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IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 54 marks

Question 1

6 marks

A chord and a major sector in a circle

Geometry and trigonometry

(a)

3 marks

Use the cosine rule, or split the isosceles triangle in half.

A B = 2 (5) sin \frac{1.9}{2} = 10 sin 0.95 \approx 8.13 m

(b)

3 marks

The angle of the shaded sector is the reflex angle.

θ = 2 π - 1.9 \approx 4.383, A = \frac{1}{2} (5)^{2} (4.383) \approx 54.8 m^{2}

Question 2

5 marks

Recovering a function from its derivative

Number and algebra

5 marks

Integrate term by term.
Use $g (0) = 4$ .

g (x) = \int (3 x^{2} + 5 e^{x}) d x = x^{3} + 5 e^{x} + C 4 = 0 + 5 + C \Rightarrow C = - 1 g (x) = x^{3} + 5 e^{x} - 1

Question 3

6 marks

Independent events with one probability three times the other

Number and algebra

6 marks

Let $b = P (B)$ , so $P (A) = 3 b$ . For independent events $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ .
A probability can't exceed 1, so

3 b + b - 3 b^{2} = 0.68 \Rightarrow 3 b^{2} - 4 b + 0.68 = 0 b = \frac{4 \pm 16 - 8.16}{6} = \frac{4 \pm 2.8}{6} \Rightarrow b = 0.2 or b = 1.13 \dots P (B) = 0.2

Question 4

6 marks

Reaction times after sleeping well: a box plot, an outlier test and what it suggests

Statistics and probability

(a)

1 mark

Read the line inside the box.

median = 0.28 s

(b)

3 marks

From the box plot, $Q_{1} = 0.27$ and $Q_{3} = 0.35$ . The upper fence is $Q_{3} + 1.5 \times IQR$ .
Since $0.46 < 0.47$ , the measurement of 0.46 s is not an outlier.

IQR = 0.35 - 0.27 = 0.08, 0.35 + 1.5 (0.08) = 0.47

(c)

1 mark

The median is close to the lower quartile and the upper whisker and box are longer. The distribution is positively skewed, so a few large values pull the mean above the median.

(d)

1 mark

The box plot for not sleeping well is centred at higher times: its median equals the upper quartile of the "slept well" data. That is consistent with slower reactions after a bad night, but it doesn't prove cause, and a sample of nine is too small for strong conclusions.

Question 5

7 marks

A particle's velocity $v = \frac{( t ^{2} + 1 ) cos t}{4}$ : direction change, acceleration and greatest speed

Calculus

(a)

2 marks

The particle changes direction when $v = 0$ . Since $t^{2} + 1 > 0$ , this needs $cos t = 0$ .

t = \frac{π}{2} \approx 1.57 s

(b)

3 marks

The acceleration is $a = v^{'} (t) = \frac{2 t cos t - ( t ^{2} + 1 ) sin t}{4}$ . Solve $a (t) = - 1.9$ on the calculator for $0 \leq t \leq 3$ .

t \approx 2.26 or 2.96 s

(c)

2 marks

Compare $∣ v ∣$ at the end points and the turning points. $v (0) = 0.25$ and $v (3) = \frac{10 cos 3}{4} \approx - 2.47$ , which is the largest size. So the speed is greatest at $t = 3$ .

a (3) = \frac{6 cos 3 - 10 sin 3}{4} \approx - 1.84 m s^{- 2}

Question 6

5 marks

The base height of a bowl from the volume of revolution of an ellipse

Number and algebra

5 marks

From the ellipse $\frac{x ^{2}}{36} + \frac{( y - 4 ) ^{2}}{16} = 1$ we get $x^{2} = 36 (1 - \frac{( y - 4 ) ^{2}}{16})$ . The interior volume is the volume of revolution from $y = h$ to $y = 4$ .
Set the volume to 285 and solve for $h$ on the calculator.

