=-\ln44\)	A1	Correct use of logarithms to reach \(-0.35t=-\ln44\).
\(t=\frac{\ln44}{0.35}\)	AG
Part (e)(ii)
\(t=10.8\) weeks	A1	10.8119…
Part (f)(i)
increasing S-shaped curve, levelling off below the asymptote and close to it at \(t=30\)	A1	\(A(30)=179.8\).
intercept \((0,\ 4)\) and asymptote \(A=180\) shown	A1
point of fastest growth marked at \((10.8,\ 90)\), where the curve is steepest and changes from bending up to bending down	A1	Allow ECF from (e)(ii).
Part (f)(ii)
15.75	A1	Accept 15.7 or 15.8.
at 10.8 weeks the area covered is increasing at about 15.8 ha per week, the greatest rate of growth	R1	Units (ha per week) required.
Part (g)(i)
horizontal stretch, scale factor 7	A1
followed by a horizontal translation of 14 units to the right	A1	Accept a translation of 2 units right followed by a stretch of scale factor 7. The order must be consistent with the size of the translation.
vertical stretch, scale factor 0.01	A1	Accept \(\frac{1}{100}\).
Part (g)(ii)
\(B(s)=\frac{1.8}{1+44e^{-0.05(s-14)}}\)	A1	Accept equivalent forms, e.g. \(0.01A\left(\frac{s-14}{7}\right)\) written out in full, or \(\frac{1.8}{1+44e^{-0.35(s-14)/7}}\). Allow ECF from (a).
domain \(s\ge14\)	A1
asymptote \(B=1.8\)	A1
Part (h)
in the first 5 weeks the growth is almost exponential and shows no sign of levelling off, so the limiting value (180) cannot be estimated reliably from these data	R1
\(t=20\) is far outside the data range, so the prediction is an extrapolation and could be badly wrong; the model should be refitted as more data arrive	R1
Part (i)
the limiting value is 120 ha, half of the lake; \(A_{R}(t)\lt120\) for all \(t\) (the asymptote is never reached)	R1	Must refer to the limiting value 120 ha being equal to (not more than) half of the 240 ha lake.
\(t=\frac{\ln29}{0.35}=9.62\) weeks	A1	Or solving \(A_{R}(t)=60\) on the GDC. Accept 9.6 weeks.

Answers: (a) \(C=44\) · (b) \(A=180\) · (c) 77.3 ha; 12.8 weeks · (d) \(E(4)=16.2\) ha, \(A(4)=15.2\) ha · (e)(ii) 10.8 weeks · (f)(ii) 15.75 ha per week · (g)(ii) \(B(s)=\dfrac{1.8}{1+44e^{-0.05(s-14)}}\), \(s\ge14\), asymptote \(B=1.8\) · (i) 9.62 weeks

Common error (c) Using half of 180 for "half of the lake". The lake is 240 ha, so half of it is 120 ha; 180 ha is only the limit of the plant's growth.

Exam tip (g)(ii) Follow one key point through the transformations to check your equation: \((0,\ 4)\) on the graph of \(A\) should map to \((14,\ 0.04)\) on the graph of \(B\).

More 2.6 questions in the question bank

Topic 3 · Geometry and trigonometry

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	3.1	3.2	3.3	3.4	3.5	3.6	3.7	3.8	Other
A coastguard search: bearings and the sine and cosine rules	●	○							1.1 2.1
A grain silo: volume, surface area and a cone angle	○	●							2.1 1.1
Electric car chargers in a town: a Voronoi diagram	○		●						2.1 1.1
Air quality sensors in a city: Voronoi cells and nearest-neighbour estimates	○		●						2.1 4.1
Solar panels on a sloping roof: scalar and vector products	○	○		●					2.3
Two delivery drones: vector kinematics and closest approach	○			○	●				5.2
A regional airline: directed graphs and counting routes						●	○		1.6
A university campus: laying fibre cable and planning a security patrol						○	●		1.6 2.1
Moving a video game spaceship: rotations, reflections and determinants	○							●	1.6 1.5
Drawing a fractal fern: repeated matrix transformations								●	1.6 1.7 1.3

3.1 Right-angled trigonometry, bearings and the sine and cosine rules · 3.2 Volume and surface area in contextmerged group

+ Add to test✓ In your testCC-AI-31-32-125 marks~30 min

A coastguard search: bearings and the sine and cosine rules

3.1 Right-angled trigonometry, bearings and the sine and cosine rules1.1 Approximation, percentage error and standard form2.1 Linear models and piecewise models3.2 Volume and surface area in contextSL

A coastguard lifeboat is sent from a harbour \(H\) to find a kayaker who has reported that they are tired and drifting. The lifeboat sails 12.6 km on a bearing of 038° to a navigation buoy \(A\). It then turns and sails 8.4 km on a bearing of 157° to reach the kayak at \(K\). Assume the sea is a horizontal plane.

This information is shown in Figure 1.

Figure 1: the route of the lifeboat from the harbour H, via the buoy A, to the kayak K

(a)

Show that angle \(H\hat{A}K=61^\circ\).

PROOFNotes

Show solution

bearing of \(H\) from \(A\) is \(038^\circ+180^\circ=218^\circ\)	M1	Or the angle between north at \(A\) and \(AH\) is \(180^\circ-38^\circ=142^\circ\) measured anticlockwise, from parallel north lines (co-interior angles).
\(218^\circ-157^\circ\)	A1	Or \(360^\circ-142^\circ-157^\circ\).
\(H\hat{A}K=61^\circ\)	AG

Answer: \(61^\circ\)

(2)

(b)

Find the direct distance \(HK\).

ROUTINENotes

Show solution

\(HK^2=12.6^2+8.4^2-2(12.6)(8.4)\cos61^\circ\)	M1	Correct substitution into the cosine rule.
\(HK^2=126.69\ldots\)	A1
\(HK=11.3\) km	A1	Accept 11.256 km. Award M0A0A0 for 18.2 km (from \(\cos119^\circ\)): the interior angle \(61^\circ\) must be used.

Answer: 11.3 km (11.256 km)

(3)

(c)

Explain why the cosine rule, and not the sine rule, was needed in part (b).

CONCEPTNotes

Show solution

two sides and the included angle are known, but no side together with its opposite angle, so the sine rule has no matching pair to start from

Accept equivalent wording. Both ideas (two sides with the included angle; no side-angle pair) are needed.

(1)

(d)

Before the lifeboat set off, the skipper estimated the direct distance \(HK\) from a chart as 11 km.

Find the percentage error in this estimate.

ROUTINENotes

Show solution

\(\left\|\frac{11-11.2559}{11.2559}\right\|\times100\%\)	M1	Exact value in the denominator. Allow ECF from (b).
\(2.27\%\)	A1	Using the rounded 11.3 gives 2.65%: award M1A0.

Answer: 2.27%

(2)

(e)

Find the area of the triangular sea region \(HAK\).

ROUTINENotes

Show solution

\(\frac12(12.6)(8.4)\sin61^\circ\)	M1	Angle must be the one between the two given sides.
\(46.3\text{ km}^2\)	A1	Accept 46.28 km².

Answer: 46.3 km²

(2)

(f)(i)

Find angle \(A\hat{H}K\).

ROUTINENotes

Show solution

\(\frac{\sin A\hat{H}K}{8.4}=\frac{\sin61^\circ}{11.2559}\)	M1	Allow ECF from (b). Or the cosine rule with three sides.
\(A\hat{H}K=40.7^\circ\)	A1	The angle is acute because it is opposite the shortest side, so \(180^\circ-40.7^\circ\) is rejected.

Answer: \(40.7^\circ\)

(2)

(f)(ii)

Once the kayaker is on board, the lifeboat sails directly back to \(H\).

Show that the bearing of \(H\) from \(K\) is \(259^\circ\), correct to the nearest degree.

PROOFNotes

Show solution

bearing of \(K\) from \(H\) \(=38^\circ+40.746^\circ=78.746^\circ\)	M1	Adding their angle from (f)(i) to 038°.
\(78.746^\circ+180^\circ=258.75^\circ\)	A1	Must see a value that rounds to 259, from unrounded working (258.7° from 40.7° is acceptable).
\(259^\circ\) to the nearest degree	AG

Answer: \(259^\circ\)

(2)

(g)(i)

Figure 2 shows the distance \(d\) km travelled by the lifeboat \(t\) minutes after it left \(H\), up to the moment it reached \(K\). The graph consists of two straight line segments.

Use Figure 2 to find the speed of the lifeboat on the first leg, \(HA\), in km per hour.

Figure 2: distance \(d\) travelled by the lifeboat against time \(t\)

GRAPHNotes

Show solution

\(\frac{12.6}{18}=0.7\) km per minute	(M1)	Reading 12.6 km and 18 minutes from the graph.
\(42\text{ km h}^{-1}\)	A1

Answer: 42 km h⁻¹

(2)

(g)(ii)

Find the equation of the second line segment, in the form \(d=mt+c\), for \(18\le t\le39\).

MODELNotes

Show solution

\(m=\frac{21-12.6}{39-18}\)	M1
\(m=0.4\)	A1
\(d=0.4t+5.4\)	A1	Substituting \((18,\,12.6)\) or \((39,\,21)\). Accept equivalent forms.

Answer: \(d=0.4t+5.4\)

(3)

(g)(iii)

Interpret the value of \(m\) in this context.

INTERPRETNotes

Show solution

on the second leg (\(A\) to \(K\)) the lifeboat travels 0.4 km every minute (24 km h⁻¹)

Must refer to the second leg and include units. Allow ECF from (g)(ii).

(1)

(h)

When the lifeboat reaches \(K\), the crew find that the kayak has drifted 2.0 km due east, to a point \(K'\). The lifeboat sails to \(K'\), picks up the kayaker and returns directly to \(H\).

Find the distance \(HK'\).

MULTINotes

Show solution

\(H\hat{K}K'=258.75^\circ-90^\circ=168.75^\circ\)	M1	Angle between the direction of \(H\) (bearing 258.75°) and due east (090°) at \(K\). Allow ECF from (f)(ii).
\(HK'^2=11.2559^2+2^2-2(11.2559)(2)\cos168.75^\circ\)	M1	Cosine rule with their angle and their \(HK\).
\(HK'=13.2\) km	A1	Accept 13.22 km. Alternative: coordinates \(K(11.04,\,2.197)\), \(K'(13.04,\,2.197)\), then Pythagoras.

Answer: 13.2 km

(3)

(i)

To mark the position where the kayak was found, the crew drop a cylindrical float of diameter 40 cm and height 65 cm.

Find the volume of the float in litres.

ROUTINENotes

Show solution

\(\pi(0.2)^2(0.65)\) or \(\pi(20)^2(65)\)	M1	Radius, not diameter, in consistent units: \(0.08168\text{ m}^3\) or \(81\,681\text{ cm}^3\).
81.7 litres	A1	Award M1A0 for 327 litres (diameter used as radius) or 0.0817 without conversion.

Answer: 81.7 L

(2)

Show full solution

Working	Mark	Notes
Part (a)
bearing of \(H\) from \(A\) is \(038^\circ+180^\circ=218^\circ\)	M1	Or the angle between north at \(A\) and \(AH\) is \(180^\circ-38^\circ=142^\circ\) measured anticlockwise, from parallel north lines (co-interior angles).
\(218^\circ-157^\circ\)	A1	Or \(360^\circ-142^\circ-157^\circ\).
\(H\hat{A}K=61^\circ\)	AG
Part (b)
\(HK^2=12.6^2+8.4^2-2(12.6)(8.4)\cos61^\circ\)	M1	Correct substitution into the cosine rule.
\(HK^2=126.69\ldots\)	A1
\(HK=11.3\) km	A1	Accept 11.256 km. Award M0A0A0 for 18.2 km (from \(\cos119^\circ\)): the interior angle \(61^\circ\) must be used.
Part (c)
two sides and the included angle are known, but no side together with its opposite angle, so the sine rule has no matching pair to start from	R1	Accept equivalent wording. Both ideas (two sides with the included angle; no side-angle pair) are needed.
Part (d)
\(\left\|\frac{11-11.2559}{11.2559}\right\|\times100\%\)	M1	Exact value in the denominator. Allow ECF from (b).
\(2.27\%\)	A1	Using the rounded 11.3 gives 2.65%: award M1A0.
Part (e)
\(\frac12(12.6)(8.4)\sin61^\circ\)	M1	Angle must be the one between the two given sides.
\(46.3\text{ km}^2\)	A1	Accept 46.28 km².
Part (f)(i)
\(\frac{\sin A\hat{H}K}{8.4}=\frac{\sin61^\circ}{11.2559}\)	M1	Allow ECF from (b). Or the cosine rule with three sides.
\(A\hat{H}K=40.7^\circ\)	A1	The angle is acute because it is opposite the shortest side, so \(180^\circ-40.7^\circ\) is rejected.
Part (f)(ii)
bearing of \(K\) from \(H\) \(=38^\circ+40.746^\circ=78.746^\circ\)	M1	Adding their angle from (f)(i) to 038°.
\(78.746^\circ+180^\circ=258.75^\circ\)	A1	Must see a value that rounds to 259, from unrounded working (258.7° from 40.7° is acceptable).
\(259^\circ\) to the nearest degree	AG
Part (g)(i)
\(\frac{12.6}{18}=0.7\) km per minute	(M1)	Reading 12.6 km and 18 minutes from the graph.
\(42\text{ km h}^{-1}\)	A1
Part (g)(ii)
\(m=\frac{21-12.6}{39-18}\)	M1
\(m=0.4\)	A1
\(d=0.4t+5.4\)	A1	Substituting \((18,\,12.6)\) or \((39,\,21)\). Accept equivalent forms.
Part (g)(iii)
on the second leg (\(A\) to \(K\)) the lifeboat travels 0.4 km every minute (24 km h⁻¹)	R1	Must refer to the second leg and include units. Allow ECF from (g)(ii).
Part (h)
\(H\hat{K}K'=258.75^\circ-90^\circ=168.75^\circ\)	M1	Angle between the direction of \(H\) (bearing 258.75°) and due east (090°) at \(K\). Allow ECF from (f)(ii).
\(HK'^2=11.2559^2+2^2-2(11.2559)(2)\cos168.75^\circ\)	M1	Cosine rule with their angle and their \(HK\).
\(HK'=13.2\) km	A1	Accept 13.22 km. Alternative: coordinates \(K(11.04,\,2.197)\), \(K'(13.04,\,2.197)\), then Pythagoras.
Part (i)
\(\pi(0.2)^2(0.65)\) or \(\pi(20)^2(65)\)	M1	Radius, not diameter, in consistent units: \(0.08168\text{ m}^3\) or \(81\,681\text{ cm}^3\).
81.7 litres	A1	Award M1A0 for 327 litres (diameter used as radius) or 0.0817 without conversion.

Answers: (a) \(61^\circ\) · (b) 11.3 km (11.256 km) · (d) 2.27% · (e) 46.3 km² · (f)(i) \(40.7^\circ\) · (f)(ii) \(259^\circ\) · (g)(i) 42 km h⁻¹ · (g)(ii) \(d=0.4t+5.4\) · (h) 13.2 km · (i) 81.7 L

Common error (b) A common error is to use a bearing (038° or 157°) inside the triangle. Only the interior angle at \(A\), 61°, belongs in the cosine rule.

Exam tip (d) Use the unrounded value of \(HK\) from your GDC. Using the rounded 11.3 km changes the second significant figure of a small percentage error.

More 3.1 questions in the question bank

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A grain silo: volume, surface area and a cone angle

3.2 Volume and surface area in context3.1 Right-angled trigonometry, bearings and the sine and cosine rules2.1 Linear models and piecewise models1.1 Approximation, percentage error and standard formSL

A farm stores wheat in a steel silo. The silo is a vertical cylinder of internal diameter 5.0 m and height 14.0 m, with a flat roof. Below the cylinder is a hopper in the shape of an inverted cone with the same diameter and a vertical depth of 3.0 m. Grain leaves through a small outlet at the bottom of the hopper.

The silo is shown in Figure 1. Ignore the thickness of the steel.

Figure 1: the silo and its hopper (not to scale)

(a)

Show that the total capacity of the silo, cylinder and hopper together, is \(295\text{ m}^3\), correct to 3 significant figures.

PROOFNotes

Show solution

cylinder: \(\pi(2.5)^2(14)\) \((=274.89\ldots)\)	M1	Radius 2.5 m, not 5.0 m.
hopper: \(\frac13\pi(2.5)^2(3)\) \((=19.635\ldots)\)	M1	Vertical depth 3.0 m used.
total \(=294.52\ldots\text{ m}^3\)	A1	Or \(93.75\pi\). Must see a value to more than 3 s.f.
\(\approx295\text{ m}^3\)	AG

Answer: \(294.52\ldots\text{ m}^3\)

(3)

(b)

Find the capacity of the silo in litres. Give your answer in the form \(a\times10^k\), where \(1\le a<10\) and \(k\in\mathbb{Z}\).

ROUTINENotes

Show solution

\(294.52\times1000=294\,524\) litres	(M1)	Using \(1\text{ m}^3=1000\) litres.
\(2.95\times10^{5}\) litres	A1	Accept \(2.94524\times10^{5}\). Award A0 for \(295\times10^{3}\).

Answer: \(2.95\times10^{5}\) L

(2)

(c)

The farmer had estimated the capacity by treating the whole silo as a single cylinder of diameter 5.0 m and height 17.0 m.

Find the percentage error in the farmer's estimate.

ROUTINENotes

Show solution

estimate \(=\pi(2.5)^2(17)=333.79\ldots\); \(\left\|\frac{333.79-294.52}{294.52}\right\|\times100\%\)	M1	Exact capacity in the denominator.
\(13.3\%\)	A1	Accept \(13\tfrac13\%\).

Answer: 13.3%

(2)

(d)

The curved outer surfaces of the cylinder and of the hopper are to be painted.

Find the total area to be painted.

MULTINotes

Show solution

slant height \(l=\sqrt{2.5^2+3^2}\)	M1
\(l=3.905\ldots\) m	A1
\(2\pi(2.5)(14)+\pi(2.5)(3.905)\)	M1	Sum of the two curved areas with their \(l\); roof and joint not included.
\(251\text{ m}^2\)	A1	Accept 250.58 m². Using the vertical depth 3.0 m in place of \(l\) gives 243.5 m²: award M0A0M1A0.

Answer: 251 m²

(4)

(e)

Wheat flows freely out of a hopper only if the sloping wall makes an angle of at least \(45^\circ\) with the horizontal.

Determine whether the wheat will flow freely out of this hopper.

MULTINotes

Show solution

\(\tan\theta=\frac{3}{2.5}\)	M1	Right-angled triangle formed by the axis, a radius and the slant edge. Or \(\cos\theta=\frac{2.5}{3.905}\).
\(\theta=50.2^\circ\)	A1
\(50.2^\circ\ge45^\circ\), so the wheat will flow freely	R1	Conclusion consistent with their angle. Do not award R1 without an angle.

Answer: \(50.2^\circ\); yes

(3)

(f)

A temperature cable is fixed in a straight line from the point \(P\) on the top rim of the cylinder to the point \(Q\) on the bottom rim, diametrically opposite \(P\), as shown in Figure 1.

Find the angle between the cable and the horizontal plane through \(Q\).

ROUTINENotes

Show solution

\(\tan\alpha=\frac{14}{5}\)	M1	Vertical 14.0 m opposite, a full diameter 5.0 m adjacent.
\(\alpha=70.3^\circ\)	A1	Award M1A0 for 79.9° (radius used in place of the diameter).

Answer: \(70.3^\circ\)

(2)

(g)(i)

After harvest the empty silo is filled with wheat. Let \(h\) metres be the depth of wheat in the cylindrical section \(t\) minutes after the hopper is full. For the first 50 minutes one conveyor is used; then a second conveyor is switched on as well. Figure 2 shows the graph of \(h\) against \(t\) until the silo is full.

Find the gradient of the second line segment and interpret it in context.

Figure 2: depth \(h\) of grain in the cylindrical section against time \(t\)

GRAPHNotes

Show solution

\(\frac{14-4}{130-50}\)	(M1)	Reading \((50,\,4)\) and \((130,\,14)\) from the graph.
\(0.125\)	A1
the depth of wheat rises by 0.125 m every minute while both conveyors run	R1	Units and context needed.

Answer: 0.125 m per minute

(3)

(g)(ii)

Find an equation for \(h\) in terms of \(t\) for \(50\le t\le130\).

MODELNotes

Show solution

\(h-4=0.125(t-50)\)	M1	Point-gradient form with a point on the segment and their gradient.
\(h=0.125t-2.25\)	A1	Allow ECF from (g)(i).

Answer: \(h=0.125t-2.25\)

(2)

(g)(iii)

Find the rate at which wheat enters the silo while both conveyors are running, in \(\text{m}^3\) per minute.

MULTINotes

Show solution

\(\pi(2.5)^2\times0.125\)	M1	Cross-sectional area times the rate of rise of the depth. Allow ECF from (g)(i).
\(2.45\text{ m}^3\) per minute	A1	Accept 2.454.

Answer: 2.45 m³ per minute

(2)

(h)

This year's wheat harvest is 240 tonnes. One cubic metre of stored wheat has a mass of 0.78 tonnes.

Determine whether the silo can hold the whole harvest. Justify your answer.

INTERPRETNotes

Show solution

\(\frac{240}{0.78}\) or \(294.52\times0.78\)	M1
\(307.7\text{ m}^3\) needed, or the silo holds \(229.7\) tonnes	A1
\(307.7>294.5\) (or \(229.7<240\)), so the silo cannot hold the whole harvest	R1	Conclusion must follow from a correct comparison.

Answer: No: 307.7 m³ needed but capacity 294.5 m³

(3)

(i)

The farm's next silo will have the same heights but its diameter will be 10% larger.

Find the capacity of the new silo, and state whether it could hold this year's harvest.

MULTINotes

Show solution

\(\pi(2.75)^2(14)+\frac13\pi(2.75)^2(3)\) or \(1.1^2\times294.52\)	M1	Award M0 for \(1.1\times294.52\).
\(356\text{ m}^3\); \(356>307.7\), so yes	A1	Accept 356.37. Conclusion required; allow ECF from (h).

Answer: 356 m³; yes

(2)

Show full solution

Working	Mark	Notes
Part (a)
cylinder: \(\pi(2.5)^2(14)\) \((=274.89\ldots)\)	M1	Radius 2.5 m, not 5.0 m.
hopper: \(\frac13\pi(2.5)^2(3)\) \((=19.635\ldots)\)	M1	Vertical depth 3.0 m used.
total \(=294.52\ldots\text{ m}^3\)	A1	Or \(93.75\pi\). Must see a value to more than 3 s.f.
\(\approx295\text{ m}^3\)	AG
Part (b)
\(294.52\times1000=294\,524\) litres	(M1)	Using \(1\text{ m}^3=1000\) litres.
\(2.95\times10^{5}\) litres	A1	Accept \(2.94524\times10^{5}\). Award A0 for \(295\times10^{3}\).
Part (c)
estimate \(=\pi(2.5)^2(17)=333.79\ldots\); \(\left\|\frac{333.79-294.52}{294.52}\right\|\times100\%\)	M1	Exact capacity in the denominator.
\(13.3\%\)	A1	Accept \(13\tfrac13\%\).
Part (d)
slant height \(l=\sqrt{2.5^2+3^2}\)	M1
\(l=3.905\ldots\) m	A1
\(2\pi(2.5)(14)+\pi(2.5)(3.905)\)	M1	Sum of the two curved areas with their \(l\); roof and joint not included.
\(251\text{ m}^2\)	A1	Accept 250.58 m². Using the vertical depth 3.0 m in place of \(l\) gives 243.5 m²: award M0A0M1A0.
Part (e)
\(\tan\theta=\frac{3}{2.5}\)	M1	Right-angled triangle formed by the axis, a radius and the slant edge. Or \(\cos\theta=\frac{2.5}{3.905}\).
\(\theta=50.2^\circ\)	A1
\(50.2^\circ\ge45^\circ\), so the wheat will flow freely	R1	Conclusion consistent with their angle. Do not award R1 without an angle.
Part (f)
\(\tan\alpha=\frac{14}{5}\)	M1	Vertical 14.0 m opposite, a full diameter 5.0 m adjacent.
\(\alpha=70.3^\circ\)	A1	Award M1A0 for 79.9° (radius used in place of the diameter).
Part (g)(i)
\(\frac{14-4}{130-50}\)	(M1)	Reading \((50,\,4)\) and \((130,\,14)\) from the graph.
\(0.125\)	A1
the depth of wheat rises by 0.125 m every minute while both conveyors run	R1	Units and context needed.
Part (g)(ii)
\(h-4=0.125(t-50)\)	M1	Point-gradient form with a point on the segment and their gradient.
\(h=0.125t-2.25\)	A1	Allow ECF from (g)(i).
Part (g)(iii)
\(\pi(2.5)^2\times0.125\)	M1	Cross-sectional area times the rate of rise of the depth. Allow ECF from (g)(i).
\(2.45\text{ m}^3\) per minute	A1	Accept 2.454.
Part (h)
\(\frac{240}{0.78}\) or \(294.52\times0.78\)	M1
\(307.7\text{ m}^3\) needed, or the silo holds \(229.7\) tonnes	A1
\(307.7>294.5\) (or \(229.7<240\)), so the silo cannot hold the whole harvest	R1	Conclusion must follow from a correct comparison.
Part (i)
\(\pi(2.75)^2(14)+\frac13\pi(2.75)^2(3)\) or \(1.1^2\times294.52\)	M1	Award M0 for \(1.1\times294.52\).
\(356\text{ m}^3\); \(356>307.7\), so yes	A1	Accept 356.37. Conclusion required; allow ECF from (h).

Answers: (a) \(294.52\ldots\text{ m}^3\) · (b) \(2.95\times10^{5}\) L · (c) 13.3% · (d) 251 m² · (e) \(50.2^\circ\); yes · (f) \(70.3^\circ\) · (g)(i) 0.125 m per minute · (g)(ii) \(h=0.125t-2.25\) · (g)(iii) 2.45 m³ per minute · (h) No: 307.7 m³ needed but capacity 294.5 m³ · (i) 356 m³; yes

Common error (d) The curved surface of a cone uses the slant height \(l\), not the vertical depth. Find \(l\) with Pythagoras first.

Exam tip (i) Only the radius changes, so both volumes are multiplied by \(1.1^2=1.21\), not by 1.1.

More 3.2 questions in the question bank

3.3 Voronoi diagrams

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Electric car chargers in a town: a Voronoi diagram

3.3 Voronoi diagrams2.1 Linear models and piecewise models3.1 Right-angled trigonometry, bearings and the sine and cosine rules1.1 Approximation, percentage error and standard formSL

A town council has four public rapid chargers for electric cars. On a map of the town, positions are given as \((x,\,y)\), where \(x\) and \(y\) are the distances in km east and north of the south-west corner of the town. The town is the rectangle \(0\le x\le14\), \(0\le y\le11\).

The chargers are at \(A(2,\,5)\), \(B(6,\,2)\), \(C(12,\,6)\) and \(D(6,\,9)\). Drivers always use the nearest charger.

Figure 1 shows the chargers and part of the Voronoi diagram for these four sites.

Figure 1: the four chargers and part of the Voronoi diagram; the edge between the cells of A and B is missing

(a)

Find the equation of the perpendicular bisector of \(AB\). Give your answer in the form \(y=mx+c\).

ALGEBRANotes

Show solution

midpoint \((4,\,3.5)\)	B1
gradient of \(AB=\frac{2-5}{6-2}=-\frac34\)	B1
\(y-3.5=\frac43(x-4)\)	M1	Negative reciprocal of their gradient, through their midpoint.
\(y=\frac43x-\frac{11}{6}\)	A1	Accept \(y=1.33x-1.83\).

Answer: \(y=\frac43x-\frac{11}{6}\)

(4)

(b)

The edge between the cells of \(A\) and \(D\) lies on the line \(y=-x+11\).

By solving your equation from part (a) and \(y=-x+11\) simultaneously, show that the cells of \(A\), \(B\) and \(D\) meet at the vertex \(V(5.5,\,5.5)\).

PROOFNotes

Show solution

\(\frac43x-\frac{11}{6}=-x+11\)	M1	Equating their equation from (a) with \(y=-x+11\).
\(\frac73x=\frac{77}{6}\Rightarrow x=5.5\), \(y=-5.5+11=5.5\)	A1	Both coordinates found from the equations, not read from Figure 1. Reading or substituting the given point only: M0A0.
\(V(5.5,\,5.5)\)	AG

Answer: \((5.5,\,5.5)\)

(2)

(c)

Show that \(V\) is the same distance from \(A\), \(B\) and \(D\).

PROOFNotes

Show solution

\(VA=\sqrt{3.5^2+0.5^2}\)	M1	Pythagoras for at least one distance.
\(VA=VB=VD=\sqrt{12.5}\;(=3.54\text{ km})\)	A1	All three seen: \(VB=\sqrt{0.5^2+3.5^2}\), \(VD=\sqrt{0.5^2+3.5^2}\).
so \(V\) is equidistant from \(A\), \(B\) and \(D\)	AG

Answer: \(\sqrt{12.5}\) km

(2)

(d)(i)

On Figure 1, draw the missing edge between the cells of \(A\) and \(B\).

GRAPHNotes

Show solution

straight edge from \(V(5.5,\,5.5)\) along \(y=\frac43x-\frac{11}{6}\) down to the boundary at about \((1.4,\,0)\)

Must start at \(V\), pass through \((4,\,3.5)\) and stop at the edge of the town.

Answer: edge from \((5.5,\,5.5)\) to \((1.375,\,0)\)

(1)

(d)(ii)

A new housing estate is planned at \(E(3.5,\,2.5)\).

Use your diagram to state which charger the residents of the estate will use, and verify your answer by calculation.

ROUTINENotes

Show solution

\(EA=\sqrt{8.5}=2.92\) km and \(EB=\sqrt{6.5}=2.55\) km	M1	Or substitute \(x=3.5\) into the bisector: \(y=2.83>2.5\), so \(E\) is on the same side as \(B\).
charger \(B\)	A1

Answer: \(B\)

(2)

(e)(i)

The other vertex of the diagram is \(W(8,\,5.5)\).

Find the distance from \(W\) to the nearest charger.

ROUTINENotes

Show solution

\(WB=\sqrt{(8-6)^2+(5.5-2)^2}\)	M1	Or \(WC\) or \(WD\), which are equal.
\(4.03\) km	A1	Accept \(\sqrt{16.25}\) or 4.031 km.

Answer: 4.03 km

(2)

(e)(ii)

The council will build a fifth charger inside the quadrilateral \(ABCD\), including its sides, at the point that is furthest from its nearest existing charger. The edges of the diagram cross the sides of \(ABCD\) at the midpoints of those sides, and the furthest of these midpoints from its nearest charger is 3.61 km away.

Explain why the new charger should be built at \(W\).

INTERPRETNotes

Show solution

the furthest point must be at a vertex inside \(ABCD\), where an edge meets a side, or at a corner (the corners are chargers, distance 0)	R1
\(W\) gives 4.03 km, more than \(V\) (3.54 km) and the midpoints (at most 3.61 km), so \(W\) is furthest from any charger	R1	Must compare \(W\) with \(V\). Allow ECF from (c) and (e)(i).

(2)

(f)

A council leaflet states that the new charger will be 4 km from each of \(B\), \(C\) and \(D\).

Find the percentage error in the leaflet's distance.

ROUTINENotes

Show solution

\(\left\|\frac{4-4.0311}{4.0311}\right\|\times100\%\)	M1	Exact distance in the denominator. Allow ECF from (e)(i).
\(0.772\%\)	A1	Using 4.03 gives 0.744%: award M1A0.

Answer: 0.772%

(2)

(g)

Instead of a new charger being built, charger \(C\) is closed permanently.

Describe how the Voronoi diagram changes, and explain why the cell of \(A\) does not change.

CONCEPTNotes

Show solution

the edges between \(B\) and \(C\) and between \(C\) and \(D\) disappear, and \(W\) is no longer a vertex	B1
the edge between \(B\) and \(D\) now continues from \(V\) along \(y=5.5\) to the east side of the town (\(x=14\))	B1	Accept 'the cell of \(C\) is shared between \(B\) (below \(y=5.5\)) and \(D\) (above)'.
the cell of \(A\) does not border the cell of \(C\), so the region that was served by \(C\) is shared only between its neighbours \(B\) and \(D\); no point becomes nearer to \(A\), so the cell of \(A\) is unchanged	R1

(3)

Show full solution

Working	Mark	Notes
Part (a)
midpoint \((4,\,3.5)\)	B1
gradient of \(AB=\frac{2-5}{6-2}=-\frac34\)	B1
\(y-3.5=\frac43(x-4)\)	M1	Negative reciprocal of their gradient, through their midpoint.
\(y=\frac43x-\frac{11}{6}\)	A1	Accept \(y=1.33x-1.83\).
Part (b)
\(\frac43x-\frac{11}{6}=-x+11\)	M1	Equating their equation from (a) with \(y=-x+11\).
\(\frac73x=\frac{77}{6}\Rightarrow x=5.5\), \(y=-5.5+11=5.5\)	A1	Both coordinates found from the equations, not read from Figure 1. Reading or substituting the given point only: M0A0.
\(V(5.5,\,5.5)\)	AG
Part (c)
\(VA=\sqrt{3.5^2+0.5^2}\)	M1	Pythagoras for at least one distance.
\(VA=VB=VD=\sqrt{12.5}\;(=3.54\text{ km})\)	A1	All three seen: \(VB=\sqrt{0.5^2+3.5^2}\), \(VD=\sqrt{0.5^2+3.5^2}\).
so \(V\) is equidistant from \(A\), \(B\) and \(D\)	AG
Part (d)(i)
straight edge from \(V(5.5,\,5.5)\) along \(y=\frac43x-\frac{11}{6}\) down to the boundary at about \((1.4,\,0)\)	A1	Must start at \(V\), pass through \((4,\,3.5)\) and stop at the edge of the town.
Part (d)(ii)
\(EA=\sqrt{8.5}=2.92\) km and \(EB=\sqrt{6.5}=2.55\) km	M1	Or substitute \(x=3.5\) into the bisector: \(y=2.83>2.5\), so \(E\) is on the same side as \(B\).
charger \(B\)	A1
Part (e)(i)
\(WB=\sqrt{(8-6)^2+(5.5-2)^2}\)	M1	Or \(WC\) or \(WD\), which are equal.
\(4.03\) km	A1	Accept \(\sqrt{16.25}\) or 4.031 km.
Part (e)(ii)
the furthest point must be at a vertex inside \(ABCD\), where an edge meets a side, or at a corner (the corners are chargers, distance 0)	R1
\(W\) gives 4.03 km, more than \(V\) (3.54 km) and the midpoints (at most 3.61 km), so \(W\) is furthest from any charger	R1	Must compare \(W\) with \(V\). Allow ECF from (c) and (e)(i).
Part (f)
\(\left\|\frac{4-4.0311}{4.0311}\right\|\times100\%\)	M1	Exact distance in the denominator. Allow ECF from (e)(i).
\(0.772\%\)	A1	Using 4.03 gives 0.744%: award M1A0.
Part (g)
the edges between \(B\) and \(C\) and between \(C\) and \(D\) disappear, and \(W\) is no longer a vertex	B1
the edge between \(B\) and \(D\) now continues from \(V\) along \(y=5.5\) to the east side of the town (\(x=14\))	B1	Accept 'the cell of \(C\) is shared between \(B\) (below \(y=5.5\)) and \(D\) (above)'.
the cell of \(A\) does not border the cell of \(C\), so the region that was served by \(C\) is shared only between its neighbours \(B\) and \(D\); no point becomes nearer to \(A\), so the cell of \(A\) is unchanged	R1

Answers: (a) \(y=\frac43x-\frac{11}{6}\) · (b) \((5.5,\,5.5)\) · (c) \(\sqrt{12.5}\) km · (d)(i) edge from \((5.5,\,5.5)\) to \((1.375,\,0)\) · (d)(ii) \(B\) · (e)(i) 4.03 km · (f) 0.772%

Common error (a) A common error is to use the gradient of \(AB\) itself, \(-\frac34\). The edge is perpendicular to \(AB\), so its gradient is \(\frac43\).

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Air quality sensors in a city: Voronoi cells and nearest-neighbour estimates

3.3 Voronoi diagrams2.1 Linear models and piecewise models4.1 Sampling and descriptive statistics3.1 Right-angled trigonometry, bearings and the sine and cosine rulesSL

A city monitors fine dust (PM2.5) with five sensors. Positions are given as \((x,\,y)\), where \(x\) and \(y\) are the distances in km east and north of the south-west corner of the city, which is the rectangle \(0\le x\le14\), \(0\le y\le12\). The positive \(y\)-direction is north.

The table shows the position of each sensor and its daily mean reading, in micrograms per cubic metre (µg m⁻³).

Sensor	A	B	C	D	E
Position	\((3,\,1)\)	\((10,\,2)\)	\((12,\,10)\)	\((3,\,7)\)	\((7,\,5)\)
Reading (µg m⁻³)	18	31	12	15	24

A primary school is at \(S(9.5,\,4)\). Figure 1 shows the sensors, the school and part of the Voronoi diagram for the five sensors.

Figure 1: the five sensors, the school \(S\) and part of the Voronoi diagram; the edge between the cells of \(D\) and \(E\) is missing

(a)(i)

By finding the distances from \(S\) to \(B\) and to \(E\), determine which sensor's cell contains the school.

ROUTINENotes

Show solution

\(SB=\sqrt{0.5^2+2^2}=2.06\) km and \(SE=\sqrt{2.5^2+1^2}=2.69\) km	M1	Both distances needed.
the cell of \(B\)	A1	Accept \(\sqrt{4.25}\) and \(\sqrt{7.25}\).

Answer: \(B\)

(2)

(a)(ii)

Hence write down the nearest-neighbour estimate of the PM2.5 reading at the school.

ROUTINENotes

Show solution

31 µg m⁻³

Allow ECF from (a)(i).

Answer: 31 µg m⁻³

(1)

(b)

Find the equation of the line on which the missing edge between the cells of \(D\) and \(E\) lies. Give your answer in the form \(y=mx+c\).

ALGEBRANotes

Show solution

midpoint of \(DE\) is \((5,\,6)\)	B1
gradient of \(DE=\frac{5-7}{7-3}=-\frac12\)	B1
\(y-6=2(x-5)\)	M1	Perpendicular gradient \(2\) through their midpoint.
\(y=2x-4\)	A1

Answer: \(y=2x-4\)

(4)

(c)

Show that \(P(11,\,6)\) is a vertex of the Voronoi diagram.

PROOFNotes

Show solution

\(PB=\sqrt{1^2+4^2}\), \(PC=\sqrt{1^2+4^2}\), \(PE=\sqrt{4^2+1^2}\)	M1	Pythagoras for the distances to \(B\), \(C\) and \(E\).
\(PB=PC=PE=\sqrt{17}=4.12\) km	A1
\(PA=\sqrt{89}=9.43\) km and \(PD=\sqrt{65}=8.06\) km are both greater than \(\sqrt{17}\), so no other sensor lies inside the circle	R1	Both other distances (or a clear statement that they are larger) needed.
so \(P\) is a vertex	AG

Answer: \(PB=PC=PE=\sqrt{17}\)

(3)

(d)

Figure 2 shows the complete Voronoi diagram within the city boundary. The area of the part of the cell of \(A\) inside the city is \(23.0\text{ km}^2\).

Use Figure 2 to estimate the area of the part of the cell of \(D\) inside the city, and comment on what the two areas suggest about the sensor network.

Figure 2: the complete Voronoi diagram within the city boundary

GRAPHNotes

Show solution

rectangle \(0\le x\le4\), \(4\le y\le12\) gives 32, plus the region to its right (about 14), or counting squares	(M1)
\(46\text{ km}^2\)	A1	Accept answers from 44 to 48 km². The exact value is \(46\tfrac13\text{ km}^2\).
the reading at \(D\) is used for about twice the area served by \(A\), so coverage is uneven: the north-west of the city is monitored less closely	R1	Any sensible comment comparing the two areas.

Answer: about 46 km²

(3)

(e)

Find the mean and the standard deviation of the five readings.

ROUTINENotes

Show solution

mean \(=20\) µg m⁻³	A1
\(\sigma=6.78\) µg m⁻³	A1	Accept \(\sqrt{46}\). Award A0 for 7.58 (the \(s_x\) value).

Answer: 20; 6.78 µg m⁻³

(2)

(f)

A maintenance drone flies in a straight line from sensor \(D\) to sensor \(B\).

Find the distance it flies and the bearing on which it flies.

MULTINotes

Show solution

\(\sqrt{7^2+5^2}=8.60\) km	B1	Accept \(\sqrt{74}\).
\(\tan^{-1}\frac57=35.5^\circ\) below due east, or \(\tan^{-1}\frac75=54.5^\circ\) east of due south	M1	Right-angled triangle with sides 7 km (east) and 5 km (south).
bearing \(125.5^\circ\)	A1	Accept \(126^\circ\).

Answer: 8.60 km on a bearing of 125.5°

(3)

(g)

Discuss the reliability of your estimate in part (a)(ii).

INTERPRETNotes

Show solution

the school is close to the edge between the cells of \(B\) and \(E\) (2.06 km against 2.69 km), so the estimate would change from 31 to 24 just across the edge	R1
PM2.5 varies gradually, so the true value is likely to lie between the readings of nearby sensors; using a single sensor makes the estimate unreliable here	R1	Accept a comment that the estimate is well above the mean of 20 because it ignores every sensor except \(B\).

(2)

(h)(i)

A sixth sensor \(F\) is installed at \((9,\,3.5)\), near the school.

Show that \(P(11,\,6)\) is no longer a vertex of the Voronoi diagram.

CONCEPTNotes

Show solution

\(PF=\sqrt{2^2+2.5^2}=3.20\) km	M1
\(3.20<\sqrt{17}=4.12\), so \(F\) is inside the empty circle of \(P\) (nearer to \(P\) than \(B\), \(C\) and \(E\)), so \(P\) is not a vertex	R1	Allow ECF from (c).
so \(P\) is no longer a vertex	AG

Answer: \(PF=3.20\) km

(2)

(h)(ii)

Sensor \(F\) records a reading of 27 µg m⁻³. State, with a reason, the new nearest-neighbour estimate at the school.

ROUTINENotes

Show solution

27 µg m⁻³, because \(SF=\sqrt{0.5^2+0.5^2}=0.707\) km is less than 2.06 km, so the school is now in the cell of \(F\)

Reason required.

Answer: 27 µg m⁻³

(1)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(SB=\sqrt{0.5^2+2^2}=2.06\) km and \(SE=\sqrt{2.5^2+1^2}=2.69\) km	M1	Both distances needed.
the cell of \(B\)	A1	Accept \(\sqrt{4.25}\) and \(\sqrt{7.25}\).
Part (a)(ii)
31 µg m⁻³	A1	Allow ECF from (a)(i).
Part (b)
midpoint of \(DE\) is \((5,\,6)\)	B1
gradient of \(DE=\frac{5-7}{7-3}=-\frac12\)	B1
\(y-6=2(x-5)\)	M1	Perpendicular gradient \(2\) through their midpoint.
\(y=2x-4\)	A1
Part (c)
\(PB=\sqrt{1^2+4^2}\), \(PC=\sqrt{1^2+4^2}\), \(PE=\sqrt{4^2+1^2}\)	M1	Pythagoras for the distances to \(B\), \(C\) and \(E\).
\(PB=PC=PE=\sqrt{17}=4.12\) km	A1
\(PA=\sqrt{89}=9.43\) km and \(PD=\sqrt{65}=8.06\) km are both greater than \(\sqrt{17}\), so no other sensor lies inside the circle	R1	Both other distances (or a clear statement that they are larger) needed.
so \(P\) is a vertex	AG
Part (d)
rectangle \(0\le x\le4\), \(4\le y\le12\) gives 32, plus the region to its right (about 14), or counting squares	(M1)
\(46\text{ km}^2\)	A1	Accept answers from 44 to 48 km². The exact value is \(46\tfrac13\text{ km}^2\).
the reading at \(D\) is used for about twice the area served by \(A\), so coverage is uneven: the north-west of the city is monitored less closely	R1	Any sensible comment comparing the two areas.
Part (e)
mean \(=20\) µg m⁻³	A1
\(\sigma=6.78\) µg m⁻³	A1	Accept \(\sqrt{46}\). Award A0 for 7.58 (the \(s_x\) value).
Part (f)
\(\sqrt{7^2+5^2}=8.60\) km	B1	Accept \(\sqrt{74}\).
\(\tan^{-1}\frac57=35.5^\circ\) below due east, or \(\tan^{-1}\frac75=54.5^\circ\) east of due south	M1	Right-angled triangle with sides 7 km (east) and 5 km (south).
bearing \(125.5^\circ\)	A1	Accept \(126^\circ\).
Part (g)
the school is close to the edge between the cells of \(B\) and \(E\) (2.06 km against 2.69 km), so the estimate would change from 31 to 24 just across the edge	R1
PM2.5 varies gradually, so the true value is likely to lie between the readings of nearby sensors; using a single sensor makes the estimate unreliable here	R1	Accept a comment that the estimate is well above the mean of 20 because it ignores every sensor except \(B\).
Part (h)(i)
\(PF=\sqrt{2^2+2.5^2}=3.20\) km	M1
\(3.20<\sqrt{17}=4.12\), so \(F\) is inside the empty circle of \(P\) (nearer to \(P\) than \(B\), \(C\) and \(E\)), so \(P\) is not a vertex	R1	Allow ECF from (c).
so \(P\) is no longer a vertex	AG
Part (h)(ii)
27 µg m⁻³, because \(SF=\sqrt{0.5^2+0.5^2}=0.707\) km is less than 2.06 km, so the school is now in the cell of \(F\)	A1	Reason required.

Answers: (a)(i) \(B\) · (a)(ii) 31 µg m⁻³ · (b) \(y=2x-4\) · (c) \(PB=PC=PE=\sqrt{17}\) · (d) about 46 km² · (e) 20; 6.78 µg m⁻³ · (f) 8.60 km on a bearing of 125.5° · (h)(i) \(PF=3.20\) km · (h)(ii) 27 µg m⁻³

Exam tip (c) Three equal distances show that \(P\) is on three bisectors. To confirm a vertex, also check that every other site is further away than that common distance.

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3.4 Vectors, scalar and vector products · 3.5 Vector kinematicsmerged group

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Solar panels on a sloping roof: scalar and vector products

3.4 Vectors, scalar and vector products3.1 Right-angled trigonometry, bearings and the sine and cosine rules3.2 Volume and surface area in context2.3 Exponential and sinusoidal modelsHL

A house has a rectangular base and a symmetrical roof with two sloping faces that meet at a horizontal ridge. A rectangular solar panel \(ABCD\) is fixed flat on the front face of the roof. Its lower edge \([AB]\) lies along the gutter and its upper edge \([DC]\) lies along the ridge.

Relative to an origin \(O\) at ground level, the \(x\)-axis points east, the \(y\)-axis points north and the \(z\)-axis points vertically up. All distances are in metres. Three corners of the panel are \(A(3,\,1,\,3)\), \(B(7,\,4,\,3)\) and \(D(0.6,\,4.2,\,5.5)\).

This information is shown in Figure 1.

Figure 1: the roof, the panel ABCD and the top T of the chimney (not to scale)

(a)

Find \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\), and hence show that the edges \([AB]\) and \([AD]\) are perpendicular.

ROUTINENotes

Show solution

\(\overrightarrow{AB}=\begin{pmatrix}4\\3\\0\end{pmatrix}\)	B1
\(\overrightarrow{AD}=\begin{pmatrix}-2.4\\3.2\\2.5\end{pmatrix}\)	B1
\(\overrightarrow{AB}\cdot\overrightarrow{AD}=4(-2.4)+3(3.2)+0(2.5)\)	M1	Correct substitution of their two vectors into the scalar product.
\(=-9.6+9.6+0=0\), so \([AB]\) and \([AD]\) are perpendicular	AG	Both the value \(0\) and the conclusion must be seen.

Answer: \(\overrightarrow{AB}=\begin{pmatrix}4\\3\\0\end{pmatrix}\), \(\overrightarrow{AD}=\begin{pmatrix}-2.4\\3.2\\2.5\end{pmatrix}\), scalar product \(0\)

(3)

(b)

Show that the roof makes an angle of \(32^\circ\) with the horizontal, correct to the nearest degree.

MULTINotes

Show solution

horizontal distance from \(A\) to \(D\): \(\sqrt{(-2.4)^2+3.2^2}=4\)	M1	The vertical rise \(5.5-3=2.5\) must also be seen or implied.
\(\tan\theta=\frac{2.5}{4}\)	M1	Or \(\sin\theta=\frac{2.5}{\sqrt{22.25}}\), or \(\cos\theta=\frac{4}{\sqrt{22.25}}\).
\(\theta=32.005\ldots^\circ\)	A1	A value to more than 2 significant figures must be seen before rounding.
\(\theta\approx32^\circ\)	AG

Answer: \(32.0^\circ\)

(3)

(c)(i)

Find \(\overrightarrow{AB}\times\overrightarrow{AD}\).

ROUTINENotes

Show solution

\(\begin{pmatrix}(3)(2.5)-(0)(3.2)\\{}(0)(-2.4)-(4)(2.5)\\{}(4)(3.2)-(3)(-2.4)\end{pmatrix}\)	M1	Correct method seen, or at least two correct components.
\(=\begin{pmatrix}7.5\\-10\\20\end{pmatrix}\)	A1

Answer: \(\begin{pmatrix}7.5\\-10\\20\end{pmatrix}\)

(2)

(c)(ii)

Hence find the area of the panel.

ROUTINENotes

Show solution

\(\sqrt{7.5^2+(-10)^2+20^2}=\sqrt{556.25}\)	M1	Allow ECF from (c)(i). Also accept \(\|\overrightarrow{AB}\|\|\overrightarrow{AD}\|=5\times4.716\ldots\)
\(23.6\ \text{m}^2\) (\(23.584\ldots\))	A1

Answer: \(23.6\ \text{m}^2\)

(2)

(d)

Both faces of the roof make the same angle with the horizontal, and both gutters are at a height of \(3\) m. The house is \(12\) m long, measured along the ridge. The space under the roof is a prism with a triangular cross-section.

Find the volume of this space.

MULTINotes

Show solution

height of the triangle \(5.5-3=2.5\) and base \(2\times4=8\)	B1	The base is twice the horizontal distance found in (b), by symmetry.
\(V=\frac{1}{2}\times8\times2.5\times12\)	M1	Area of the triangular cross-section multiplied by the length, with their values.
\(V=120\ \text{m}^3\)	A1

Answer: \(120\ \text{m}^3\)

(3)

(e)

The electrical power output \(P\) kW of the panel is modelled by \(P=0.18\,S\cos\theta\), where \(S\ \text{m}^2\) is the area of the panel and \(\theta\) is the angle between the upward normal to the panel and the direction of the sun.

At 11:00 the direction of the sun is given by the vector \(\begin{pmatrix}2\\-1\\2\end{pmatrix}\).

Find the power output at 11:00.

MULTINotes

Show solution

\(\cos\theta=\frac{(7.5)(2)+(-10)(-1)+(20)(2)}{\sqrt{556.25}\times\sqrt{9}}=\frac{65}{70.75\ldots}\)	M1	Their normal from (c)(i), or any positive multiple of it such as \((3,-4,8)\).
\(\cos\theta=0.9187\ldots\) (\(\theta=23.3^\circ\))	A1
\(P=0.18\times23.58\ldots\times0.9187\ldots=3.90\) kW	A1	Allow ECF from (c)(ii). The exact value is \(3.9\).

Answer: \(3.90\) kW

(3)

(f)

The top of the chimney is at \(T(0,\,1,\,6)\). At 16:00 the direction of the sun is given by the vector \(\begin{pmatrix}-7\\-3\\3\end{pmatrix}\).

By writing down a vector equation of the line through \(B\) in the direction of the sun, determine whether the shadow of \(T\) falls on the corner \(B\) of the panel at 16:00.

ALGEBRANotes

Show solution

\(\mathbf r=\begin{pmatrix}7\\4\\3\end{pmatrix}+\lambda\begin{pmatrix}-7\\-3\\3\end{pmatrix}\)	B1	Accept \(\mathbf r=\ldots\) or \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\ldots\); a different parameter letter is fine.
\(7-7\lambda=0\Rightarrow\lambda=1\)	M1	Setting one component equal to the matching coordinate of \(T\) and solving.
\(y=4-3=1\) and \(z=3+3=6\), so \(T\) lies on the line, on the sun's side of \(B\) (\(\lambda>0\)): the shadow of \(T\) falls on \(B\)	R1	Both remaining components must be checked. Do not award for a conclusion based on one component only.

Answer: Yes: \(T\) is on the line at \(\lambda=1\)

(3)

(g)(i)

On a clear day, the power output \(P\) kW is modelled by \(P(t)=a\sin\left(b(t-c)\right)+d\), where \(t\) is the time in hours after midnight, the angle \(b(t-c)\) is in degrees, \(a,b,d>0\) and \(0<c<12\). The model has a period of 24 hours.

Figure 2 shows the graph of the model for \(6\le t\le20\). The maximum point is \((13,\,4.2)\), and the points \((7,\,1.2)\) and \((19,\,1.2)\) lie on the graph.

Find the values of \(a\), \(b\), \(c\) and \(d\).

Figure 2: graph of the model for the power output, \(6\le t\le20\)

GRAPHNotes

Show solution

\(b=\frac{360}{24}=15\)	B1
\(d=1.2\)	B1	The points at \(t=7\) and \(t=19\) are 6 hours (a quarter of a period) either side of the maximum, so they lie on the principal axis.
\(a=4.2-1.2=3\)	B1	Allow ECF from their \(d\).
\(c=7\)	B1	The graph rises through the principal axis at \(t=7\).

Answer: \(a=3,\ b=15,\ c=7,\ d=1.2\)

(4)

(g)(ii)

Use the model to find the power output at 11:00. Compare your answer with your answer to part (e), and suggest one reason for any difference.

INTERPRETNotes

Show solution

\(P(11)=3\sin(60^\circ)+1.2=3.80\) kW (\(3.798\ldots\))	A1	Allow ECF from (g)(i).
the values are close (they differ by about 0.1 kW, about 3%), so the two models agree well; e.g. the sinusoidal model is a smooth curve fitted to a whole day, while the vector model uses the actual direction of the sun at that moment and assumes the factor 0.18 is exact	R1	Any sensible comparison together with a sensible reason.

Answer: \(3.80\) kW, close to \(3.90\) kW

(2)

(h)

A bracket is now fitted that tilts the panel about its lower edge \([AB]\) so that the panel makes an angle of \(45^\circ\) with the horizontal. The upward normal to the tilted panel is in the direction of \(\begin{pmatrix}3\\-4\\5\end{pmatrix}\).

Find the power output at 11:00 with the bracket fitted, and state whether the bracket increases the output at that time.

MULTINotes

Show solution

\(\cos\theta=\frac{(3)(2)+(-4)(-1)+(5)(2)}{\sqrt{50}\times3}=\frac{20}{21.21\ldots}\)	M1
\(\cos\theta=0.9428\ldots\) (\(\theta=19.5^\circ\))	A1
\(P=0.18\times23.58\ldots\times0.9428\ldots=4.00\) kW, greater than \(3.90\) kW, so the output increases	A1	The area of the panel does not change when it is tilted. Allow ECF from (c)(ii) and (e).

Answer: \(4.00\) kW; the output increases

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\overrightarrow{AB}=\begin{pmatrix}4\\3\\0\end{pmatrix}\)	B1
\(\overrightarrow{AD}=\begin{pmatrix}-2.4\\3.2\\2.5\end{pmatrix}\)	B1
\(\overrightarrow{AB}\cdot\overrightarrow{AD}=4(-2.4)+3(3.2)+0(2.5)\)	M1	Correct substitution of their two vectors into the scalar product.
\(=-9.6+9.6+0=0\), so \([AB]\) and \([AD]\) are perpendicular	AG	Both the value \(0\) and the conclusion must be seen.
Part (b)
horizontal distance from \(A\) to \(D\): \(\sqrt{(-2.4)^2+3.2^2}=4\)	M1	The vertical rise \(5.5-3=2.5\) must also be seen or implied.
\(\tan\theta=\frac{2.5}{4}\)	M1	Or \(\sin\theta=\frac{2.5}{\sqrt{22.25}}\), or \(\cos\theta=\frac{4}{\sqrt{22.25}}\).
\(\theta=32.005\ldots^\circ\)	A1	A value to more than 2 significant figures must be seen before rounding.
\(\theta\approx32^\circ\)	AG
Part (c)(i)
\(\begin{pmatrix}(3)(2.5)-(0)(3.2)\\{}(0)(-2.4)-(4)(2.5)\\{}(4)(3.2)-(3)(-2.4)\end{pmatrix}\)	M1	Correct method seen, or at least two correct components.
\(=\begin{pmatrix}7.5\\-10\\20\end{pmatrix}\)	A1
Part (c)(ii)
\(\sqrt{7.5^2+(-10)^2+20^2}=\sqrt{556.25}\)	M1	Allow ECF from (c)(i). Also accept \(\|\overrightarrow{AB}\|\|\overrightarrow{AD}\|=5\times4.716\ldots\)
\(23.6\ \text{m}^2\) (\(23.584\ldots\))	A1
Part (d)
height of the triangle \(5.5-3=2.5\) and base \(2\times4=8\)	B1	The base is twice the horizontal distance found in (b), by symmetry.
\(V=\frac{1}{2}\times8\times2.5\times12\)	M1	Area of the triangular cross-section multiplied by the length, with their values.
\(V=120\ \text{m}^3\)	A1
Part (e)
\(\cos\theta=\frac{(7.5)(2)+(-10)(-1)+(20)(2)}{\sqrt{556.25}\times\sqrt{9}}=\frac{65}{70.75\ldots}\)	M1	Their normal from (c)(i), or any positive multiple of it such as \((3,-4,8)\).
\(\cos\theta=0.9187\ldots\) (\(\theta=23.3^\circ\))	A1
\(P=0.18\times23.58\ldots\times0.9187\ldots=3.90\) kW	A1	Allow ECF from (c)(ii). The exact value is \(3.9\).
Part (f)
\(\mathbf r=\begin{pmatrix}7\\4\\3\end{pmatrix}+\lambda\begin{pmatrix}-7\\-3\\3\end{pmatrix}\)	B1	Accept \(\mathbf r=\ldots\) or \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\ldots\); a different parameter letter is fine.
\(7-7\lambda=0\Rightarrow\lambda=1\)	M1	Setting one component equal to the matching coordinate of \(T\) and solving.
\(y=4-3=1\) and \(z=3+3=6\), so \(T\) lies on the line, on the sun's side of \(B\) (\(\lambda>0\)): the shadow of \(T\) falls on \(B\)	R1	Both remaining components must be checked. Do not award for a conclusion based on one component only.
Part (g)(i)
\(b=\frac{360}{24}=15\)	B1
\(d=1.2\)	B1	The points at \(t=7\) and \(t=19\) are 6 hours (a quarter of a period) either side of the maximum, so they lie on the principal axis.
\(a=4.2-1.2=3\)	B1	Allow ECF from their \(d\).
\(c=7\)	B1	The graph rises through the principal axis at \(t=7\).
Part (g)(ii)
\(P(11)=3\sin(60^\circ)+1.2=3.80\) kW (\(3.798\ldots\))	A1	Allow ECF from (g)(i).
the values are close (they differ by about 0.1 kW, about 3%), so the two models agree well; e.g. the sinusoidal model is a smooth curve fitted to a whole day, while the vector model uses the actual direction of the sun at that moment and assumes the factor 0.18 is exact	R1	Any sensible comparison together with a sensible reason.
Part (h)
\(\cos\theta=\frac{(3)(2)+(-4)(-1)+(5)(2)}{\sqrt{50}\times3}=\frac{20}{21.21\ldots}\)	M1
\(\cos\theta=0.9428\ldots\) (\(\theta=19.5^\circ\))	A1
\(P=0.18\times23.58\ldots\times0.9428\ldots=4.00\) kW, greater than \(3.90\) kW, so the output increases	A1	The area of the panel does not change when it is tilted. Allow ECF from (c)(ii) and (e).

Answers: (a) \(\overrightarrow{AB}=\begin{pmatrix}4\\3\\0\end{pmatrix}\), \(\overrightarrow{AD}=\begin{pmatrix}-2.4\\3.2\\2.5\end{pmatrix}\), scalar product \(0\) · (b) \(32.0^\circ\) · (c)(i) \(\begin{pmatrix}7.5\\-10\\20\end{pmatrix}\) · (c)(ii) \(23.6\ \text{m}^2\) · (d) \(120\ \text{m}^3\) · (e) \(3.90\) kW · (f) Yes: \(T\) is on the line at \(\lambda=1\) · (g)(i) \(a=3,\ b=15,\ c=7,\ d=1.2\) · (g)(ii) \(3.80\) kW, close to \(3.90\) kW · (h) \(4.00\) kW; the output increases

Exam tip (b) In a “show that” with a rounded target, write the unrounded value (here \(32.005\ldots^\circ\)) before the rounded one. Quoting \(32^\circ\) on its own does not show anything.

Common error (e) The angle in the model is measured from the normal, not from the surface of the panel. Using \(\overrightarrow{AB}\) or \(\overrightarrow{AD}\) instead of \(\overrightarrow{AB}\times\overrightarrow{AD}\) gives the wrong angle.

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Two delivery drones: vector kinematics and closest approach

3.5 Vector kinematics3.4 Vectors, scalar and vector products5.2 Optimisation in context3.1 Right-angled trigonometry, bearings and the sine and cosine rulesHL

Two delivery drones fly in straight lines at constant velocity. Relative to an origin \(O\) at ground level, the \(x\)-axis points east, the \(y\)-axis points north and the \(z\)-axis points vertically up. Distances are in metres and the time \(t\) is in seconds, for \(0\le t\le25\).

Drone A leaves a depot at the point \((0,\,0,\,30)\) at \(t=0\) with velocity \(\begin{pmatrix}4\\3\\0\end{pmatrix}\) m s⁻¹. Drone B passes through the point \((80,\,65,\,10)\) at \(t=0\) with velocity \(\begin{pmatrix}-2\\-2\\2\end{pmatrix}\) m s⁻¹.

Figure 1 shows the flight paths seen from above.

Figure 1: the flight paths of the drones for \(0\le t\le25\), seen from above (heights not shown)

(a)

Find the speed of each drone.

ROUTINENotes

Show solution

A: \(\sqrt{4^2+3^2+0^2}=5\) m s⁻¹	B1
B: \(\sqrt{(-2)^2+(-2)^2+2^2}=\sqrt{12}=3.46\) m s⁻¹	B1

Answer: A: \(5\) m s⁻¹; B: \(3.46\) m s⁻¹

(2)

(b)(i)

Find the bearing on which drone A is flying.

ROUTINENotes

Show solution

angle from north \(=\tan^{-1}\left(\frac{4}{3}\right)\)	M1	Using the east and north components with the angle measured from north.
\(053^\circ\) (\(053.1^\circ\))	A1	Accept \(53.1^\circ\). \(036.9^\circ\) is A0.

Answer: \(053^\circ\)

(2)

(b)(ii)

Find the angle between the flight path of drone B and the horizontal.

ROUTINENotes

Show solution

horizontal speed \(\sqrt{(-2)^2+(-2)^2}=\sqrt8\) and \(\tan\alpha=\frac{2}{\sqrt8}\)	M1	Or \(\sin\alpha=\frac{2}{\sqrt{12}}\).
\(\alpha=35.3^\circ\)	A1

Answer: \(35.3^\circ\)

(2)

(c)(i)

Show that the flight paths of the two drones intersect.

PROOFNotes

Show solution

\(\begin{pmatrix}4\lambda\\3\lambda\\30\end{pmatrix}=\begin{pmatrix}80-2\mu\\65-2\mu\\10+2\mu\end{pmatrix}\)	M1	Equating the two paths with different parameters. Using \(t\) for both tests a collision, not an intersection: M0.
\(z\): \(30=10+2\mu\Rightarrow\mu=10\); \(x\): \(4\lambda=80-20\Rightarrow\lambda=15\)	A1	Any two of the equations solved.
\(y\): \(3(15)=45\) and \(65-2(10)=45\), so the third equation is satisfied	R1	The check in the third equation must be shown.
so the flight paths intersect	AG

Answer: \(\lambda=15,\ \mu=10\) satisfy all three equations

(3)

(c)(ii)

Write down the coordinates of the point of intersection.

ROUTINENotes

Show solution

\((60,\,45,\,30)\)

Answer: \((60,\,45,\,30)\)

(1)

(d)

Let \(d\) m be the distance between the drones at time \(t\).

Show that \(d^2=65t^2-1690t+11025\).

PROOFNotes

Show solution

\(\mathbf r_A-\mathbf r_B=\begin{pmatrix}4t\\3t\\30\end{pmatrix}-\begin{pmatrix}80-2t\\65-2t\\10+2t\end{pmatrix}=\begin{pmatrix}6t-80\\5t-65\\20-2t\end{pmatrix}\)	M1	Both position vectors written with the same \(t\) and subtracted (or \(\mathbf r_B-\mathbf r_A\)).
\(d^2=(6t-80)^2+(5t-65)^2+(20-2t)^2\)	M1	Sum of the squares of their components.
\(=36t^2-960t+6400+25t^2-650t+4225+4t^2-80t+400\)	A1	All three brackets expanded correctly.
\(=65t^2-1690t+11025\)	AG

(3)

(e)

For safety, the drones must never be closer than \(10\) m to each other.

By differentiating \(d^2\) with respect to \(t\), find the least distance between the drones. Hence determine whether the safety rule is broken.

MULTINotes

Show solution

\(\frac{d}{dt}\left(d^2\right)=130t-1690=0\)	M1
\(t=13\)	A1	This is a minimum because the second derivative is \(130>0\) (or the coefficient of \(t^2\) is positive).
\(d^2=65(13)^2-1690(13)+11025=40\), so \(d=\sqrt{40}=6.32\) m	A1
\(6.32<10\), so the safety rule is broken	R1	Allow ECF from their least distance.

Answer: \(6.32\) m at \(t=13\) s; the rule is broken

(4)

(f)(i)

On the axes in Figure 2, sketch the graph of \(d\) against \(t\) for \(0\le t\le25\). Label the minimum point and the end points of the graph with their coordinates.

Figure 2: axes for the graph of \(d\) against \(t\)

GRAPHNotes

Show solution

smooth curve that decreases to a minimum and then increases, with a rounded minimum above the \(t\)-axis	A1	Not a V shape, and the curve must not touch the \(t\)-axis.
minimum labelled \((13,\,6.32)\)	A1	Allow ECF from (e).
end points labelled \((0,\,105)\) and \((25,\,97.0)\)	A1	Accept \((25,\,96.9)\) or \((25,\,97)\).

(3)

(f)(ii)

Find the length of time for which the drones are closer than \(10\) m to each other.

MULTINotes

Show solution

\(65t^2-1690t+11025=100\) (or \(d=10\)) gives \(t=12.039\ldots\) and \(t=13.960\ldots\)	(M1)	Solving on the GDC, or the intersections of the graph with \(d=10\).
\(1.92\) s (\(1.9215\ldots\))	A1

Answer: \(1.92\) s

(2)

(g)

Explain why the drones do not collide, even though their flight paths intersect.

INTERPRETNotes

Show solution

drone A reaches \((60,\,45,\,30)\) at \(t=15\), but drone B is there at \(t=10\)	R1	From the parameter values in (c).
they are never at the same point at the same time; the least distance between them is \(6.32\) m, not \(0\)	R1

(2)

(h)

Drone B is rerouted. It still passes through \((80,\,65,\,10)\) at \(t=0\), but its velocity is now \(\begin{pmatrix}-3\\1\\1\end{pmatrix}\) m s⁻¹.

Find the acute angle between the new flight path of drone B and the flight path of drone A.

ROUTINENotes

Show solution

\(\cos\theta=\frac{(4)(-3)+(3)(1)+(0)(1)}{5\times\sqrt{11}}=\frac{-9}{16.58\ldots}\)	M1
\(\theta=122.9^\circ\)	A1	Or \(\cos\theta=\frac{\|-9\|}{5\sqrt{11}}=0.5427\ldots\) directly.
acute angle \(=180^\circ-122.9^\circ=57.1^\circ\)	A1

Answer: \(57.1^\circ\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
A: \(\sqrt{4^2+3^2+0^2}=5\) m s⁻¹	B1
B: \(\sqrt{(-2)^2+(-2)^2+2^2}=\sqrt{12}=3.46\) m s⁻¹	B1
Part (b)(i)
angle from north \(=\tan^{-1}\left(\frac{4}{3}\right)\)	M1	Using the east and north components with the angle measured from north.
\(053^\circ\) (\(053.1^\circ\))	A1	Accept \(53.1^\circ\). \(036.9^\circ\) is A0.
Part (b)(ii)
horizontal speed \(\sqrt{(-2)^2+(-2)^2}=\sqrt8\) and \(\tan\alpha=\frac{2}{\sqrt8}\)	M1	Or \(\sin\alpha=\frac{2}{\sqrt{12}}\).
\(\alpha=35.3^\circ\)	A1
Part (c)(i)
\(\begin{pmatrix}4\lambda\\3\lambda\\30\end{pmatrix}=\begin{pmatrix}80-2\mu\\65-2\mu\\10+2\mu\end{pmatrix}\)	M1	Equating the two paths with different parameters. Using \(t\) for both tests a collision, not an intersection: M0.
\(z\): \(30=10+2\mu\Rightarrow\mu=10\); \(x\): \(4\lambda=80-20\Rightarrow\lambda=15\)	A1	Any two of the equations solved.
\(y\): \(3(15)=45\) and \(65-2(10)=45\), so the third equation is satisfied	R1	The check in the third equation must be shown.
so the flight paths intersect	AG
Part (c)(ii)
\((60,\,45,\,30)\)	A1
Part (d)
\(\mathbf r_A-\mathbf r_B=\begin{pmatrix}4t\\3t\\30\end{pmatrix}-\begin{pmatrix}80-2t\\65-2t\\10+2t\end{pmatrix}=\begin{pmatrix}6t-80\\5t-65\\20-2t\end{pmatrix}\)	M1	Both position vectors written with the same \(t\) and subtracted (or \(\mathbf r_B-\mathbf r_A\)).
\(d^2=(6t-80)^2+(5t-65)^2+(20-2t)^2\)	M1	Sum of the squares of their components.
\(=36t^2-960t+6400+25t^2-650t+4225+4t^2-80t+400\)	A1	All three brackets expanded correctly.
\(=65t^2-1690t+11025\)	AG
Part (e)
\(\frac{d}{dt}\left(d^2\right)=130t-1690=0\)	M1
\(t=13\)	A1	This is a minimum because the second derivative is \(130>0\) (or the coefficient of \(t^2\) is positive).
\(d^2=65(13)^2-1690(13)+11025=40\), so \(d=\sqrt{40}=6.32\) m	A1
\(6.32<10\), so the safety rule is broken	R1	Allow ECF from their least distance.
Part (f)(i)
smooth curve that decreases to a minimum and then increases, with a rounded minimum above the \(t\)-axis	A1	Not a V shape, and the curve must not touch the \(t\)-axis.
minimum labelled \((13,\,6.32)\)	A1	Allow ECF from (e).
end points labelled \((0,\,105)\) and \((25,\,97.0)\)	A1	Accept \((25,\,96.9)\) or \((25,\,97)\).
Part (f)(ii)
\(65t^2-1690t+11025=100\) (or \(d=10\)) gives \(t=12.039\ldots\) and \(t=13.960\ldots\)	(M1)	Solving on the GDC, or the intersections of the graph with \(d=10\).
\(1.92\) s (\(1.9215\ldots\))	A1
Part (g)
drone A reaches \((60,\,45,\,30)\) at \(t=15\), but drone B is there at \(t=10\)	R1	From the parameter values in (c).
they are never at the same point at the same time; the least distance between them is \(6.32\) m, not \(0\)	R1
Part (h)
\(\cos\theta=\frac{(4)(-3)+(3)(1)+(0)(1)}{5\times\sqrt{11}}=\frac{-9}{16.58\ldots}\)	M1
\(\theta=122.9^\circ\)	A1	Or \(\cos\theta=\frac{\|-9\|}{5\sqrt{11}}=0.5427\ldots\) directly.
acute angle \(=180^\circ-122.9^\circ=57.1^\circ\)	A1

Answers: (a) A: \(5\) m s⁻¹; B: \(3.46\) m s⁻¹ · (b)(i) \(053^\circ\) · (b)(ii) \(35.3^\circ\) · (c)(i) \(\lambda=15,\ \mu=10\) satisfy all three equations · (c)(ii) \((60,\,45,\,30)\) · (e) \(6.32\) m at \(t=13\) s; the rule is broken · (f)(ii) \(1.92\) s · (h) \(57.1^\circ\)

Exam tip (e) The time that minimises \(d^2\) also minimises \(d\), and \(d^2\) is a quadratic, so it is much easier to differentiate. Remember to take the square root at the end.

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3.6 Graph theory and adjacency matrices · 3.7 Minimum spanning trees, Chinese postman and travelling salesmanmerged group

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A regional airline: directed graphs and counting routes

3.6 Graph theory and adjacency matrices1.6 Matrices and matrix algebra3.7 Minimum spanning trees, Chinese postman and travelling salesmanHL

A small airline flies between six island airports \(A\), \(B\), \(C\), \(D\), \(E\) and \(F\). Some routes are flown in both directions and some in one direction only.

Flown in both directions: between \(A\) and \(F\); between \(D\) and \(F\).
Flown in one direction only: \(A\) to \(C\); \(B\) to \(A\); \(C\) to \(B\); \(C\) to \(D\); \(D\) to \(E\); \(E\) to \(C\).

A journey is a sequence of scheduled flights, each one leaving from the airport where the previous one landed. Figure 1 shows the positions of the airports.

Figure 1: the positions of the six airports (not to scale)

(a)

On Figure 1, draw a directed graph to represent the scheduled routes.

GRAPHNotes

Show solution

arrows in both directions (or a double-headed edge) between \(A\) and \(F\) and between \(D\) and \(F\)	A1
single arrows \(A\to C\), \(B\to A\), \(C\to B\), \(C\to D\), \(D\to E\), \(E\to C\), and no other edges	A1	All six directions must be correct.

(2)

(b)

Write down the adjacency matrix \(\mathbf M\) of the directed graph, with the vertices in the order \(A,B,C,D,E,F\) and each row showing the flights that leave an airport.

ROUTINENotes

Show solution

at least four rows of \(\mathbf M\) correct	A1	Award A1A0 for the transpose of the correct matrix.
\(\mathbf M=\begin{pmatrix}0&0&1&0&0&1\\1&0&0&0&0&0\\0&1&0&1&0&0\\0&0&0&0&1&1\\0&0&1&0&0&0\\1&0&0&1&0&0\end{pmatrix}\)	A1	Fully correct matrix, rows = departures.

Answer: \(\mathbf M=\begin{pmatrix}0&0&1&0&0&1\\1&0&0&0&0&0\\0&1&0&1&0&0\\0&0&0&0&1&1\\0&0&1&0&0&0\\1&0&0&1&0&0\end{pmatrix}\)

(2)

(c)

Find \(\mathbf M^2\). Hence show that there are exactly two journeys consisting of two flights from \(A\) to \(D\), and list them.

ROUTINENotes

Show solution

\(\mathbf M^2=\begin{pmatrix}1&1&0&2&0&0\\0&0&1&0&0&1\\1&0&0&0&1&1\\1&0&1&1&0&0\\0&1&0&1&0&0\\0&0&1&0&1&2\end{pmatrix}\)	(M1)	Correct matrix seen, or at least row \(A\) correct.
the entry in row \(A\), column \(D\) is \(2\)	A1	Must state which entry is read.
\(A\to C\to D\) and \(A\to F\to D\)	A1
so there are exactly two such journeys	AG

Answer: \(A\to C\to D\) and \(A\to F\to D\)

(3)

(d)

Let \(\mathbf S=\mathbf M+\mathbf M^2+\mathbf M^3\). Find \(\mathbf S\). State the position of the only zero entry of \(\mathbf S\) and interpret it in context.

INTERPRETNotes

Show solution

\(\mathbf S=\begin{pmatrix}2&1&2&2&2&4\\2&1&1&2&0&1\\2&1&2&2&1&2\\1&1&2&2&2&3\\1&1&1&1&1&1\\3&1&2&4&1&2\end{pmatrix}\)	(M1)	Found on the GDC; a correct matrix implies the mark.
zero in row \(B\), column \(E\)	A1
there is no journey from \(B\) to \(E\) using three flights or fewer	R1	Must refer to the airports and to “three or fewer” (or “at most three”) flights.

Answer: Row \(B\), column \(E\): no journey from \(B\) to \(E\) with three or fewer flights

(3)

(e)

A network of routes is strongly connected if it is possible to travel from every airport to every other airport by a journey.

Explain why the zero entry in \(\mathbf S\) does not show that the network is not strongly connected, and write down a journey from \(B\) to \(E\).

CONCEPTNotes

Show solution

\(\mathbf S\) only counts journeys of at most three flights; a longer journey may still exist	R1
\(B\to A\to C\to D\to E\) (or \(B\to A\to F\to D\to E\))	A1	A journey of four flights.

Answer: e.g. \(B\to A\to C\to D\to E\)

(2)

(f)

The route from \(D\) to \(E\) is suspended.

State the change to \(\mathbf M\), and explain why the network is no longer strongly connected.

INTERPRETNotes

Show solution

the entry in row \(D\), column \(E\) changes from \(1\) to \(0\)	A1
column \(E\) of the new matrix is all zeros, so the in-degree of \(E\) is \(0\)	R1
no flight lands at \(E\), so no journey from another airport can reach \(E\)	R1	Award for a correct conclusion in context, e.g. “passengers cannot fly to \(E\) at all”.

Answer: Row \(D\), column \(E\) becomes \(0\); \(E\) can no longer be reached

(3)

(g)

The airline also operates a charter plane, which can fly directly between any two of the airports, in either direction. The flight times, in minutes, are shown in the table.

	A	B	C	D	E	F
A	–	25	45	50	55	30
B	25	–	30	65	60	45
C	45	30	–	45	40	35
D	50	65	45	–	30	35
E	55	60	40	30	–	50
F	30	45	35	35	50	–

An engineer, using the charter plane, must visit every airport exactly once, starting and finishing at \(A\).

The directed graph of the scheduled routes contains exactly one Hamiltonian cycle that starts at \(A\). Write down this cycle, and use the table to find its total flight time.

ROUTINENotes

Show solution

\(A\to F\to D\to E\to C\to B\to A\)	B1
\(30+35+30+40+30+25=190\) minutes	B1	Allow ECF from their cycle.

Answer: \(AFDECBA\), \(190\) minutes

(2)

(h)

Use the nearest-neighbour algorithm, starting at \(A\), to find another upper bound for the engineer's total flight time.

ROUTINENotes

Show solution

\(A\to B\to C\to F\to D\to E\to A\)	M1	The order must be seen; from \(C\) the nearest unvisited airport is \(F\) (35).
\(25+30+35+35+30+55=210\) minutes	A1

Answer: \(ABCFDEA\), \(210\) minutes

(2)

(i)

By deleting vertex \(A\), find a lower bound for the engineer's total flight time.

MULTINotes

Show solution

minimum spanning tree of \(B,C,D,E,F\): \(BC\ 30\), \(DE\ 30\), \(CF\ 35\), \(DF\ 35\); weight \(130\)	M1	Kruskal's or Prim's algorithm on the five remaining airports; \(CE\ 40\) is rejected (cycle \(C\,F\,D\,E\,C\)).
two shortest edges at \(A\): \(AB\ 25\) and \(AF\ 30\)	M1
lower bound \(=130+25+30=185\) minutes	A1

Answer: \(185\) minutes

(3)

(j)

Write down the best interval for the shortest possible total flight time that can be found from your answers to parts (g), (h) and (i). Justify your choice of upper bound.

INTERPRETNotes

Show solution

\(185\le\text{shortest time}\le190\) minutes	A1	Allow ECF from (g), (h) and (i).
both \(190\) and \(210\) are times of actual tours, so the shortest time is at most the smaller one, \(190\)	R1

Answer: \(185\le\text{time}\le190\) minutes

(2)

Show full solution

Working	Mark	Notes
Part (a)
arrows in both directions (or a double-headed edge) between \(A\) and \(F\) and between \(D\) and \(F\)	A1
single arrows \(A\to C\), \(B\to A\), \(C\to B\), \(C\to D\), \(D\to E\), \(E\to C\), and no other edges	A1	All six directions must be correct.
Part (b)
at least four rows of \(\mathbf M\) correct	A1	Award A1A0 for the transpose of the correct matrix.
\(\mathbf M=\begin{pmatrix}0&0&1&0&0&1\\1&0&0&0&0&0\\0&1&0&1&0&0\\0&0&0&0&1&1\\0&0&1&0&0&0\\1&0&0&1&0&0\end{pmatrix}\)	A1	Fully correct matrix, rows = departures.
Part (c)
\(\mathbf M^2=\begin{pmatrix}1&1&0&2&0&0\\0&0&1&0&0&1\\1&0&0&0&1&1\\1&0&1&1&0&0\\0&1&0&1&0&0\\0&0&1&0&1&2\end{pmatrix}\)	(M1)	Correct matrix seen, or at least row \(A\) correct.
the entry in row \(A\), column \(D\) is \(2\)	A1	Must state which entry is read.
\(A\to C\to D\) and \(A\to F\to D\)	A1
so there are exactly two such journeys	AG
Part (d)
\(\mathbf S=\begin{pmatrix}2&1&2&2&2&4\\2&1&1&2&0&1\\2&1&2&2&1&2\\1&1&2&2&2&3\\1&1&1&1&1&1\\3&1&2&4&1&2\end{pmatrix}\)	(M1)	Found on the GDC; a correct matrix implies the mark.
zero in row \(B\), column \(E\)	A1
there is no journey from \(B\) to \(E\) using three flights or fewer	R1	Must refer to the airports and to “three or fewer” (or “at most three”) flights.
Part (e)
\(\mathbf S\) only counts journeys of at most three flights; a longer journey may still exist	R1
\(B\to A\to C\to D\to E\) (or \(B\to A\to F\to D\to E\))	A1	A journey of four flights.
Part (f)
the entry in row \(D\), column \(E\) changes from \(1\) to \(0\)	A1
column \(E\) of the new matrix is all zeros, so the in-degree of \(E\) is \(0\)	R1
no flight lands at \(E\), so no journey from another airport can reach \(E\)	R1	Award for a correct conclusion in context, e.g. “passengers cannot fly to \(E\) at all”.
Part (g)
\(A\to F\to D\to E\to C\to B\to A\)	B1
\(30+35+30+40+30+25=190\) minutes	B1	Allow ECF from their cycle.
Part (h)
\(A\to B\to C\to F\to D\to E\to A\)	M1	The order must be seen; from \(C\) the nearest unvisited airport is \(F\) (35).
\(25+30+35+35+30+55=210\) minutes	A1
Part (i)
minimum spanning tree of \(B,C,D,E,F\): \(BC\ 30\), \(DE\ 30\), \(CF\ 35\), \(DF\ 35\); weight \(130\)	M1	Kruskal's or Prim's algorithm on the five remaining airports; \(CE\ 40\) is rejected (cycle \(C\,F\,D\,E\,C\)).
two shortest edges at \(A\): \(AB\ 25\) and \(AF\ 30\)	M1
lower bound \(=130+25+30=185\) minutes	A1
Part (j)
\(185\le\text{shortest time}\le190\) minutes	A1	Allow ECF from (g), (h) and (i).
both \(190\) and \(210\) are times of actual tours, so the shortest time is at most the smaller one, \(190\)	R1

Answers: (b) \(\mathbf M=\begin{pmatrix}0&0&1&0&0&1\\1&0&0&0&0&0\\0&1&0&1&0&0\\0&0&0&0&1&1\\0&0&1&0&0&0\\1&0&0&1&0&0\end{pmatrix}\) · (c) \(A\to C\to D\) and \(A\to F\to D\) · (d) Row \(B\), column \(E\): no journey from \(B\) to \(E\) with three or fewer flights · (e) e.g. \(B\to A\to C\to D\to E\) · (f) Row \(D\), column \(E\) becomes \(0\); \(E\) can no longer be reached · (g) \(AFDECBA\), \(190\) minutes · (h) \(ABCFDEA\), \(210\) minutes · (i) \(185\) minutes · (j) \(185\le\text{time}\le190\) minutes

Common error (c) \(\mathbf M^2\) means \(\mathbf M\times\mathbf M\) as a matrix product. Squaring each entry of \(\mathbf M\) leaves the matrix unchanged and counts nothing.