V = π \int_{h}^{4} 36 (1 - \frac{( y - 4 ) ^{2}}{16}) d y = 36 π [(4 - h) - \frac{( 4 - h ) ^{3}}{48}] 36 π [(4 - h) - \frac{( 4 - h ) ^{3}}{48}] = 285 \Rightarrow h \approx 0.793 cm

Question 7

8 marks

A limit that exists only for one value of $k$ , and l'Hôpital's rule

Calculus

(a)

2 marks

As $x \to 0$ the denominator $x^{2} \to 0$ . For the limit to be finite the numerator must also tend to 0.

arctan (cos 0) - k = arctan 1 - k = \frac{π}{4} - k = 0 \Rightarrow k = \frac{π}{4}

(b)

6 marks

With $k = \frac{π}{4}$ the limit has the form $\frac{0}{0}$ . Differentiate the top and bottom.
Simplify to $\frac{- sin x}{2 x ( 1 + cos ^{2} x )}$ . This is $\frac{0}{0}$ again, so apply the rule a second time. The derivative of the top is $- cos x$ , and the derivative of the bottom, by the product rule, is $2 (1 + cos^{2} x) + 2 x (- 2 cos x sin x)$ .

x \to 0 lim \frac{arctan ( cos x ) - \frac{π}{4}}{x ^{2}} = x \to 0 lim \frac{- sin x / ( 1 + cos ^{2} x )}{2 x} x \to 0 lim \frac{- cos x}{2 ( 1 + cos ^{2} x ) - 4 x cos x sin x} = \frac{- 1}{2 ( 2 ) - 0} = - \frac{1}{4}

Question 8

7 marks

Two javelin throwers and who qualifies

Number and algebra

7 marks

Rachel: $R \sim N (56.5, 3^{2})$ , so $P (R \geq 60) \approx 0.1216$ . Sophia: $S \sim N (57.5, 1. 8^{2})$ , so $P (S \geq 60) \approx 0.0824$ .
In five throws, the number of throws of 60 m or more is binomial. A competitor qualifies with at least one.
Exactly one of them qualifies: Rachel only, or Sophia only.

P (Rachel qualifies) = 1 - (1 - 0.1216)^{5} \approx 0.4772 P (Sophia qualifies) = 1 - (1 - 0.0824)^{5} \approx 0.3495 0.4772 (1 - 0.3495) + 0.3495 (1 - 0.4772) \approx 0.3104 + 0.1827 \approx 0.493

Question 9

4 marks

Counting six-digit numbers with distinct and increasing digits

Statistics and probability

(a)

2 marks

The first digit has 9 choices (not 0). The remaining five places take distinct digits from the 9 left, in order.

9 \times^{9} P_{5} = 9 \times (9 \times 8 \times 7 \times 6 \times 5) = 136080

(b)

2 marks

In increasing order the first digit is the smallest, so 0 can't be used. Any choice of 6 different digits from 1–9 gives exactly one such number.

(6 9) = 84

Section B · Extended response

Questions 10–12 · 56 marks

Question 10

15 marks

Two growing plants: heights, crossing points and growth rates

Functions

(a)

3 marks

(i) At $t = 0$ for Plant B:

(ii) For Plant A, with the calculator in radians:

h_{B} (0) = 32 cm h_{A} (0) = sin 6 + 27 \approx 26.7 cm

(b)

3 marks

Solve $h_{A} (t) = h_{B} (t)$ on the calculator for $0 \leq t \leq 9$ .

t \approx 4.01, 4.70, 5.88 weeks

(c)

3 marks

Subtract the heights.
For $t > 6$ , $t - 5 > 1$ , and $sin (2 t + 6) \geq - 1$ . So the difference is greater than 0.
So for $t > 6$ Plant A was always taller than Plant B.

h_{A} (t) - h_{B} (t) = sin (2 t + 6) + t - 5 h_{A} (t) - h_{B} (t) > 1 - 1 = 0

(d)

6 marks

The rates are $h_{A}^{'} (t) = 2 cos (2 t + 6) + 9$ and $h_{B}^{'} (t) = 8$ . Plant B grows faster when $2 cos (2 t + 6) + 9 < 8$ , that is when $cos (2 t + 6) < - \frac{1}{2}$ .
Solve $h_{A}^{'} (t) = h_{B}^{'} (t)$ on the calculator: the crossings are at about 1.19, 2.24; 4.33, 5.38; and 7.47, 8.52. Between each pair, B is growing faster.

(2.236 - 1.189) + (5.378 - 4.330) + (8.519 - 7.472) \approx 3.14 weeks

Mentor tip

Exactly: each of the three intervals has length $\frac{π}{3}$ , so the total is $π$ weeks.