Exam tip (j) Every complete tour gives an upper bound, so when you have more than one, keep the smallest. The nearest-neighbour tour is not always the best one you have found.

More 3.6 questions in the question bank

+ Add to test✓ In your testCC-AI-36-37-225 marks~30 min

A university campus: laying fibre cable and planning a security patrol

3.7 Minimum spanning trees, Chinese postman and travelling salesman3.6 Graph theory and adjacency matrices1.6 Matrices and matrix algebra2.1 Linear models and piecewise modelsHL

A university campus has seven buildings: Admin \(A\), Cafeteria \(C\), Engineering \(E\), Gym \(G\), Halls of residence \(H\), Library \(L\) and Science \(S\). Figure 1 shows the footpaths between the buildings and their lengths in metres. The same information is given in the table, where a dash means that there is no direct footpath.

	A	C	E	G	H	L	S
A	–	130	–	–	–	160	–
C	130	–	–	–	170	100	150
E	–	–	–	140	–	220	120
G	–	–	140	–	210	–	190
H	–	170	–	210	–	–	–
L	160	100	220	–	–	–	–
S	–	150	120	190	–	–	–

Figure 1: footpaths between the seven buildings, with lengths in metres (not to scale)

(a)

Explain why the network of footpaths is not a tree, and write down the number of edges in a spanning tree of this network.

CONCEPTNotes

Show solution

it contains a cycle, e.g. \(A\,C\,L\,A\) (or it has \(10\) edges, more than \(7-1=6\))	R1
\(6\) edges	B1

Answer: It contains a cycle; \(6\) edges

(2)

(b)

Fibre-optic cable is to be laid along some of the footpaths so that all seven buildings are connected.

Use Kruskal's algorithm to show that the minimum total length of cable is \(810\) m. Show the order in which you consider the edges and state any edge that you reject.

ROUTINENotes

Show solution

edges in increasing order: \(LC\ 100\), \(SE\ 120\), \(AC\ 130\), \(EG\ 140\), \(CS\ 150\), \(AL\ 160\), \(CH\ 170\), …	M1	At least the first seven edges in the correct order.
accept \(LC\), \(SE\), \(AC\), \(EG\), \(CS\)	A1
reject \(AL\), because it forms the cycle \(A\,C\,L\,A\)	A1	The reason must be given.
accept \(CH\); this gives \(6\) edges, so stop	A1
\(100+120+130+140+150+170=810\) m	AG

Answer: \(LC, SE, AC, EG, CS, CH\); \(810\) m

(4)

(c)

Use Prim's algorithm on the table, starting at \(H\), to confirm your answer to part (b). List the edges in the order in which they are added.

ROUTINENotes

Show solution

first edge \(HC\ 170\)	M1	The cheapest edge from \(H\).
then \(CL\ 100\), \(CA\ 130\), \(CS\ 150\)	A1
then \(SE\ 120\), \(EG\ 140\): the same six edges, total \(810\) m	A1

Answer: \(HC, CL, CA, CS, SE, EG\)

(3)

(d)(i)

Two contractors quote for laying the cable. For \(x\) metres of cable, the cost in euros is \(40x+1500\) for contractor P and \(32x+7500\) for contractor Q. Both costs are given by the matrix product \(\begin{pmatrix}x&1\end{pmatrix}\begin{pmatrix}40&32\\1500&7500\end{pmatrix}\).

Find this product for the cable in part (b), and state which contractor is cheaper.

ROUTINENotes

Show solution

\(\begin{pmatrix}810&1\end{pmatrix}\begin{pmatrix}40&32\\1500&7500\end{pmatrix}=\begin{pmatrix}33900&33420\end{pmatrix}\)	A1
contractor Q is cheaper (by €480)	A1	Allow ECF from their product.

Answer: \(\begin{pmatrix}33900&33420\end{pmatrix}\); Q is cheaper

(2)

(d)(ii)

Find the length of cable for which the two quotes are equal, and explain what this means for the university.

MODELNotes

Show solution

\(40x+1500=32x+7500\)	M1	Equating the two costs, or the intersection of the two lines on the GDC.
\(x=750\) m	A1
for more than \(750\) m of cable Q is cheaper, because its rate per metre is lower, even though its fixed fee is higher; \(810>750\), so Q should be chosen	R1

Answer: \(750\) m

(3)

(e)

Each night a security guard must walk along every footpath at least once, starting and finishing at \(H\).

Explain why the guard's route must use some footpaths more than once.

INTERPRETNotes

Show solution

\(L\), \(S\), \(E\) and \(G\) have odd degree (\(3\))	B1
a closed route using every footpath exactly once (an Eulerian circuit) needs every vertex to have even degree	R1	Award for an equivalent argument: each visit to a building uses one footpath in and one footpath out.

(2)

(f)

Find the length of the shortest route for the guard.

MULTINotes

Show solution

the three pairings of \(L,S,E,G\) considered, using shortest paths	M1
\(LS+EG=250+140=390\); \(LE+SG=220+190=410\); \(LG+SE=360+120=480\)	A1	Shortest paths: \(L\) to \(S\) is \(250\) via \(C\); \(L\) to \(G\) is \(360\) via \(E\).
total length of all footpaths \(=1590\) m, added to their least pairing	M1
\(1590+390=1980\) m	A1

Answer: \(1980\) m

(4)

(g)

Using Figure 1, state the footpaths that the guard walks twice on this route.

GRAPHNotes

Show solution

\(LC\) and \(CS\)	A1
\(EG\)	A1

Answer: \(LC\), \(CS\) and \(EG\)

(2)

(h)

A new footpath of length \(230\) m is opened between \(L\) and \(S\). The guard must walk along this footpath as well.

Determine the effect of the new footpath on the length of the guard's shortest route.

MODELNotes

Show solution

\(L\) and \(S\) now have even degree, so only \(E\) and \(G\) are odd	R1
new length \(=1590+230+140\)	M1	Their total plus the new footpath plus the shortest path from \(E\) to \(G\).
\(=1960\) m, which is \(20\) m shorter than before	A1	Allow ECF from (f). The route is shorter although there is an extra \(230\) m of footpath to walk.

Answer: \(1960\) m: \(20\) m shorter

(3)

Show full solution

Working	Mark	Notes
Part (a)
it contains a cycle, e.g. \(A\,C\,L\,A\) (or it has \(10\) edges, more than \(7-1=6\))	R1
\(6\) edges	B1
Part (b)
edges in increasing order: \(LC\ 100\), \(SE\ 120\), \(AC\ 130\), \(EG\ 140\), \(CS\ 150\), \(AL\ 160\), \(CH\ 170\), …	M1	At least the first seven edges in the correct order.
accept \(LC\), \(SE\), \(AC\), \(EG\), \(CS\)	A1
reject \(AL\), because it forms the cycle \(A\,C\,L\,A\)	A1	The reason must be given.
accept \(CH\); this gives \(6\) edges, so stop	A1
\(100+120+130+140+150+170=810\) m	AG
Part (c)
first edge \(HC\ 170\)	M1	The cheapest edge from \(H\).
then \(CL\ 100\), \(CA\ 130\), \(CS\ 150\)	A1
then \(SE\ 120\), \(EG\ 140\): the same six edges, total \(810\) m	A1
Part (d)(i)
\(\begin{pmatrix}810&1\end{pmatrix}\begin{pmatrix}40&32\\1500&7500\end{pmatrix}=\begin{pmatrix}33900&33420\end{pmatrix}\)	A1
contractor Q is cheaper (by €480)	A1	Allow ECF from their product.
Part (d)(ii)
\(40x+1500=32x+7500\)	M1	Equating the two costs, or the intersection of the two lines on the GDC.
\(x=750\) m	A1
for more than \(750\) m of cable Q is cheaper, because its rate per metre is lower, even though its fixed fee is higher; \(810>750\), so Q should be chosen	R1
Part (e)
\(L\), \(S\), \(E\) and \(G\) have odd degree (\(3\))	B1
a closed route using every footpath exactly once (an Eulerian circuit) needs every vertex to have even degree	R1	Award for an equivalent argument: each visit to a building uses one footpath in and one footpath out.
Part (f)
the three pairings of \(L,S,E,G\) considered, using shortest paths	M1
\(LS+EG=250+140=390\); \(LE+SG=220+190=410\); \(LG+SE=360+120=480\)	A1	Shortest paths: \(L\) to \(S\) is \(250\) via \(C\); \(L\) to \(G\) is \(360\) via \(E\).
total length of all footpaths \(=1590\) m, added to their least pairing	M1
\(1590+390=1980\) m	A1
Part (g)
\(LC\) and \(CS\)	A1
\(EG\)	A1
Part (h)
\(L\) and \(S\) now have even degree, so only \(E\) and \(G\) are odd	R1
new length \(=1590+230+140\)	M1	Their total plus the new footpath plus the shortest path from \(E\) to \(G\).
\(=1960\) m, which is \(20\) m shorter than before	A1	Allow ECF from (f). The route is shorter although there is an extra \(230\) m of footpath to walk.

Answers: (a) It contains a cycle; \(6\) edges · (b) \(LC, SE, AC, EG, CS, CH\); \(810\) m · (c) \(HC, CL, CA, CS, SE, EG\) · (d)(i) \(\begin{pmatrix}33900&33420\end{pmatrix}\); Q is cheaper · (d)(ii) \(750\) m · (f) \(1980\) m · (g) \(LC\), \(CS\) and \(EG\) · (h) \(1960\) m: \(20\) m shorter

Exam tip (b) With Kruskal's algorithm, write every edge you consider, mark it accepted or rejected, and name the cycle that a rejected edge would make. Stop as soon as you have \(n-1\) edges.

Common error (f) There is no direct footpath from \(L\) to \(S\). The shortest way between them is \(L\,C\,S\) (\(250\) m), and that whole path is what gets repeated.

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3.8 Matrix transformations

+ Add to test✓ In your testCC-AI-38-128 marks~35 min

Moving a video game spaceship: rotations, reflections and determinants

3.8 Matrix transformations1.6 Matrices and matrix algebra3.1 Right-angled trigonometry, bearings and the sine and cosine rules1.5 Complex numbers in Cartesian, polar and exponential formHL

In a two-dimensional video game, the outline of a spaceship is stored as the quadrilateral \(PQRS\) with vertices \(P(6,0)\), \(Q(-3,3)\), \(R(-1,0)\) and \(S(-3,-3)\), measured in game units from the centre of the screen, which is the origin. The outline is shown in Figure 1. The area of \(PQRS\) is 21 square units.

The game engine moves the spaceship by multiplying the position vector of every vertex by a \(2\times2\) matrix. All transformations are about the origin.

Figure 1: the spaceship outline \(PQRS\)

(a)(i)

When the player presses one key, the engine applies the matrix \(\mathbf{M}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\).

Find the images of \(P\) and \(Q\) under \(\mathbf{M}\).

ROUTINENotes

Show solution

\(P'(0,-6)\)	A1	Accept column vectors.
\(Q'(3,3)\)	A1

Answer: \(P'(0,-6)\), \(Q'(3,3)\)

(2)

(a)(ii)

Describe fully the single geometric transformation represented by \(\mathbf{M}\).

CONCEPTNotes

Show solution

rotation about the origin	A1	Accept centre \((0,0)\).
through \(90^\circ\) clockwise	A1	Accept \(270^\circ\) anticlockwise or \(-90^\circ\). Do not award A1A1 for a reflection or a stretch.

Answer: rotation of \(90^\circ\) clockwise about the origin

(2)

(b)

The outline is symmetrical about the \(x\)-axis. Find the size of angle \(Q\hat{P}S\) at the nose of the spaceship.

ROUTINENotes

Show solution

right-angled triangle with legs 3 and 9, e.g. \(\tan\theta=\frac{3}{9}\) for half the angle	(M1)	Any valid trigonometric method, e.g. the cosine rule with \(PQ=PS=\sqrt{90}\) and \(QS=6\).
\(Q\hat{P}S=2\times18.43\ldots^\circ=36.9^\circ\)	A1	\(36.8698\ldots^\circ\). Answer \(18.4^\circ\) (half the angle only) scores M1A0.

Answer: \(36.9^\circ\)

(2)

(c)

A second move rotates the spaceship \(30^\circ\) anticlockwise about the origin, using the matrix \(\mathbf{R}\). A third move reflects it in the line \(y=x\), using the matrix \(\mathbf{F}\).

Write down \(\mathbf{R}\) and \(\mathbf{F}\), giving the entries of \(\mathbf{R}\) as exact values.

ROUTINENotes

Show solution

\(\mathbf{R}=\begin{pmatrix}\frac{\sqrt3}{2}&-\frac12\\\frac12&\frac{\sqrt3}{2}\end{pmatrix}\)	A1	Accept \(\cos30^\circ\) and \(\sin30^\circ\) entries evaluated exactly. Decimals only (0.866) score A0.
\(\mathbf{F}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\)	A1

Answer: \(\mathbf{R}=\begin{pmatrix}\frac{\sqrt3}{2}&-\frac12\\\frac12&\frac{\sqrt3}{2}\end{pmatrix}\), \(\mathbf{F}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\)

(2)

(d)(i)

In a combined move the spaceship is rotated using \(\mathbf{R}\) and then reflected using \(\mathbf{F}\). This combined move is represented by the single matrix \(\mathbf{C}\).

Find \(\mathbf{C}\), giving its entries as exact values.

ALGEBRANotes

Show solution

\(\mathbf{C}=\mathbf{FR}\)	(M1)	Correct order seen or implied by a correct answer. \(\mathbf{RF}\) scores M0A0, but see (d)(ii).
\(\mathbf{C}=\begin{pmatrix}\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{pmatrix}\)	A1	Allow ECF from their matrices in (c).

Answer: \(\mathbf{C}=\begin{pmatrix}\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{pmatrix}\)

(2)

(d)(ii)

Find the matrix for the reflection using \(\mathbf{F}\) followed by the rotation using \(\mathbf{R}\). Hence state whether the order of the two moves matters.

ALGEBRANotes

Show solution

\(\mathbf{RF}=\begin{pmatrix}-\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac12\end{pmatrix}\)	A1	Allow ECF from (c).
\(\mathbf{RF}\neq\mathbf{FR}\), so the order matters (the spaceship ends in a different position)	R1	Must compare with their \(\mathbf{C}\). A correct example, e.g. \(P\) goes to \((3,3\sqrt3)\) under \(\mathbf{FR}\) but to \((-3,3\sqrt3)\) under \(\mathbf{RF}\), is also acceptable.

Answer: \(\begin{pmatrix}-\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac12\end{pmatrix}\); the order matters

(2)

(e)

Find the coordinates of the images \(P'\), \(Q'\), \(R'\) and \(S'\) of the four vertices under \(\mathbf{C}\). On Figure 2, draw and label the image \(P'Q'R'S'\).

Figure 2: grid for the image of the spaceship (the original outline is dashed)

GRAPHNotes

Show solution

\(\mathbf{C}\) multiplied by at least two position vectors correctly	(M1)
\(P'(3,\ 5.20)\), \(Q'(1.10,\ -4.10)\), \(R'(-0.5,\ -0.866)\), \(S'(-4.10,\ -1.10)\)	A1	Exact forms \(P'(3,3\sqrt3)\), \(Q'\left(\frac{3\sqrt3-3}{2},-\frac{3\sqrt3+3}{2}\right)\), \(R'\left(-\frac12,-\frac{\sqrt3}{2}\right)\), \(S'\left(-\frac{3\sqrt3+3}{2},\frac{3-3\sqrt3}{2}\right)\). Allow ECF from (d)(i).
quadrilateral drawn with vertices within half a small square of their points, joined in order and labelled	A1	FT their coordinates.

Answer: \(P'(3,5.20)\), \(Q'(1.10,-4.10)\), \(R'(-0.5,-0.866)\), \(S'(-4.10,-1.10)\)

(3)

(f)

Show that \(\det\mathbf{C}=-1\).

PROOFNotes

Show solution

\(\det\mathbf{C}=\frac12\times\left(-\frac12\right)-\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}\)	M1	Correct substitution into \(ad-bc\).
\(=-\frac14-\frac34\)	A1	Accept \(\det\mathbf{F}\times\det\mathbf{R}=(-1)(1)\) with both determinants shown.
\(=-1\)	AG

Answer: \(-1\)

(2)

(g)

Interpret the value of \(\det\mathbf{C}\) for the image of the spaceship. Refer to your diagram in Figure 2.

INTERPRETNotes

Show solution

\(\|\det\mathbf{C}\|=1\), so the area of the image is unchanged at 21 square units	R1	Must mention area or size.
the negative sign means the spaceship is flipped over (a mirror image): \(P\to Q\to R\to S\) runs anticlockwise but \(P'\to Q'\to R'\to S'\) runs clockwise	R1	Accept "orientation reversed" or "reflected". Do not accept "negative area".

(2)

(h)

After the combined move \(\mathbf{C}\), the laser cannon on the spaceship is at the point \((1,\sqrt3)\).

Find \(\mathbf{C}^{-1}\) and hence find the position of the laser cannon before the move.

ALGEBRANotes

Show solution

\(\mathbf{C}^{-1}=\frac{1}{-1}\begin{pmatrix}-\frac12&-\frac{\sqrt3}{2}\\-\frac{\sqrt3}{2}&\frac12\end{pmatrix}=\begin{pmatrix}\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{pmatrix}\)	B1	Accept from the GDC. \(\mathbf{C}^{-1}=\mathbf{C}\): a reflection is undone by itself.
\(\mathbf{C}^{-1}\begin{pmatrix}1\\\sqrt3\end{pmatrix}=\begin{pmatrix}\frac12+\frac32\\\frac{\sqrt3}{2}-\frac{\sqrt3}{2}\end{pmatrix}\)	(M1)	Inverse on the left of the column vector.
\((2,\ 0)\)	A1	Allow ECF from their \(\mathbf{C}\).

Answer: \(\mathbf{C}^{-1}=\mathbf{C}\); cannon was at \((2,0)\)

(3)

(i)

A programmer suggests modelling the rotation through \(30^\circ\) anticlockwise with complex numbers instead. The vertex \(Q\) is represented by \(w=-3+3i\) on an Argand diagram.

By writing \(w\) in exponential form, find \(w\,e^{i\pi/6}\) in the form \(a+bi\), and verify that it represents the image of \(Q\) under \(\mathbf{R}\).

MULTINotes

Show solution

\(w=3\sqrt2\,e^{3\pi i/4}\), so \(w\,e^{i\pi/6}=3\sqrt2\,e^{i\left(\frac{3\pi}{4}+\frac{\pi}{6}\right)}=3\sqrt2\,e^{11\pi i/12}\)	M1	Correct modulus and argument of \(w\) and the arguments added. Accept \(4.24e^{2.36i}\times e^{0.524i}=4.24e^{2.88i}\).
\(=-4.10+1.10i\)	A1	Exact: \(-\frac{3\sqrt3+3}{2}+\frac{3\sqrt3-3}{2}i\). Expanding \((-3+3i)\left(\frac{\sqrt3}{2}+\frac12i\right)\) directly, without the exponential form, does not use the method asked for: M0, but award 1 mark for the correct value and the R1 below (at most 2 marks).
\(\mathbf{R}\begin{pmatrix}-3\\3\end{pmatrix}=\begin{pmatrix}-4.10\\1.10\end{pmatrix}\), the same point	R1	Allow ECF from their \(\mathbf{R}\).

Answer: \(-4.10+1.10i\), the same as \(\mathbf{R}Q\)

(3)

(j)

In a power-up, the engine applies \(\mathbf{C}\) and then an enlargement with centre the origin and scale factor 2.

Find the single matrix for this power-up and hence find the area of the image of the spaceship.

ROUTINENotes

Show solution

\(\begin{pmatrix}2&0\\0&2\end{pmatrix}\mathbf{C}=\begin{pmatrix}1&\sqrt3\\\sqrt3&-1\end{pmatrix}\)	B1	Allow ECF from (d)(i).
\(\det=1\times(-1)-\sqrt3\times\sqrt3=-4\), so the area scale factor is \(\|-4\|=4\)	M1	Accept \(2^2\times1\).
area \(=4\times21=84\) square units	A1	\(42\) (scale factor 2 used for area) scores M0A0.

Answer: \(\begin{pmatrix}1&\sqrt3\\\sqrt3&-1\end{pmatrix}\); 84 square units

(3)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(P'(0,-6)\)	A1	Accept column vectors.
\(Q'(3,3)\)	A1
Part (a)(ii)
rotation about the origin	A1	Accept centre \((0,0)\).
through \(90^\circ\) clockwise	A1	Accept \(270^\circ\) anticlockwise or \(-90^\circ\). Do not award A1A1 for a reflection or a stretch.
Part (b)
right-angled triangle with legs 3 and 9, e.g. \(\tan\theta=\frac{3}{9}\) for half the angle	(M1)	Any valid trigonometric method, e.g. the cosine rule with \(PQ=PS=\sqrt{90}\) and \(QS=6\).
\(Q\hat{P}S=2\times18.43\ldots^\circ=36.9^\circ\)	A1	\(36.8698\ldots^\circ\). Answer \(18.4^\circ\) (half the angle only) scores M1A0.
Part (c)
\(\mathbf{R}=\begin{pmatrix}\frac{\sqrt3}{2}&-\frac12\\\frac12&\frac{\sqrt3}{2}\end{pmatrix}\)	A1	Accept \(\cos30^\circ\) and \(\sin30^\circ\) entries evaluated exactly. Decimals only (0.866) score A0.
\(\mathbf{F}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\)	A1
Part (d)(i)
\(\mathbf{C}=\mathbf{FR}\)	(M1)	Correct order seen or implied by a correct answer. \(\mathbf{RF}\) scores M0A0, but see (d)(ii).
\(\mathbf{C}=\begin{pmatrix}\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{pmatrix}\)	A1	Allow ECF from their matrices in (c).
Part (d)(ii)
\(\mathbf{RF}=\begin{pmatrix}-\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac12\end{pmatrix}\)	A1	Allow ECF from (c).
\(\mathbf{RF}\neq\mathbf{FR}\), so the order matters (the spaceship ends in a different position)	R1	Must compare with their \(\mathbf{C}\). A correct example, e.g. \(P\) goes to \((3,3\sqrt3)\) under \(\mathbf{FR}\) but to \((-3,3\sqrt3)\) under \(\mathbf{RF}\), is also acceptable.
Part (e)
\(\mathbf{C}\) multiplied by at least two position vectors correctly	(M1)
\(P'(3,\ 5.20)\), \(Q'(1.10,\ -4.10)\), \(R'(-0.5,\ -0.866)\), \(S'(-4.10,\ -1.10)\)	A1	Exact forms \(P'(3,3\sqrt3)\), \(Q'\left(\frac{3\sqrt3-3}{2},-\frac{3\sqrt3+3}{2}\right)\), \(R'\left(-\frac12,-\frac{\sqrt3}{2}\right)\), \(S'\left(-\frac{3\sqrt3+3}{2},\frac{3-3\sqrt3}{2}\right)\). Allow ECF from (d)(i).
quadrilateral drawn with vertices within half a small square of their points, joined in order and labelled	A1	FT their coordinates.
Part (f)
\(\det\mathbf{C}=\frac12\times\left(-\frac12\right)-\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}\)	M1	Correct substitution into \(ad-bc\).
\(=-\frac14-\frac34\)	A1	Accept \(\det\mathbf{F}\times\det\mathbf{R}=(-1)(1)\) with both determinants shown.
\(=-1\)	AG
Part (g)
\(\|\det\mathbf{C}\|=1\), so the area of the image is unchanged at 21 square units	R1	Must mention area or size.
the negative sign means the spaceship is flipped over (a mirror image): \(P\to Q\to R\to S\) runs anticlockwise but \(P'\to Q'\to R'\to S'\) runs clockwise	R1	Accept "orientation reversed" or "reflected". Do not accept "negative area".
Part (h)
\(\mathbf{C}^{-1}=\frac{1}{-1}\begin{pmatrix}-\frac12&-\frac{\sqrt3}{2}\\-\frac{\sqrt3}{2}&\frac12\end{pmatrix}=\begin{pmatrix}\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{pmatrix}\)	B1	Accept from the GDC. \(\mathbf{C}^{-1}=\mathbf{C}\): a reflection is undone by itself.
\(\mathbf{C}^{-1}\begin{pmatrix}1\\\sqrt3\end{pmatrix}=\begin{pmatrix}\frac12+\frac32\\\frac{\sqrt3}{2}-\frac{\sqrt3}{2}\end{pmatrix}\)	(M1)	Inverse on the left of the column vector.
\((2,\ 0)\)	A1	Allow ECF from their \(\mathbf{C}\).
Part (i)
\(w=3\sqrt2\,e^{3\pi i/4}\), so \(w\,e^{i\pi/6}=3\sqrt2\,e^{i\left(\frac{3\pi}{4}+\frac{\pi}{6}\right)}=3\sqrt2\,e^{11\pi i/12}\)	M1	Correct modulus and argument of \(w\) and the arguments added. Accept \(4.24e^{2.36i}\times e^{0.524i}=4.24e^{2.88i}\).
\(=-4.10+1.10i\)	A1	Exact: \(-\frac{3\sqrt3+3}{2}+\frac{3\sqrt3-3}{2}i\). Expanding \((-3+3i)\left(\frac{\sqrt3}{2}+\frac12i\right)\) directly, without the exponential form, does not use the method asked for: M0, but award 1 mark for the correct value and the R1 below (at most 2 marks).
\(\mathbf{R}\begin{pmatrix}-3\\3\end{pmatrix}=\begin{pmatrix}-4.10\\1.10\end{pmatrix}\), the same point	R1	Allow ECF from their \(\mathbf{R}\).
Part (j)
\(\begin{pmatrix}2&0\\0&2\end{pmatrix}\mathbf{C}=\begin{pmatrix}1&\sqrt3\\\sqrt3&-1\end{pmatrix}\)	B1	Allow ECF from (d)(i).
\(\det=1\times(-1)-\sqrt3\times\sqrt3=-4\), so the area scale factor is \(\|-4\|=4\)	M1	Accept \(2^2\times1\).
area \(=4\times21=84\) square units	A1	\(42\) (scale factor 2 used for area) scores M0A0.

Answers: (a)(i) \(P'(0,-6)\), \(Q'(3,3)\) · (a)(ii) rotation of \(90^\circ\) clockwise about the origin · (b) \(36.9^\circ\) · (c) \(\mathbf{R}=\begin{pmatrix}\frac{\sqrt3}{2}&-\frac12\\\frac12&\frac{\sqrt3}{2}\end{pmatrix}\), \(\mathbf{F}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\) · (d)(i) \(\mathbf{C}=\begin{pmatrix}\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{pmatrix}\) · (d)(ii) \(\begin{pmatrix}-\frac12&\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac12\end{pmatrix}\); the order matters · (e) \(P'(3,5.20)\), \(Q'(1.10,-4.10)\), \(R'(-0.5,-0.866)\), \(S'(-4.10,-1.10)\) · (f) \(-1\) · (h) \(\mathbf{C}^{-1}=\mathbf{C}\); cannon was at \((2,0)\) · (i) \(-4.10+1.10i\), the same as \(\mathbf{R}Q\) · (j) \(\begin{pmatrix}1&\sqrt3\\\sqrt3&-1\end{pmatrix}\); 84 square units

Exam tip (a)(ii) To describe a matrix, read its columns: the first column is the image of \((1,0)\), the second the image of \((0,1)\). Here \((1,0)\mapsto(0,-1)\) and \((0,1)\mapsto(1,0)\), a quarter-turn clockwise.

Common error (d)(i) \(\mathbf{R}\) first, then \(\mathbf{F}\) is the product \(\mathbf{FR}\): the first transformation is written next to the point. Multiplying \(\mathbf{RF}\) gives a different, perfectly reasonable-looking matrix.

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Drawing a fractal fern: repeated matrix transformations

3.8 Matrix transformations1.6 Matrices and matrix algebra1.7 Eigenvalues and eigenvectors1.3 Arithmetic and geometric sequences in contextHL

A graphics program draws one frond of a fern-like fractal on a screen. Starting from a point \(\mathbf{x}_0\), it repeatedly applies the map

\[\mathbf{x}_{n+1}=\mathbf{A}\mathbf{x}_n+\mathbf{b},\qquad \mathbf{A}=\begin{pmatrix}0.5&0.2\\0.1&0.4\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}2\\2.4\end{pmatrix},\]

where the coordinates are in centimetres from the bottom left corner of the screen. Each step is a matrix transformation about the origin followed by a translation by the vector \(\mathbf{b}\).

(a)

The program starts at \(\mathbf{x}_0=\begin{pmatrix}0\\0\end{pmatrix}\). Find \(\mathbf{x}_1\), \(\mathbf{x}_2\) and \(\mathbf{x}_3\).

ROUTINENotes

Show solution

\(\mathbf{x}_1=\begin{pmatrix}2\\2.4\end{pmatrix}\)	B1
\(\mathbf{x}_2=\mathbf{A}\begin{pmatrix}2\\2.4\end{pmatrix}+\mathbf{b}=\begin{pmatrix}1.48+2\\1.16+2.4\end{pmatrix}\)	(M1)	Multiplying by \(\mathbf{A}\) before adding \(\mathbf{b}\).
\(\mathbf{x}_2=\begin{pmatrix}3.48\\3.56\end{pmatrix}\), \(\mathbf{x}_3=\begin{pmatrix}4.452\\4.172\end{pmatrix}\)	A1	Accept \(\mathbf{x}_3=(4.45,\ 4.17)\). Both needed.

Answer: \(\mathbf{x}_1=(2,2.4)\), \(\mathbf{x}_2=(3.48,3.56)\), \(\mathbf{x}_3=(4.452,4.172)\)

(3)

(b)

Find \(\det\mathbf{A}\) and state what it tells you about the effect of one step of the map on a shape drawn on the screen.

CONCEPTNotes

Show solution

\(\det\mathbf{A}=0.5\times0.4-0.2\times0.1=0.18\)	A1
each step multiplies area by 0.18 (the translation does not change area), and as \(\det\mathbf{A}>0\) the shape is not flipped over	R1	Both ideas needed: area scale factor and orientation preserved.

Answer: \(0.18\)

(2)

(c)

The program draws a leaf of area \(40\ \text{cm}^2\), then draws its image under the map, then the image of that image, and so on.

Explain why the areas of the leaves form a geometric sequence, and find the total area of the first six leaves drawn.

MODELNotes

Show solution

each leaf has area \(0.18\times\) the previous one, so the areas are geometric with \(u_1=40\), \(r=0.18\)	R1	Must mention multiplying by the same factor each time.
\(S_6=\frac{40(1-0.18^6)}{1-0.18}\)	(M1)	Allow ECF from (b). Adding \(40+7.2+1.296+\ldots\) also scores.
\(=48.8\ \text{cm}^2\)	A1	\(48.7788\ldots\)

Answer: \(48.8\ \text{cm}^2\)

(3)

(d)

A fixed point of the map is a point \(\mathbf{f}\) with \(\mathbf{A}\mathbf{f}+\mathbf{b}=\mathbf{f}\).

By solving \((\mathbf{I}-\mathbf{A})\mathbf{f}=\mathbf{b}\), show that the fixed point is \(F(6,5)\).

PROOFNotes

Show solution

\(\mathbf{I}-\mathbf{A}=\begin{pmatrix}0.5&-0.2\\-0.1&0.6\end{pmatrix}\)	B1
\(\mathbf{f}=(\mathbf{I}-\mathbf{A})^{-1}\mathbf{b}=\frac{1}{0.28}\begin{pmatrix}0.6&0.2\\0.1&0.5\end{pmatrix}\begin{pmatrix}2\\2.4\end{pmatrix}\)	M1	Inverse on the left, by formula or GDC. Solving the simultaneous equations \(0.5x-0.2y=2\), \(-0.1x+0.6y=2.4\) is also acceptable.
\(=\frac{1}{0.28}\begin{pmatrix}1.68\\1.4\end{pmatrix}\)	A1	Or equivalent working leading to 6 and 5.
\(=\begin{pmatrix}6\\5\end{pmatrix}\)	AG	Substituting \((6,5)\) to check only scores at most B1M0A0.

Answer: \(F(6,5)\)

(3)

(e)

To erase a frond the program has to run the map backwards. One point drawn is \((4.8,\ 4.4)\).

Use \(\mathbf{A}^{-1}\) to find the point from which it was drawn.

ALGEBRANotes

Show solution

\(\mathbf{x}_{n-1}=\mathbf{A}^{-1}\left(\begin{pmatrix}4.8\\4.4\end{pmatrix}-\mathbf{b}\right)=\frac{1}{0.18}\begin{pmatrix}0.4&-0.2\\-0.1&0.5\end{pmatrix}\begin{pmatrix}2.8\\2\end{pmatrix}\)	M1	Subtract \(\mathbf{b}\) before multiplying by \(\mathbf{A}^{-1}\). \(\mathbf{A}^{-1}\begin{pmatrix}4.8\\4.4\end{pmatrix}-\mathbf{b}\) scores M0.
\((4,\ 4)\)	A1

Answer: \((4,4)\)

(2)

(f)

Find the eigenvalues of \(\mathbf{A}\).

ROUTINENotes

Show solution

\(\det(\mathbf{A}-\lambda\mathbf{I})=(0.5-\lambda)(0.4-\lambda)-0.02=0\)	M1	Or \(\lambda^2-0.9\lambda+0.18=0\) from the trace and determinant.
\(\lambda^2-0.9\lambda+0.18=0\)	(A1)
\(\lambda=0.6\) or \(\lambda=0.3\)	A1	Check: sum 0.9 = trace, product 0.18 = determinant.

Answer: \(0.6\) and \(0.3\)

(3)

(g)

Find an eigenvector of \(\mathbf{A}\) for each eigenvalue.

ROUTINENotes

Show solution

\(\lambda=0.6\): \(-0.1x+0.2y=0\), eigenvector \(\begin{pmatrix}2\\1\end{pmatrix}\)	A1	Any non-zero multiple. Allow ECF from (f).
\(\lambda=0.3\): \(0.2x+0.2y=0\), eigenvector \(\begin{pmatrix}1\\-1\end{pmatrix}\)	A1	Any non-zero multiple.

Answer: \(\begin{pmatrix}2\\1\end{pmatrix}\) for 0.6; \(\begin{pmatrix}1\\-1\end{pmatrix}\) for 0.3

(2)

(h)

Figure 1 shows the fixed point \(F\) and the points \(\mathbf{x}_4\), \(\mathbf{x}_5\) and \(\mathbf{x}_6\).

On Figure 1, plot the points \(\mathbf{x}_0\), \(\mathbf{x}_1\), \(\mathbf{x}_2\) and \(\mathbf{x}_3\). Draw the line through \(F\) in the direction of the eigenvector for the larger eigenvalue, and describe how the points move.

Figure 1: the fixed point \(F\) and the points \(\mathbf{x}_4\), \(\mathbf{x}_5\), \(\mathbf{x}_6\)

GRAPHNotes

Show solution

\(\mathbf{x}_0\) to \(\mathbf{x}_3\) plotted correctly	A1	Within half a small square. FT from (a).
straight line through \(F(6,5)\) with gradient \(\frac12\), e.g. through \((2,3)\)	A1	FT from their eigenvector in (g).
the points get closer and closer to \(F\), and approach it along (close to) this line from the lower left	R1	Both ideas needed.

(3)

(i)

Let \(\mathbf{e}_n=\mathbf{x}_n-\mathbf{f}\), where \(\mathbf{f}=\begin{pmatrix}6\\5\end{pmatrix}\).

Show that \(\mathbf{e}_{n+1}=\mathbf{A}\mathbf{e}_n\). Hence use your answers to (f) and (g) to explain why the points approach \(F\) whatever the starting point, and why they finally approach it along the line drawn in (h).

INTERPRETNotes

Show solution

\(\mathbf{e}_{n+1}=\mathbf{A}\mathbf{x}_n+\mathbf{b}-(\mathbf{A}\mathbf{f}+\mathbf{b})=\mathbf{A}(\mathbf{x}_n-\mathbf{f})=\mathbf{A}\mathbf{e}_n\)	R1	Uses \(\mathbf{f}=\mathbf{A}\mathbf{f}+\mathbf{b}\).
writing \(\mathbf{e}_0=p\begin{pmatrix}2\\1\end{pmatrix}+q\begin{pmatrix}1\\-1\end{pmatrix}\) gives \(\mathbf{e}_n=p(0.6)^n\begin{pmatrix}2\\1\end{pmatrix}+q(0.3)^n\begin{pmatrix}1\\-1\end{pmatrix}\); both eigenvalues lie between \(-1\) and 1, so \(\mathbf{e}_n\to\mathbf{0}\) and \(\mathbf{x}_n\to\mathbf{f}\)	R1	Accept a clear statement that the distance from \(F\) is multiplied by factors 0.6 and 0.3 along the eigenvector directions.
\(0.3^n\) shrinks faster than \(0.6^n\), so the \(\begin{pmatrix}2\\1\end{pmatrix}\) term dominates and \(\mathbf{e}_n\) points along the line of gradient \(\frac12\)	R1	For \(\mathbf{x}_0=\mathbf{0}\): \(p=-\frac{11}{3}\), \(q=\frac43\) (not required).

(3)

(j)(i)

The designer replaces \(\mathbf{A}\) by \(\mathbf{B}=2.5\mathbf{A}=\begin{pmatrix}1.25&0.5\\0.25&1\end{pmatrix}\), keeping the same \(\mathbf{b}\).

Find \(\det\mathbf{B}\) and state what now happens to the areas of the leaves.

CONCEPTNotes

Show solution

\(\det\mathbf{B}=1.25-0.125=1.125\)	A1	Accept \(2.5^2\times0.18\).
each leaf is \(1.125\) times the area of the previous one, so the areas increase (by 12.5% per step) without limit	R1

Answer: \(1.125\); the areas grow

(2)

(j)(ii)

Write down the eigenvalues of \(\mathbf{B}\). Hence explain whether the points \(\mathbf{x}_n\) still approach a fixed point.

INTERPRETNotes

Show solution

\(1.5\) and \(0.75\)	A1	\(2.5\times0.6\) and \(2.5\times0.3\); allow ECF from (f).
\(1.5>1\), so the component of \(\mathbf{x}_n-\mathbf{f}\) along \(\begin{pmatrix}2\\1\end{pmatrix}\) grows by 50% each step: the points move away from the fixed point and do not approach it	R1	Unless they start exactly on the other eigenvector line. Do not accept an argument based only on \(\det\mathbf{B}>1\).

Answer: \(1.5\), \(0.75\); no, the points move away

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(\mathbf{x}_1=\begin{pmatrix}2\\2.4\end{pmatrix}\)	B1
\(\mathbf{x}_2=\mathbf{A}\begin{pmatrix}2\\2.4\end{pmatrix}+\mathbf{b}=\begin{pmatrix}1.48+2\\1.16+2.4\end{pmatrix}\)	(M1)	Multiplying by \(\mathbf{A}\) before adding \(\mathbf{b}\).
\(\mathbf{x}_2=\begin{pmatrix}3.48\\3.56\end{pmatrix}\), \(\mathbf{x}_3=\begin{pmatrix}4.452\\4.172\end{pmatrix}\)	A1	Accept \(\mathbf{x}_3=(4.45,\ 4.17)\). Both needed.
Part (b)
\(\det\mathbf{A}=0.5\times0.4-0.2\times0.1=0.18\)	A1
each step multiplies area by 0.18 (the translation does not change area), and as \(\det\mathbf{A}>0\) the shape is not flipped over	R1	Both ideas needed: area scale factor and orientation preserved.
Part (c)
each leaf has area \(0.18\times\) the previous one, so the areas are geometric with \(u_1=40\), \(r=0.18\)	R1	Must mention multiplying by the same factor each time.
\(S_6=\frac{40(1-0.18^6)}{1-0.18}\)	(M1)	Allow ECF from (b). Adding \(40+7.2+1.296+\ldots\) also scores.
\(=48.8\ \text{cm}^2\)	A1	\(48.7788\ldots\)
Part (d)
\(\mathbf{I}-\mathbf{A}=\begin{pmatrix}0.5&-0.2\\-0.1&0.6\end{pmatrix}\)	B1
\(\mathbf{f}=(\mathbf{I}-\mathbf{A})^{-1}\mathbf{b}=\frac{1}{0.28}\begin{pmatrix}0.6&0.2\\0.1&0.5\end{pmatrix}\begin{pmatrix}2\\2.4\end{pmatrix}\)	M1	Inverse on the left, by formula or GDC. Solving the simultaneous equations \(0.5x-0.2y=2\), \(-0.1x+0.6y=2.4\) is also acceptable.
\(=\frac{1}{0.28}\begin{pmatrix}1.68\\1.4\end{pmatrix}\)	A1	Or equivalent working leading to 6 and 5.
\(=\begin{pmatrix}6\\5\end{pmatrix}\)	AG	Substituting \((6,5)\) to check only scores at most B1M0A0.
Part (e)
\(\mathbf{x}_{n-1}=\mathbf{A}^{-1}\left(\begin{pmatrix}4.8\\4.4\end{pmatrix}-\mathbf{b}\right)=\frac{1}{0.18}\begin{pmatrix}0.4&-0.2\\-0.1&0.5\end{pmatrix}\begin{pmatrix}2.8\\2\end{pmatrix}\)	M1	Subtract \(\mathbf{b}\) before multiplying by \(\mathbf{A}^{-1}\). \(\mathbf{A}^{-1}\begin{pmatrix}4.8\\4.4\end{pmatrix}-\mathbf{b}\) scores M0.
\((4,\ 4)\)	A1
Part (f)
\(\det(\mathbf{A}-\lambda\mathbf{I})=(0.5-\lambda)(0.4-\lambda)-0.02=0\)	M1	Or \(\lambda^2-0.9\lambda+0.18=0\) from the trace and determinant.
\(\lambda^2-0.9\lambda+0.18=0\)	(A1)
\(\lambda=0.6\) or \(\lambda=0.3\)	A1	Check: sum 0.9 = trace, product 0.18 = determinant.
Part (g)
\(\lambda=0.6\): \(-0.1x+0.2y=0\), eigenvector \(\begin{pmatrix}2\\1\end{pmatrix}\)	A1	Any non-zero multiple. Allow ECF from (f).
\(\lambda=0.3\): \(0.2x+0.2y=0\), eigenvector \(\begin{pmatrix}1\\-1\end{pmatrix}\)	A1	Any non-zero multiple.
Part (h)
\(\mathbf{x}_0\) to \(\mathbf{x}_3\) plotted correctly	A1	Within half a small square. FT from (a).
straight line through \(F(6,5)\) with gradient \(\frac12\), e.g. through \((2,3)\)	A1	FT from their eigenvector in (g).
the points get closer and closer to \(F\), and approach it along (close to) this line from the lower left	R1	Both ideas needed.
Part (i)
\(\mathbf{e}_{n+1}=\mathbf{A}\mathbf{x}_n+\mathbf{b}-(\mathbf{A}\mathbf{f}+\mathbf{b})=\mathbf{A}(\mathbf{x}_n-\mathbf{f})=\mathbf{A}\mathbf{e}_n\)	R1	Uses \(\mathbf{f}=\mathbf{A}\mathbf{f}+\mathbf{b}\).
writing \(\mathbf{e}_0=p\begin{pmatrix}2\\1\end{pmatrix}+q\begin{pmatrix}1\\-1\end{pmatrix}\) gives \(\mathbf{e}_n=p(0.6)^n\begin{pmatrix}2\\1\end{pmatrix}+q(0.3)^n\begin{pmatrix}1\\-1\end{pmatrix}\); both eigenvalues lie between \(-1\) and 1, so \(\mathbf{e}_n\to\mathbf{0}\) and \(\mathbf{x}_n\to\mathbf{f}\)	R1	Accept a clear statement that the distance from \(F\) is multiplied by factors 0.6 and 0.3 along the eigenvector directions.
\(0.3^n\) shrinks faster than \(0.6^n\), so the \(\begin{pmatrix}2\\1\end{pmatrix}\) term dominates and \(\mathbf{e}_n\) points along the line of gradient \(\frac12\)	R1	For \(\mathbf{x}_0=\mathbf{0}\): \(p=-\frac{11}{3}\), \(q=\frac43\) (not required).
Part (j)(i)
\(\det\mathbf{B}=1.25-0.125=1.125\)	A1	Accept \(2.5^2\times0.18\).
each leaf is \(1.125\) times the area of the previous one, so the areas increase (by 12.5% per step) without limit	R1
Part (j)(ii)
\(1.5\) and \(0.75\)	A1	\(2.5\times0.6\) and \(2.5\times0.3\); allow ECF from (f).
\(1.5>1\), so the component of \(\mathbf{x}_n-\mathbf{f}\) along \(\begin{pmatrix}2\\1\end{pmatrix}\) grows by 50% each step: the points move away from the fixed point and do not approach it	R1	Unless they start exactly on the other eigenvector line. Do not accept an argument based only on \(\det\mathbf{B}>1\).

Answers: (a) \(\mathbf{x}_1=(2,2.4)\), \(\mathbf{x}_2=(3.48,3.56)\), \(\mathbf{x}_3=(4.452,4.172)\) · (b) \(0.18\) · (c) \(48.8\ \text{cm}^2\) · (d) \(F(6,5)\) · (e) \((4,4)\) · (f) \(0.6\) and \(0.3\) · (g) \(\begin{pmatrix}2\\1\end{pmatrix}\) for 0.6; \(\begin{pmatrix}1\\-1\end{pmatrix}\) for 0.3 · (j)(i) \(1.125\); the areas grow · (j)(ii) \(1.5\), \(0.75\); no, the points move away

Exam tip (i) In \(\mathbf{x}_{n+1}=\mathbf{A}\mathbf{x}_n+\mathbf{b}\), measure positions from the fixed point. The translation then disappears and the long-term behaviour is decided by the eigenvalues of \(\mathbf{A}\) alone.

More 3.8 questions in the question bank

Topic 4 · Statistics and probability

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	4.1	4.2	4.3	4.4	4.5	4.6	4.7	4.8	4.9	Other
How much do students sleep? Sampling and summary statistics	●	○								1.1
Expected goals in a football league: correlation and regression	○	●								2.1 1.1
A penalty shootout: tree diagrams, the binomial model and a chi-squared test	○		●	○						—
A blood pressure screening day: normal model, binomial counts and independence	○		○	●						1.1
Do carbon-plate shoes make runners faster? Confidence intervals and t-tests	○			○	●		○			—
Arrivals at an emergency department: Poisson models and combined variables			○	○		●	○			—
A music streaming service: subscribers moving between plans			○					●		1.6 1.7
How much power does a cyclist need? Power regression and the coefficient of determination		○							●	2.5 2.2

4.1 Sampling and descriptive statistics · 4.2 Correlation and regressionmerged group

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How much do students sleep? Sampling and summary statistics

4.1 Sampling and descriptive statistics4.2 Correlation and regression1.1 Approximation, percentage error and standard formSL

The student council of a secondary school wants to know how many hours students sleep on a school night. The school has 960 students: 288 in Grade 9, 264 in Grade 10, 216 in Grade 11 and 192 in Grade 12.

(a)(i)

The council takes a stratified sample of 50 students, using the grades as strata.

Find the number of students that should be chosen from each grade.

ROUTINENotes

Show solution

\(\frac{288}{960}\times50=15\), \(\frac{264}{960}\times50=13.75\), \(\frac{216}{960}\times50=11.25\), \(\frac{192}{960}\times50=10\)	(M1)	At least two correct proportional calculations.
Grade 9: 15, Grade 10: 14, Grade 11: 11, Grade 12: 10	A1	Whole numbers with total 50 required.

Answer: 15, 14, 11, 10

(2)

(a)(ii)

One council member suggests posting the question as an online poll on the school's social media page at 11 pm instead. Give one reason why the results of this poll could be biased.

CONCEPTNotes

Show solution

e.g. the students who reply are those online at 11 pm, who are likely to sleep less than typical students; or replies are self-selected / not random, so some groups are over-represented

Must refer to this context. "It is not random" alone is not enough.

(1)

(b)(i)

The results for the 50 students in the stratified sample are shown in the table.

Hours of sleep, \(h\)	Number of students
\(4\le h<5\)	1
\(5\le h<6\)	3
\(6\le h<7\)	9
\(7\le h<8\)	18
\(8\le h<9\)	14
\(9\le h<10\)	5

Find an estimate of the mean and an estimate of the standard deviation of the number of hours of sleep.

ROUTINENotes

Show solution

mean \(\approx7.62\) hours	A1	Mid-interval values \(4.5, 5.5, \ldots, 9.5\). \(\frac{381}{50}\).
standard deviation \(\approx1.12\) hours	A1	\(1.12498\ldots\) Accept 1.125.

Answer: mean \(\approx7.62\) h; \(\sigma\approx1.12\) h

(2)

(b)(ii)

The mean of the 50 exact recorded values was 7.59 hours. Find the percentage error in your estimate of the mean.

ROUTINENotes

Show solution

\(\left\|\frac{7.62-7.59}{7.59}\right\|\times100\%\)	M1	Exact value 7.59 in the denominator. Allow ECF from (b)(i).
\(0.395\%\)	A1	\(0.39525\ldots\%\)

Answer: \(0.395\%\)

(2)

(c)

On the grid in Figure 1, draw a cumulative frequency graph for the data in the table.

Figure 1: grid for the cumulative frequency graph

GRAPHNotes

Show solution

cumulative frequencies \(1,\,4,\,13,\,31,\,45,\,50\)	B1	May be seen on the graph (correct heights of the plotted points).
points plotted at the upper class boundaries \(5, 6, \ldots, 10\), and starting at \((4,0)\)	A1	Within half a small square.
points joined with a smooth curve or straight line segments	A1

(3)

(d)

Use your graph to estimate the median and the interquartile range of the hours of sleep.

GRAPHNotes

Show solution

median \(\approx7.7\) hours	B1	Reading at 25. Accept 7.6 to 7.8.
readings at 12.5 and 37.5: \(Q_1\approx6.9\), \(Q_3\approx8.5\)	(M1)	Accept \(Q_1\) from 6.8 to 7.1 and \(Q_3\) from 8.3 to 8.6. (Straight line segments give \(Q_1=6.94\), \(Q_3=8.46\).)
\(\text{IQR}\approx1.5\) hours	A1	Accept 1.3 to 1.8. FT from their graph.

Answer: median \(\approx7.7\) h; IQR \(\approx1.5\) h

(3)

(e)

Twelve of the students in the sample also recorded their average daily screen time outside lessons.

Screen time, \(x\) (hours)	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6	7
Sleep, \(y\) (hours)	8.9	8.6	8.7	8.1	8	7.9	7.4	7.6	7	6.9	6.5	4.6

Show that the sleep value 4.6 hours is an outlier among these twelve sleep values.

ROUTINENotes

Show solution

\(Q_1=6.95\), \(Q_3=8.35\), \(\text{IQR}=1.4\)	B1	From one-variable statistics on the twelve values.
lower fence \(=6.95-1.5\times1.4=4.85\)	M1
\(4.6<4.85\), so 4.6 is an outlier	AG

Answer: lower fence 4.85

(2)

(f)

On Figure 2, draw a box-and-whisker plot of the twelve sleep values, showing the outlier.

Figure 2: scale for the box-and-whisker plot

GRAPHNotes

Show solution

box from \(Q_1=6.95\) to \(Q_3=8.35\) with median \(7.75\)	A1	Within half a small square.
whiskers to 6.5 and 8.9, with 4.6 marked separately as a cross	A1	A whisker drawn to 4.6 scores A0.

(2)

(g)

The scatter diagram of the twelve students is shown in Figure 3.

Find the value of Pearson's correlation coefficient, \(r\), and the equation of the regression line of \(y\) on \(x\). Give all values correct to 3 significant figures.

Figure 3: sleep against daily screen time for twelve students

ROUTINENotes

Show solution

\(r=-0.940\)	A1	\(-0.93956\ldots\) Accept \(-0.94\).
\(y=-0.593x+\ldots\)	A1	\(-0.59279\ldots\)
\(y=\ldots+9.76\)	A1	\(9.7643\ldots\) If the coefficients are not given to 3 s.f. (e.g. \(y=-0.6x+9.8\)), award only one of the two A1 marks for the equation.

Answer: \(r=-0.940\); \(y=-0.593x+9.76\)

(3)

(h)

Interpret the gradient of the regression line in context, and use the line to estimate the sleep of a student whose daily screen time is 3.2 hours.

INTERPRETNotes

Show solution

each extra hour of screen time is associated with about 0.593 hours (about 36 minutes) less sleep, on average	R1	Must mention context and "on average" or equivalent. FT their gradient.
\(-0.59279\times3.2+9.7643=7.87\) hours	A1	Accept 7.86 to 7.87 from rounded coefficients.

Answer: 0.593 h less sleep per extra hour; about 7.87 h

(2)

(i)

A parent concludes that screen time causes students to sleep less. Discuss whether the data support this conclusion.

INTERPRETNotes

Show solution

the strong negative correlation shows association, not cause	R1
e.g. a third factor such as homework load or a late part-time job could reduce sleep and increase evening screen use; or the direction may be reversed (students who cannot sleep use screens); or only 12 students, including an outlier	R1	One relevant explanation in context.

(2)

(j)

The council decides to report sleep in minutes, so every value of \(h\) in the table is multiplied by 60.

Write down the new estimates of the mean and the standard deviation.

ROUTINENotes

Show solution

mean \(\approx457\) minutes	A1	\(60\times7.62=457.2\). Allow ECF from (b)(i).
standard deviation \(\approx67.5\) minutes	A1	\(60\times1.12498\ldots=67.499\ldots\). Accept 67.2 from 1.12.

Answer: mean \(\approx457\) min; \(\sigma\approx67.5\) min

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(\frac{288}{960}\times50=15\), \(\frac{264}{960}\times50=13.75\), \(\frac{216}{960}\times50=11.25\), \(\frac{192}{960}\times50=10\)	(M1)	At least two correct proportional calculations.
Grade 9: 15, Grade 10: 14, Grade 11: 11, Grade 12: 10	A1	Whole numbers with total 50 required.
Part (a)(ii)
e.g. the students who reply are those online at 11 pm, who are likely to sleep less than typical students; or replies are self-selected / not random, so some groups are over-represented	R1	Must refer to this context. "It is not random" alone is not enough.
Part (b)(i)
mean \(\approx7.62\) hours	A1	Mid-interval values \(4.5, 5.5, \ldots, 9.5\). \(\frac{381}{50}\).
standard deviation \(\approx1.12\) hours	A1	\(1.12498\ldots\) Accept 1.125.
Part (b)(ii)
\(\left\|\frac{7.62-7.59}{7.59}\right\|\times100\%\)	M1	Exact value 7.59 in the denominator. Allow ECF from (b)(i).
\(0.395\%\)	A1	\(0.39525\ldots\%\)
Part (c)
cumulative frequencies \(1,\,4,\,13,\,31,\,45,\,50\)	B1	May be seen on the graph (correct heights of the plotted points).
points plotted at the upper class boundaries \(5, 6, \ldots, 10\), and starting at \((4,0)\)	A1	Within half a small square.
points joined with a smooth curve or straight line segments	A1
Part (d)
median \(\approx7.7\) hours	B1	Reading at 25. Accept 7.6 to 7.8.
readings at 12.5 and 37.5: \(Q_1\approx6.9\), \(Q_3\approx8.5\)	(M1)	Accept \(Q_1\) from 6.8 to 7.1 and \(Q_3\) from 8.3 to 8.6. (Straight line segments give \(Q_1=6.94\), \(Q_3=8.46\).)
\(\text{IQR}\approx1.5\) hours	A1	Accept 1.3 to 1.8. FT from their graph.
Part (e)
\(Q_1=6.95\), \(Q_3=8.35\), \(\text{IQR}=1.4\)	B1	From one-variable statistics on the twelve values.
lower fence \(=6.95-1.5\times1.4=4.85\)	M1
\(4.6<4.85\), so 4.6 is an outlier	AG
Part (f)
box from \(Q_1=6.95\) to \(Q_3=8.35\) with median \(7.75\)	A1	Within half a small square.
whiskers to 6.5 and 8.9, with 4.6 marked separately as a cross	A1	A whisker drawn to 4.6 scores A0.
Part (g)
\(r=-0.940\)	A1	\(-0.93956\ldots\) Accept \(-0.94\).
\(y=-0.593x+\ldots\)	A1	\(-0.59279\ldots\)
\(y=\ldots+9.76\)	A1	\(9.7643\ldots\) If the coefficients are not given to 3 s.f. (e.g. \(y=-0.6x+9.8\)), award only one of the two A1 marks for the equation.
Part (h)
each extra hour of screen time is associated with about 0.593 hours (about 36 minutes) less sleep, on average	R1	Must mention context and "on average" or equivalent. FT their gradient.
\(-0.59279\times3.2+9.7643=7.87\) hours	A1	Accept 7.86 to 7.87 from rounded coefficients.
Part (i)
the strong negative correlation shows association, not cause	R1
e.g. a third factor such as homework load or a late part-time job could reduce sleep and increase evening screen use; or the direction may be reversed (students who cannot sleep use screens); or only 12 students, including an outlier	R1	One relevant explanation in context.
Part (j)
mean \(\approx457\) minutes	A1	\(60\times7.62=457.2\). Allow ECF from (b)(i).
standard deviation \(\approx67.5\) minutes	A1	\(60\times1.12498\ldots=67.499\ldots\). Accept 67.2 from 1.12.

Answers: (a)(i) 15, 14, 11, 10 · (b)(i) mean \(\approx7.62\) h; \(\sigma\approx1.12\) h · (b)(ii) \(0.395\%\) · (d) median \(\approx7.7\) h; IQR \(\approx1.5\) h · (e) lower fence 4.85 · (g) \(r=-0.940\); \(y=-0.593x+9.76\) · (h) 0.593 h less sleep per extra hour; about 7.87 h · (j) mean \(\approx457\) min; \(\sigma\approx67.5\) min

Exam tip (a)(i) After rounding each stratum to a whole number, add them up: the total must still equal the sample size.

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Expected goals in a football league: correlation and regression

4.2 Correlation and regression4.1 Sampling and descriptive statistics2.1 Linear models and piecewise models1.1 Approximation, percentage error and standard formSL

In football, the expected goals (xG) of a shot is the probability that a shot from that position and situation is scored. Adding these over a season gives a club's total xG, a measure of the quality of the chances it created.

The table shows the total xG, \(x\), and the number of goals actually scored, \(y\), for the ten clubs in a league over one season.

Club	A	B	C	D	E	F	G	H	I	J
Expected goals, \(x\)	38.2	41.5	44.6	47.0	52.3	55.1	58.4	63.0	69.6	76.8
Goals scored, \(y\)	36	44	41	68	50	61	57	66	74	81

(a)

On the grid in Figure 1, draw a scatter diagram of the data.

Figure 1: grid for the scatter diagram

GRAPHNotes

Show solution

at least 8 points plotted correctly	A1	Within half a small square.
all 10 points plotted correctly	A1

(2)

(b)

Find the mean and the standard deviation of

the expected goals;
the goals scored.

ROUTINENotes

Show solution

\(\bar{x}=54.65\), \(\sigma_x=11.9\)	A1	\(11.8603\ldots\)
\(\bar{y}=57.8\), \(\sigma_y=14.1\)	A1	\(14.1124\ldots\) Using \(s_{n-1}\) (12.5 and 14.9) scores A0 once only.

Answer: \(\bar{x}=54.65\), \(\sigma_x\approx11.9\); \(\bar{y}=57.8\), \(\sigma_y\approx14.1\)

(2)

(c)

Find Pearson's correlation coefficient, \(r\), and describe the correlation between \(x\) and \(y\).

ROUTINENotes

Show solution

\(r=0.880\)	A1	\(0.87956\ldots\)
strong positive (linear) correlation	A1	Both "strong" and "positive" needed.

Answer: \(r\approx0.880\); strong positive

(2)

(d)

Find the equation of the regression line of \(y\) on \(x\).

ROUTINENotes

Show solution

\(y=1.05x+\ldots\)	A1	\(1.04657\ldots\)
\(y=\ldots+0.605\)	A1	\(0.60473\ldots\)

Answer: \(y=1.05x+0.605\)

(2)

(e)

Show that the regression line passes through the mean point \((\bar{x},\bar{y})\).

PROOFNotes

Show solution

substituting \(x=54.65\): \(1.04657\ldots\times54.65+0.60473\ldots\)	M1	Unrounded or at least 4 s.f. coefficients.
\(=57.8\)	A1	With 1.05 and 0.605 the value is 57.99; award M1A0 unless they comment on rounding.
\(=\bar{y}\), so \((\bar{x},\bar{y})\) lies on the line	AG

(2)

(f)

Interpret, in context, the gradient and the \(y\)-intercept of the regression line. Comment on whether the \(y\)-intercept is meaningful.

INTERPRETNotes

Show solution

gradient: for each extra 1 xG a club creates, it scores about 1.05 more goals, on average	R1
intercept: a club with xG of 0 would be predicted to score about 0.6 goals; \(x=0\) is far outside the data (38.2 to 76.8), so this is extrapolation and not meaningful	R1	Both the meaning and the comment needed.

(2)

(g)

Find Spearman's rank correlation coefficient, \(r_s\), for the data and compare it with your value of \(r\).

MULTINotes

Show solution

ranks of \(x\): 1 to 10 in order; ranks of \(y\): \(1, 3, 2, 8, 4, 6, 5, 7, 9, 10\)	(M1)	Ranking from the largest instead is fine if done for both variables.
\(r_s=0.855\)	A1	\(0.85454\ldots=\frac{47}{55}\)
both show a strong positive relationship: clubs with higher xG tend to score more goals; \(r_s\) is slightly smaller because club D is ranked well out of order	R1	Any sensible comparison of the two values in context.

Answer: \(r_s\approx0.855\)

(3)

(h)(i)

In the next season a different club in this league had a total xG of 74.0.

Use the regression line to predict the number of goals it scored, and comment on the reliability of your prediction.

INTERPRETNotes

Show solution

\(1.04657\times74.0+0.60473=78.1\)	A1	Accept 78 goals.
74.0 lies within the range of the data (38.2 to 76.8) and \(r\) is close to 1, so the prediction is fairly reliable (interpolation)	R1	Accept a comment that the line comes from a different season, so reliability is limited.

Answer: about 78 goals; interpolation, fairly reliable

(2)

(h)(ii)

The club actually scored 70 goals. Find the percentage error in the prediction.

ROUTINENotes

Show solution

\(\left\|\frac{78.05-70}{70}\right\|\times100\%\)	M1	Actual value in the denominator. Allow ECF from (h)(i).
\(11.5\%\)	A1	\(11.50\ldots\%\); 11.6% from 78.1 and 11.4% from 78 are accepted.

Answer: \(11.5\%\)

(2)

(i)

A journalist writes: "The strong correlation proves that a high xG causes a club to score more goals." Explain why this conclusion is not justified.

INTERPRETNotes

Show solution

correlation measures association only and does not show cause	R1
both variables depend on a third factor: the number and quality of chances (shots) a club creates, and the skill of its players; xG is a measurement of those chances, not something that makes the ball go in	R1	Any sensible third-variable explanation in context.

(2)

(j)

Club D is removed from the data because its finishing was unusually successful.

Find the new value of \(r\) and comment on the effect of removing club D.

MULTINotes

Show solution

data without D entered (9 clubs)	(M1)
\(r=0.983\)	A1	\(0.98261\ldots\)
\(r\) increases (from 0.880): D lies far above the trend, so without it the points lie much closer to a straight line and the correlation is stronger	R1	Must refer to D being far from the line.

Answer: \(r\approx0.983\); the correlation becomes stronger

(3)

Show full solution

Working	Mark	Notes
Part (a)
at least 8 points plotted correctly	A1	Within half a small square.
all 10 points plotted correctly	A1
Part (b)
\(\bar{x}=54.65\), \(\sigma_x=11.9\)	A1	\(11.8603\ldots\)
\(\bar{y}=57.8\), \(\sigma_y=14.1\)	A1	\(14.1124\ldots\) Using \(s_{n-1}\) (12.5 and 14.9) scores A0 once only.
Part (c)
\(r=0.880\)	A1	\(0.87956\ldots\)
strong positive (linear) correlation	A1	Both "strong" and "positive" needed.
Part (d)
\(y=1.05x+\ldots\)	A1	\(1.04657\ldots\)
\(y=\ldots+0.605\)	A1	\(0.60473\ldots\)
Part (e)
substituting \(x=54.65\): \(1.04657\ldots\times54.65+0.60473\ldots\)	M1	Unrounded or at least 4 s.f. coefficients.
\(=57.8\)	A1	With 1.05 and 0.605 the value is 57.99; award M1A0 unless they comment on rounding.
\(=\bar{y}\), so \((\bar{x},\bar{y})\) lies on the line	AG
Part (f)
gradient: for each extra 1 xG a club creates, it scores about 1.05 more goals, on average	R1
intercept: a club with xG of 0 would be predicted to score about 0.6 goals; \(x=0\) is far outside the data (38.2 to 76.8), so this is extrapolation and not meaningful	R1	Both the meaning and the comment needed.
Part (g)
ranks of \(x\): 1 to 10 in order; ranks of \(y\): \(1, 3, 2, 8, 4, 6, 5, 7, 9, 10\)	(M1)	Ranking from the largest instead is fine if done for both variables.
\(r_s=0.855\)	A1	\(0.85454\ldots=\frac{47}{55}\)
both show a strong positive relationship: clubs with higher xG tend to score more goals; \(r_s\) is slightly smaller because club D is ranked well out of order	R1	Any sensible comparison of the two values in context.
Part (h)(i)
\(1.04657\times74.0+0.60473=78.1\)	A1	Accept 78 goals.
74.0 lies within the range of the data (38.2 to 76.8) and \(r\) is close to 1, so the prediction is fairly reliable (interpolation)	R1	Accept a comment that the line comes from a different season, so reliability is limited.
Part (h)(ii)
\(\left\|\frac{78.05-70}{70}\right\|\times100\%\)	M1	Actual value in the denominator. Allow ECF from (h)(i).
\(11.5\%\)	A1	\(11.50\ldots\%\); 11.6% from 78.1 and 11.4% from 78 are accepted.
Part (i)
correlation measures association only and does not show cause	R1
both variables depend on a third factor: the number and quality of chances (shots) a club creates, and the skill of its players; xG is a measurement of those chances, not something that makes the ball go in	R1	Any sensible third-variable explanation in context.
Part (j)
data without D entered (9 clubs)	(M1)
\(r=0.983\)	A1	\(0.98261\ldots\)
\(r\) increases (from 0.880): D lies far above the trend, so without it the points lie much closer to a straight line and the correlation is stronger	R1	Must refer to D being far from the line.

Answers: (b) \(\bar{x}=54.65\), \(\sigma_x\approx11.9\); \(\bar{y}=57.8\), \(\sigma_y\approx14.1\) · (c) \(r\approx0.880\); strong positive · (d) \(y=1.05x+0.605\) · (g) \(r_s\approx0.855\) · (h)(i) about 78 goals; interpolation, fairly reliable · (h)(ii) \(11.5\%\) · (j) \(r\approx0.983\); the correlation becomes stronger

Common error (j) Removing an outlier is not automatic. Report the analysis with and without the point and say why it was removed.

More 4.2 questions in the question bank

4.3 Probability and the binomial distribution · 4.4 Normal distribution and chi-squared testsmerged group

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A penalty shootout: tree diagrams, the binomial model and a chi-squared test

4.3 Probability and the binomial distribution4.4 Normal distribution and chi-squared tests4.1 Sampling and descriptive statisticsSL

A football academy studies the penalty kicks of its striker, Dani. For each penalty, Dani shoots to the left (\(L\)) with probability 0.55, and otherwise to the right (\(R\)). When Dani shoots to the left, the probability of scoring a goal (\(G\)) is 0.8; when Dani shoots to the right, it is 0.7. A penalty that does not give a goal is recorded as \(N\).

The information is shown on the tree diagram in Figure 1.

Figure 1: direction of the shot, then the outcome of the penalty

(a)

Write down the values of \(a\), \(b\) and \(c\).

ROUTINENotes

Show solution

\(a=0.45\)	A1
\(b=0.2\) and \(c=0.3\)	A1	Both needed for this mark.

Answer: \(a=0.45\), \(b=0.2\), \(c=0.3\)

(2)

(b)

Show that the probability that Dani scores a goal with a randomly chosen penalty is 0.755.

ROUTINENotes

Show solution

\(P(G)=0.55\times0.8+0.45\times0.7\)	M1	Multiplies along both paths that end in \(G\) and adds.
\(=0.44+0.315\)	A1
\(=0.755\)	AG

(2)

(c)

One of Dani's penalties does not give a goal. Find the probability that it was shot to the left.

MULTINotes

Show solution

\(P(N)=0.55\times0.2+0.45\times0.3=0.245\)	(M1)	Or \(1-0.755=0.245\).
\(P(L\mid N)=\frac{P(L\cap N)}{P(N)}=\frac{0.11}{0.245}\)	M1	Numerator \(0.55\times0.2\) over their \(P(N)\).
\(=0.449\)	A1	Accept \(\frac{22}{49}\). Allow ECF from their \(b\) and \(c\).

Answer: \(\frac{22}{49}\approx0.449\)

(3)

(d)(i)

In a training session Dani takes 5 penalties. Let \(X\) be the number of goals scored.

Explain why \(X\) can be modelled by \(X\sim\text{B}(5,\,0.755)\), stating one assumption that must be made about the penalties.

MODELNotes

Show solution

a fixed number of trials (5 penalties), each with two outcomes (goal or no goal) and the same probability of a goal, 0.755	R1	Must refer to the context; a list of the four conditions with no context earns R0.
the penalties must be independent: the outcome of one penalty does not affect the next (e.g. a miss does not affect Dani's confidence), and Dani's direction and accuracy do not change during the session	R1	Accept independence or constant probability, explained in context.

(2)

(d)(ii)

Find the probability that Dani scores at least 4 goals.

ROUTINENotes

Show solution

\(P(X\ge4)=1-P(X\le3)\)	(M1)	Or \(P(X=4)+P(X=5)=0.398+0.245\).
\(=0.643\)	A1

Answer: \(0.643\)

(2)

(e)

Figure 2 shows the probability distribution of \(X\).

Write down the most likely number of goals. Describe the shape of the distribution and explain it in terms of the probability of a goal.

Figure 2: the probability distribution of \(X\sim\text{B}(5,\,0.755)\)

GRAPHNotes

Show solution

4 goals	A1	Read from the tallest bar.
the distribution is skewed to the left (negatively skewed), because \(p=0.755\) is large (greater than 0.5), so most of the probability is on the larger numbers of goals with a tail towards 0	R1	Needs both the skew and the reason.

Answer: 4 goals; skewed to the left

(2)

(f)

The coach repeats the session of 5 penalties under match pressure, with crowd noise played through speakers. Under these conditions each of the scoring probabilities 0.8 and 0.7 falls by 0.1. The probabilities of shooting to the left and to the right are unchanged.

Find the probability that Dani scores at least 4 goals in this session.

MULTINotes

Show solution

new probability of a goal \(=0.55\times0.7+0.45\times0.6\ (=0.655)\)	M1
\(Y\sim\text{B}(5,\,0.655)\): \(P(Y\ge4)=1-P(Y\le3)\)	M1	Or \(P(Y=4)+P(Y=5)\). Allow their new probability.
\(=0.438\)	A1	Allow ECF from an incorrect new probability.

Answer: \(0.438\)

(3)

(g)(i)

The academy also analyses 200 penalties taken in one season of a professional league. For each penalty it records the direction of the shot (left, centre or right) and whether a goal was scored. The results are shown in the table.

	Left	Centre	Right	Total
Goal	68	28	53	149
No goal	27	2	22	51
Total	95	30	75	200

State the type of data given by the direction of the shot, and the type of data given by \(X\) in part (d).

CONCEPTNotes

Show solution

direction of the shot: categorical	A1	Accept qualitative.
\(X\): discrete (numerical)	A1

Answer: categorical; discrete

(2)

(g)(ii)

A \(\chi^2\) test for independence is carried out at the 5% significance level. Write down the null and alternative hypotheses.

ROUTINENotes

Show solution

\(H_0\): the outcome of a penalty is independent of the direction of the shot; \(H_1\): the outcome is not independent of the direction

Both needed, in context. Accept 'not associated' / 'associated'.

(1)

(g)(iii)

Find the expected frequency of penalties shot to the centre that did not give a goal.

ROUTINENotes

Show solution

\(E=\frac{30\times51}{200}\)	M1	Row total times column total over grand total.
\(=7.65\)	A1

Answer: \(7.65\)

(2)

(g)(iv)

Carry out the test. State the degrees of freedom, the value of \(\chi^2_{\text{calc}}\), the \(p\)-value and the conclusion in context.

MULTINotes

Show solution

\(\nu=(2-1)(3-1)=2\)	A1
\(\chi^2_{\text{calc}}=6.61\)	A1	From the GDC matrix of observed counts (6.6080…).
\(p=0.0367\)	A1
\(0.0367\lt0.05\), so reject \(H_0\): there is evidence at the 5% level that whether a goal is scored depends on the direction of the shot	R1	Accept \(6.61\gt5.99\). Do not accept 'accept \(H_1\)' with no context. Allow ECF from their \(p\)-value.

Answer: \(\nu=2\), \(\chi^2_{\text{calc}}=6.61\), \(p=0.0367\); reject \(H_0\)

(4)

(h)

The largest contribution to \(\chi^2_{\text{calc}}\) comes from the penalties shot to the centre that did not give a goal.

Interpret this for a goalkeeper facing a penalty, and give one reason why the academy should be careful about applying the result to Dani.

INTERPRETNotes

Show solution

only 2 shots to the centre failed against 7.65 expected, so shots to the centre were scored more often than independence predicts; a goalkeeper who dives to one side leaves the centre open and could consider staying in the centre more often	R1	Must use the observed and expected counts (or 'fewer than expected').
the 200 penalties come from one league in one season and from many different players, not a random sample of Dani's penalties, so the pattern may not apply to Dani	R1	Accept another valid sampling point, e.g. only 30 shots went to the centre, so that estimate is based on few penalties.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(a=0.45\)	A1
\(b=0.2\) and \(c=0.3\)	A1	Both needed for this mark.
Part (b)
\(P(G)=0.55\times0.8+0.45\times0.7\)	M1	Multiplies along both paths that end in \(G\) and adds.
\(=0.44+0.315\)	A1
\(=0.755\)	AG
Part (c)
\(P(N)=0.55\times0.2+0.45\times0.3=0.245\)	(M1)	Or \(1-0.755=0.245\).
\(P(L\mid N)=\frac{P(L\cap N)}{P(N)}=\frac{0.11}{0.245}\)	M1	Numerator \(0.55\times0.2\) over their \(P(N)\).
\(=0.449\)	A1	Accept \(\frac{22}{49}\). Allow ECF from their \(b\) and \(c\).
Part (d)(i)
a fixed number of trials (5 penalties), each with two outcomes (goal or no goal) and the same probability of a goal, 0.755	R1	Must refer to the context; a list of the four conditions with no context earns R0.
the penalties must be independent: the outcome of one penalty does not affect the next (e.g. a miss does not affect Dani's confidence), and Dani's direction and accuracy do not change during the session	R1	Accept independence or constant probability, explained in context.
Part (d)(ii)
\(P(X\ge4)=1-P(X\le3)\)	(M1)	Or \(P(X=4)+P(X=5)=0.398+0.245\).
\(=0.643\)	A1
Part (e)
4 goals	A1	Read from the tallest bar.
the distribution is skewed to the left (negatively skewed), because \(p=0.755\) is large (greater than 0.5), so most of the probability is on the larger numbers of goals with a tail towards 0	R1	Needs both the skew and the reason.
Part (f)
new probability of a goal \(=0.55\times0.7+0.45\times0.6\ (=0.655)\)	M1
\(Y\sim\text{B}(5,\,0.655)\): \(P(Y\ge4)=1-P(Y\le3)\)	M1	Or \(P(Y=4)+P(Y=5)\). Allow their new probability.
\(=0.438\)	A1	Allow ECF from an incorrect new probability.
Part (g)(i)
direction of the shot: categorical	A1	Accept qualitative.
\(X\): discrete (numerical)	A1
Part (g)(ii)
\(H_0\): the outcome of a penalty is independent of the direction of the shot; \(H_1\): the outcome is not independent of the direction	A1	Both needed, in context. Accept 'not associated' / 'associated'.
Part (g)(iii)
\(E=\frac{30\times51}{200}\)	M1	Row total times column total over grand total.
\(=7.65\)	A1
Part (g)(iv)
\(\nu=(2-1)(3-1)=2\)	A1
\(\chi^2_{\text{calc}}=6.61\)	A1	From the GDC matrix of observed counts (6.6080…).
\(p=0.0367\)	A1
\(0.0367\lt0.05\), so reject \(H_0\): there is evidence at the 5% level that whether a goal is scored depends on the direction of the shot	R1	Accept \(6.61\gt5.99\). Do not accept 'accept \(H_1\)' with no context. Allow ECF from their \(p\)-value.
Part (h)
only 2 shots to the centre failed against 7.65 expected, so shots to the centre were scored more often than independence predicts; a goalkeeper who dives to one side leaves the centre open and could consider staying in the centre more often	R1	Must use the observed and expected counts (or 'fewer than expected').
the 200 penalties come from one league in one season and from many different players, not a random sample of Dani's penalties, so the pattern may not apply to Dani	R1	Accept another valid sampling point, e.g. only 30 shots went to the centre, so that estimate is based on few penalties.

Answers: (a) \(a=0.45\), \(b=0.2\), \(c=0.3\) · (c) \(\frac{22}{49}\approx0.449\) · (d)(ii) \(0.643\) · (e) 4 goals; skewed to the left · (f) \(0.438\) · (g)(i) categorical; discrete · (g)(iii) \(7.65\) · (g)(iv) \(\nu=2\), \(\chi^2_{\text{calc}}=6.61\), \(p=0.0367\); reject \(H_0\)

Common error (c) \(P(L\mid N)\) is not the branch value 0.2, which is \(P(N\mid L)\). The given event, \(N\), goes in the denominator: add every path that ends in \(N\).

Exam tip (g)(iv) Write every step: hypotheses in context, \(\nu\), \(\chi^2_{\text{calc}}\), the \(p\)-value, the comparison with 0.05 and a sentence that answers the question asked.

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A blood pressure screening day: normal model, binomial counts and independence

4.4 Normal distribution and chi-squared tests4.3 Probability and the binomial distribution4.1 Sampling and descriptive statistics1.1 Approximation, percentage error and standard formSL

A pharmacy runs a free blood pressure screening day. The systolic blood pressure \(X\) mmHg of an adult who attends is modelled by \(X\sim N(122,\,14^2)\). A reading above 140 mmHg is classed as high.

(a)(i)

On the axes in Figure 1, sketch the distribution of \(X\). Shade the region that represents the probability that a reading is high.

Figure 1: axes for the sketch of the distribution of \(X\)

GRAPHNotes

Show solution

symmetric bell-shaped curve with its peak at 122, approaching the axis at both ends	A1	The curve should be close to the axis by about 80 and 164 (three standard deviations from the mean).
region to the right of 140 shaded	A1	140 must be clearly to the right of the peak.

(2)

(a)(ii)

Find the probability that a reading is high.

ROUTINENotes

Show solution

\(P(X\gt140)\): normal cdf with lower bound 140, upper bound \(10^{99}\), \(\mu=122\), \(\sigma=14\)	(M1)	Or \(z=\frac{140-122}{14}=1.286\).
\(=0.0993\)	A1

Answer: \(0.0993\)

(2)

(b)

Find the reading that is exceeded by 5% of the adults who attend.

ROUTINENotes

Show solution

inverse normal with area 0.95 to the left, \(\mu=122\), \(\sigma=14\)	(M1)	Using area 0.05 (giving 99.0) earns M0A0.
\(145\) mmHg	A1	Accept 145.0.

Answer: \(145\) mmHg (145.0)

(2)

(c)

At a second pharmacy, the readings are normally distributed with mean 128 mmHg, and 20% of the readings are high.

Find the standard deviation of the readings at the second pharmacy.

MULTINotes

Show solution

inverse normal with area 0.8, \(\mu=0\), \(\sigma=1\): \(z=0.8416\)	(M1)
\(\frac{140-128}{\sigma}=0.8416\)	M1	Or \(140=128+0.8416\sigma\).
\(\sigma=14.3\) mmHg	A1	14.258…

Answer: \(14.3\) mmHg

(3)

(d)

Twenty adults are chosen at random from those who attend the first pharmacy. Let \(Y\) be the number of them with a high reading.

Show that the probability that at most 2 of them have a high reading is 0.68, correct to 2 decimal places.

MULTINotes

Show solution

\(Y\sim\text{B}(20,\,0.0993)\)	M1	Uses their probability from (a)(ii).
\(P(Y\le2)\) using binomial cdf with \(k=2\)	M1	Or \(P(Y=0)+P(Y=1)+P(Y=2)\).
\(=0.681\)	A1	Must see a value to at least 3 s.f.
\(\approx0.68\)	AG

(3)

(e)

The pharmacy lowers the threshold for a high reading from 140 mmHg to 130 mmHg.

Find the new value of \(P(Y\le2)\), and explain why it is smaller than the value in part (d).

MULTINotes

Show solution

\(P(X\gt130)=0.284\)	A1	0.28385…
\(Y\sim\text{B}(20,\,0.284)\): \(P(Y\le2)=0.0488\)	A1	Allow ECF from their probability.
a larger proportion of readings is now classed as high, so the success probability increases and the distribution of \(Y\) moves to larger values; 2 or fewer high readings among 20 adults becomes much less likely	R1	Must link the larger probability of a high reading to the shift in \(Y\).

Answer: \(0.0488\)

(3)

(f)

The pharmacy's automatic monitor can differ from a reference measurement. For one adult the monitor displays 137 mmHg, while a reference device records 142 mmHg.

Taking the reference reading as the exact value, calculate the percentage error of the monitor reading. Comment on the consequence of this error for the screening, using the threshold of 140 mmHg.

CONCEPTNotes

Show solution

\(\varepsilon=\left\|\frac{137-142}{142}\right\|\times100\%\)	M1	Exact value 142 in the denominator.
\(=3.52\%\)	A1
although the percentage error is small, it moves the reading from above 140 to below it, so this adult would wrongly be classed as not having a high reading	R1	Must refer to the threshold.

Answer: \(3.52\%\)

(3)

(g)(i)

During the day, 240 adults are classified by age group and by whether their reading is high.

	18–39	40–59	60 and over	Total
High	6	14	16	36
Not high	74	86	44	204
Total	80	100	60	240

A \(\chi^2\) test for independence is carried out at the 5% significance level. Write down the null hypothesis.

ROUTINENotes

Show solution

\(H_0\): whether a reading is high is independent of age group

Must be in context.

(1)

(g)(ii)

Find the expected frequency of adults aged 60 and over with a high reading.

ROUTINENotes

Show solution

\(E=\frac{60\times36}{240}\)	M1
\(=9\)	A1

Answer: \(9\)

(2)

(g)(iii)

Carry out the test. State the degrees of freedom, the value of \(\chi^2_{\text{calc}}\), the \(p\)-value and the conclusion in context.