Question 11

20 marks

Two airplanes: bearing, speeds, angle between flight paths and closest approach

Geometry and trigonometry

(a)

2 marks

Airplane B moves 4 km east and 2 km north per minute, so its bearing is the angle from north, measured clockwise.

tan^{- 1} \frac{4}{2} \approx 63. 4^{\circ} \Rightarrow bearing 06 3^{\circ}

(b)

2 marks

The speed is the length of the direction vector.
Since $7.48 > 4.90$ , airplane A travels faster.

∣ b_{A} ∣ = (- 6)^{2} + 2^{2} + 4^{2} = 56 \approx 7.48 km min^{- 1}, ∣ b_{B} ∣ = 4^{2} + 2^{2} + (- 2)^{2} = 24 \approx 4.90 km min^{- 1}

(c)

4 marks

Use the scalar product of the direction vectors $- 6 24$ and $42 - 2$ .
The acute angle between the lines is the supplement.

cos θ = \frac{- 24 + 4 - 8}{56 24} = \frac{- 28}{1344} \approx - 0.764 \Rightarrow θ \approx 139. 8^{\circ} 18 0^{\circ} - 139. 8^{\circ} \approx 40. 2^{\circ}

(d)

7 marks

(i) Equate positions with separate times $t_{1}$ and $t_{2}$ .

Solve the second and third equations: $t_{1} = 2$ and $t_{2} = \frac{3}{2}$ . Check the first: $19 - 12 = 7 = 1 + 6$ . Then

(ii) Airplane A reaches P at $t = 2$ and B at $t = 1.5$ .

19 - 6 t_{1} = 1 + 4 t_{2}, - 1 + 2 t_{1} = 2 t_{2}, 1 + 4 t_{1} = 12 - 2 t_{2} P (7, 3, 9) 2 - 1.5 = 0.5 minutes (30 seconds)

(e)

5 marks

The distance between the planes at time $t$ is the length of $r_{B} - r_{A}$ . Find each component: east $= (1 + 4 t) - (19 - 6 t) = 10 t - 18$ , north $= 2 t - (- 1 + 2 t) = 1$ , height $= (12 - 2 t) - (1 + 4 t) = 11 - 6 t$ .
Minimise $D (t)$ on the calculator, or set $D^{'} (t) = 0$ .

r_{B} - r_{A} = 10 t - 18 1 11 - 6 t, D (t) = (10 t - 18)^{2} + 1^{2} + (11 - 6 t)^{2} t \approx 1.81, D_{m i n} \approx 1.01 km

Question 12

21 marks

A logistic population model: maximum growth rate and an exact solution

Calculus

(a)

1 mark

$\frac{d P}{d t}$ is the rate at which the population of marsupials is changing: its growth rate, in animals per year.

(b)

4 marks

Differentiate $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ with the product rule, using the chain rule for $\frac{d P}{d t}$ .
Replace $\frac{d P}{d t}$ by the original expression.

\frac{d ^{2} P}{d t ^{2}} = k \frac{d P}{d t} (1 - \frac{P}{N}) - \frac{k}{N} P \frac{d P}{d t} = k \frac{d P}{d t} (1 - \frac{2 P}{N}) \frac{d ^{2} P}{d t ^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})

(c)

5 marks

For $0 < P < N$ the factors $P$ and $1 - \frac{P}{N}$ are positive, so the sign of $\frac{d ^{2} P}{d t ^{2}}$ is the sign of $1 - \frac{2 P}{N}$ .
For $P < \frac{N}{2}$ that is positive (so the growth rate is increasing), for $P > \frac{N}{2}$ it is negative (so the growth rate is decreasing). The growth rate changes from increasing to decreasing at $P = \frac{N}{2}$ .
So the population grows at its maximum rate when $P = \frac{N}{2}$ .

(d)

2 marks

Substitute $P = \frac{N}{2}$ into $\frac{d P}{d t}$ .

\frac{d P}{d t}_{m a x} = k \cdot \frac{N}{2} (1 - \frac{1}{2}) = \frac{k N}{4}

(e)

7 marks

Separate the variables: $\frac{N}{P ( N - P )} d P = k d t$ . Use partial fractions.
At $t = 0$ , $P = P_{0}$ , so $C = ln \frac{P _{0}}{N - P _{0}}$ . Combine the logarithms.