MULTINotes

Show solution

\(\nu=(2-1)(3-1)=2\)	A1
\(\chi^2_{\text{calc}}=10.0\)	A1	10.013…
\(p=0.00669\)	A1
\(0.00669\lt0.05\), so reject \(H_0\): there is evidence that whether a reading is high depends on age group	R1	Accept \(10.0\gt5.99\). Allow ECF from their \(p\)-value.

Answer: \(\nu=2\), \(\chi^2_{\text{calc}}=10.0\), \(p=0.00669\); reject \(H_0\)

(4)

(h)

Describe the association shown by the data, and comment on whether the conclusion of part (g)(iii) can be applied to all adults in the town.

INTERPRETNotes

Show solution

adults aged 60 and over had more high readings than expected (16 against 9), and adults aged 18–39 fewer (6 against 12), so high readings are more common in the older age group	R1	Must compare observed with expected counts.
the adults chose to attend the screening day, so this is not a random sample of the town: people who are worried about their blood pressure may be more likely to attend, so the result may not apply to all adults	R1	Accept any valid source of bias in this context.

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
symmetric bell-shaped curve with its peak at 122, approaching the axis at both ends	A1	The curve should be close to the axis by about 80 and 164 (three standard deviations from the mean).
region to the right of 140 shaded	A1	140 must be clearly to the right of the peak.
Part (a)(ii)
\(P(X\gt140)\): normal cdf with lower bound 140, upper bound \(10^{99}\), \(\mu=122\), \(\sigma=14\)	(M1)	Or \(z=\frac{140-122}{14}=1.286\).
\(=0.0993\)	A1
Part (b)
inverse normal with area 0.95 to the left, \(\mu=122\), \(\sigma=14\)	(M1)	Using area 0.05 (giving 99.0) earns M0A0.
\(145\) mmHg	A1	Accept 145.0.
Part (c)
inverse normal with area 0.8, \(\mu=0\), \(\sigma=1\): \(z=0.8416\)	(M1)
\(\frac{140-128}{\sigma}=0.8416\)	M1	Or \(140=128+0.8416\sigma\).
\(\sigma=14.3\) mmHg	A1	14.258…
Part (d)
\(Y\sim\text{B}(20,\,0.0993)\)	M1	Uses their probability from (a)(ii).
\(P(Y\le2)\) using binomial cdf with \(k=2\)	M1	Or \(P(Y=0)+P(Y=1)+P(Y=2)\).
\(=0.681\)	A1	Must see a value to at least 3 s.f.
\(\approx0.68\)	AG
Part (e)
\(P(X\gt130)=0.284\)	A1	0.28385…
\(Y\sim\text{B}(20,\,0.284)\): \(P(Y\le2)=0.0488\)	A1	Allow ECF from their probability.
a larger proportion of readings is now classed as high, so the success probability increases and the distribution of \(Y\) moves to larger values; 2 or fewer high readings among 20 adults becomes much less likely	R1	Must link the larger probability of a high reading to the shift in \(Y\).
Part (f)
\(\varepsilon=\left\|\frac{137-142}{142}\right\|\times100\%\)	M1	Exact value 142 in the denominator.
\(=3.52\%\)	A1
although the percentage error is small, it moves the reading from above 140 to below it, so this adult would wrongly be classed as not having a high reading	R1	Must refer to the threshold.
Part (g)(i)
\(H_0\): whether a reading is high is independent of age group	A1	Must be in context.
Part (g)(ii)
\(E=\frac{60\times36}{240}\)	M1
\(=9\)	A1
Part (g)(iii)
\(\nu=(2-1)(3-1)=2\)	A1
\(\chi^2_{\text{calc}}=10.0\)	A1	10.013…
\(p=0.00669\)	A1
\(0.00669\lt0.05\), so reject \(H_0\): there is evidence that whether a reading is high depends on age group	R1	Accept \(10.0\gt5.99\). Allow ECF from their \(p\)-value.
Part (h)
adults aged 60 and over had more high readings than expected (16 against 9), and adults aged 18–39 fewer (6 against 12), so high readings are more common in the older age group	R1	Must compare observed with expected counts.
the adults chose to attend the screening day, so this is not a random sample of the town: people who are worried about their blood pressure may be more likely to attend, so the result may not apply to all adults	R1	Accept any valid source of bias in this context.

Answers: (a)(ii) \(0.0993\) · (b) \(145\) mmHg (145.0) · (c) \(14.3\) mmHg · (e) \(0.0488\) · (f) \(3.52\%\) · (g)(ii) \(9\) · (g)(iii) \(\nu=2\), \(\chi^2_{\text{calc}}=10.0\), \(p=0.00669\); reject \(H_0\)

Common error (b) Inverse normal always works with the area to the left. 'Exceeded by 5%' means 95% of the area lies to the left of the value.

More 4.4 questions in the question bank

4.5 T-tests and confidence intervals · 4.6 The Poisson distribution · 4.7 Linear combinations of random variablesmerged group

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Do carbon-plate shoes make runners faster? Confidence intervals and t-tests

4.5 T-tests and confidence intervals4.7 Linear combinations of random variables4.1 Sampling and descriptive statistics4.4 Normal distribution and chi-squared testsHL

A running club investigates whether carbon-plate shoes make its members faster. Ten members are chosen at random. Each runs the same 5 km course twice, one week apart: once in standard shoes and once in carbon-plate shoes, in a random order. Their times, in seconds, are shown in the table.

Runner	Standard shoes (s)	Carbon-plate shoes (s)
A	1252	1234
B	1386	1381
C	1174	1152
D	1503	1507
E	1297	1282
F	1428	1417
G	1341	1332
H	1216	1190
I	1465	1462
J	1319	1307

For each runner, let \(d\) = (time in standard shoes) \(-\) (time in carbon-plate shoes), so that a positive value of \(d\) is an improvement. The values of \(d\) are treated as a random sample from the population of differences for all members of the club, which has mean \(\mu\).

(a)

Find the ten values of \(d\), and hence find unbiased estimates of the mean and the variance of the population of differences.

HL onlyROUTINENotes

Show solution

\(d\): 18, 5, 22, \(-4\), 15, 11, 9, 26, 3, 12	(M1)	Allow one slip in the list for this mark.
\(\bar{d}=11.7\) s	A1
\(s_{n-1}^2=81.8\) s\(^2\)	A1	81.788…; from \(s_{n-1}=9.0437\). Using \(\sigma_n^2=73.6\) earns A0.

Answer: \(\bar{d}=11.7\) s, \(s_{n-1}^2=81.8\) s\(^2\)

(3)

(b)

The differences are modelled by \(D\sim\text{N}(\mu,\sigma^2)\). Write down the distribution of the sample mean \(\bar{D}\) for samples of 10 runners. Hence, using your estimate of \(\sigma^2\) from part (a), show that the standard deviation of \(\bar{D}\) is approximately 2.86 seconds.

HL onlyROUTINENotes

Show solution

\(\bar{D}\sim\text{N}\left(\mu,\frac{\sigma^2}{10}\right)\)	A1
\(\sqrt{\frac{81.79}{10}}\) or \(\frac{9.0437}{\sqrt{10}}=2.8599\)	M1	Must see the substitution and a value to at least 3 s.f.
\(\approx2.86\)	AG

(2)

(c)

Find a 95% confidence interval for \(\mu\).

HL onlyMULTINotes

Show solution

inverse \(t\) with area 0.975, \(\nu=9\): \(t=2.262\)	(M1)
\(11.7\pm2.262\times2.860\)	M1	Or GDC \(t\)-interval with the data list of \(d\).
\((5.23,\ 18.2)\) seconds	A1	Accept \(5.23\lt\mu\lt18.2\). Using \(z=1.96\) earns M1A0.

Answer: \((5.23,\ 18.2)\) s

(3)

(d)

A runner says: “There is a 95% probability that \(\mu\) lies in the interval found in part (c).”

Explain why this statement is not correct, and give a correct interpretation of the interval.

HL onlyINTERPRETNotes

Show solution

\(\mu\) is a fixed (unknown) value, so this particular interval either contains it or does not; it is the interval that varies from sample to sample	R1
if many samples of 10 runners were taken, about 95% of the intervals built in this way would contain \(\mu\); we are 95% confident that the mean improvement for club members is between 5.23 s and 18.2 s	R1	Either version, in context.

(2)

(e)

Explain why the club analyses the differences \(d\) with a one-sample method, rather than comparing the two columns of times with a two-sample pooled \(t\)-test.

HL onlyCONCEPTNotes

Show solution

the two sets of times come from the same ten runners, so they are not independent samples, which the pooled test requires	R1
the runners' times vary a lot (from about 1150 s to 1500 s); working with each runner's difference removes this variation, so the effect of the shoes is easier to detect	R1	Accept 'each runner acts as their own control'.

(2)

(f)

Carry out a suitable \(t\)-test, at the 5% significance level, to decide whether carbon-plate shoes reduce the mean time of club members. State your hypotheses, the value of the test statistic, the \(p\)-value and your conclusion in context.

HL onlyMULTINotes

Show solution

\(H_0:\mu=0\); \(H_1:\mu\gt0\)	A1	Accept in words: the mean improvement is zero / is greater than zero.
\(t=\frac{11.7-0}{2.860}=4.09\)	A1	4.0911…; from GDC one-sample \(t\)-test on \(d\).
\(p=0.00136\)	A1	A two-tailed value 0.00271 earns A0.
\(0.00136\lt0.05\), so reject \(H_0\)	R1
there is sufficient evidence at the 5% level that carbon-plate shoes reduce the mean 5 km time of club members	A1	Allow ECF from their \(p\)-value.

Answer: \(t=4.09\), \(p=0.00136\); reject \(H_0\)

(5)

(g)

Figure 1 shows the \(t\)-distribution used in part (f). The shaded region, to the right of \(t=c\), has area 0.05.

Find the value of \(c\). Explain, with reference to Figure 1, how the value of the test statistic in part (f) leads to the same conclusion.

Figure 1: the \(t\)-distribution with \(\nu=9\); the shaded critical region has area 0.05

HL onlyGRAPHNotes

Show solution

\(c=1.83\) (inverse \(t\) with area 0.95, \(\nu=9\))	A1	1.8331…
\(t=4.09\gt1.83\), so the test statistic lies in the shaded critical region and \(H_0\) is rejected	R1	Allow ECF from their \(t\).

Answer: \(c=1.83\)

(2)

(h)(i)

Using a normal model with the estimates found in part (a), estimate the proportion of club members who would be slower in carbon-plate shoes.

MODELNotes

Show solution

\(P(D\lt0)\) with \(D\sim N(11.7,\,9.044^2)\)	(M1)	Standard deviation \(\sqrt{81.79}\), not the variance; or \(z=\frac{0-11.7}{9.044}=-1.294\).
\(0.0979\)	A1	Accept 9.79%.

Answer: \(0.0979\) (about 10%)

(2)

(h)(ii)

State the assumption about the population of differences that is needed for parts (c) and (f). Comment, with reference to the ten values of \(d\), on whether this assumption is reasonable.

HL onlyINTERPRETNotes

Show solution

the differences are normally distributed	R1
the values are roughly symmetric (mean 11.7 close to median 11.5) with no extreme values, so the assumption is reasonable, although a sample of 10 is too small to check it with confidence	R1	Must refer to the data.

(2)

(i)

The club repeats the trial with 20 randomly chosen members. Assume that the new sample has the same mean and standard deviation of \(d\) as before.

Predict how the width of the 95% confidence interval changes, and estimate the new width.

HL onlyCONCEPTNotes

Show solution

the width \(2t_\nu\frac{s_{n-1}}{\sqrt n}\) decreases: \(\sqrt n\) increases and \(t_{19}=2.093\) is smaller than \(t_9=2.262\)	R1
new width \(=2\times2.093\times\frac{9.044}{\sqrt{20}}=8.47\) s (from 12.9 s), about 65% of the original width	A1	Accept 8.46 to 8.47. Allow 8.5.

Answer: about \(8.47\) s (from \(12.9\) s)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(d\): 18, 5, 22, \(-4\), 15, 11, 9, 26, 3, 12	(M1)	Allow one slip in the list for this mark.
\(\bar{d}=11.7\) s	A1
\(s_{n-1}^2=81.8\) s\(^2\)	A1	81.788…; from \(s_{n-1}=9.0437\). Using \(\sigma_n^2=73.6\) earns A0.
Part (b)
\(\bar{D}\sim\text{N}\left(\mu,\frac{\sigma^2}{10}\right)\)	A1
\(\sqrt{\frac{81.79}{10}}\) or \(\frac{9.0437}{\sqrt{10}}=2.8599\)	M1	Must see the substitution and a value to at least 3 s.f.
\(\approx2.86\)	AG
Part (c)
inverse \(t\) with area 0.975, \(\nu=9\): \(t=2.262\)	(M1)
\(11.7\pm2.262\times2.860\)	M1	Or GDC \(t\)-interval with the data list of \(d\).
\((5.23,\ 18.2)\) seconds	A1	Accept \(5.23\lt\mu\lt18.2\). Using \(z=1.96\) earns M1A0.
Part (d)
\(\mu\) is a fixed (unknown) value, so this particular interval either contains it or does not; it is the interval that varies from sample to sample	R1
if many samples of 10 runners were taken, about 95% of the intervals built in this way would contain \(\mu\); we are 95% confident that the mean improvement for club members is between 5.23 s and 18.2 s	R1	Either version, in context.
Part (e)
the two sets of times come from the same ten runners, so they are not independent samples, which the pooled test requires	R1
the runners' times vary a lot (from about 1150 s to 1500 s); working with each runner's difference removes this variation, so the effect of the shoes is easier to detect	R1	Accept 'each runner acts as their own control'.
Part (f)
\(H_0:\mu=0\); \(H_1:\mu\gt0\)	A1	Accept in words: the mean improvement is zero / is greater than zero.
\(t=\frac{11.7-0}{2.860}=4.09\)	A1	4.0911…; from GDC one-sample \(t\)-test on \(d\).
\(p=0.00136\)	A1	A two-tailed value 0.00271 earns A0.
\(0.00136\lt0.05\), so reject \(H_0\)	R1
there is sufficient evidence at the 5% level that carbon-plate shoes reduce the mean 5 km time of club members	A1	Allow ECF from their \(p\)-value.
Part (g)
\(c=1.83\) (inverse \(t\) with area 0.95, \(\nu=9\))	A1	1.8331…
\(t=4.09\gt1.83\), so the test statistic lies in the shaded critical region and \(H_0\) is rejected	R1	Allow ECF from their \(t\).
Part (h)(i)
\(P(D\lt0)\) with \(D\sim N(11.7,\,9.044^2)\)	(M1)	Standard deviation \(\sqrt{81.79}\), not the variance; or \(z=\frac{0-11.7}{9.044}=-1.294\).
\(0.0979\)	A1	Accept 9.79%.
Part (h)(ii)
the differences are normally distributed	R1
the values are roughly symmetric (mean 11.7 close to median 11.5) with no extreme values, so the assumption is reasonable, although a sample of 10 is too small to check it with confidence	R1	Must refer to the data.
Part (i)
the width \(2t_\nu\frac{s_{n-1}}{\sqrt n}\) decreases: \(\sqrt n\) increases and \(t_{19}=2.093\) is smaller than \(t_9=2.262\)	R1
new width \(=2\times2.093\times\frac{9.044}{\sqrt{20}}=8.47\) s (from 12.9 s), about 65% of the original width	A1	Accept 8.46 to 8.47. Allow 8.5.

Answers: (a) \(\bar{d}=11.7\) s, \(s_{n-1}^2=81.8\) s\(^2\) · (c) \((5.23,\ 18.2)\) s · (f) \(t=4.09\), \(p=0.00136\); reject \(H_0\) · (g) \(c=1.83\) · (h)(i) \(0.0979\) (about 10%) · (i) about \(8.47\) s (from \(12.9\) s)

Common error (a) For an estimate of a population variance use \(s_{n-1}\) (the GDC's \(s_x\)), not \(\sigma_x\). The divisor-\(n\) value is too small on average.

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Arrivals at an emergency department: Poisson models and combined variables

4.6 The Poisson distribution4.7 Linear combinations of random variables4.3 Probability and the binomial distribution4.4 Normal distribution and chi-squared testsHL

Records from the previous year show that, during weekday daytime hours, patients arrive at the emergency department of a city hospital on foot (walk-in patients) at a mean rate of 4.5 per hour, and by ambulance at a mean rate of 1.2 per hour.

Let \(W\) and \(A\) be the numbers of walk-in patients and ambulance patients who arrive in one hour. They are modelled by independent distributions \(W\sim\text{Po}(4.5)\) and \(A\sim\text{Po}(1.2)\).

(a)

Give one reason why a Poisson model is reasonable for \(W\), and one reason why the arrivals of walk-in patients might not be independent.

HL onlyINTERPRETNotes

Show solution

patients usually arrive separately, at random, with no upper limit on the number in an hour, and the mean rate is roughly constant during weekday daytime hours	R1	One valid reason in context is enough.
an incident such as a road accident, a fire or an outbreak of food poisoning can bring several patients at about the same time, so one arrival makes others more likely	R1	Accept people arriving together in groups (not singly). 'Arrivals are random' alone earns R0.

(2)

(b)

Find the probability that at least 3 walk-in patients arrive in a 20-minute interval.

HL onlyROUTINENotes

Show solution

\(\lambda=\frac{20}{60}\times4.5=1.5\); \(P(X\ge3)=1-P(X\le2)\)	(M1)	Using 4.5 for 20 minutes earns M0.
\(=0.191\)	A1

Answer: \(0.191\)

(2)

(c)

Let \(T=W+A\) be the total number of patients who arrive in one hour.

Show that the probability that more than 10 patients arrive in one hour is 0.031, correct to 2 significant figures.

HL onlyMULTINotes

Show solution

\(T\sim\text{Po}(4.5+1.2)=\text{Po}(5.7)\), since \(W\) and \(A\) are independent	M1
\(P(T\gt10)=1-P(T\le10)\)	M1	Using \(1-P(T\le11)\) earns M0.
\(=1-0.96856=0.0314\)	A1	Must see a value to at least 3 s.f.
\(\approx0.031\)	AG

(3)

(d)

An hour is called busy if 8 or more patients arrive. The numbers of arrivals in different hours are independent.

Find the probability that at least 3 hours of an 8-hour shift are busy.

HL onlyMULTINotes

Show solution

\(P(T\ge8)=1-P(T\le7)\)	(M1)
\(=0.216\)	A1	0.21585…
\(B\sim\text{B}(8,\,0.216)\); \(P(B\ge3)=1-P(B\le2)\)	M1	Binomial with their probability of a busy hour.
\(=0.239\)	A1	Allow ECF from their probability.

Answer: \(0.239\)

(4)

(e)(i)

To check the model for \(T\), the total number of arrivals was recorded in 120 randomly chosen weekday daytime hours this year. The table shows the observed frequencies and the expected frequencies for \(T\sim\text{Po}(5.7)\), correct to 2 decimal places.

Arrivals in one hour	\(\le2\)	3	4	5	6	7	8	\(\ge9\)
Observed frequency	14	8	15	19	22	17	8	17
Expected frequency	\(q\)	12.39	17.66	20.13	19.13	15.57	11.10	14.81

Find the value of \(q\).

HL onlyROUTINENotes

Show solution

\(q=120\times P(T\le2)\)	(M1)
\(=9.21\)	A1	9.2128…

Answer: \(q=9.21\)

(2)

(e)(ii)

Figure 1 shows the observed frequencies and the expected frequencies for 3 or more arrivals.

With reference to Figure 1 and your answer to part (e)(i), describe how the observed frequencies compare with those predicted by the model.

Figure 1: observed frequencies (bars) and expected frequencies for \(T\sim\text{Po}(5.7)\) (points), for 3 or more arrivals

HL onlyGRAPHNotes

Show solution

there were more quiet hours than predicted (14 hours with 2 or fewer arrivals against 9.21), and fewer hours with exactly 3 or exactly 8 arrivals than predicted	R1	Must quote at least one category with numbers.
otherwise the observed and expected frequencies are close and have a similar shape, with most hours having 4 to 7 arrivals	R1	Accept 'the differences are small and in both directions'.

(2)

(e)(iii)

Carry out a \(\chi^2\) goodness-of-fit test at the 5% significance level to decide whether \(\text{Po}(5.7)\) is a suitable model for \(T\). State your hypotheses, the degrees of freedom, the value of \(\chi^2_{\text{calc}}\), the \(p\)-value and your conclusion in context.

MULTINotes

Show solution

\(H_0\): \(T\) follows the distribution \(\text{Po}(5.7)\); \(H_1\): it does not	A1
\(\nu=8-1=7\)	A1	All expected frequencies are at least 5, so no categories are combined; \(\lambda\) is not estimated from these data.
\(\chi^2_{\text{calc}}=6.26\), \(p=0.510\)	A1	Both values needed.
\(0.510\gt0.05\), so do not reject \(H_0\): there is insufficient evidence that \(\text{Po}(5.7)\) is unsuitable for the number of arrivals in an hour	R1	'Accept \(H_0\)' or 'proves the model' earns R0. Allow ECF from their \(p\)-value.

Answer: \(\nu=7\), \(\chi^2_{\text{calc}}=6.26\), \(p=0.510\); do not reject \(H_0\)

(4)

(f)(i)

The hospital estimates that the first assessment of a patient costs €300 for a walk-in patient and €500 for an ambulance patient. The total cost, in euros, of first assessments for the patients who arrive in one hour is \(C=300W+500A\).

Find the mean and the standard deviation of \(C\).

HL onlyMULTINotes

Show solution

\(\operatorname{E}(C)=300(4.5)+500(1.2)=1950\)	A1
\(\operatorname{Var}(C)=300^2(4.5)+500^2(1.2)=705\,000\)	M1	Uses \(\operatorname{Var}(W)=4.5\) and \(\operatorname{Var}(A)=1.2\) with squared coefficients.
standard deviation \(=\sqrt{705\,000}=840\) euros	A1	839.64…

Answer: mean €1950, standard deviation €840

(3)

(f)(ii)

Explain why \(C\) does not follow a Poisson distribution.

HL onlyCONCEPTNotes

Show solution

for a Poisson variable the mean equals the variance, but \(\operatorname{E}(C)=1950\ne\operatorname{Var}(C)=705\,000\)

Accept: \(C\) only takes multiples of 100, not every non-negative integer.

(1)

(g)

During a city festival the mean rate of walk-in patients rises to 7.5 per hour. The ambulance rate does not change.

Find the probability that more than 10 patients arrive in one hour during the festival, and comment on your answer.

HL onlyMODELNotes

Show solution

\(T\sim\text{Po}(7.5+1.2)=\text{Po}(8.7)\)	M1
\(P(T\gt10)=1-P(T\le10)=0.259\)	A1
more than 10 arrivals in an hour becomes about 8 times as likely as usual (0.259 against 0.031), about one hour in four, so the department should plan extra staff for the festival	R1	Must compare with part (c).

Answer: \(0.259\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
patients usually arrive separately, at random, with no upper limit on the number in an hour, and the mean rate is roughly constant during weekday daytime hours	R1	One valid reason in context is enough.
an incident such as a road accident, a fire or an outbreak of food poisoning can bring several patients at about the same time, so one arrival makes others more likely	R1	Accept people arriving together in groups (not singly). 'Arrivals are random' alone earns R0.
Part (b)
\(\lambda=\frac{20}{60}\times4.5=1.5\); \(P(X\ge3)=1-P(X\le2)\)	(M1)	Using 4.5 for 20 minutes earns M0.
\(=0.191\)	A1
Part (c)
\(T\sim\text{Po}(4.5+1.2)=\text{Po}(5.7)\), since \(W\) and \(A\) are independent	M1
\(P(T\gt10)=1-P(T\le10)\)	M1	Using \(1-P(T\le11)\) earns M0.
\(=1-0.96856=0.0314\)	A1	Must see a value to at least 3 s.f.
\(\approx0.031\)	AG
Part (d)
\(P(T\ge8)=1-P(T\le7)\)	(M1)
\(=0.216\)	A1	0.21585…
\(B\sim\text{B}(8,\,0.216)\); \(P(B\ge3)=1-P(B\le2)\)	M1	Binomial with their probability of a busy hour.
\(=0.239\)	A1	Allow ECF from their probability.
Part (e)(i)
\(q=120\times P(T\le2)\)	(M1)
\(=9.21\)	A1	9.2128…
Part (e)(ii)
there were more quiet hours than predicted (14 hours with 2 or fewer arrivals against 9.21), and fewer hours with exactly 3 or exactly 8 arrivals than predicted	R1	Must quote at least one category with numbers.
otherwise the observed and expected frequencies are close and have a similar shape, with most hours having 4 to 7 arrivals	R1	Accept 'the differences are small and in both directions'.
Part (e)(iii)
\(H_0\): \(T\) follows the distribution \(\text{Po}(5.7)\); \(H_1\): it does not	A1
\(\nu=8-1=7\)	A1	All expected frequencies are at least 5, so no categories are combined; \(\lambda\) is not estimated from these data.
\(\chi^2_{\text{calc}}=6.26\), \(p=0.510\)	A1	Both values needed.
\(0.510\gt0.05\), so do not reject \(H_0\): there is insufficient evidence that \(\text{Po}(5.7)\) is unsuitable for the number of arrivals in an hour	R1	'Accept \(H_0\)' or 'proves the model' earns R0. Allow ECF from their \(p\)-value.
Part (f)(i)
\(\operatorname{E}(C)=300(4.5)+500(1.2)=1950\)	A1
\(\operatorname{Var}(C)=300^2(4.5)+500^2(1.2)=705\,000\)	M1	Uses \(\operatorname{Var}(W)=4.5\) and \(\operatorname{Var}(A)=1.2\) with squared coefficients.
standard deviation \(=\sqrt{705\,000}=840\) euros	A1	839.64…
Part (f)(ii)
for a Poisson variable the mean equals the variance, but \(\operatorname{E}(C)=1950\ne\operatorname{Var}(C)=705\,000\)	R1	Accept: \(C\) only takes multiples of 100, not every non-negative integer.
Part (g)
\(T\sim\text{Po}(7.5+1.2)=\text{Po}(8.7)\)	M1
\(P(T\gt10)=1-P(T\le10)=0.259\)	A1
more than 10 arrivals in an hour becomes about 8 times as likely as usual (0.259 against 0.031), about one hour in four, so the department should plan extra staff for the festival	R1	Must compare with part (c).

Answers: (b) \(0.191\) · (d) \(0.239\) · (e)(i) \(q=9.21\) · (e)(iii) \(\nu=7\), \(\chi^2_{\text{calc}}=6.26\), \(p=0.510\); do not reject \(H_0\) · (f)(i) mean €1950, standard deviation €840 · (g) \(0.259\)

Common error (b) Rescale \(\lambda\) to the interval in the question before calculating: 20 minutes is \(\frac13\) of an hour, so \(\lambda=1.5\), not 4.5.

More 4.6 questions in the question bank

4.8 Markov chains and transition matrices · 4.9 Non-linear regression and the coefficient of determinationmerged group

+ Add to test✓ In your testCC-AI-48-49-128 marks~35 min

A music streaming service: subscribers moving between plans

4.8 Markov chains and transition matrices1.6 Matrices and matrix algebra1.7 Eigenvalues and eigenvectors4.3 Probability and the binomial distributionHL

A music streaming service offers three plans: Free (F), Premium (P) and Group (G). Each subscriber is on exactly one plan at the start of each month. The plan a subscriber is on from one month to the next is modelled as a Markov chain, with the monthly transition probabilities shown in Figure 1.

At the start of January (month 0) the service has 60 000 Free, 30 000 Premium and 10 000 Group subscribers. Assume that the total number of subscribers stays at 100 000.

Figure 1: monthly transition diagram for the three plans

(a)(i)

Write down the transition matrix \(\mathbf T\), with the states in the order F, P, G, so that \(\mathbf s_{n+1}=\mathbf T\mathbf s_n\).

HL onlyROUTINENotes

Show solution

each column holds the probabilities of moving from one plan, e.g. first column \((0.80,\ 0.15,\ 0.05)\)	A1	At least one column correct and placed as a column.
\(\mathbf T=\begin{pmatrix}0.80&0.10&0.05\\0.15&0.85&0.05\\0.05&0.05&0.90\end{pmatrix}\)	A1	All nine entries correct. Award A0A0 for the transpose (rows summing to 1).

Answer: \(\mathbf T=\begin{pmatrix}0.80&0.10&0.05\\0.15&0.85&0.05\\0.05&0.05&0.90\end{pmatrix}\)

(2)

(a)(ii)

Explain why the entries in each column of \(\mathbf T\) add up to 1.

HL onlyCONCEPTNotes

Show solution

a column lists the probabilities of every possible plan next month for a subscriber on one plan this month; each subscriber must be on exactly one plan, so these probabilities sum to 1

Accept 'the subscriber has to go somewhere (including staying)'.

(1)

(b)

Find the number of subscribers on each plan at the start of July (month 6).

HL onlyROUTINENotes

Show solution

\(\mathbf s_6=\mathbf T^6\begin{pmatrix}60\,000\\30\,000\\10\,000\end{pmatrix}\)	(M1)	Award for a correct product with the power on \(\mathbf T\) only.
Free 33 587, Premium 41 880, Group 24 533	A1	Accept 33 600, 41 900, 24 500. Allow ECF from (a)(i).

Answer: Free 33 587, Premium 41 880, Group 24 533

(2)

(c)(i)

A subscriber is on the Premium plan at the start of January.

Find the probability that this subscriber is on the Premium plan at the start of each of February, March and April.

ROUTINENotes

Show solution

\(0.85\times0.85\times0.85\)	M1	Multiplies three 'stay on Premium' probabilities.
\(=0.614\)	A1	Accept 0.614125.

Answer: \(0.614\)

(2)

(c)(ii)

Find the probability that this subscriber is on the Premium plan at the start of April, and explain why this is greater than your answer to (c)(i).

HL onlyCONCEPTNotes

Show solution

entry in row P, column P of \(\mathbf T^3\): \(0.659\)	A1	Accept 0.65875. Allow ECF from (a)(i).
this also counts subscribers who leave Premium and return before April (e.g. P to F to P to P), whereas (c)(i) needs them to stay on Premium every month	R1

Answer: \(0.659\)

(2)

(d)(i)

The long-term distribution of subscribers is \(\mathbf s=\begin{pmatrix}a\\b\\c\end{pmatrix}\), where \(a+b+c=1\).

By solving \(\mathbf T\mathbf s=\mathbf s\), show that in the long term \(\dfrac{7}{18}\) of the subscribers are on the Premium plan.

HL onlyALGEBRANotes

Show solution

at least two equations from \(\mathbf T\mathbf s=\mathbf s\), e.g. \(0.80a+0.10b+0.05c=a\) and \(0.05a+0.05b+0.90c=c\)	M1
\(4a=2b+c\) and \(2c=a+b\)	A1	Or equivalent simplified forms, e.g. \(3b=3a+c\) from the second row.
using \(a+b+c=1\): \(2c=1-c\), so \(c=\frac{1}{3}\)	M1	Uses the condition that the entries sum to 1.
\(4a=2\left(\frac{2}{3}-a\right)+\frac{1}{3}\), so \(a=\frac{5}{18}\)	A1
\(b=1-\frac{5}{18}-\frac{6}{18}=\frac{7}{18}\)	AG	A GDC eigenvector or a high power of the matrix alone does not 'solve' the system: award at most M1.

Answer: \(\mathbf s=\left(\frac{5}{18},\ \frac{7}{18},\ \frac{1}{3}\right)^T\)

(4)

(d)(ii)

\(\mathbf s\) is an eigenvector of \(\mathbf T\). Write down the corresponding eigenvalue.

The other two eigenvalues of \(\mathbf T\) are \(0.85\) and \(\lambda\). Use the trace of \(\mathbf T\) to find \(\lambda\).

HL onlyCONCEPTNotes

Show solution

eigenvalue \(1\)	A1
\(\operatorname{tr}\mathbf T=0.80+0.85+0.90=2.55=1+0.85+\lambda\)	M1	Sum of the eigenvalues equated to the trace.
\(\lambda=0.7\)	A1	A value from the GDC eigenvalue function without the trace method: A1 only.

Answer: \(1\); \(\lambda=0.7\)

(3)

(e)(i)

Let \(p_n\) be the proportion of subscribers on the Premium plan at the start of month \(n\). Figure 2 shows \(p_n\) for \(0\le n\le24\).

By drawing a suitable horizontal line on Figure 2 and using your GDC, find the smallest value of \(N\) such that \(p_n\) differs from \(\dfrac{7}{18}\) by less than 0.01 for every \(n\ge N\).

Figure 2: the proportion of subscribers on the Premium plan at the start of month \(n\), for \(0\le n\le24\)

HL onlyGRAPHNotes

Show solution

draws the line \(p=\frac{7}{18}+0.01\approx0.399\) and sees that the points fall below it from about \(n=16\) or \(17\); then checks with the GDC, e.g. \(p_{16}=0.39963\) (difference \(0.0107\)) and \(p_{17}=0.39814\) (difference \(0.0093\))	(M1)	Accept a GDC table of \(\mathbf T^n\mathbf s_0\). The graph alone cannot separate \(n=16\) from \(n=17\) (the values differ from the line by less than 0.001).
\(N=17\)	A1	\(N=2\) scores M0A0.

Answer: \(N=17\)

(2)

(e)(ii)

A manager notices that \(p_2=0.3815\), which is within 0.01 of \(\dfrac{7}{18}\), and concludes that the Premium share has already settled at its long-term value.

Use Figure 2 to comment on this conclusion.

HL onlyINTERPRETNotes

Show solution

the conclusion is wrong: after month 2 the proportion keeps rising, to about 0.419 at month 6, well above \(\frac{7}{18}\)	R1	Must refer to the overshoot shown in the graph.
\(p_2\) is close to \(\frac{7}{18}\) only because the graph is passing through that level on the way up; it only stays within 0.01 from month 17 onwards	R1	Allow ECF from (e)(i).

(2)

(f)

State two assumptions of this Markov chain model that may not be realistic for the streaming service, giving a brief reason for each.

HL onlyMODELNotes

Show solution

e.g. the transition probabilities stay the same every month, but special offers, price changes or the time of year can change them	R1
e.g. the next plan depends only on the current plan, but a subscriber's history (such as how long they have been on Premium) may matter; or the total stays at 100 000, but new users join and others cancel	R1	Any two different assumptions, each with a reason.

(2)

(g)(i)

The service raises the price of the Premium plan from €10 to €12 per month. Now, each month, 80% of Premium subscribers stay on Premium, 15% move to Free and 5% move to Group. The other transition probabilities are unchanged.

Find the new long-term distribution of subscribers.

HL onlyMULTINotes

Show solution

new matrix with second column \((0.15,\ 0.80,\ 0.05)\): \(\begin{pmatrix}0.80&0.15&0.05\\0.15&0.80&0.05\\0.05&0.05&0.90\end{pmatrix}\)	M1
attempt to find the steady state, e.g. solving \(\mathbf T\mathbf s=\mathbf s\) with entries summing to 1, an eigenvector for eigenvalue 1, or a high power such as \(\mathbf T^{100}\)	M1
\(\left(\frac{1}{3},\ \frac{1}{3},\ \frac{1}{3}\right)^T\): one third on each plan	A1	Accept 0.333 for each.

Answer: \(\left(\frac{1}{3},\ \frac{1}{3},\ \frac{1}{3}\right)^T\)

(3)

(g)(ii)

Considering only the income from Premium subscriptions, determine whether the price rise increases the long-term monthly income from the 100 000 subscribers.

HL onlyMODELNotes

Show solution

long-term income = long-term number on Premium \(\times\) price, e.g. \(\frac{7}{18}\times100\,000\times10\)	M1	For either situation.
before: €388 889; after: \(\frac{1}{3}\times100\,000\times12=\) €400 000	A1	Accept €389 000. Allow ECF from (g)(i).
yes: the long-term Premium income rises, by about €11 100 per month (about 2.9%), even though fewer subscribers stay on Premium	R1	Conclusion must follow from their values.

Answer: Yes: €400 000 compared with €388 889 per month

(3)

Show full solution

Working	Mark	Notes
Part (a)(i)
each column holds the probabilities of moving from one plan, e.g. first column \((0.80,\ 0.15,\ 0.05)\)	A1	At least one column correct and placed as a column.
\(\mathbf T=\begin{pmatrix}0.80&0.10&0.05\\0.15&0.85&0.05\\0.05&0.05&0.90\end{pmatrix}\)	A1	All nine entries correct. Award A0A0 for the transpose (rows summing to 1).
Part (a)(ii)
a column lists the probabilities of every possible plan next month for a subscriber on one plan this month; each subscriber must be on exactly one plan, so these probabilities sum to 1	R1	Accept 'the subscriber has to go somewhere (including staying)'.
Part (b)
\(\mathbf s_6=\mathbf T^6\begin{pmatrix}60\,000\\30\,000\\10\,000\end{pmatrix}\)	(M1)	Award for a correct product with the power on \(\mathbf T\) only.
Free 33 587, Premium 41 880, Group 24 533	A1	Accept 33 600, 41 900, 24 500. Allow ECF from (a)(i).
Part (c)(i)
\(0.85\times0.85\times0.85\)	M1	Multiplies three 'stay on Premium' probabilities.
\(=0.614\)	A1	Accept 0.614125.
Part (c)(ii)
entry in row P, column P of \(\mathbf T^3\): \(0.659\)	A1	Accept 0.65875. Allow ECF from (a)(i).
this also counts subscribers who leave Premium and return before April (e.g. P to F to P to P), whereas (c)(i) needs them to stay on Premium every month	R1
Part (d)(i)
at least two equations from \(\mathbf T\mathbf s=\mathbf s\), e.g. \(0.80a+0.10b+0.05c=a\) and \(0.05a+0.05b+0.90c=c\)	M1
\(4a=2b+c\) and \(2c=a+b\)	A1	Or equivalent simplified forms, e.g. \(3b=3a+c\) from the second row.
using \(a+b+c=1\): \(2c=1-c\), so \(c=\frac{1}{3}\)	M1	Uses the condition that the entries sum to 1.
\(4a=2\left(\frac{2}{3}-a\right)+\frac{1}{3}\), so \(a=\frac{5}{18}\)	A1
\(b=1-\frac{5}{18}-\frac{6}{18}=\frac{7}{18}\)	AG	A GDC eigenvector or a high power of the matrix alone does not 'solve' the system: award at most M1.
Part (d)(ii)
eigenvalue \(1\)	A1
\(\operatorname{tr}\mathbf T=0.80+0.85+0.90=2.55=1+0.85+\lambda\)	M1	Sum of the eigenvalues equated to the trace.
\(\lambda=0.7\)	A1	A value from the GDC eigenvalue function without the trace method: A1 only.
Part (e)(i)
draws the line \(p=\frac{7}{18}+0.01\approx0.399\) and sees that the points fall below it from about \(n=16\) or \(17\); then checks with the GDC, e.g. \(p_{16}=0.39963\) (difference \(0.0107\)) and \(p_{17}=0.39814\) (difference \(0.0093\))	(M1)	Accept a GDC table of \(\mathbf T^n\mathbf s_0\). The graph alone cannot separate \(n=16\) from \(n=17\) (the values differ from the line by less than 0.001).
\(N=17\)	A1	\(N=2\) scores M0A0.
Part (e)(ii)
the conclusion is wrong: after month 2 the proportion keeps rising, to about 0.419 at month 6, well above \(\frac{7}{18}\)	R1	Must refer to the overshoot shown in the graph.
\(p_2\) is close to \(\frac{7}{18}\) only because the graph is passing through that level on the way up; it only stays within 0.01 from month 17 onwards	R1	Allow ECF from (e)(i).
Part (f)
e.g. the transition probabilities stay the same every month, but special offers, price changes or the time of year can change them	R1
e.g. the next plan depends only on the current plan, but a subscriber's history (such as how long they have been on Premium) may matter; or the total stays at 100 000, but new users join and others cancel	R1	Any two different assumptions, each with a reason.
Part (g)(i)
new matrix with second column \((0.15,\ 0.80,\ 0.05)\): \(\begin{pmatrix}0.80&0.15&0.05\\0.15&0.80&0.05\\0.05&0.05&0.90\end{pmatrix}\)	M1
attempt to find the steady state, e.g. solving \(\mathbf T\mathbf s=\mathbf s\) with entries summing to 1, an eigenvector for eigenvalue 1, or a high power such as \(\mathbf T^{100}\)	M1
\(\left(\frac{1}{3},\ \frac{1}{3},\ \frac{1}{3}\right)^T\): one third on each plan	A1	Accept 0.333 for each.
Part (g)(ii)
long-term income = long-term number on Premium \(\times\) price, e.g. \(\frac{7}{18}\times100\,000\times10\)	M1	For either situation.
before: €388 889; after: \(\frac{1}{3}\times100\,000\times12=\) €400 000	A1	Accept €389 000. Allow ECF from (g)(i).
yes: the long-term Premium income rises, by about €11 100 per month (about 2.9%), even though fewer subscribers stay on Premium	R1	Conclusion must follow from their values.

Answers: (a)(i) \(\mathbf T=\begin{pmatrix}0.80&0.10&0.05\\0.15&0.85&0.05\\0.05&0.05&0.90\end{pmatrix}\) · (b) Free 33 587, Premium 41 880, Group 24 533 · (c)(i) \(0.614\) · (c)(ii) \(0.659\) · (d)(i) \(\mathbf s=\left(\frac{5}{18},\ \frac{7}{18},\ \frac{1}{3}\right)^T\) · (d)(ii) \(1\); \(\lambda=0.7\) · (e)(i) \(N=17\) · (g)(i) \(\left(\frac{1}{3},\ \frac{1}{3},\ \frac{1}{3}\right)^T\) · (g)(ii) Yes: €400 000 compared with €388 889 per month

Common error (a)(i) Writing each state's outgoing probabilities as a row gives the transpose of \(\mathbf T\). With \(\mathbf s_{n+1}=\mathbf T\mathbf s_n\) each column is a 'from' state, so every column must add up to 1.