\frac{N}{P ( N - P )} = \frac{1}{P} + \frac{1}{N - P} \Rightarrow \int (\frac{1}{P} + \frac{1}{N - P}) d P = \int k d t ln P - ln (N - P) = k t + C ln \frac{P}{N - P} - ln \frac{P _{0}}{N - P _{0}} = k t \Rightarrow k t = ln (\frac{P ( N - P _{0} )}{P _{0} ( N - P )})

(f)

2 marks

Put $N = 4 P_{0}$ , $P = 3 P_{0}$ and $t = 10$ into the formula.

10 k = ln \frac{3 P _{0} \cdot 3 P _{0}}{P _{0} \cdot P _{0}} = ln 9 \Rightarrow k = \frac{ln 9}{10} \approx 0.220

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus36 marks · 33%
Q5, Q7, Q12
Geometry and trigonometry26 marks · 24%
Q1, Q11
Number and algebra23 marks · 21%
Q2, Q3, Q6, Q8
Functions15 marks · 14%
Q10
Statistics and probability10 marks · 9%
Q4, Q9

Mentor insights

Traps to watch for

01
Check the end of the interval · Q5
The greatest speed on $0 \leq t \leq 3$ might be at an end point. Compare $∣ v ∣$ at the turning points of $v$ and at $t = 0$ and $t = 3$ .
02
Zero over zero first · Q7
A finite limit with denominator $x^{2} \to 0$ needs the numerator to tend to 0 too. That forces $k$ . Then l'Hôpital's rule applies, and may need to be used twice.
03
Rates are derivatives · Q10
In (d) compare the derivatives $h_{A}^{'} (t)$ and $h_{B}^{'} (t)$ . The condition is a cosine inequality, and each period contributes the same length of time.
04
Bearing is clockwise from north · Q11
Bearing uses north and east components only: $tan^{- 1} \frac{east}{north}$ . Speed uses the whole direction vector, including height.
05
Rate of growth maximum is the peak of $\frac{d P}{d t}$ · Q12
The maximum of $\frac{d P}{d t}$ is where its derivative, $\frac{d ^{2} P}{d t ^{2}}$ , is zero and changes sign from positive to negative.

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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 54 marks

Question 1

6 marks

A chord and a major sector in a circle

Geometry and trigonometry

(a)

3 marks

Use the cosine rule, or split the isosceles triangle in half.

A B = 2 (5) sin \frac{1.9}{2} = 10 sin 0.95 \approx 8.13 m

(b)

3 marks

The angle of the shaded sector is the reflex angle.

θ = 2 π - 1.9 \approx 4.383, A = \frac{1}{2} (5)^{2} (4.383) \approx 54.8 m^{2}

Question 2

5 marks

Recovering a function from its derivative

Number and algebra

5 marks

Integrate term by term.
Use $g (0) = 4$ .

g (x) = \int (3 x^{2} + 5 e^{x}) d x = x^{3} + 5 e^{x} + C 4 = 0 + 5 + C \Rightarrow C = - 1 g (x) = x^{3} + 5 e^{x} - 1

Question 3

6 marks

Independent events with one probability three times the other

Number and algebra

6 marks

Let $b = P (B)$ , so $P (A) = 3 b$ . For independent events $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ .
A probability can't exceed 1, so

3 b + b - 3 b^{2} = 0.68 \Rightarrow 3 b^{2} - 4 b + 0.68 = 0 b = \frac{4 \pm 16 - 8.16}{6} = \frac{4 \pm 2.8}{6} \Rightarrow b = 0.2 or b = 1.13 \dots P (B) = 0.2

Question 4

6 marks

Reaction times after sleeping well: a box plot, an outlier test and what it suggests

Statistics and probability

(a)

1 mark

Read the line inside the box.

median = 0.28 s

(b)

3 marks

From the box plot, $Q_{1} = 0.27$ and $Q_{3} = 0.35$ . The upper fence is $Q_{3} + 1.5 \times IQR$ .
Since $0.46 < 0.47$ , the measurement of 0.46 s is not an outlier.

IQR = 0.35 - 0.27 = 0.08, 0.35 + 1.5 (0.08) = 0.47

(c)

1 mark

The median is close to the lower quartile and the upper whisker and box are longer. The distribution is positively skewed, so a few large values pull the mean above the median.