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How much power does a cyclist need? Power regression and the coefficient of determination

4.9 Non-linear regression and the coefficient of determination2.5 Logarithmic scales and linearising data2.2 Quadratic and cubic models4.2 Correlation and regressionHL

A cyclist rides on a flat, sheltered road in still air. A power meter on the bicycle records the mean power \(P\) watts (W) needed to ride at each of eight steady speeds \(v\) km/h.

Speed \(v\) (km/h)	16	20	24	28	32	36	40	44
Power \(P\) (W)	29	52	84	125	179	238	305	414

Figure 1: power needed against speed

(a)

Use your GDC to write down the coefficient of determination \(R^2\) for each of the following models:

the linear model \(P=mv+c\)
the quadratic model \(P=av^2+bv+c\)
the power model \(P=av^n\).

HL onlyROUTINENotes

Show solution

linear: \(R^2=0.953\)	A1	Linear model \(P=13.3v-221\).
quadratic: \(R^2=0.998\)	A1	\(P=0.362v^2-8.38v+73.7\); accept 0.9978.
power: \(R^2=0.9997\)	A1	\(P=0.0215v^{2.60}\). Some GDCs compute \(R^2\) for a power fit differently; accept 0.998 (0.9983).

Answer: \(0.953\), \(0.998\), \(0.9997\)

(3)

(b)

Suggest which of the three models is the most appropriate for these data. Give two reasons, at least one of which refers to the context.

HL onlyMODELNotes

Show solution

power model, e.g. it has the highest \(R^2\) and only two parameters	R1	The choice of model must be stated.
context, e.g. the power model gives \(P=0\) when \(v=0\); the linear model predicts a negative power (about \(-8\) W) at 16 km/h; the quadratic predicts its least power at about 12 km/h and about 74 W when the bike is not moving	R1	A reason based on \(R^2\) alone scores R1 R0.

Answer: the power model

(2)

(c)

The table gives \(\ln v\) and \(\ln P\) for the data, correct to 3 decimal places.

\(\ln v\)	2.773	2.996	3.178	3.332	3.466	3.584	3.689	3.784
\(\ln P\)	3.367	3.951		4.828	5.187	5.472		6.026

Complete the table.

HL onlyROUTINENotes

Show solution

\(v=24\): \(\ln 84=4.431\)	A1
\(v=40\): \(\ln 305=5.720\)	A1	Accept 5.72. Do not accept log base 10 values (1.924, 2.484).

Answer: \(4.431\) and \(5.720\)

(2)

(d)

Figure 2 shows six of the eight points \((\ln v,\ \ln P)\).

Plot the two remaining points on Figure 2, and state what the pattern of all eight points suggests about the relationship between \(P\) and \(v\).

Figure 2: six of the eight points \((\ln v,\ \ln P)\)

HL onlyGRAPHNotes

Show solution

\((3.178,\ 4.431)\) plotted	A1	Within half a small square. Allow ECF from (c).
\((3.689,\ 5.720)\) plotted	A1
the points lie close to a straight line, so \(\ln P\) is (approximately) linear in \(\ln v\) and a power model \(P=av^n\) is suitable	R1	Must refer to the straight line and to a power relationship.

(3)

(e)(i)

Find the equation of the regression line of \(\ln P\) on \(\ln v\), and the value of Pearson's product-moment correlation coefficient \(r\).

ROUTINENotes

Show solution

gradient \(2.60\)	A1	From the table values the GDC gives 2.601 (2.6014); from unrounded logarithms 2.601 (2.6007).
\(\ln P=2.60\ln v-3.84\)	A1	Intercept \(-3.84\) (accept \(-3.843\) from the table values or \(-3.840\) from unrounded logarithms).
\(r=0.9998\)	A1	Accept 1.00 (3 s.f.) if 0.9998 seen earlier, or 0.99983.

Answer: \(\ln P=2.60\ln v-3.84\); \(r=0.9998\)

(3)

(e)(ii)

Hence show that the data can be modelled by \(P=av^n\), where \(n=2.60\) correct to 3 significant figures, and find the value of \(a\).

HL onlyALGEBRANotes

Show solution

\(P=e^{2.60\ln v-3.84}=e^{-3.84}\times e^{2.60\ln v}\)	M1	Takes exponentials of both sides of their line.
\(e^{2.60\ln v}=e^{\ln v^{2.60}}=v^{2.60}\), so \(P=e^{-3.84}v^{2.60}\) with \(n=2.60\)	M1	Uses \(k\ln v=\ln v^k\). Or compares \(\ln P=\ln a+n\ln v\) term by term.
\(a=e^{-3.84}=0.0215\)	A1	Accept 0.0214 or 0.0215.
\(n=2.60\)	AG

Answer: \(a=0.0215\)

(3)

(f)

Use the model to estimate the power needed to ride at 50 km/h, and comment on the reliability of your estimate.

HL onlyMODELNotes

Show solution

\(0.0215\times50^{2.60}\)	M1	Allow ECF from (e)(ii).
\(=562\) W	A1	Accept answers from 559 W to 564 W (\(a=0.0214\) gives 559 W).
50 km/h lies outside the data (16 to 44 km/h), so this is extrapolation and less reliable (although only a little beyond the data)	R1	Must mention extrapolation or the data range.

Answer: about 562 W

(3)

(g)

A classmate says: “\(R^2\) for the power model is so close to 1 that this proves the power needed depends on speed exactly by \(P=av^n\).”

Explain why this statement is not correct.

HL onlyINTERPRETNotes

Show solution

\(R^2\) only gives the proportion of the variation in \(P\) that the model explains for these eight data points	R1	Do not accept 'the model is 99.97% accurate'.
other models can also give \(R^2\) close to 1 (the quadratic gave 0.998), so a high \(R^2\) cannot show which formula is the true one, or that it holds outside 16 to 44 km/h	R1	Accept a reference to measurement error or to checking the residuals.

(2)

(h)

The cyclist repeats the test wearing an aerodynamic helmet. At every speed the power needed is 6% lower than before.

Describe, with reasons, the effect on the values of \(a\) and \(n\) in the power model.

HL onlyCONCEPTNotes

Show solution

each new power is \(0.94P\), so \(\ln(0.94P)=\ln P+\ln0.94=\ln P-0.0619\): every point on Figure 2 moves down by the same amount	M1
the gradient of the line is unchanged, so \(n\) stays \(2.60\)	A1
the intercept decreases by \(0.0619\), so \(a\) is multiplied by \(0.94\): \(a=0.0202\)	A1	Allow ECF from (e)(ii).

Answer: \(n\) unchanged; \(a\) becomes \(0.94\times0.0215=0.0202\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
linear: \(R^2=0.953\)	A1	Linear model \(P=13.3v-221\).
quadratic: \(R^2=0.998\)	A1	\(P=0.362v^2-8.38v+73.7\); accept 0.9978.
power: \(R^2=0.9997\)	A1	\(P=0.0215v^{2.60}\). Some GDCs compute \(R^2\) for a power fit differently; accept 0.998 (0.9983).
Part (b)
power model, e.g. it has the highest \(R^2\) and only two parameters	R1	The choice of model must be stated.
context, e.g. the power model gives \(P=0\) when \(v=0\); the linear model predicts a negative power (about \(-8\) W) at 16 km/h; the quadratic predicts its least power at about 12 km/h and about 74 W when the bike is not moving	R1	A reason based on \(R^2\) alone scores R1 R0.
Part (c)
\(v=24\): \(\ln 84=4.431\)	A1
\(v=40\): \(\ln 305=5.720\)	A1	Accept 5.72. Do not accept log base 10 values (1.924, 2.484).
Part (d)
\((3.178,\ 4.431)\) plotted	A1	Within half a small square. Allow ECF from (c).
\((3.689,\ 5.720)\) plotted	A1
the points lie close to a straight line, so \(\ln P\) is (approximately) linear in \(\ln v\) and a power model \(P=av^n\) is suitable	R1	Must refer to the straight line and to a power relationship.
Part (e)(i)
gradient \(2.60\)	A1	From the table values the GDC gives 2.601 (2.6014); from unrounded logarithms 2.601 (2.6007).
\(\ln P=2.60\ln v-3.84\)	A1	Intercept \(-3.84\) (accept \(-3.843\) from the table values or \(-3.840\) from unrounded logarithms).
\(r=0.9998\)	A1	Accept 1.00 (3 s.f.) if 0.9998 seen earlier, or 0.99983.
Part (e)(ii)
\(P=e^{2.60\ln v-3.84}=e^{-3.84}\times e^{2.60\ln v}\)	M1	Takes exponentials of both sides of their line.
\(e^{2.60\ln v}=e^{\ln v^{2.60}}=v^{2.60}\), so \(P=e^{-3.84}v^{2.60}\) with \(n=2.60\)	M1	Uses \(k\ln v=\ln v^k\). Or compares \(\ln P=\ln a+n\ln v\) term by term.
\(a=e^{-3.84}=0.0215\)	A1	Accept 0.0214 or 0.0215.
\(n=2.60\)	AG
Part (f)
\(0.0215\times50^{2.60}\)	M1	Allow ECF from (e)(ii).
\(=562\) W	A1	Accept answers from 559 W to 564 W (\(a=0.0214\) gives 559 W).
50 km/h lies outside the data (16 to 44 km/h), so this is extrapolation and less reliable (although only a little beyond the data)	R1	Must mention extrapolation or the data range.
Part (g)
\(R^2\) only gives the proportion of the variation in \(P\) that the model explains for these eight data points	R1	Do not accept 'the model is 99.97% accurate'.
other models can also give \(R^2\) close to 1 (the quadratic gave 0.998), so a high \(R^2\) cannot show which formula is the true one, or that it holds outside 16 to 44 km/h	R1	Accept a reference to measurement error or to checking the residuals.
Part (h)
each new power is \(0.94P\), so \(\ln(0.94P)=\ln P+\ln0.94=\ln P-0.0619\): every point on Figure 2 moves down by the same amount	M1
the gradient of the line is unchanged, so \(n\) stays \(2.60\)	A1
the intercept decreases by \(0.0619\), so \(a\) is multiplied by \(0.94\): \(a=0.0202\)	A1	Allow ECF from (e)(ii).

Answers: (a) \(0.953\), \(0.998\), \(0.9997\) · (b) the power model · (c) \(4.431\) and \(5.720\) · (e)(i) \(\ln P=2.60\ln v-3.84\); \(r=0.9998\) · (e)(ii) \(a=0.0215\) · (f) about 562 W · (h) \(n\) unchanged; \(a\) becomes \(0.94\times0.0215=0.0202\)

Exam tip (b) When choosing between models, back up the \(R^2\) comparison with a reason from the context, such as what each model predicts at the ends of the data range or at \(v=0\).

More 4.9 questions in the question bank

Topic 5 · Calculus

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	5.1	5.2	5.3	5.4	5.5	5.6	5.7	5.8	5.9	Other
A roller coaster section: gradients, tangents and the steepest drop	●	○								2.2 3.1
Fencing a riverside solar farm: minimising cost	○	●								2.2 1.1
Water flowing into a reservoir after a storm: rates, totals and the trapezoidal rule	○		●							2.2 1.1
Measuring a river's flow: cross-section area and the trapezoidal rule	○		●							3.2 2.2
A wave-energy buoy: derivatives of sine, cosine and exponential functions	○	○		●						2.3
A car going over a speed bump: product, chain and quotient rules for damped motion				○	●					2.3 1.4
Designing a 3D-printed vase: volumes of revolution			○			●				3.2 2.2
An express lift in a skyscraper: velocity, acceleration and distance			○	○		○	●			—
Harvesting a fish farm: slope fields and Euler's method	○							●		2.6 2.3
Foxes and rabbits on an island: a coupled system and its eigenvalues								○	●	1.7 1.5

5.1 Gradients, tangents and rates of change · 5.2 Optimisation in contextmerged group

+ Add to test✓ In your testCC-AI-51-52-128 marks~35 min

A roller coaster section: gradients, tangents and the steepest drop

5.1 Gradients, tangents and rates of change2.2 Quadratic and cubic models5.2 Optimisation in context3.1 Right-angled trigonometry, bearings and the sine and cosine rulesSL

A section of a roller coaster track is modelled by

\[h(x)=0.002x^3-0.15x^2+1.35x+64,\qquad 0\le x\le60,\]

where \(h\) metres is the height of the track above horizontal ground and \(x\) metres is the horizontal distance from the start of the section.

(a)

Find \(h'(x)\).

ROUTINENotes

Show solution

\(0.006x^2-0.3x\)	A1	Two correct terms.
\(h'(x)=0.006x^2-0.3x+1.35\)	A1	Fully correct, with the constant 64 differentiated to 0.

Answer: \(h'(x)=0.006x^2-0.3x+1.35\)

(2)

(b)

Find \(h'(0)\) and interpret your answer in context.

CONCEPTNotes

Show solution

\(h'(0)=1.35\)	A1
at the start of the section the track is climbing: its height increases by about 1.35 m for each metre moved horizontally	R1	Must refer to height increasing (rising) and to the rate per metre of horizontal distance.

Answer: \(1.35\)

(2)

(c)

Find the coordinates of the top of the hill and of the bottom of the dip.

MULTINotes

Show solution

\(0.006x^2-0.3x+1.35=0\)	M1	Sets their derivative equal to zero.
\(x=5\) or \(x=45\)	A1	Equivalent to \(x^2-50x+225=0\); GDC or formula.
top of the hill \((5,\ 67.25)\)	A1	Accept 67.3.
bottom of the dip \((45,\ 3.25)\)	A1	Points must be correctly identified as top and bottom.

Answer: top \((5,\ 67.25)\); bottom \((45,\ 3.25)\)

(4)

(d)

Sketch the graph of \(y=h(x)\) for \(0\le x\le60\) on the axes in Figure 1. Label the top of the hill, the bottom of the dip and both end points with their coordinates.

Figure 1: axes for the graph of \(y=h(x)\)

GRAPHNotes

Show solution

cubic shape on \(0\le x\le60\): rises to a maximum, falls to a minimum, then rises, within the domain only	A1
maximum \((5,\ 67.25)\) and minimum \((45,\ 3.25)\) labelled in approximately the correct positions	A1	Allow ECF from (c).
end points \((0,\ 64)\) and \((60,\ 37)\) labelled	A1

(3)

(e)(i)

Show that the track is steepest on the way down when \(x=25\).

ALGEBRANotes

Show solution

the steepest descent is where \(h'(x)\) takes its least (most negative) value	R1
\(h'(x)\) is a quadratic with a positive coefficient of \(x^2\), so its least value is at its vertex, \(x=-\frac{b}{2a}\)	M1	Accept the minimum of the graph of \(y=h'(x)\) found on a GDC, with a sketch or the GDC output.
\(x=\frac{0.3}{2(0.006)}=\frac{0.3}{0.012}\)	A1
\(x=25\)	AG

(3)

(e)(ii)

Find the angle between the track and the horizontal at \(x=25\).

ROUTINENotes

Show solution

\(h'(25)=-2.4\), so \(\tan\theta=2.4\)	M1	Uses the size of the gradient as the tangent of the angle.
\(\theta=67.4^\circ\)	A1	Accept 67.38°.

Answer: \(67.4^\circ\)

(2)

(f)(i)

A straight support strut meets the track at the point A, where \(x=35\).

Find the equation of the tangent to the track at A.

ROUTINENotes

Show solution

\(h'(35)=0.006(35)^2-0.3(35)+1.35=-1.8\)	M1	Substitutes \(x=35\) into their \(h'(x)\).
\(h(35)=13.25\), so \(y-13.25=-1.8(x-35)\)	A1	Correct point and gradient. Allow ECF from (a).
\(y=-1.8x+76.25\)	A1	Accept any equivalent form.

Answer: \(y=-1.8x+76.25\)

(3)

(f)(ii)

The strut is perpendicular to the track at A and runs in a straight line down to the ground, as shown in Figure 2.

Find the length of the strut.

Figure 2: the strut meets the track at right angles at A

MULTINotes

Show solution

gradient of the strut \(=-\frac{1}{-1.8}=\frac{5}{9}\ (0.556)\)	M1	Negative reciprocal of their gradient from (f)(i).
\(y-13.25=\frac{5}{9}(x-35)\); at the ground \(y=0\): \(x=35-13.25\times1.8=11.15\)	A1	Or angle with the ground \(\theta=\tan^{-1}\frac{5}{9}=29.1^\circ\).
length \(=\sqrt{23.85^2+13.25^2}\)	M1	Or \(\frac{13.25}{\sin29.05^\circ}\).
\(=27.3\) m	A1	Allow ECF from (f)(i).

Answer: 27.3 m

(4)

(g)

Explain what the sign of \(h'(x)\) tells a rider about the track for \(5<x<45\) and for \(45<x\le60\).

INTERPRETNotes

Show solution

for \(5<x<45\), \(h'(x)<0\): the height decreases as the rider moves forward, so the rider is going downhill	R1
for \(45<x\le60\), \(h'(x)>0\): the height increases, so the rider is climbing again	R1	Both the sign and its meaning are needed for each mark.

(2)

(h)

The designers raise the whole section by 6 m, so that the new height is \(h(x)+6\).

State, with a reason, which of your answers to (c), (e)(ii) and (f)(i) change, and give any new values.

CONCEPTNotes

Show solution

the derivative of \(h(x)+6\) is still \(h'(x)\) (the constant differentiates to 0), so all gradients are unchanged	R1
the \(x\)-coordinates in (c) and the angle \(67.4^\circ\) in (e)(ii) do not change	A1
the heights increase by 6: top \((5,\ 73.25)\), bottom \((45,\ 9.25)\); the tangent becomes \(y=-1.8x+82.25\)	A1	Allow ECF from (c) and (f)(i).

Answer: \((5,\ 73.25)\), \((45,\ 9.25)\), \(y=-1.8x+82.25\); angle unchanged

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(0.006x^2-0.3x\)	A1	Two correct terms.
\(h'(x)=0.006x^2-0.3x+1.35\)	A1	Fully correct, with the constant 64 differentiated to 0.
Part (b)
\(h'(0)=1.35\)	A1
at the start of the section the track is climbing: its height increases by about 1.35 m for each metre moved horizontally	R1	Must refer to height increasing (rising) and to the rate per metre of horizontal distance.
Part (c)
\(0.006x^2-0.3x+1.35=0\)	M1	Sets their derivative equal to zero.
\(x=5\) or \(x=45\)	A1	Equivalent to \(x^2-50x+225=0\); GDC or formula.
top of the hill \((5,\ 67.25)\)	A1	Accept 67.3.
bottom of the dip \((45,\ 3.25)\)	A1	Points must be correctly identified as top and bottom.
Part (d)
cubic shape on \(0\le x\le60\): rises to a maximum, falls to a minimum, then rises, within the domain only	A1
maximum \((5,\ 67.25)\) and minimum \((45,\ 3.25)\) labelled in approximately the correct positions	A1	Allow ECF from (c).
end points \((0,\ 64)\) and \((60,\ 37)\) labelled	A1
Part (e)(i)
the steepest descent is where \(h'(x)\) takes its least (most negative) value	R1
\(h'(x)\) is a quadratic with a positive coefficient of \(x^2\), so its least value is at its vertex, \(x=-\frac{b}{2a}\)	M1	Accept the minimum of the graph of \(y=h'(x)\) found on a GDC, with a sketch or the GDC output.
\(x=\frac{0.3}{2(0.006)}=\frac{0.3}{0.012}\)	A1
\(x=25\)	AG
Part (e)(ii)
\(h'(25)=-2.4\), so \(\tan\theta=2.4\)	M1	Uses the size of the gradient as the tangent of the angle.
\(\theta=67.4^\circ\)	A1	Accept 67.38°.
Part (f)(i)
\(h'(35)=0.006(35)^2-0.3(35)+1.35=-1.8\)	M1	Substitutes \(x=35\) into their \(h'(x)\).
\(h(35)=13.25\), so \(y-13.25=-1.8(x-35)\)	A1	Correct point and gradient. Allow ECF from (a).
\(y=-1.8x+76.25\)	A1	Accept any equivalent form.
Part (f)(ii)
gradient of the strut \(=-\frac{1}{-1.8}=\frac{5}{9}\ (0.556)\)	M1	Negative reciprocal of their gradient from (f)(i).
\(y-13.25=\frac{5}{9}(x-35)\); at the ground \(y=0\): \(x=35-13.25\times1.8=11.15\)	A1	Or angle with the ground \(\theta=\tan^{-1}\frac{5}{9}=29.1^\circ\).
length \(=\sqrt{23.85^2+13.25^2}\)	M1	Or \(\frac{13.25}{\sin29.05^\circ}\).
\(=27.3\) m	A1	Allow ECF from (f)(i).
Part (g)
for \(5<x<45\), \(h'(x)<0\): the height decreases as the rider moves forward, so the rider is going downhill	R1
for \(45<x\le60\), \(h'(x)>0\): the height increases, so the rider is climbing again	R1	Both the sign and its meaning are needed for each mark.
Part (h)
the derivative of \(h(x)+6\) is still \(h'(x)\) (the constant differentiates to 0), so all gradients are unchanged	R1
the \(x\)-coordinates in (c) and the angle \(67.4^\circ\) in (e)(ii) do not change	A1
the heights increase by 6: top \((5,\ 73.25)\), bottom \((45,\ 9.25)\); the tangent becomes \(y=-1.8x+82.25\)	A1	Allow ECF from (c) and (f)(i).

Answers: (a) \(h'(x)=0.006x^2-0.3x+1.35\) · (b) \(1.35\) · (c) top \((5,\ 67.25)\); bottom \((45,\ 3.25)\) · (e)(ii) \(67.4^\circ\) · (f)(i) \(y=-1.8x+76.25\) · (f)(ii) 27.3 m · (h) \((5,\ 73.25)\), \((45,\ 9.25)\), \(y=-1.8x+82.25\); angle unchanged

Common error (e)(ii) The gradient \(h'(25)=-2.4\) is a ratio (vertical rise per horizontal metre), not an angle. The angle with the horizontal is \(\tan^{-1}(2.4)\); the negative sign only says that the track is going down.

More 5.1 questions in the question bank

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Fencing a riverside solar farm: minimising cost

5.2 Optimisation in context5.1 Gradients, tangents and rates of change2.2 Quadratic and cubic models1.1 Approximation, percentage error and standard formSL

A solar farm is to be built on a rectangular plot of area 24 000 m² beside a straight river, as shown in Figure 1. The side along the river needs no fence.

The side opposite the river has length \(x\) metres and uses security fencing costing €45 per metre. The two sides perpendicular to the river each have length \(y\) metres and use standard fencing costing €24 per metre. Let \(C\) be the total cost of the fencing, in euros.

Figure 1: the plot beside the river (not to scale)

(a)

Show that \(C(x)=45x+\dfrac{1\,152\,000}{x}\).

ALGEBRANotes

Show solution

\(xy=24\,000\), so \(y=\frac{24\,000}{x}\)	B1
\(C=45x+24y+24y=45x+48y\)	M1	Cost of all three fenced sides.
\(C=45x+\frac{48\times24\,000}{x}\)	A1	Substitution seen.
\(C(x)=45x+\frac{1\,152\,000}{x}\)	AG

(3)

(b)

Find \(C'(x)\).

ROUTINENotes

Show solution

\(C(x)=45x+1\,152\,000x^{-1}\), derivative of \(45x\) is \(45\)	A1
\(C'(x)=45-1\,152\,000x^{-2}\)	A1	Accept \(45-\frac{1\,152\,000}{x^2}\).

Answer: \(C'(x)=45-\dfrac{1\,152\,000}{x^2}\)

(2)

(c)(i)

Find the value of \(x\) that minimises the cost of the fencing, and the corresponding value of \(y\).

MULTINotes

Show solution

\(45-\frac{1\,152\,000}{x^2}=0\)	M1	Sets their derivative equal to zero.
\(x^2=25\,600\), \(x=160\)	A1	Reject \(x=-160\) (a length cannot be negative).
\(y=\frac{24\,000}{160}=150\)	A1

Answer: \(x=160\), \(y=150\)

(3)

(c)(ii)

Use the sign of \(C'(x)\) to justify that this value of \(x\) gives a minimum.

CONCEPTNotes

Show solution

values either side, e.g. \(C'(150)=-6.2\) and \(C'(170)=5.14\)	M1	Any values with \(x\) a little below and a little above 160.
\(C'\) changes from negative to positive at \(x=160\), so the cost has a minimum there	R1

(2)

(c)(iii)

Write down the minimum cost of the fencing.

ROUTINENotes

Show solution

€14 400

Answer: €14 400

(1)

(d)

Before any calculus was used, a planner proposed a square plot of area 24 000 m².

Find the cost of fencing this square plot, and the percentage error when this cost is used as an estimate of the minimum cost.

ROUTINENotes

Show solution

side \(=\sqrt{24\,000}=154.919\ldots\) m, cost \(=45(154.919\ldots)+48(154.919\ldots)\)	M1	Or \(C(\sqrt{24\,000})\).
€14 407.50	A1	Accept €14 408. Do not accept 14 400 (3 s.f.) here.
\(\varepsilon=\left\|\frac{14\,407.50-14\,400}{14\,400}\right\|\times100\%=0.0521\%\)	A1	Allow ECF from (c)(iii).

Answer: €14 407.50; 0.0521%

(3)

(e)(i)

The plot can extend at most 160 m from the river, and the land along the river is 300 m long.

Show that \(150\le x\le300\).

CONCEPTNotes

Show solution

\(y\le160\) means \(\frac{24\,000}{x}\le160\), so \(x\ge150\); and \(x\le300\) from the length of land along the river	R1	Both bounds must be justified.
\(150\le x\le300\)	AG

(1)

(e)(ii)

Sketch the graph of \(y=C(x)\) for \(150\le x\le300\) on the axes in Figure 2. Label the minimum point and both end points with their coordinates.

Figure 2: axes for the graph of \(y=C(x)\)

GRAPHNotes

Show solution

curve drawn only for \(150\le x\le300\), falling slightly then rising, concave up	A1
minimum \((160,\ 14\,400)\) labelled	A1	Allow ECF from (c).
end points \((150,\ 14\,430)\) and \((300,\ 17\,340)\) labelled	A1

(3)

(f)

Use your answers to (d) and (e)(ii) to comment on how much the choice of dimensions near the minimum affects the cost, and suggest one practical reason why the operators might not use the exact minimum-cost dimensions.

INTERPRETNotes

Show solution

the cost changes very little near the minimum: the square plot costs only about €7.50 (0.05%) more, and \(x=150\) only €30 more	R1	Must use values from (d) or (e)(ii).
e.g. a layout that suits the panel rows or the access roads, the shape of the land, gate positions, or fencing sold in fixed lengths matters more than such a small saving	R1	Any sensible practical reason.

(2)

(g)(i)

The plot found in (c)(i) is built. Rows of panels are installed parallel to the river. With \(n\) rows, the annual income from the farm, in euros, is modelled by \(I(n)=n(9000-180n)\).

Find the number of rows that gives the greatest annual income, and this income.

ROUTINENotes

Show solution

\(I(n)=-180n^2+9000n\), vertex at \(n=-\frac{9000}{2(-180)}\)	M1	Or a GDC maximum, or \(I'(n)=9000-360n=0\).
\(n=25\) rows, \(I=\) €112 500	A1	Both values needed.

Answer: 25 rows, €112 500

(2)

(g)(ii)

A maintenance track takes up some of the plot, so that at most 20 rows can be installed.

Find the greatest annual income now, giving a reason for your answer.

CONCEPTNotes

Show solution

\(I\) is increasing for \(n<25\) (the vertex is at \(n=25\)), so the greatest value for \(n\le20\) is at \(n=20\)	R1
\(I(20)=20(9000-3600)=\) €108 000	A1

Answer: €108 000 (20 rows)

(2)

(h)

Suppose instead that, before the plot was chosen, the price of the security fencing had risen to €60 per metre. The price of the standard fencing is unchanged, and the plot must satisfy \(150\le x\le300\).

Find the dimensions of the plot that now minimise the cost of the fencing, and the minimum cost.

MULTINotes

Show solution

\(C(x)=60x+\frac{1\,152\,000}{x}\), \(C'(x)=60-\frac{1\,152\,000}{x^2}\)	M1
\(C'(x)=0\) when \(x=\sqrt{19\,200}=138.6\), which is outside the domain	A1
\(C'(x)>0\) for \(150\le x\le300\) (e.g. \(C'(150)=8.8\)), so \(C\) is increasing and the minimum is at the end point \(x=150\)	R1
plot 150 m by 160 m; minimum cost €16 680	A1	€16 628 (from \(x=138.6\)) scores at most M1A1R0A0.

Answer: 150 m by 160 m; €16 680

(4)

Show full solution

Working	Mark	Notes
Part (a)
\(xy=24\,000\), so \(y=\frac{24\,000}{x}\)	B1
\(C=45x+24y+24y=45x+48y\)	M1	Cost of all three fenced sides.
\(C=45x+\frac{48\times24\,000}{x}\)	A1	Substitution seen.
\(C(x)=45x+\frac{1\,152\,000}{x}\)	AG
Part (b)
\(C(x)=45x+1\,152\,000x^{-1}\), derivative of \(45x\) is \(45\)	A1
\(C'(x)=45-1\,152\,000x^{-2}\)	A1	Accept \(45-\frac{1\,152\,000}{x^2}\).
Part (c)(i)
\(45-\frac{1\,152\,000}{x^2}=0\)	M1	Sets their derivative equal to zero.
\(x^2=25\,600\), \(x=160\)	A1	Reject \(x=-160\) (a length cannot be negative).
\(y=\frac{24\,000}{160}=150\)	A1
Part (c)(ii)
values either side, e.g. \(C'(150)=-6.2\) and \(C'(170)=5.14\)	M1	Any values with \(x\) a little below and a little above 160.
\(C'\) changes from negative to positive at \(x=160\), so the cost has a minimum there	R1
Part (c)(iii)
€14 400	A1
Part (d)
side \(=\sqrt{24\,000}=154.919\ldots\) m, cost \(=45(154.919\ldots)+48(154.919\ldots)\)	M1	Or \(C(\sqrt{24\,000})\).
€14 407.50	A1	Accept €14 408. Do not accept 14 400 (3 s.f.) here.
\(\varepsilon=\left\|\frac{14\,407.50-14\,400}{14\,400}\right\|\times100\%=0.0521\%\)	A1	Allow ECF from (c)(iii).
Part (e)(i)
\(y\le160\) means \(\frac{24\,000}{x}\le160\), so \(x\ge150\); and \(x\le300\) from the length of land along the river	R1	Both bounds must be justified.
\(150\le x\le300\)	AG
Part (e)(ii)
curve drawn only for \(150\le x\le300\), falling slightly then rising, concave up	A1
minimum \((160,\ 14\,400)\) labelled	A1	Allow ECF from (c).
end points \((150,\ 14\,430)\) and \((300,\ 17\,340)\) labelled	A1
Part (f)
the cost changes very little near the minimum: the square plot costs only about €7.50 (0.05%) more, and \(x=150\) only €30 more	R1	Must use values from (d) or (e)(ii).
e.g. a layout that suits the panel rows or the access roads, the shape of the land, gate positions, or fencing sold in fixed lengths matters more than such a small saving	R1	Any sensible practical reason.
Part (g)(i)
\(I(n)=-180n^2+9000n\), vertex at \(n=-\frac{9000}{2(-180)}\)	M1	Or a GDC maximum, or \(I'(n)=9000-360n=0\).
\(n=25\) rows, \(I=\) €112 500	A1	Both values needed.
Part (g)(ii)
\(I\) is increasing for \(n<25\) (the vertex is at \(n=25\)), so the greatest value for \(n\le20\) is at \(n=20\)	R1
\(I(20)=20(9000-3600)=\) €108 000	A1
Part (h)
\(C(x)=60x+\frac{1\,152\,000}{x}\), \(C'(x)=60-\frac{1\,152\,000}{x^2}\)	M1
\(C'(x)=0\) when \(x=\sqrt{19\,200}=138.6\), which is outside the domain	A1
\(C'(x)>0\) for \(150\le x\le300\) (e.g. \(C'(150)=8.8\)), so \(C\) is increasing and the minimum is at the end point \(x=150\)	R1
plot 150 m by 160 m; minimum cost €16 680	A1	€16 628 (from \(x=138.6\)) scores at most M1A1R0A0.

Answers: (b) \(C'(x)=45-\dfrac{1\,152\,000}{x^2}\) · (c)(i) \(x=160\), \(y=150\) · (c)(iii) €14 400 · (d) €14 407.50; 0.0521% · (g)(i) 25 rows, €112 500 · (g)(ii) €108 000 (20 rows) · (h) 150 m by 160 m; €16 680

Exam tip (h) After solving \(C'(x)=0\), always check that the solution lies in the domain. If it does not, the least value on a closed interval is at one of the end points: compare their values.

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5.3 Integration and the trapezoidal rule

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Water flowing into a reservoir after a storm: rates, totals and the trapezoidal rule

5.3 Integration and the trapezoidal rule5.1 Gradients, tangents and rates of change2.2 Quadratic and cubic models1.1 Approximation, percentage error and standard formSL

After a heavy storm, streams carry water into a mountain reservoir. The rate at which water flows into the reservoir is modelled by

\[r(t)=-0.6t^2+7.2t+4,\qquad 0\le t\le12,\]

where \(r\) is measured in thousand m\(^3\) per hour and \(t\) is the time in hours after the rain stops. The graph of \(r\) is shown in Figure 1.

Figure 1: graph of the inflow rate \(r(t)\) for \(0\le t\le12\)

(a)

Find the time at which the inflow rate is greatest, and the greatest inflow rate.

ROUTINENotes

Show solution

\(r'(t)=-1.2t+7.2=0\)	M1	Differentiating and setting the derivative to zero.
\(t=6\) hours	A1	Accept \(t=-\frac{b}{2a}=\frac{-7.2}{2(-0.6)}=6\) or a GDC maximum. Allow ECF from their \(r'(t)\).
\(r(6)=25.6\) thousand m\(^3\) per hour	A1	Accept 25 600 m\(^3\) per hour.

Answer: \(t=6\) hours; \(25.6\) thousand m\(^3\) per hour

(3)

(b)

Show that, at \(t=12\), the inflow rate has returned to its value at \(t=0\).

PROOFNotes

Show solution

\(r(12)=-0.6(144)+7.2(12)+4=-86.4+86.4+4\)	M1	Substitution must be seen.
\(r(12)=4\) and \(r(0)=4\)	A1	Both values seen.
\(r(12)=r(0)\)	AG	Alternative: the axis of symmetry is \(t=6\) and 12 is as far from 6 as 0 is (M1A1).

Answer: \(r(12)=r(0)=4\)

(2)

(c)

By integrating, find the total volume of water that flows into the reservoir during the first 10 hours. Show your working.

ROUTINENotes

Show solution

\(\int_0^{10}(-0.6t^2+7.2t+4)\,dt\)	M1	Integral with correct limits seen.
\(\left[-0.2t^3+3.6t^2+4t\right]_0^{10}\)	A1
\(=-200+360+40=200\) thousand m\(^3\)	A1	Accept \(2.00\times10^5\) m\(^3\). A GDC value with no antiderivative seen: M1A0A1.

Answer: \(200\) thousand m\(^3\) \((2.00\times10^5\text{ m}^3)\)

(3)

(d)(i)

A hydrologist evaluates the model every 2 hours.

\(t\) (hours)	0	2	4	6	8	10
\(r(t)\)	4	16	\(p\)	25.6	\(q\)	16

Write down the value of \(p\) and the value of \(q\).

ROUTINENotes

Show solution

\(p=23.2,\ q=23.2\)

Both values needed.

Answer: \(p=q=23.2\)

(1)

(d)(ii)

Use the trapezoidal rule with the six values in the table to estimate the volume of water that flows into the reservoir during the first 10 hours.

ROUTINENotes

Show solution

\(\frac{2}{2}\left[4+16+2(16+23.2+25.6+23.2)\right]\)	M1	Correct \(h=2\) and correct substitution into the rule. Allow ECF from (d)(i).
\(=196\) thousand m\(^3\)	A1

Answer: \(196\) thousand m\(^3\)

(2)

(d)(iii)

With reference to the shape of the graph in Figure 1, explain why your answer to (d)(ii) is smaller than your answer to (c).

GRAPHNotes

Show solution

the curve is concave down (it bends downwards) for \(0\le t\le10\)	R1	Accept a sketch of a chord below the arc with a comment.
so the straight tops of the trapezia (chords) lie below the curve and the trapezia miss some of the area under it, giving an underestimate	R1	A bare statement "underestimate" with no reason based on the shape: R0R0.

(2)

(d)(iv)

Find the percentage error in the estimate from (d)(ii), taking your answer to (c) as the exact value.

ROUTINENotes

Show solution

\(\varepsilon=\left\|\frac{196-200}{200}\right\|\times100\%\)	M1	The exact value 200 must be the denominator. Allow ECF from (c) and (d)(ii).
\(=2\%\)	A1

Answer: \(2\%\)

(2)

(e)

Before the spillway is used, all the water flowing in is stored, so the volume \(V\) thousand m\(^3\) in the reservoir satisfies \(V'(t)=r(t)\). Five hours after the rain stops, the reservoir holds 2015 thousand m\(^3\).

Find an expression for \(V(t)\).

MODELNotes

Show solution

\(V(t)=\int(-0.6t^2+7.2t+4)\,dt=-0.2t^3+3.6t^2+4t+c\)	M1	Integrating term by term with a constant of integration; at least two terms correct.
\(-0.2(125)+3.6(25)+4(5)+c=2015\), so \(85+c=2015\)	M1	Substituting \(t=5,\ V=2015\) into their integrated expression.
\(V(t)=-0.2t^3+3.6t^2+4t+1930\)	A1	Fully correct expression, with \(c=1930\).

Answer: \(V(t)=-0.2t^3+3.6t^2+4t+1930\)

(3)

(f)

An engineer suggests using the model for \(r\) beyond \(t=12\).

Find the value of \(t\) for which \(r(t)=0\), and explain why the model should not be used after this time.

INTERPRETNotes

Show solution

\(t=12.5\) hours \((12.53\ldots)\)	A1	Reject the negative root. Accept 12.532.
after this time \(r(t)\lt0\), which would mean water flowing out of the reservoir up the streams; an inflow rate cannot be negative (the flow would stop at zero instead)	R1	Must refer to a negative inflow being impossible in context.

Answer: \(t=12.5\) hours

(2)

(g)(i)

The reservoir is now managed differently. From \(t=0\), a spillway removes water from the reservoir at a constant rate of 5 thousand m\(^3\) per hour, while the inflow is still given by \(r(t)\).

Find the change in the volume of water stored during the first 10 hours.

MODELNotes

Show solution

\(\int_0^{10}(r(t)-5)\,dt\) or \(200-5\times10\)	M1	Allow ECF from (c).
\(150\) thousand m\(^3\) (increase)	A1

Answer: an increase of \(150\) thousand m\(^3\)

(2)

(g)(ii)

Find the time in \(0\le t\le12\) at which the volume of water stored is greatest. Justify your answer.

MULTINotes

Show solution

rate of change of volume \(=r(t)-5\); \(-0.6t^2+7.2t-1=0\)	M1	Setting the net rate (or \(r(t)=5\)) to zero.
\(t=11.9\) hours \((11.859\ldots)\)	A1	The other root, \(t=0.141\), must be rejected or classified as a minimum.
the net rate changes from positive (volume increasing) to negative (volume decreasing) at \(t=11.9\), and it is positive for all of \(0.141\lt t\lt11.9\), so the volume there is greater than at \(t=0\)	R1	Accept a sign diagram or graph of \(r(t)-5\) with a comment.

Answer: \(t=11.9\) hours

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(r'(t)=-1.2t+7.2=0\)	M1	Differentiating and setting the derivative to zero.
\(t=6\) hours	A1	Accept \(t=-\frac{b}{2a}=\frac{-7.2}{2(-0.6)}=6\) or a GDC maximum. Allow ECF from their \(r'(t)\).
\(r(6)=25.6\) thousand m\(^3\) per hour	A1	Accept 25 600 m\(^3\) per hour.
Part (b)
\(r(12)=-0.6(144)+7.2(12)+4=-86.4+86.4+4\)	M1	Substitution must be seen.
\(r(12)=4\) and \(r(0)=4\)	A1	Both values seen.
\(r(12)=r(0)\)	AG	Alternative: the axis of symmetry is \(t=6\) and 12 is as far from 6 as 0 is (M1A1).
Part (c)
\(\int_0^{10}(-0.6t^2+7.2t+4)\,dt\)	M1	Integral with correct limits seen.
\(\left[-0.2t^3+3.6t^2+4t\right]_0^{10}\)	A1
\(=-200+360+40=200\) thousand m\(^3\)	A1	Accept \(2.00\times10^5\) m\(^3\). A GDC value with no antiderivative seen: M1A0A1.
Part (d)(i)
\(p=23.2,\ q=23.2\)	A1	Both values needed.
Part (d)(ii)
\(\frac{2}{2}\left[4+16+2(16+23.2+25.6+23.2)\right]\)	M1	Correct \(h=2\) and correct substitution into the rule. Allow ECF from (d)(i).
\(=196\) thousand m\(^3\)	A1
Part (d)(iii)
the curve is concave down (it bends downwards) for \(0\le t\le10\)	R1	Accept a sketch of a chord below the arc with a comment.
so the straight tops of the trapezia (chords) lie below the curve and the trapezia miss some of the area under it, giving an underestimate	R1	A bare statement "underestimate" with no reason based on the shape: R0R0.
Part (d)(iv)
\(\varepsilon=\left\|\frac{196-200}{200}\right\|\times100\%\)	M1	The exact value 200 must be the denominator. Allow ECF from (c) and (d)(ii).
\(=2\%\)	A1
Part (e)
\(V(t)=\int(-0.6t^2+7.2t+4)\,dt=-0.2t^3+3.6t^2+4t+c\)	M1	Integrating term by term with a constant of integration; at least two terms correct.
\(-0.2(125)+3.6(25)+4(5)+c=2015\), so \(85+c=2015\)	M1	Substituting \(t=5,\ V=2015\) into their integrated expression.
\(V(t)=-0.2t^3+3.6t^2+4t+1930\)	A1	Fully correct expression, with \(c=1930\).
Part (f)
\(t=12.5\) hours \((12.53\ldots)\)	A1	Reject the negative root. Accept 12.532.
after this time \(r(t)\lt0\), which would mean water flowing out of the reservoir up the streams; an inflow rate cannot be negative (the flow would stop at zero instead)	R1	Must refer to a negative inflow being impossible in context.
Part (g)(i)
\(\int_0^{10}(r(t)-5)\,dt\) or \(200-5\times10\)	M1	Allow ECF from (c).
\(150\) thousand m\(^3\) (increase)	A1
Part (g)(ii)
rate of change of volume \(=r(t)-5\); \(-0.6t^2+7.2t-1=0\)	M1	Setting the net rate (or \(r(t)=5\)) to zero.
\(t=11.9\) hours \((11.859\ldots)\)	A1	The other root, \(t=0.141\), must be rejected or classified as a minimum.
the net rate changes from positive (volume increasing) to negative (volume decreasing) at \(t=11.9\), and it is positive for all of \(0.141\lt t\lt11.9\), so the volume there is greater than at \(t=0\)	R1	Accept a sign diagram or graph of \(r(t)-5\) with a comment.

Answers: (a) \(t=6\) hours; \(25.6\) thousand m\(^3\) per hour · (b) \(r(12)=r(0)=4\) · (c) \(200\) thousand m\(^3\) \((2.00\times10^5\text{ m}^3)\) · (d)(i) \(p=q=23.2\) · (d)(ii) \(196\) thousand m\(^3\) · (d)(iv) \(2\%\) · (e) \(V(t)=-0.2t^3+3.6t^2+4t+1930\) · (f) \(t=12.5\) hours · (g)(i) an increase of \(150\) thousand m\(^3\) · (g)(ii) \(t=11.9\) hours

Common error (d)(ii) Five strips need six values: the end values 4 and 16 are used once and the four interior values twice. Doubling every value gives 392, twice the correct total.

Exam tip (d)(iii) Justify over- or underestimates with the shape: concave down means the chords lie below the curve, so the rule underestimates.

More 5.3 questions in the question bank

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Measuring a river's flow: cross-section area and the trapezoidal rule

5.3 Integration and the trapezoidal rule5.1 Gradients, tangents and rates of change3.2 Volume and surface area in context2.2 Quadratic and cubic modelsSL

A river flows through a town in a channel between two vertical concrete walls 24 m apart. To estimate the flow, hydrologists measure the depth of the water, \(d\) metres, at distances \(x\) metres from the left wall, every 4 m across the channel (Figure 1). At both walls the water depth is zero.

\(x\) (m)	0	4	8	12	16	20	24
\(d\) (m)	0	1.9	3.3	3.6	3.1	2.1	0

When the depth is plotted upwards against \(x\), the area under the graph equals the area of the cross-section of the water.

Figure 1: cross-section of the channel, with the depth measured every 4 m (not to scale)

(a)

On the axes in Figure 2, plot the seven measured depths and join them with a smooth curve.

Figure 2: axes for the depth \(d\) against the distance \(x\) from the left wall

GRAPHNotes

Show solution

all seven points plotted correctly	A1	Within half a small square.
smooth curve through (or very close to) the points, starting and ending on the \(x\)-axis, bending downwards	A1	Not straight-line segments.

(2)

(b)

Use the trapezoidal rule with the data in the table to estimate the area of the cross-section.

ROUTINENotes

Show solution

\(h=\frac{24}{6}=4\)	B1	May be implied by correct substitution.
\(\frac{4}{2}\left[0+0+2(1.9+3.3+3.6+3.1+2.1)\right]\)	M1	Ends once, the five interior values twice.
\(=56.0\) m\(^2\)	A1

Answer: \(56.0\) m\(^2\)

(3)

(c)

A hydrologist models the depth by \(d(x)=ax(24-x)\), \(0\le x\le24\), where \(a\) is a constant. The model is chosen to give the measured depth at the centre of the channel.

Show that \(a=0.025\).

PROOFNotes

Show solution

\(a(12)(24-12)=3.6\)	M1	Substituting \(x=12,\ d=3.6\).
\(144a=3.6\Rightarrow a=\frac{3.6}{144}\)	A1
\(a=0.025\)	AG

Answer: \(a=0.025\)

(2)

(d)(i)

Find \(d'(x)\). Hence show that, according to the model, the deepest point of the channel is at its centre, and write down the greatest depth.

ROUTINENotes

Show solution

\(d(x)=0.6x-0.025x^2\Rightarrow d'(x)=0.6-0.05x\)	A1	Expansion may be implied.
\(0.6-0.05x=0\Rightarrow x=12\) (the centre)	M1	Setting their \(d'(x)=0\).
\(d'(x)\gt0\) for \(x\lt12\) and \(d'(x)\lt0\) for \(x\gt12\), so this is a maximum (the deepest point)	R1	Accept: the coefficient of \(x^2\) is negative, so the graph is a parabola with a maximum at its vertex. A stationary point with no reason why it is the deepest: R0.
greatest depth \(d(12)=3.6\) m	A1

Answer: \(d'(x)=0.6-0.05x\); deepest at \(x=12\), depth \(3.6\) m

(4)

(d)(ii)

Find \(d'(20)\) and interpret its value in context.

INTERPRETNotes

Show solution

\(d'(20)=-0.4\)	A1	Allow ECF from (d)(i).
at \(x=20\), the depth decreases by 0.4 m for each metre moved towards the right wall (the bed rises steeply there)	R1	Must mention a rate per metre across the channel and a decrease.

Answer: \(-0.4\) m per m

(2)

(e)

Show that the model gives the area of the cross-section as exactly 57.6 m\(^2\).

PROOFNotes

Show solution

\(\int_0^{24}(0.6x-0.025x^2)\,dx\)	M1	Integral with correct limits.
\(\left[0.3x^2-\frac{0.025x^3}{3}\right]_0^{24}\)	A1	Accept \(0.3x^2-\frac{x^3}{120}\).
\(=172.8-115.2\)	A1	Both values seen.
\(=57.6\) m\(^2\)	AG	A GDC value alone earns no marks in a "show that".

Answer: \(57.6\) m\(^2\)

(3)

(f)(i)

With reference to the shape of the graph of \(d\), explain why the trapezoidal estimate in (b) is smaller than the area given by the model.

INTERPRETNotes

Show solution

the graph of \(d\) is concave down (bends downwards)	R1	Accept: \(d\) is a quadratic with negative \(x^2\) coefficient.
so each chord lies below the curve and the trapezia leave out small regions under the curve: an underestimate	R1

(2)

(f)(ii)

Suggest how the hydrologists could obtain a more accurate trapezoidal estimate from new measurements.

INTERPRETNotes

Show solution

measure the depth at smaller intervals (for example every 1 m or 2 m), so that more, narrower strips are used

(1)

(g)

The water flows along the channel at a constant speed of 0.8 m s\(^{-1}\). In each hour, the water passing a point forms a prism whose cross-section is the cross-section of the channel.

Using the area from the model, find the volume of water that flows past a point in one hour. Give your answer in m\(^3\) and in litres.

MULTINotes

Show solution

length of prism \(=0.8\times3600=2880\) m, \(V=57.6\times2880\)	M1	Accept \(57.6\times0.8=46.08\) m\(^3\) per second, then \(\times3600\).
\(V=165\,888\approx166\,000\) m\(^3\)	A1	Accept \(1.66\times10^5\) m\(^3\).
\(1.66\times10^8\) litres	A1	Using \(1\text{ m}^3=1000\) litres. Allow ECF from their volume in m\(^3\).

Answer: \(1.66\times10^5\) m\(^3\) \(=1.66\times10^8\) litres

(3)

(h)(i)

After a night of rain, the water level rises by 0.5 m, so that the depth at every point across the channel is 0.5 m greater than before. The walls are high enough that the width stays 24 m.

Find the new area of the cross-section, according to the model.

MODELNotes

Show solution

\(57.6+0.5\times24\) or \(\int_0^{24}(d(x)+0.5)\,dx\)	M1	Adding the rectangle 24 m by 0.5 m.
\(69.6\) m\(^2\)	A1

Answer: \(69.6\) m\(^2\)

(2)

(h)(ii)

The water now flows at 1.1 m s\(^{-1}\). Find the volume of water that flows past a point in one hour.

MODELNotes

Show solution

\(69.6\times1.1\times3600\)	M1	Allow ECF from (h)(i).
\(275\,616\approx276\,000\) m\(^3\)	A1	Accept \(2.76\times10^5\) m\(^3\).

Answer: \(2.76\times10^5\) m\(^3\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
all seven points plotted correctly	A1	Within half a small square.
smooth curve through (or very close to) the points, starting and ending on the \(x\)-axis, bending downwards	A1	Not straight-line segments.
Part (b)
\(h=\frac{24}{6}=4\)	B1	May be implied by correct substitution.
\(\frac{4}{2}\left[0+0+2(1.9+3.3+3.6+3.1+2.1)\right]\)	M1	Ends once, the five interior values twice.
\(=56.0\) m\(^2\)	A1
Part (c)
\(a(12)(24-12)=3.6\)	M1	Substituting \(x=12,\ d=3.6\).
\(144a=3.6\Rightarrow a=\frac{3.6}{144}\)	A1
\(a=0.025\)	AG
Part (d)(i)
\(d(x)=0.6x-0.025x^2\Rightarrow d'(x)=0.6-0.05x\)	A1	Expansion may be implied.
\(0.6-0.05x=0\Rightarrow x=12\) (the centre)	M1	Setting their \(d'(x)=0\).
\(d'(x)\gt0\) for \(x\lt12\) and \(d'(x)\lt0\) for \(x\gt12\), so this is a maximum (the deepest point)	R1	Accept: the coefficient of \(x^2\) is negative, so the graph is a parabola with a maximum at its vertex. A stationary point with no reason why it is the deepest: R0.
greatest depth \(d(12)=3.6\) m	A1
Part (d)(ii)
\(d'(20)=-0.4\)	A1	Allow ECF from (d)(i).
at \(x=20\), the depth decreases by 0.4 m for each metre moved towards the right wall (the bed rises steeply there)	R1	Must mention a rate per metre across the channel and a decrease.
Part (e)
\(\int_0^{24}(0.6x-0.025x^2)\,dx\)	M1	Integral with correct limits.
\(\left[0.3x^2-\frac{0.025x^3}{3}\right]_0^{24}\)	A1	Accept \(0.3x^2-\frac{x^3}{120}\).
\(=172.8-115.2\)	A1	Both values seen.
\(=57.6\) m\(^2\)	AG	A GDC value alone earns no marks in a "show that".
Part (f)(i)
the graph of \(d\) is concave down (bends downwards)	R1	Accept: \(d\) is a quadratic with negative \(x^2\) coefficient.
so each chord lies below the curve and the trapezia leave out small regions under the curve: an underestimate	R1
Part (f)(ii)
measure the depth at smaller intervals (for example every 1 m or 2 m), so that more, narrower strips are used	R1
Part (g)
length of prism \(=0.8\times3600=2880\) m, \(V=57.6\times2880\)	M1	Accept \(57.6\times0.8=46.08\) m\(^3\) per second, then \(\times3600\).
\(V=165\,888\approx166\,000\) m\(^3\)	A1	Accept \(1.66\times10^5\) m\(^3\).
\(1.66\times10^8\) litres	A1	Using \(1\text{ m}^3=1000\) litres. Allow ECF from their volume in m\(^3\).
Part (h)(i)
\(57.6+0.5\times24\) or \(\int_0^{24}(d(x)+0.5)\,dx\)	M1	Adding the rectangle 24 m by 0.5 m.
\(69.6\) m\(^2\)	A1
Part (h)(ii)
\(69.6\times1.1\times3600\)	M1	Allow ECF from (h)(i).
\(275\,616\approx276\,000\) m\(^3\)	A1	Accept \(2.76\times10^5\) m\(^3\).

Answers: (b) \(56.0\) m\(^2\) · (c) \(a=0.025\) · (d)(i) \(d'(x)=0.6-0.05x\); deepest at \(x=12\), depth \(3.6\) m · (d)(ii) \(-0.4\) m per m · (e) \(57.6\) m\(^2\) · (g) \(1.66\times10^5\) m\(^3\) \(=1.66\times10^8\) litres · (h)(i) \(69.6\) m\(^2\) · (h)(ii) \(2.76\times10^5\) m\(^3\)

Exam tip (f)(i) In a "too big or too small" question, name the concavity and say where the chords lie. "Underestimate" on its own earns nothing.

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5.4 Derivatives of trigonometric, exponential and logarithmic functions · 5.5 Chain, product and quotient rulesmerged group

+ Add to test✓ In your testCC-AI-54-55-126 marks~30 min

A wave-energy buoy: derivatives of sine, cosine and exponential functions

5.4 Derivatives of trigonometric, exponential and logarithmic functions2.3 Exponential and sinusoidal models5.1 Gradients, tangents and rates of change5.2 Optimisation in contextHL

A wave-energy buoy floats on the sea and moves up and down as waves pass. On a calm day its height above the seabed is modelled by

\[h(t)=8-1.2\cos t,\qquad t\ge0,\]

where \(h\) is in metres, \(t\) is the time in seconds and the angle is in radians. The graph of \(h\) is shown in Figure 1.

Figure 1: height of the buoy, \(h(t)=8-1.2\cos t\), for \(0\le t\le2\pi\); the dashed line is \(h=8\)

(a)

Write down the amplitude and the period of the motion, and the greatest height of the buoy above the seabed.

ROUTINENotes

Show solution

amplitude \(1.2\) m	A1
period \(2\pi\approx6.28\) s; greatest height \(9.2\) m	A1	Both needed.

Answer: \(1.2\) m; \(2\pi\approx6.28\) s; \(9.2\) m

(2)

(b)

Find expressions for \(h'(t)\) and \(h''(t)\).

HL onlyROUTINENotes

Show solution

\(h'(t)=1.2\sin t\)	A1	\(-1.2\times(-\sin t)\); sign must be correct.
\(h''(t)=1.2\cos t\)	A1	Allow ECF from their \(h'(t)\).

Answer: \(h'(t)=1.2\sin t,\ h''(t)=1.2\cos t\)

(2)

(c)

Show that \(h''(t)=-\big(h(t)-8\big)\) for all \(t\).

HL onlyPROOFNotes

Show solution

\(h(t)-8=-1.2\cos t\)	M1
\(-\big(h(t)-8\big)=1.2\cos t=h''(t)\)	A1	Must use their correct \(h''(t)\) from (b).
\(h''(t)=-\big(h(t)-8\big)\)	AG

(2)

(d)

Find the greatest vertical speed of the buoy, the first time \(t\gt0\) at which it occurs, and the height of the buoy at that time.

HL onlyMULTINotes

Show solution

\(h''(t)=0\Rightarrow\cos t=0\), or \(\|\sin t\|=1\)	M1	Any valid method for the extreme of \(h'\).
greatest speed \(1.2\) m s\(^{-1}\) at \(t=\frac{\pi}{2}\approx1.57\) s	A1
height \(h=8\) m	A1	The buoy is at its mean level.

Answer: \(1.2\) m s\(^{-1}\) at \(t=\frac{\pi}{2}\approx1.57\) s, when \(h=8\) m

(3)

(e)(i)

On the axes in Figure 2, sketch the graph of \(h'(t)\) for \(0\le t\le2\pi\).

Figure 2: axes for \(h'(t)\), with the same time scale as Figure 1

HL onlyGRAPHNotes

Show solution

sine-shaped curve starting at the origin, with maximum \(1.2\) at \(t=\frac{\pi}{2}\) and minimum \(-1.2\) at \(t=\frac{3\pi}{2}\)	A1
crosses the \(t\)-axis at \(t=\pi\) and ends at \((2\pi,0)\)	A1	Allow ECF from their \(h'(t)\) if a sinusoid of the correct period.

(2)

(e)(ii)

By comparing Figure 1 and your sketch, explain how the position of the buoy is related to the times at which \(h'(t)=0\) and the times at which \(h'(t)\) is greatest.

HL onlyINTERPRETNotes

Show solution

\(h'(t)=0\) at \(t=0,\pi,2\pi\), when the buoy is at its lowest or highest point (turning points of \(h\))	R1
\(h'(t)\) is greatest at \(t=\frac{\pi}{2}\), when the buoy passes the mean level \(h=8\) going up; the graph of \(h'\) is a quarter of a period (\(\frac{\pi}{2}\) s) ahead of the graph of \(h\)	R1	Accept: the peaks of \(h'\) are a quarter period before the peaks of \(h\).

(2)

(f)

Find the equation of the tangent to the graph of \(h\) at \(t=1\). Give your answer in the form \(y=mt+c\).

HL onlyROUTINENotes

Show solution

gradient \(h'(1)=1.2\sin1=1.0097\ldots\)	M1	Substituting \(t=1\) into their \(h'(t)\) (radians). Allow ECF from (b).
point \(h(1)=8-1.2\cos1=7.3516\ldots\); \(y-7.3516=1.0098(t-1)\)	A1	Correct point and gradient in a line equation.
\(y=1.01t+6.34\)	A1	Accept \(y=1.0098t+6.3419\).

Answer: \(y=1.01t+6.34\)

(3)

(g)

A student evaluates \(h'(1)\) with a calculator set to degrees and obtains 0.0209.

Explain why this value is incorrect.

HL onlyINTERPRETNotes

Show solution

the result \(\frac{d}{dt}\cos t=-\sin t\) (so \(h'(t)=1.2\sin t\)) is valid only when \(t\) is in radians	R1
in degree mode the calculator finds \(1.2\sin(1^\circ)=0.0209\), the sine of 1 degree instead of 1 radian; the correct value is \(1.2\sin1=1.01\) m s\(^{-1}\)	R1	Accept any clear statement that the model's angle \(t\) is in radians, so degree mode evaluates the wrong angle.

(2)

(h)(i)

During a storm, the buoy's generator is protected by a brake that dissipates power. The power generated is modelled by \(G(x)=60x+20\) kW and the power dissipated by the brake by \(B(x)=4e^{x}\) kW, where \(x\) is the time in hours after the storm begins, \(0\le x\le4\).

Find \(B'(2)\) and interpret your answer in context.

HL onlyMODELNotes

Show solution

\(B'(2)=4e^{2}=29.6\)	A1	Accept 29.556.
two hours after the storm begins, the power dissipated by the brake is increasing at 29.6 kW per hour	R1	Must include the rate of increase with correct units.

Answer: \(29.6\) kW per hour

(2)

(h)(ii)

The net power output is \(N(x)=G(x)-B(x)\).

Find the time at which \(N\) is greatest, justify that it is a maximum, and find the greatest net power output.

HL onlyMULTINotes

Show solution

\(N'(x)=60-4e^{x}=0\)	M1
\(e^{x}=15\Rightarrow x=\ln15=2.71\) hours	A1	Accept 2.708 hours, or 2 hours 42 minutes.
\(N''(x)=-4e^{x}\lt0\) for all \(x\), so a maximum	R1	Or sign change of \(N'\): positive before, negative after. Endpoint values \(N(0)=16\), \(N(4)=41.6\) are smaller.
\(N(\ln15)=60\ln15+20-60=122\) kW	A1	Accept 122.48 kW.

Answer: \(x=\ln15\approx2.71\) hours; \(122\) kW

(4)

(i)

On a stormier day the waves are bigger, and the height of the buoy is modelled by \(h(t)=8-2.4\cos t\).

State, with a reason, the effect of this change on the greatest vertical speed of the buoy and on the result of part (c).

HL onlyINTERPRETNotes

Show solution

\(h'(t)=2.4\sin t\), so the greatest speed doubles to \(2.4\) m s\(^{-1}\)	R1
\(h''(t)=2.4\cos t=-\big(h(t)-8\big)\) again, so the result of (c) is unchanged (the period \(2\pi\) s does not change)	R1

(2)

Show full solution

Working	Mark	Notes
Part (a)
amplitude \(1.2\) m	A1
period \(2\pi\approx6.28\) s; greatest height \(9.2\) m	A1	Both needed.
Part (b)
\(h'(t)=1.2\sin t\)	A1	\(-1.2\times(-\sin t)\); sign must be correct.
\(h''(t)=1.2\cos t\)	A1	Allow ECF from their \(h'(t)\).
Part (c)
\(h(t)-8=-1.2\cos t\)	M1
\(-\big(h(t)-8\big)=1.2\cos t=h''(t)\)	A1	Must use their correct \(h''(t)\) from (b).
\(h''(t)=-\big(h(t)-8\big)\)	AG
Part (d)
\(h''(t)=0\Rightarrow\cos t=0\), or \(\|\sin t\|=1\)	M1	Any valid method for the extreme of \(h'\).
greatest speed \(1.2\) m s\(^{-1}\) at \(t=\frac{\pi}{2}\approx1.57\) s	A1
height \(h=8\) m	A1	The buoy is at its mean level.
Part (e)(i)
sine-shaped curve starting at the origin, with maximum \(1.2\) at \(t=\frac{\pi}{2}\) and minimum \(-1.2\) at \(t=\frac{3\pi}{2}\)	A1
crosses the \(t\)-axis at \(t=\pi\) and ends at \((2\pi,0)\)	A1	Allow ECF from their \(h'(t)\) if a sinusoid of the correct period.
Part (e)(ii)
\(h'(t)=0\) at \(t=0,\pi,2\pi\), when the buoy is at its lowest or highest point (turning points of \(h\))	R1
\(h'(t)\) is greatest at \(t=\frac{\pi}{2}\), when the buoy passes the mean level \(h=8\) going up; the graph of \(h'\) is a quarter of a period (\(\frac{\pi}{2}\) s) ahead of the graph of \(h\)	R1	Accept: the peaks of \(h'\) are a quarter period before the peaks of \(h\).
Part (f)
gradient \(h'(1)=1.2\sin1=1.0097\ldots\)	M1	Substituting \(t=1\) into their \(h'(t)\) (radians). Allow ECF from (b).
point \(h(1)=8-1.2\cos1=7.3516\ldots\); \(y-7.3516=1.0098(t-1)\)	A1	Correct point and gradient in a line equation.
\(y=1.01t+6.34\)	A1	Accept \(y=1.0098t+6.3419\).
Part (g)
the result \(\frac{d}{dt}\cos t=-\sin t\) (so \(h'(t)=1.2\sin t\)) is valid only when \(t\) is in radians	R1
in degree mode the calculator finds \(1.2\sin(1^\circ)=0.0209\), the sine of 1 degree instead of 1 radian; the correct value is \(1.2\sin1=1.01\) m s\(^{-1}\)	R1	Accept any clear statement that the model's angle \(t\) is in radians, so degree mode evaluates the wrong angle.
Part (h)(i)
\(B'(2)=4e^{2}=29.6\)	A1	Accept 29.556.
two hours after the storm begins, the power dissipated by the brake is increasing at 29.6 kW per hour	R1	Must include the rate of increase with correct units.
Part (h)(ii)
\(N'(x)=60-4e^{x}=0\)	M1
\(e^{x}=15\Rightarrow x=\ln15=2.71\) hours	A1	Accept 2.708 hours, or 2 hours 42 minutes.
\(N''(x)=-4e^{x}\lt0\) for all \(x\), so a maximum	R1	Or sign change of \(N'\): positive before, negative after. Endpoint values \(N(0)=16\), \(N(4)=41.6\) are smaller.
\(N(\ln15)=60\ln15+20-60=122\) kW	A1	Accept 122.48 kW.
Part (i)
\(h'(t)=2.4\sin t\), so the greatest speed doubles to \(2.4\) m s\(^{-1}\)	R1
\(h''(t)=2.4\cos t=-\big(h(t)-8\big)\) again, so the result of (c) is unchanged (the period \(2\pi\) s does not change)	R1

Answers: (a) \(1.2\) m; \(2\pi\approx6.28\) s; \(9.2\) m · (b) \(h'(t)=1.2\sin t,\ h''(t)=1.2\cos t\) · (d) \(1.2\) m s\(^{-1}\) at \(t=\frac{\pi}{2}\approx1.57\) s, when \(h=8\) m · (f) \(y=1.01t+6.34\) · (h)(i) \(29.6\) kW per hour · (h)(ii) \(x=\ln15\approx2.71\) hours; \(122\) kW

Common error (b) The derivative of \(\cos t\) is \(-\sin t\), so \(-1.2\cos t\) differentiates to \(+1.2\sin t\). A wrong sign here makes the buoy start by moving down, which contradicts Figure 1.

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A car going over a speed bump: product, chain and quotient rules for damped motion

5.5 Chain, product and quotient rules5.4 Derivatives of trigonometric, exponential and logarithmic functions2.3 Exponential and sinusoidal models1.4 Exponents and logarithmsHL

After a car's front wheels pass over a speed bump, the body of the car bounces up and down while the shock absorbers slow the motion. The vertical displacement of the body above its normal resting position is modelled by

\[y(t)=6e^{-1.5t}\cos(4t),\qquad t\ge0,\]

where \(y\) is in centimetres, \(t\) is the time in seconds after the bump, and the angle is in radians. The curves \(y=\pm6e^{-1.5t}\) are called the envelope curves.

(a)

Use the product rule and the chain rule to show that \(y'(t)=-6e^{-1.5t}\big(1.5\cos(4t)+4\sin(4t)\big)\).

HL onlyALGEBRANotes

Show solution

\(u=6e^{-1.5t},\ v=\cos(4t)\); \(y'=uv'+vu'\)	M1	Product rule attempted with two terms.
\(u'=-9e^{-1.5t}\)	A1	Chain rule on the exponential.
\(v'=-4\sin(4t)\), so \(y'=6e^{-1.5t}(-4\sin4t)+\cos(4t)(-9e^{-1.5t})\)	A1	Chain rule on the cosine, substituted correctly.
\(y'(t)=-6e^{-1.5t}\big(1.5\cos(4t)+4\sin(4t)\big)\)	AG

(3)

(b)

Show that \(y'(t)=0\) when \(\tan(4t)=-0.375\).

HL onlyPROOFNotes

Show solution

\(y'(t)=0\Rightarrow1.5\cos(4t)+4\sin(4t)=0\), because \(e^{-1.5t}\gt0\) is never zero	M1	The reason that the exponential factor cannot be zero must be stated.
\(4\sin(4t)=-1.5\cos(4t)\Rightarrow\frac{\sin4t}{\cos4t}=-\frac{1.5}{4}\)	A1	Dividing by \(\cos(4t)\) (or equivalent).
\(\tan(4t)=-0.375\)	AG

(2)

(c)

Find \(y'(0)\) and interpret your answer in context.

HL onlyMODELNotes

Show solution

\(y'(0)=-6(1.5)=-9\)	A1
just after the bump the car body is moving downwards at \(9\) cm s\(^{-1}\)	R1	Must refer to direction (downwards) and units.

Answer: \(-9\) cm s\(^{-1}\)

(2)

(d)

Find the first time \(t\gt0\) at which the car body is stationary, and its displacement at that time.

HL onlyMULTINotes

Show solution

\(4t=\pi+\arctan(-0.375)=2.7828\ldots\)	M1	Or a GDC zero of \(y'\). The GDC value \(\arctan(-0.375)\lt0\) must be adjusted by \(\pi\).
\(t=0.696\) s	A1	Accept 0.6957.
\(y=6e^{-1.5(0.6957)}\cos(2.7828)=-1.98\) cm	A1	1.98 cm below the resting position.

Answer: \(t=0.696\) s; \(y=-1.98\) cm

(3)

(e)

The motion is considered to have settled once the upper envelope \(6e^{-1.5t}\) is less than 0.5 cm.

Use logarithms to find the time after which the motion has settled.

HL onlyALGEBRANotes

Show solution

\(6e^{-1.5t}=0.5\Rightarrow e^{-1.5t}=\frac{1}{12}\)	M1	Isolating the exponential.
\(-1.5t=\ln\frac{1}{12}\)	M1
\(t=\frac{\ln12}{1.5}=1.66\) s	A1	Accept 1.6566. A GDC intersection with no logarithm seen earns no marks (the question asks for logarithms).

Answer: \(t=1.66\) s

(3)

(f)(i)

Write down the period of \(\cos(4t)\).

ROUTINENotes

Show solution

\(\frac{2\pi}{4}=\frac{\pi}{2}\approx1.57\) s

Answer: \(\frac{\pi}{2}\approx1.57\) s

(1)

(f)(ii)

Figure 1 shows the envelope curves. On Figure 1, sketch the graph of \(y(t)\) for \(0\le t\le2\).

Figure 1: the envelope curves \(y=6e^{-1.5t}\) and \(y=-6e^{-1.5t}\) for \(0\le t\le2\)

HL onlyGRAPHNotes

Show solution

starts at \((0,6)\) on the upper envelope and decreases	A1
oscillates between the envelope curves, touching them, with decreasing amplitude	A1
crosses the \(t\)-axis near \(t=0.39,\ 1.18,\ 1.96\) (every \(\frac{\pi}{4}\) s), with a minimum near \((0.70,\,-1.98)\)	A1	Allow ECF from (d) for the minimum.

(3)

(g)(i)

When the shock absorbers are worn, the motion is modelled by \(w(t)=\dfrac{6\cos(4t)}{e^{0.6t}}\).

Use the quotient rule to show that \(w'(t)=-\dfrac{6\big(0.6\cos(4t)+4\sin(4t)\big)}{e^{0.6t}}\).

HL onlyALGEBRANotes

Show solution

\(u=6\cos(4t),\ v=e^{0.6t}\); \(w'=\frac{vu'-uv'}{v^2}\)	M1	Correct order in the numerator.
\(u'=-24\sin(4t),\ v'=0.6e^{0.6t}\)	A1
\(w'=\frac{e^{0.6t}(-24\sin4t)-6\cos4t(0.6e^{0.6t})}{e^{1.2t}}\)	A1	Correct substitution; then cancel \(e^{0.6t}\).
\(w'(t)=-\frac{6(0.6\cos4t+4\sin4t)}{e^{0.6t}}\)	AG

(3)

(g)(ii)

Using the same settling condition as in (e), find how long the motion takes to settle with worn shock absorbers, and compare it with your answer to (e).

HL onlyMODELNotes

Show solution

\(6e^{-0.6t}=0.5\Rightarrow e^{-0.6t}=\frac{1}{12}\Rightarrow t=\frac{\ln12}{0.6}\)	M1	Or \(\frac{6}{e^{0.6t}}=0.5\). A GDC intersection is acceptable here.
\(t=4.14\) s	A1	Accept 4.1415.
about 2.5 times as long as the 1.66 s with good shock absorbers	R1	Allow ECF from (e). Accept "2.48 s longer".

Answer: \(4.14\) s, about 2.5 times as long

(3)

(h)

Explain what the constant 1.5 in \(e^{-1.5t}\) represents physically, and why it does not change the time between successive upward bounces.

INTERPRETNotes

Show solution

it controls how quickly the size of the bounces dies away: a larger constant makes the envelope decay faster, so the car settles sooner (stronger damping by the shock absorbers)	R1
the time between upward bounces is set by \(\cos(4t)\) only (period \(\frac{\pi}{2}\) s), which does not contain the constant	R1	Accept reference to the worn model in (g) having the same period.

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(u=6e^{-1.5t},\ v=\cos(4t)\); \(y'=uv'+vu'\)	M1	Product rule attempted with two terms.
\(u'=-9e^{-1.5t}\)	A1	Chain rule on the exponential.
\(v'=-4\sin(4t)\), so \(y'=6e^{-1.5t}(-4\sin4t)+\cos(4t)(-9e^{-1.5t})\)	A1	Chain rule on the cosine, substituted correctly.
\(y'(t)=-6e^{-1.5t}\big(1.5\cos(4t)+4\sin(4t)\big)\)	AG
Part (b)
\(y'(t)=0\Rightarrow1.5\cos(4t)+4\sin(4t)=0\), because \(e^{-1.5t}\gt0\) is never zero	M1	The reason that the exponential factor cannot be zero must be stated.
\(4\sin(4t)=-1.5\cos(4t)\Rightarrow\frac{\sin4t}{\cos4t}=-\frac{1.5}{4}\)	A1	Dividing by \(\cos(4t)\) (or equivalent).
\(\tan(4t)=-0.375\)	AG
Part (c)
\(y'(0)=-6(1.5)=-9\)	A1
just after the bump the car body is moving downwards at \(9\) cm s\(^{-1}\)	R1	Must refer to direction (downwards) and units.
Part (d)
\(4t=\pi+\arctan(-0.375)=2.7828\ldots\)	M1	Or a GDC zero of \(y'\). The GDC value \(\arctan(-0.375)\lt0\) must be adjusted by \(\pi\).
\(t=0.696\) s	A1	Accept 0.6957.
\(y=6e^{-1.5(0.6957)}\cos(2.7828)=-1.98\) cm	A1	1.98 cm below the resting position.
Part (e)
\(6e^{-1.5t}=0.5\Rightarrow e^{-1.5t}=\frac{1}{12}\)	M1	Isolating the exponential.
\(-1.5t=\ln\frac{1}{12}\)	M1
\(t=\frac{\ln12}{1.5}=1.66\) s	A1	Accept 1.6566. A GDC intersection with no logarithm seen earns no marks (the question asks for logarithms).
Part (f)(i)
\(\frac{2\pi}{4}=\frac{\pi}{2}\approx1.57\) s	A1
Part (f)(ii)
starts at \((0,6)\) on the upper envelope and decreases	A1
oscillates between the envelope curves, touching them, with decreasing amplitude	A1
crosses the \(t\)-axis near \(t=0.39,\ 1.18,\ 1.96\) (every \(\frac{\pi}{4}\) s), with a minimum near \((0.70,\,-1.98)\)	A1	Allow ECF from (d) for the minimum.
Part (g)(i)
\(u=6\cos(4t),\ v=e^{0.6t}\); \(w'=\frac{vu'-uv'}{v^2}\)	M1	Correct order in the numerator.
\(u'=-24\sin(4t),\ v'=0.6e^{0.6t}\)	A1
\(w'=\frac{e^{0.6t}(-24\sin4t)-6\cos4t(0.6e^{0.6t})}{e^{1.2t}}\)	A1	Correct substitution; then cancel \(e^{0.6t}\).
\(w'(t)=-\frac{6(0.6\cos4t+4\sin4t)}{e^{0.6t}}\)	AG
Part (g)(ii)
\(6e^{-0.6t}=0.5\Rightarrow e^{-0.6t}=\frac{1}{12}\Rightarrow t=\frac{\ln12}{0.6}\)	M1	Or \(\frac{6}{e^{0.6t}}=0.5\). A GDC intersection is acceptable here.
\(t=4.14\) s	A1	Accept 4.1415.
about 2.5 times as long as the 1.66 s with good shock absorbers	R1	Allow ECF from (e). Accept "2.48 s longer".
Part (h)
it controls how quickly the size of the bounces dies away: a larger constant makes the envelope decay faster, so the car settles sooner (stronger damping by the shock absorbers)	R1
the time between upward bounces is set by \(\cos(4t)\) only (period \(\frac{\pi}{2}\) s), which does not contain the constant	R1	Accept reference to the worn model in (g) having the same period.

Answers: (c) \(-9\) cm s\(^{-1}\) · (d) \(t=0.696\) s; \(y=-1.98\) cm · (e) \(t=1.66\) s · (f)(i) \(\frac{\pi}{2}\approx1.57\) s · (g)(ii) \(4.14\) s, about 2.5 times as long

Common error (a) Missing the inside derivatives: \(\frac{d}{dt}\cos(4t)=-4\sin(4t)\), not \(-\sin(4t)\), and \(\frac{d}{dt}e^{-1.5t}=-1.5e^{-1.5t}\).

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5.6 Integration by substitution and volumes of revolution · 5.7 Kinematics with calculusmerged group

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Designing a 3D-printed vase: volumes of revolution

5.6 Integration by substitution and volumes of revolution3.2 Volume and surface area in context2.2 Quadratic and cubic models5.3 Integration and the trapezoidal ruleHL

A designer prints a plastic vase that is a solid of revolution about a vertical axis. Heights \(h\) are measured in cm from the bottom of the vase, and \(r\) cm is the distance from the axis.

The outer surface is formed by rotating the curve \(r=8-0.02(h-10)^2\), \(0\le h\le25\), through \(360^\circ\) about the \(h\)-axis.

The inner surface, which holds the water, is formed by rotating the curve \(r=7.5-0.02(h-10)^2\), \(1\le h\le25\), through \(360^\circ\) about the same axis. The vase is open at the top and its base is a solid disc 1 cm thick.

(a)

Write down the external diameter of the vase at its widest point, and the height \(h\) at which this occurs.

ROUTINENotes

Show solution

widest point at the vertex of the parabola, \(h=10\) cm	A1	Read from the vertex form \(8-0.02(h-10)^2\).
external diameter \(2\times8=16\) cm	A1	A radius of 8 cm with no doubling scores A0 for this mark.

Answer: \(16\) cm, at \(h=10\) cm

(2)

(b)

On Figure 1, sketch the curve \(r=8-0.02(h-10)^2\) for \(0\le h\le25\), labelling the coordinates \((r,h)\) of its end points. Shade the region that is rotated to form the solid enclosed by the outer surface.

Figure 1: axes for the profile of the vase (radius \(r\) horizontal, height \(h\) vertical)

GRAPHNotes

Show solution

smooth curve with \(r\) increasing from \(h=0\) to a maximum at \((8,10)\), then decreasing to \(h=25\)	A1	The curve bulges away from the \(h\)-axis; the maximum point must be in the correct place.
end points \((6,0)\) and \((3.5,25)\) labelled	A1	Accept end points correctly placed on the grid with coordinates written.
region between the curve and the \(h\)-axis, for \(0\le h\le25\), shaded	A1	The region must be bounded by \(h=0\) and \(h=25\). Do not award for shading on the wrong side of the curve.

(3)

(c)

Using the substitution \(u=h-10\), show that the volume of the solid enclosed by the outer surface is \(3780\) cm³, correct to 3 significant figures.

PROOFNotes

Show solution

\(V=\pi\displaystyle\int_0^{25}\left(8-0.02(h-10)^2\right)^2dh\)	M1	Must include \(\pi\), the square and the limits 0 and 25.
\(du=dh\); when \(h=0\), \(u=-10\) and when \(h=25\), \(u=15\)	M1	New limits must be seen. Keeping the limits 0 and 25 in \(u\) scores M0.
\(\pi\displaystyle\int_{-10}^{15}\left(64-0.32u^2+0.0004u^4\right)du=\pi\left[64u-\frac{0.32u^3}{3}+0.00008u^5\right]_{-10}^{15}\)	A1	Correct expansion of the square and correct antiderivative.
\(=\pi(660.75+541.33\ldots)=1202.08\ldots\pi=3776.4\ldots\)	A1	Exact value \(\frac{14425\pi}{12}\). A GDC value alone, without the working above, scores at most M1.
\(V\approx3780\) cm³	AG

Answer: \(V=\frac{14425\pi}{12}=3776.4\ldots\approx3780\) cm³

(4)

(d)

Find the capacity of the vase, giving your answer in litres.

MULTINotes

Show solution

\(\pi\displaystyle\int_1^{25}\left(7.5-0.02(h-10)^2\right)^2dh\)	M1	Inner curve with limits 1 and 25. Limits 0 and 25 score M0.
\(=3157.5\ldots\) cm³	A1	Exact value \(\frac{3140856\pi}{3125}\).
\(3.16\) litres	A1	Allow ECF from an incorrect volume for the conversion (divide by 1000).

Answer: \(3.16\) litres (\(3160\) cm³)

(3)

(e)

Find the volume of plastic needed to print the vase.

ROUTINENotes

Show solution

\(3776.45\ldots-3157.53\ldots\)	M1	Outer solid minus inner solid. Allow ECF from (d).
\(619\) cm³	A1	\(618.92\ldots\) cm³.

Answer: \(619\) cm³

(2)

(f)

Before the design was finished, the capacity was estimated by treating the inside of the vase as a cylinder with the greatest internal radius, 7.5 cm, and the internal height, 24 cm.

Find this estimate, and find by how many litres it exceeds the capacity found in (d).

ROUTINENotes

Show solution

\(\pi(7.5)^2(24)\)	M1
\(4240\) cm³ (\(4241.15\ldots\))	A1	Exact \(1350\pi\).
\(4.241\ldots-3.157\ldots=1.08\) litres	A1	Allow ECF from (d).

Answer: \(4240\) cm³; it exceeds the capacity by \(1.08\) litres

(3)

(g)

A potter makes a ceramic copy of the vase by hand. She measures the internal radius \(r\) cm at heights \(h\) cm above the bottom of her vase.

\(h\) (cm)	1	5	9	13	17	21	25
\(r\) (cm)	5.9	6.9	7.4	7.4	6.6	5.2	3.2

Use the trapezoidal rule with all the values in the table to estimate the capacity of the ceramic vase, in cm³.

MULTINotes

Show solution

capacity \(=\pi\displaystyle\int_1^{25}r^2\,dh\), so the ordinates are \(\pi r^2\): \(109.4,\ 149.6,\ 172.0,\ 172.0,\ 136.8,\ 85.0,\ 32.2\)	M1	Squaring the radii is required. Applying the rule to \(r\) itself scores M0 for this mark.
\(\frac{4}{2}\big[109.36+2(149.57+172.03+172.03+136.85+84.95)+32.17\big]\)	M1	Strip width 4 and the correct pattern of coefficients. Equivalent: \(2\pi\big[34.81+2(227.73)+10.24\big]\).
\(3140\) cm³ (\(3144.8\ldots\))	A1	Exact form \(1001.02\pi\).

Answer: \(3140\) cm³

(3)

(h)

Explain why the method in (g) gives a much better estimate of a vase's capacity than the cylinder method in (f).

INTERPRETNotes

Show solution

the cylinder uses the greatest radius at every height, but the vase is narrower near the base and much narrower near the top, so the cylinder includes space outside the vase (a large overestimate)	R1
the trapezoidal rule uses the actual cross-sectional areas at seven heights, so it follows the changing radius; its only error comes from joining neighbouring values with straight lines over 4 cm intervals	R1	Accept: more measurements (smaller strips) would make it more accurate still.

(2)

(i)

A larger version of the printed vase is made by multiplying every length in the design by 1.5. Find the capacity of the larger vase, in litres.

ROUTINENotes

Show solution

\(1.5^3\times3157.5\ldots\)	M1	Volume scale factor \(1.5^3=3.375\). Allow a correct new integral. Multiplying by 1.5 or \(1.5^2\) scores M0.
\(10.7\) litres (\(10657\) cm³)	A1	Allow ECF from (d).

Answer: \(10.7\) litres

(2)

Show full solution

Working	Mark	Notes
Part (a)
widest point at the vertex of the parabola, \(h=10\) cm	A1	Read from the vertex form \(8-0.02(h-10)^2\).
external diameter \(2\times8=16\) cm	A1	A radius of 8 cm with no doubling scores A0 for this mark.
Part (b)
smooth curve with \(r\) increasing from \(h=0\) to a maximum at \((8,10)\), then decreasing to \(h=25\)	A1	The curve bulges away from the \(h\)-axis; the maximum point must be in the correct place.
end points \((6,0)\) and \((3.5,25)\) labelled	A1	Accept end points correctly placed on the grid with coordinates written.
region between the curve and the \(h\)-axis, for \(0\le h\le25\), shaded	A1	The region must be bounded by \(h=0\) and \(h=25\). Do not award for shading on the wrong side of the curve.
Part (c)
\(V=\pi\displaystyle\int_0^{25}\left(8-0.02(h-10)^2\right)^2dh\)	M1	Must include \(\pi\), the square and the limits 0 and 25.
\(du=dh\); when \(h=0\), \(u=-10\) and when \(h=25\), \(u=15\)	M1	New limits must be seen. Keeping the limits 0 and 25 in \(u\) scores M0.
\(\pi\displaystyle\int_{-10}^{15}\left(64-0.32u^2+0.0004u^4\right)du=\pi\left[64u-\frac{0.32u^3}{3}+0.00008u^5\right]_{-10}^{15}\)	A1	Correct expansion of the square and correct antiderivative.
\(=\pi(660.75+541.33\ldots)=1202.08\ldots\pi=3776.4\ldots\)	A1	Exact value \(\frac{14425\pi}{12}\). A GDC value alone, without the working above, scores at most M1.
\(V\approx3780\) cm³	AG
Part (d)
\(\pi\displaystyle\int_1^{25}\left(7.5-0.02(h-10)^2\right)^2dh\)	M1	Inner curve with limits 1 and 25. Limits 0 and 25 score M0.
\(=3157.5\ldots\) cm³	A1	Exact value \(\frac{3140856\pi}{3125}\).
\(3.16\) litres	A1	Allow ECF from an incorrect volume for the conversion (divide by 1000).
Part (e)
\(3776.45\ldots-3157.53\ldots\)	M1	Outer solid minus inner solid. Allow ECF from (d).
\(619\) cm³	A1	\(618.92\ldots\) cm³.
Part (f)
\(\pi(7.5)^2(24)\)	M1
\(4240\) cm³ (\(4241.15\ldots\))	A1	Exact \(1350\pi\).
\(4.241\ldots-3.157\ldots=1.08\) litres	A1	Allow ECF from (d).
Part (g)
capacity \(=\pi\displaystyle\int_1^{25}r^2\,dh\), so the ordinates are \(\pi r^2\): \(109.4,\ 149.6,\ 172.0,\ 172.0,\ 136.8,\ 85.0,\ 32.2\)	M1	Squaring the radii is required. Applying the rule to \(r\) itself scores M0 for this mark.
\(\frac{4}{2}\big[109.36+2(149.57+172.03+172.03+136.85+84.95)+32.17\big]\)	M1	Strip width 4 and the correct pattern of coefficients. Equivalent: \(2\pi\big[34.81+2(227.73)+10.24\big]\).
\(3140\) cm³ (\(3144.8\ldots\))	A1	Exact form \(1001.02\pi\).
Part (h)
the cylinder uses the greatest radius at every height, but the vase is narrower near the base and much narrower near the top, so the cylinder includes space outside the vase (a large overestimate)	R1
the trapezoidal rule uses the actual cross-sectional areas at seven heights, so it follows the changing radius; its only error comes from joining neighbouring values with straight lines over 4 cm intervals	R1	Accept: more measurements (smaller strips) would make it more accurate still.
Part (i)
\(1.5^3\times3157.5\ldots\)	M1	Volume scale factor \(1.5^3=3.375\). Allow a correct new integral. Multiplying by 1.5 or \(1.5^2\) scores M0.
\(10.7\) litres (\(10657\) cm³)	A1	Allow ECF from (d).

Answers: (a) \(16\) cm, at \(h=10\) cm · (c) \(V=\frac{14425\pi}{12}=3776.4\ldots\approx3780\) cm³ · (d) \(3.16\) litres (\(3160\) cm³) · (e) \(619\) cm³ · (f) \(4240\) cm³; it exceeds the capacity by \(1.08\) litres · (g) \(3140\) cm³ · (i) \(10.7\) litres

Common error (c) Square the whole radius: \(\left(8-0.02u^2\right)^2=64-0.32u^2+0.0004u^4\), not \(64+0.0004u^4\).

Exam tip (i) When every length is multiplied by \(k\), every volume is multiplied by \(k^3\); there is no need to set up a new integral.

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An express lift in a skyscraper: velocity, acceleration and distance

5.7 Kinematics with calculus5.3 Integration and the trapezoidal rule5.6 Integration by substitution and volumes of revolution5.4 Derivatives of trigonometric, exponential and logarithmic functionsHL

An express lift in a skyscraper starts from rest at ground level and travels upwards. Its velocity \(v\) m s\(^{-1}\), \(t\) seconds after it starts, is modelled by

\(v(t)=9t\,e^{-0.04t^2},\quad 0\le t\le20.\)

Upwards is the positive direction. It can be shown that the acceleration of the lift is \(a(t)=9e^{-0.04t^2}\left(1-0.08t^2\right)\) m s\(^{-2}\).

(a)

Find the velocity and the acceleration of the lift when \(t=2\), and state, with a reason, whether the lift is speeding up or slowing down at that moment.

ROUTINENotes

Show solution

\(v(2)=18e^{-0.16}=15.3\) m s\(^{-1}\)	A1	\(15.338\ldots\)
\(a(2)=9e^{-0.16}(0.68)=5.22\) m s\(^{-2}\)	A1	\(5.2151\ldots\)
speeding up, because \(v\) and \(a\) are both positive (same sign)	R1	Do not award R1 for 'a is positive' alone.

Answer: \(v=15.3\) m s\(^{-1}\), \(a=5.22\) m s\(^{-2}\); speeding up

(3)

(b)

Find the exact time at which the velocity of the lift is greatest, justify that it is a maximum, and find the greatest velocity.

MULTINotes

Show solution

\(a=0\Rightarrow1-0.08t^2=0\) (since \(e^{-0.04t^2}\gt0\))	M1
\(t=\sqrt{12.5}=\frac{5\sqrt2}{2}\) s \((3.54\) s\()\)	A1	Accept \(\frac{5}{\sqrt2}\). Reject the negative root, outside the domain.
\(a\gt0\) for \(0\le t\lt\sqrt{12.5}\) and \(a\lt0\) for \(t\gt\sqrt{12.5}\), so \(v\) increases then decreases: a maximum	R1	Accept a sign test with values either side, or a GDC graph of \(v\) described.
\(v_{\max}=19.3\) m s\(^{-1}\)	A1	\(19.2996\ldots\); exact \(\frac{45\sqrt2}{2}e^{-0.5}\).

Answer: \(t=\sqrt{12.5}=\frac{5\sqrt2}{2}\) s; \(v_{\max}=19.3\) m s\(^{-1}\)

(4)

(c)

The height of the lift above ground level at time \(T\) seconds is \(s(T)\) metres. Use the substitution \(u=-0.04t^2\) to show that

\(s(T)=112.5\left(1-e^{-0.04T^2}\right).\)

PROOFNotes

Show solution

\(s(T)=\displaystyle\int_0^T9t\,e^{-0.04t^2}\,dt\)	M1	Lift starts at ground level, so no constant is needed.
\(du=-0.08t\,dt\), so \(9t\,dt=-112.5\,du\)	M1
when \(t=0\), \(u=0\); when \(t=T\), \(u=-0.04T^2\): \(\displaystyle\int_0^{-0.04T^2}-112.5e^u\,du\)	A1	Accept returning to \(t\) before using the limits 0 and \(T\).
\(=-112.5\left[e^u\right]_0^{-0.04T^2}=-112.5\left(e^{-0.04T^2}-1\right)\)	A1
\(s(T)=112.5\left(1-e^{-0.04T^2}\right)\)	AG

(4)

(d)(i)

Hence write down the height of the lift when \(t=20\), correct to one decimal place.

ROUTINENotes

Show solution

\(112.5\) m

\(112.49998\ldots\); \(e^{-16}\) is negligible.

Answer: \(112.5\) m

(1)

(d)(ii)

Find the exact time at which the lift is 100 m above ground level.

ALGEBRANotes

Show solution

\(112.5\left(1-e^{-0.04T^2}\right)=100\)	M1
\(e^{-0.04T^2}=\frac19\Rightarrow0.04T^2=\ln9\)	M1	Correct use of logarithms.
\(T=5\sqrt{\ln9}\) s \((7.41\) s\()\)	A1	Accept \(\sqrt{25\ln9}\). A decimal answer only scores M1M1A0.

Answer: \(T=5\sqrt{\ln9}\approx7.41\) s

(3)

(e)(i)

On a different trip, a service lift moves in the same shaft. Its velocity–time graph for \(0\le t\le28\), with upwards positive, is shown in Figure 1.

Find the total distance travelled by the service lift.

Figure 1: velocity–time graph of the service lift (upwards positive)

GRAPHNotes

Show solution

areas: \(\frac12(9+15)(6)=72\) and \(\frac12(4+8)(4)=24\)	M1	Either area correct, from trapezia or triangles plus rectangles.
\(96\) m	A1

Answer: \(96\) m

(2)

(e)(ii)

The service lift starts at a platform 80 m above ground level. Find its least height above ground level during these 28 seconds, and its height above ground level when \(t=28\).

GRAPHNotes

Show solution

signed areas: \(-72\) for \(0\le t\le15\) and \(+24\) for \(20\le t\le28\)	M1	The area below the axis counts as negative. Allow ECF from their areas in (e)(i).
least height \(80-72=8\) m (for \(15\le t\le20\))	A1
height at \(t=28\): \(80-72+24=32\) m	A1	Allow ECF from their least height.

Answer: least height \(8\) m; \(32\) m above ground level at \(t=28\)

(3)

(f)

Passenger lifts are usually designed so that the magnitude of the acceleration does not exceed 1.5 m s\(^{-2}\). Comment on whether the model for the express lift is realistic.

INTERPRETNotes

Show solution

greatest magnitude of acceleration is \(a(0)=9\) m s\(^{-2}\) (the deceleration reaches about \(4.0\) m s\(^{-2}\))	A1	Accept any value of \(\|a\|\gt1.5\) found from the model with its time, e.g. \(a(2)=5.22\).
this is far above 1.5 m s⁻², so the model is not realistic for passengers, especially at the start, where a real lift's acceleration builds up gradually from zero	R1	Do not award R1 without a value from the model.

(2)

(g)(i)

The lift's speed is now limited to 8 m s\(^{-1}\). During the part of the journey where the model's velocity exceeds 8 m s\(^{-1}\), the lift instead travels at a constant 8 m s\(^{-1}\) and covers the same distance. All other parts of the journey are unchanged.

Find the two times at which the model's velocity is 8 m s\(^{-1}\).

ROUTINENotes

Show solution

\(t_1=0.919\) s	A1	\(0.91946\ldots\) (GDC: intersection of \(v\) with \(v=8\)).
\(t_2=7.24\) s	A1	\(7.24159\ldots\)

Answer: \(0.919\) s and \(7.24\) s

(2)

(g)(ii)

Find how much longer the journey takes with the speed limit.

MULTINotes

Show solution

distance between the times: \(s(7.2416)-s(0.9195)=94.95\) m	M1	Use of (c) or \(\int_{0.919}^{7.24}v\,dt\). Allow ECF from (g)(i).
time at 8 m s\(^{-1}\): \(\frac{94.95}{8}=11.87\) s	M1
extra time \(=11.87-(7.2416-0.9195)=5.55\) s	A1	\(5.5466\ldots\) s.

Answer: \(5.55\) s longer

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(v(2)=18e^{-0.16}=15.3\) m s\(^{-1}\)	A1	\(15.338\ldots\)
\(a(2)=9e^{-0.16}(0.68)=5.22\) m s\(^{-2}\)	A1	\(5.2151\ldots\)
speeding up, because \(v\) and \(a\) are both positive (same sign)	R1	Do not award R1 for 'a is positive' alone.
Part (b)
\(a=0\Rightarrow1-0.08t^2=0\) (since \(e^{-0.04t^2}\gt0\))	M1
\(t=\sqrt{12.5}=\frac{5\sqrt2}{2}\) s \((3.54\) s\()\)	A1	Accept \(\frac{5}{\sqrt2}\). Reject the negative root, outside the domain.
\(a\gt0\) for \(0\le t\lt\sqrt{12.5}\) and \(a\lt0\) for \(t\gt\sqrt{12.5}\), so \(v\) increases then decreases: a maximum	R1	Accept a sign test with values either side, or a GDC graph of \(v\) described.
\(v_{\max}=19.3\) m s\(^{-1}\)	A1	\(19.2996\ldots\); exact \(\frac{45\sqrt2}{2}e^{-0.5}\).
Part (c)
\(s(T)=\displaystyle\int_0^T9t\,e^{-0.04t^2}\,dt\)	M1	Lift starts at ground level, so no constant is needed.
\(du=-0.08t\,dt\), so \(9t\,dt=-112.5\,du\)	M1
when \(t=0\), \(u=0\); when \(t=T\), \(u=-0.04T^2\): \(\displaystyle\int_0^{-0.04T^2}-112.5e^u\,du\)	A1	Accept returning to \(t\) before using the limits 0 and \(T\).
\(=-112.5\left[e^u\right]_0^{-0.04T^2}=-112.5\left(e^{-0.04T^2}-1\right)\)	A1
\(s(T)=112.5\left(1-e^{-0.04T^2}\right)\)	AG
Part (d)(i)
\(112.5\) m	A1	\(112.49998\ldots\); \(e^{-16}\) is negligible.
Part (d)(ii)
\(112.5\left(1-e^{-0.04T^2}\right)=100\)	M1
\(e^{-0.04T^2}=\frac19\Rightarrow0.04T^2=\ln9\)	M1	Correct use of logarithms.
\(T=5\sqrt{\ln9}\) s \((7.41\) s\()\)	A1	Accept \(\sqrt{25\ln9}\). A decimal answer only scores M1M1A0.
Part (e)(i)
areas: \(\frac12(9+15)(6)=72\) and \(\frac12(4+8)(4)=24\)	M1	Either area correct, from trapezia or triangles plus rectangles.
\(96\) m	A1
Part (e)(ii)
signed areas: \(-72\) for \(0\le t\le15\) and \(+24\) for \(20\le t\le28\)	M1	The area below the axis counts as negative. Allow ECF from their areas in (e)(i).
least height \(80-72=8\) m (for \(15\le t\le20\))	A1
height at \(t=28\): \(80-72+24=32\) m	A1	Allow ECF from their least height.
Part (f)
greatest magnitude of acceleration is \(a(0)=9\) m s\(^{-2}\) (the deceleration reaches about \(4.0\) m s\(^{-2}\))	A1	Accept any value of \(\|a\|\gt1.5\) found from the model with its time, e.g. \(a(2)=5.22\).
this is far above 1.5 m s⁻², so the model is not realistic for passengers, especially at the start, where a real lift's acceleration builds up gradually from zero	R1	Do not award R1 without a value from the model.
Part (g)(i)
\(t_1=0.919\) s	A1	\(0.91946\ldots\) (GDC: intersection of \(v\) with \(v=8\)).
\(t_2=7.24\) s	A1	\(7.24159\ldots\)
Part (g)(ii)
distance between the times: \(s(7.2416)-s(0.9195)=94.95\) m	M1	Use of (c) or \(\int_{0.919}^{7.24}v\,dt\). Allow ECF from (g)(i).
time at 8 m s\(^{-1}\): \(\frac{94.95}{8}=11.87\) s	M1
extra time \(=11.87-(7.2416-0.9195)=5.55\) s	A1	\(5.5466\ldots\) s.

Answers: (a) \(v=15.3\) m s\(^{-1}\), \(a=5.22\) m s\(^{-2}\); speeding up · (b) \(t=\sqrt{12.5}=\frac{5\sqrt2}{2}\) s; \(v_{\max}=19.3\) m s\(^{-1}\) · (d)(i) \(112.5\) m · (d)(ii) \(T=5\sqrt{\ln9}\approx7.41\) s · (e)(i) \(96\) m · (e)(ii) least height \(8\) m; \(32\) m above ground level at \(t=28\) · (g)(i) \(0.919\) s and \(7.24\) s · (g)(ii) \(5.55\) s longer

Exam tip (g)(ii) Write down the integral or equation you give the GDC, such as \(9t\,e^{-0.04t^2}=8\), not only the answers it returns.

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5.8 Slope fields and Euler's method · 5.9 Coupled and second-order differential equationsmerged group

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Harvesting a fish farm: slope fields and Euler's method

5.8 Slope fields and Euler's method2.6 Transformations of graphs and logistic models2.3 Exponential and sinusoidal models5.1 Gradients, tangents and rates of changeHL

A fish farm stocks a large lake with young trout. The number of trout, \(N\) thousand, \(t\) years after stocking is modelled by the differential equation

\(\dfrac{dN}{dt}=0.6N\left(1-\dfrac{N}{12}\right),\quad N\ge0.\)

Initially 2000 trout are released, so \(N(0)=2\).

(a)

Find the values of \(N\) for which the number of trout does not change, and interpret the non-zero value in context.

MODELNotes

Show solution

\(0.6N\left(1-\frac{N}{12}\right)=0\)	M1
\(N=0\) or \(N=12\)	A1
12 000 trout is the largest population the lake can support in the long term (its carrying capacity); the population levels off there	R1	Accept 'the population tends to 12 000 trout'. Must refer to 12 000 trout, not 12.

Answer: \(N=0\) or \(N=12\); 12 000 trout is the carrying capacity

(3)

(b)(i)

Complete the table of gradients.

\(N\)	0	3	6	9	12	15
\(\dfrac{dN}{dt}\)	0		1.8		0

ROUTINENotes

Show solution

\(N=3\): \(1.35\) and \(N=9\): \(1.35\)	B1	Both values needed.
\(N=15\): \(-2.25\)	B1

Answer: \(1.35,\ 1.35,\ -2.25\)

(2)

(b)(ii)

Figure 1 shows part of the slope field. On Figure 1:

draw the missing segments at the hollow points on the lines \(N=3\), \(N=9\) and \(N=15\);
sketch the solution curve with \(N(0)=2\) and the solution curve with \(N(0)=15\).

Figure 1: part of the slope field for \(\dfrac{dN}{dt}=0.6N\left(1-\dfrac{N}{12}\right)\); the hollow points mark where segments are missing

GRAPHNotes

Show solution

segments drawn with the same gradient along each row: equal positive gradients at \(N=3\) and \(N=9\), steep negative gradients at \(N=15\)	A1	Allow ECF from (b)(i).
curve from \((0,2)\) increasing, steepest near \(N=6\), approaching \(N=12\) from below without reaching it	A1	S-shaped; must follow the segments.
curve from \((0,15)\) decreasing, approaching \(N=12\) from above	A1	The two curves must not cross the line \(N=12\) or each other.

(3)

(c)

Use Euler's method with a step length of 0.5 years to estimate \(N(2)\).

ROUTINENotes

Show solution

\(N_{n+1}=N_n+0.5\times0.6N_n\left(1-\frac{N_n}{12}\right)\), with \(N_0=2\)	M1	Correct formula seen or implied by a correct first step.
\(N_1=2.5\), \(N_2=3.09375\), \(N_3=3.78259\ldots\)	A1	At least two correct intermediate values.
\(N(2)\approx4.56\)	A1	\(4.55967\ldots\) (about 4560 trout).

Answer: \(N(2)\approx4.56\) (4560 trout)

(3)

(d)(i)

It can be shown that the solution of the differential equation with \(N(0)=2\) is \(N(t)=\dfrac{12}{1+5e^{-0.6t}}\).

Find \(N(2)\) using this solution, and hence find the error in your estimate from (c).

ROUTINENotes

Show solution

\(N(2)=\frac{12}{1+5e^{-1.2}}=4.79\)	B1	\(4.78856\ldots\)
error \(=4.7886-4.5597=0.229\) (about 229 trout)	B1	Allow ECF from (c). Accept \(-0.229\).

Answer: \(4.79\); error \(0.229\) thousand

(2)

(d)(ii)

Explain why Euler's method gives an underestimate of \(N(2)\), and describe the effect of using a step length of 0.25 years instead.

INTERPRETNotes

Show solution

for \(0\le t\le2\), \(N\lt6\), so as \(N\) increases the gradient \(\frac{dN}{dt}\) increases: the solution curve is concave up, and each tangent step lies below the curve	R1	Must refer to the gradient increasing or concavity, not just 'Euler is inaccurate'.
a step of 0.25 gives a closer estimate (still too low), with the error roughly halved	R1	Accept 'more, shorter steps follow the curve more closely'. (With step 0.25 the estimate is \(4.67\).)

(2)

(e)(i)

When \(N\) is small, the model is approximately \(\dfrac{dN}{dt}=0.6N\).

Write down the solution of this equation with \(N(0)=2\), and find the annual percentage increase that it predicts.

MODELNotes

Show solution

\(N=2e^{0.6t}\)	B1
annual growth factor \(\frac{N(t+1)}{N(t)}=e^{0.6}\)	M1	Or compare \(N(1)=3.644\ldots\) with \(N(0)=2\). Allow ECF from their solution.
\(e^{0.6}=1.822\), so \(82.2\%\) per year	A1	Not 60%.

Answer: \(N=2e^{0.6t}\); \(82.2\%\) per year

(3)

(e)(ii)

Find the number of trout predicted after 4 years by the solution in (e)(i) and by the solution in (d)(i). Hence explain why the model in (e)(i) is not suitable for long-term predictions.

INTERPRETNotes

Show solution

\(2e^{2.4}=22.0\) (22 000 trout) and \(\frac{12}{1+5e^{-2.4}}=8.26\) (8260 trout)	A1	Both values needed. Allow ECF from (e)(i).
the exponential model grows without limit and already exceeds the carrying capacity of 12 000; it ignores the limited food and space in the lake	R1

Answer: 22 000 and 8260 trout

(2)

(f)(i)

The farm now harvests trout each year in proportion to the number in the lake, so that

\(\dfrac{dN}{dt}=0.6N\left(1-\dfrac{N}{12}\right)-hN,\)

where the constant \(h\), \(0\lt h\lt0.6\), is the harvesting rate per year.

For \(h=0.15\), find the values of \(N\) for which the number of trout does not change, and determine, with a reason, whether the non-zero value is stable.

MODELNotes

Show solution

\(0.6N\left(1-\frac{N}{12}\right)-0.15N=0\) (or \(N(0.45-0.05N)=0\))	M1
\(N=0\) or \(N=9\)	A1
\(\frac{dN}{dt}\gt0\) for \(0\lt N\lt9\) and \(\frac{dN}{dt}\lt0\) for \(N\gt9\), so the population moves towards 9: \(N=9\) is stable	R1	Test values may be used, e.g. at \(N=5\) the rate is \(1\) and at \(N=10\) it is \(-0.5\).

Answer: \(N=0\) or \(N=9\); \(N=9\) is stable

(3)

(f)(ii)

The yield \(Y\) is the number of trout, in thousands, harvested per year when the population is at its non-zero equilibrium. Show that \(Y=12h-20h^2\).

PROOFNotes

Show solution

\(0.6\left(1-\frac{N}{12}\right)=h\Rightarrow N=12-20h\)	M1	Non-zero equilibrium in terms of \(h\); check: \(h=0.15\) gives 9.
\(Y=hN=h(12-20h)\)	A1	Yield = rate \(\times\) population must be seen.
\(Y=12h-20h^2\)	AG

(2)

(f)(iii)

By differentiating \(Y\) with respect to \(h\), find the harvesting rate that gives the greatest yield, and find this greatest yield.

MULTINotes

Show solution

\(\frac{dY}{dh}=12-40h\)	M1	Setting the derivative equal to zero may be seen in the next line.
\(12-40h=0\Rightarrow h=0.3\)	A1	It is a maximum, since \(Y\) is a quadratic with negative \(h^2\) coefficient.
\(Y=12(0.3)-20(0.3)^2=1.8\) (1800 trout per year)	A1	Allow ECF from their \(h\) with \(0\lt h\lt0.6\).

Answer: \(h=0.3\); greatest yield \(1.8\) thousand (1800 trout) per year

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(0.6N\left(1-\frac{N}{12}\right)=0\)	M1
\(N=0\) or \(N=12\)	A1
12 000 trout is the largest population the lake can support in the long term (its carrying capacity); the population levels off there	R1	Accept 'the population tends to 12 000 trout'. Must refer to 12 000 trout, not 12.
Part (b)(i)
\(N=3\): \(1.35\) and \(N=9\): \(1.35\)	B1	Both values needed.
\(N=15\): \(-2.25\)	B1
Part (b)(ii)
segments drawn with the same gradient along each row: equal positive gradients at \(N=3\) and \(N=9\), steep negative gradients at \(N=15\)	A1	Allow ECF from (b)(i).
curve from \((0,2)\) increasing, steepest near \(N=6\), approaching \(N=12\) from below without reaching it	A1	S-shaped; must follow the segments.
curve from \((0,15)\) decreasing, approaching \(N=12\) from above	A1	The two curves must not cross the line \(N=12\) or each other.
Part (c)
\(N_{n+1}=N_n+0.5\times0.6N_n\left(1-\frac{N_n}{12}\right)\), with \(N_0=2\)	M1	Correct formula seen or implied by a correct first step.
\(N_1=2.5\), \(N_2=3.09375\), \(N_3=3.78259\ldots\)	A1	At least two correct intermediate values.
\(N(2)\approx4.56\)	A1	\(4.55967\ldots\) (about 4560 trout).
Part (d)(i)
\(N(2)=\frac{12}{1+5e^{-1.2}}=4.79\)	B1	\(4.78856\ldots\)
error \(=4.7886-4.5597=0.229\) (about 229 trout)	B1	Allow ECF from (c). Accept \(-0.229\).
Part (d)(ii)
for \(0\le t\le2\), \(N\lt6\), so as \(N\) increases the gradient \(\frac{dN}{dt}\) increases: the solution curve is concave up, and each tangent step lies below the curve	R1	Must refer to the gradient increasing or concavity, not just 'Euler is inaccurate'.
a step of 0.25 gives a closer estimate (still too low), with the error roughly halved	R1	Accept 'more, shorter steps follow the curve more closely'. (With step 0.25 the estimate is \(4.67\).)
Part (e)(i)
\(N=2e^{0.6t}\)	B1
annual growth factor \(\frac{N(t+1)}{N(t)}=e^{0.6}\)	M1	Or compare \(N(1)=3.644\ldots\) with \(N(0)=2\). Allow ECF from their solution.
\(e^{0.6}=1.822\), so \(82.2\%\) per year	A1	Not 60%.
Part (e)(ii)
\(2e^{2.4}=22.0\) (22 000 trout) and \(\frac{12}{1+5e^{-2.4}}=8.26\) (8260 trout)	A1	Both values needed. Allow ECF from (e)(i).
the exponential model grows without limit and already exceeds the carrying capacity of 12 000; it ignores the limited food and space in the lake	R1
Part (f)(i)
\(0.6N\left(1-\frac{N}{12}\right)-0.15N=0\) (or \(N(0.45-0.05N)=0\))	M1
\(N=0\) or \(N=9\)	A1
\(\frac{dN}{dt}\gt0\) for \(0\lt N\lt9\) and \(\frac{dN}{dt}\lt0\) for \(N\gt9\), so the population moves towards 9: \(N=9\) is stable	R1	Test values may be used, e.g. at \(N=5\) the rate is \(1\) and at \(N=10\) it is \(-0.5\).
Part (f)(ii)
\(0.6\left(1-\frac{N}{12}\right)=h\Rightarrow N=12-20h\)	M1	Non-zero equilibrium in terms of \(h\); check: \(h=0.15\) gives 9.
\(Y=hN=h(12-20h)\)	A1	Yield = rate \(\times\) population must be seen.
\(Y=12h-20h^2\)	AG
Part (f)(iii)
\(\frac{dY}{dh}=12-40h\)	M1	Setting the derivative equal to zero may be seen in the next line.
\(12-40h=0\Rightarrow h=0.3\)	A1	It is a maximum, since \(Y\) is a quadratic with negative \(h^2\) coefficient.
\(Y=12(0.3)-20(0.3)^2=1.8\) (1800 trout per year)	A1	Allow ECF from their \(h\) with \(0\lt h\lt0.6\).

Answers: (a) \(N=0\) or \(N=12\); 12 000 trout is the carrying capacity · (b)(i) \(1.35,\ 1.35,\ -2.25\) · (c) \(N(2)\approx4.56\) (4560 trout) · (d)(i) \(4.79\); error \(0.229\) thousand · (e)(i) \(N=2e^{0.6t}\); \(82.2\%\) per year · (e)(ii) 22 000 and 8260 trout · (f)(i) \(N=0\) or \(N=9\); \(N=9\) is stable · (f)(iii) \(h=0.3\); greatest yield \(1.8\) thousand (1800 trout) per year

Exam tip (d)(ii) Justify the direction of an Euler error with concavity: a concave-up solution makes every tangent step land below the curve.

Common error (f)(iii) The greatest yield does not come from the largest equilibrium population: harvesting hard lowers the population, so \(Y=hN\) has a maximum in between.

More 5.8 questions in the question bank

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Foxes and rabbits on an island: a coupled system and its eigenvalues

5.9 Coupled and second-order differential equations1.7 Eigenvalues and eigenvectors5.8 Slope fields and Euler's method1.5 Complex numbers in Cartesian, polar and exponential formHL

On an island, populations of rabbits and foxes have been stable for many years. After a disturbance, ecologists model the deviations from the stable values:

\(x\) is the number of rabbits above their stable value, in hundreds;
\(y\) is the number of foxes above their stable value, in tens,

\(t\) months after the disturbance. Negative values mean below the stable value. Near equilibrium,

\(\dfrac{dx}{dt}=0.1x-0.5y,\qquad\dfrac{dy}{dt}=0.4x-0.1y.\)

A mild winter leaves 400 extra rabbits while the foxes are at their stable value, so \(x(0)=4\) and \(y(0)=0\).

(a)

Interpret, in context, the sign of the coefficient of \(y\) in the first equation and the sign of the coefficient of \(x\) in the second equation.

MODELNotes

Show solution

\(-0.5y\): more foxes than usual makes the rabbit numbers decrease (foxes eat rabbits)	R1	Must refer to the effect of one species on the rate of change of the other.
\(+0.4x\): more rabbits than usual makes the fox numbers increase (more food for foxes)	R1

(2)

(b)(i)

Use Euler's method with a step length of 0.5 months to estimate \(x(1)\) and \(y(1)\).

ROUTINENotes

Show solution

\(x_{n+1}=x_n+0.5(0.1x_n-0.5y_n)\), \(y_{n+1}=y_n+0.5(0.4x_n-0.1y_n)\), both using the old values	M1	Using \(x_{n+1}\) in the \(y\) update scores M0.
\(x_1=4.2\), \(y_1=0.8\)	A1
\(x(1)\approx4.21\), \(y(1)\approx1.60\)	A1	Second step rates: \(0.02\) and \(1.6\).

Answer: \(x(1)\approx4.21\), \(y(1)\approx1.60\)

(3)

(b)(ii)

The exact values are \(x(1)=4.01\) and \(y(1)=1.55\), correct to 3 significant figures. Comment on the accuracy of your estimates, and state how they could be improved.

CONCEPTNotes

Show solution

both estimates are too large (the rabbit estimate by about 0.2, i.e. 20 rabbits), so the step of 0.5 months is too coarse	R1	Accept a comparison of either value with a sensible comment.
use a smaller step length (more steps); the error is roughly proportional to the step length	R1

(2)

(c)(i)

The system can be written as \(\dfrac{d\mathbf X}{dt}=M\mathbf X\), where \(\mathbf X=\begin{pmatrix}x\\y\end{pmatrix}\). Write down the matrix \(M\).

ROUTINENotes

Show solution

\(M=\begin{pmatrix}0.1&-0.5\\0.4&-0.1\end{pmatrix}\)

Answer: \(M=\begin{pmatrix}0.1&-0.5\\0.4&-0.1\end{pmatrix}\)

(1)

(c)(ii)

Show that the eigenvalues of \(M\) are \(\pm i\sqrt{0.19}\).

PROOFNotes

Show solution

\(\det(M-\lambda I)=(0.1-\lambda)(-0.1-\lambda)-(-0.5)(0.4)=0\)	M1	Or trace \(0\) and determinant \(0.19\) in \(\lambda^2-(\operatorname{tr}M)\lambda+\det M=0\).
\(\lambda^2-0.01+0.2=0\Rightarrow\lambda^2+0.19=0\)	A1
\(\lambda=\pm\sqrt{-0.19}=\pm\sqrt{0.19}\,i\)	A1	The step from the negative square root to \(i\) must be seen.
\(\lambda=\pm i\sqrt{0.19}\)	AG

(3)

(c)(iii)

Hence state the type of equilibrium at the origin, and describe the shape of the trajectories in the phase plane.

CONCEPTNotes

Show solution

centre	A1
the eigenvalues are purely imaginary (real part 0), so the trajectories are closed loops around the origin that neither grow nor shrink	R1

Answer: a centre: closed loops around the origin

(2)

(d)

On Figure 1, sketch the trajectory that starts at \((4,0)\). Show the direction of motion with arrows.

Figure 1: axes for the phase portrait (\(x\): rabbits, in hundreds; \(y\): foxes, in tens; both measured from equilibrium)

GRAPHNotes

Show solution

closed loop around the origin passing through \((4,0)\)	A1	A spiral scores A0.
anticlockwise direction (at \((4,0)\), \(\frac{dy}{dt}=1.6\gt0\), so the motion is upwards)	A1
loop symmetric about the origin, crossing the axes at about \((\pm4,0)\) and \((0,\pm3.6)\)	A1	Accept a slightly tilted oval with \(y\)-intercepts between 3 and 4 in size.

(3)

(e)

Find the time taken for the populations to complete one cycle.

ROUTINENotes

Show solution

period \(=\frac{2\pi}{\sqrt{0.19}}\)	M1	Uses the imaginary part of the eigenvalues.
\(14.4\) months	A1	\(14.414\ldots\)

Answer: \(14.4\) months

(2)

(f)

Use your trajectory to describe how the numbers of rabbits and foxes change during one cycle, starting at \(t=0\).

INTERPRETNotes

Show solution

rabbit numbers rise slightly and then fall while fox numbers rise; the foxes reach their peak after the rabbits have peaked	R1	The idea that the fox peak lags the rabbit peak is required.
both then fall below their stable values (rabbits first), recover, and the cycle repeats every 14.4 months with the same size; the populations never settle at equilibrium	R1	Allow ECF from (e).

(2)

(g)(i)

A disease affects both species, and the model becomes

\(\dfrac{dx}{dt}=-0.05x-0.5y,\qquad\dfrac{dy}{dt}=0.4x-0.15y.\)

Find the eigenvalues of the new matrix, giving your answers in the form \(a+bi\).

ROUTINENotes

Show solution

\(\lambda^2+0.2\lambda+0.2075=0\)	M1	Trace \(-0.2\), determinant \(0.0075+0.2=0.2075\).
\(\lambda=-0.1\pm0.444i\)	A1	Exact \(-0.1\pm\sqrt{0.1975}\,i\).

Answer: \(-0.1\pm0.444i\)

(2)

(g)(ii)

Describe the long-term behaviour of the rabbit and fox populations under the new model.

INTERPRETNotes

Show solution

complex eigenvalues with negative real part, so the origin is a stable spiral	R1	Allow ECF from (g)(i).
both populations still oscillate, but the oscillations shrink and the numbers settle at their stable values	R1

Answer: stable spiral: damped oscillations back to equilibrium

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(-0.5y\): more foxes than usual makes the rabbit numbers decrease (foxes eat rabbits)	R1	Must refer to the effect of one species on the rate of change of the other.
\(+0.4x\): more rabbits than usual makes the fox numbers increase (more food for foxes)	R1
Part (b)(i)
\(x_{n+1}=x_n+0.5(0.1x_n-0.5y_n)\), \(y_{n+1}=y_n+0.5(0.4x_n-0.1y_n)\), both using the old values	M1	Using \(x_{n+1}\) in the \(y\) update scores M0.
\(x_1=4.2\), \(y_1=0.8\)	A1
\(x(1)\approx4.21\), \(y(1)\approx1.60\)	A1	Second step rates: \(0.02\) and \(1.6\).
Part (b)(ii)
both estimates are too large (the rabbit estimate by about 0.2, i.e. 20 rabbits), so the step of 0.5 months is too coarse	R1	Accept a comparison of either value with a sensible comment.
use a smaller step length (more steps); the error is roughly proportional to the step length	R1
Part (c)(i)
\(M=\begin{pmatrix}0.1&-0.5\\0.4&-0.1\end{pmatrix}\)	A1
Part (c)(ii)
\(\det(M-\lambda I)=(0.1-\lambda)(-0.1-\lambda)-(-0.5)(0.4)=0\)	M1	Or trace \(0\) and determinant \(0.19\) in \(\lambda^2-(\operatorname{tr}M)\lambda+\det M=0\).
\(\lambda^2-0.01+0.2=0\Rightarrow\lambda^2+0.19=0\)	A1
\(\lambda=\pm\sqrt{-0.19}=\pm\sqrt{0.19}\,i\)	A1	The step from the negative square root to \(i\) must be seen.
\(\lambda=\pm i\sqrt{0.19}\)	AG
Part (c)(iii)
centre	A1
the eigenvalues are purely imaginary (real part 0), so the trajectories are closed loops around the origin that neither grow nor shrink	R1
Part (d)
closed loop around the origin passing through \((4,0)\)	A1	A spiral scores A0.
anticlockwise direction (at \((4,0)\), \(\frac{dy}{dt}=1.6\gt0\), so the motion is upwards)	A1
loop symmetric about the origin, crossing the axes at about \((\pm4,0)\) and \((0,\pm3.6)\)	A1	Accept a slightly tilted oval with \(y\)-intercepts between 3 and 4 in size.
Part (e)
period \(=\frac{2\pi}{\sqrt{0.19}}\)	M1	Uses the imaginary part of the eigenvalues.
\(14.4\) months	A1	\(14.414\ldots\)
Part (f)
rabbit numbers rise slightly and then fall while fox numbers rise; the foxes reach their peak after the rabbits have peaked	R1	The idea that the fox peak lags the rabbit peak is required.
both then fall below their stable values (rabbits first), recover, and the cycle repeats every 14.4 months with the same size; the populations never settle at equilibrium	R1	Allow ECF from (e).
Part (g)(i)
\(\lambda^2+0.2\lambda+0.2075=0\)	M1	Trace \(-0.2\), determinant \(0.0075+0.2=0.2075\).
\(\lambda=-0.1\pm0.444i\)	A1	Exact \(-0.1\pm\sqrt{0.1975}\,i\).
Part (g)(ii)
complex eigenvalues with negative real part, so the origin is a stable spiral	R1	Allow ECF from (g)(i).
both populations still oscillate, but the oscillations shrink and the numbers settle at their stable values	R1

Answers: (b)(i) \(x(1)\approx4.21\), \(y(1)\approx1.60\) · (c)(i) \(M=\begin{pmatrix}0.1&-0.5\\0.4&-0.1\end{pmatrix}\) · (c)(iii) a centre: closed loops around the origin · (e) \(14.4\) months · (g)(i) \(-0.1\pm0.444i\) · (g)(ii) stable spiral: damped oscillations back to equilibrium

Common error (g)(ii) The type depends on the real part of complex eigenvalues: negative means the loops shrink (stable spiral), zero means closed loops (centre).

More 5.9 questions in the question bank

	A	B	C	D	E	F
A	–	25	45	50	55	30
B	25	–	30	65	60	45
C	45	30	–	45	40	35
D	50	65	45	–	30	35
E	55	60	40	30	–	50
F	30	45	35	35	50	–

	A	C	E	G	H	L	S
A	–	130	–	–	–	160	–
C	130	–	–	–	170	100	150
E	–	–	–	140	–	220	120
G	–	–	140	–	210	–	190
H	–	170	–	210	–	–	–
L	160	100	220	–	–	–	–
S	–	150	120	190	–	–	–

	A	B	C	D	E	F
A	–	25	45	50	55	30
B	25	–	30	65	60	45
C	45	30	–	45	40	35
D	50	65	45	–	30	35
E	55	60	40	30	–	50
F	30	45	35	35	50	–

	A	C	E	G	H	L	S
A	–	130	–	–	–	160	–
C	130	–	–	–	170	100	150
E	–	–	–	140	–	220	120
G	–	–	140	–	210	–	190
H	–	170	–	210	–	–	–
L	160	100	220	–	–	–	–
S	–	150	120	190	–	–	–

	A	B	C	D	E	F
A	–	25	45	50	55	30
B	25	–	30	65	60	45
C	45	30	–	45	40	35
D	50	65	45	–	30	35
E	55	60	40	30	–	50
F	30	45	35	35	50	–

	A	C	E	G	H	L	S
A	–	130	–	–	–	160	–
C	130	–	–	–	170	100	150
E	–	–	–	140	–	220	120
G	–	–	140	–	210	–	190
H	–	170	–	210	–	–	–
L	160	100	220	–	–	–	–
S	–	150	120	190	–	–	–