(d)

1 mark

Question 5

7 marks

A particle's velocity $v = \frac{( t ^{2} + 1 ) cos t}{4}$ : direction change, acceleration and greatest speed

Calculus

(a)

2 marks

The particle changes direction when $v = 0$ . Since $t^{2} + 1 > 0$ , this needs $cos t = 0$ .

t = \frac{π}{2} \approx 1.57 s

(b)

3 marks

The acceleration is $a = v^{'} (t) = \frac{2 t cos t - ( t ^{2} + 1 ) sin t}{4}$ . Solve $a (t) = - 1.9$ on the calculator for $0 \leq t \leq 3$ .

t \approx 2.26 or 2.96 s

(c)

2 marks

Compare $∣ v ∣$ at the end points and the turning points. $v (0) = 0.25$ and $v (3) = \frac{10 cos 3}{4} \approx - 2.47$ , which is the largest size. So the speed is greatest at $t = 3$ .

a (3) = \frac{6 cos 3 - 10 sin 3}{4} \approx - 1.84 m s^{- 2}

Question 6

5 marks

The base height of a bowl from the volume of revolution of an ellipse

Number and algebra

5 marks

From the ellipse $\frac{x ^{2}}{36} + \frac{( y - 4 ) ^{2}}{16} = 1$ we get $x^{2} = 36 (1 - \frac{( y - 4 ) ^{2}}{16})$ . The interior volume is the volume of revolution from $y = h$ to $y = 4$ .
Set the volume to 285 and solve for $h$ on the calculator.

V = π \int_{h}^{4} 36 (1 - \frac{( y - 4 ) ^{2}}{16}) d y = 36 π [(4 - h) - \frac{( 4 - h ) ^{3}}{48}] 36 π [(4 - h) - \frac{( 4 - h ) ^{3}}{48}] = 285 \Rightarrow h \approx 0.793 cm

Question 7

8 marks

A limit that exists only for one value of $k$ , and l'Hôpital's rule

Calculus

(a)

2 marks

As $x \to 0$ the denominator $x^{2} \to 0$ . For the limit to be finite the numerator must also tend to 0.

arctan (cos 0) - k = arctan 1 - k = \frac{π}{4} - k = 0 \Rightarrow k = \frac{π}{4}

(b)

6 marks

With $k = \frac{π}{4}$ the limit has the form $\frac{0}{0}$ . Differentiate the top and bottom.
Simplify to $\frac{- sin x}{2 x ( 1 + cos ^{2} x )}$ . This is $\frac{0}{0}$ again, so apply the rule a second time. The derivative of the top is $- cos x$ , and the derivative of the bottom, by the product rule, is $2 (1 + cos^{2} x) + 2 x (- 2 cos x sin x)$ .

x \to 0 lim \frac{arctan ( cos x ) - \frac{π}{4}}{x ^{2}} = x \to 0 lim \frac{- sin x / ( 1 + cos ^{2} x )}{2 x} x \to 0 lim \frac{- cos x}{2 ( 1 + cos ^{2} x ) - 4 x cos x sin x} = \frac{- 1}{2 ( 2 ) - 0} = - \frac{1}{4}

Question 8

7 marks

Two javelin throwers and who qualifies

Number and algebra

7 marks

Rachel: $R \sim N (56.5, 3^{2})$ , so $P (R \geq 60) \approx 0.1216$ . Sophia: $S \sim N (57.5, 1. 8^{2})$ , so $P (S \geq 60) \approx 0.0824$ .
In five throws, the number of throws of 60 m or more is binomial. A competitor qualifies with at least one.
Exactly one of them qualifies: Rachel only, or Sophia only.

P (Rachel qualifies) = 1 - (1 - 0.1216)^{5} \approx 0.4772 P (Sophia qualifies) = 1 - (1 - 0.0824)^{5} \approx 0.3495 0.4772 (1 - 0.3495) + 0.3495 (1 - 0.4772) \approx 0.3104 + 0.1827 \approx 0.493

Question 9

4 marks

Counting six-digit numbers with distinct and increasing digits

Statistics and probability

(a)

2 marks

The first digit has 9 choices (not 0). The remaining five places take distinct digits from the 9 left, in order.

9 \times^{9} P_{5} = 9 \times (9 \times 8 \times 7 \times 6 \times 5) = 136080

(b)

2 marks

In increasing order the first digit is the smallest, so 0 can't be used. Any choice of 6 different digits from 1–9 gives exactly one such number.

(6 9) = 84

Section B · Extended response

Questions 10–12 · 56 marks

Question 10

15 marks

Two growing plants: heights, crossing points and growth rates

Functions

(a)

3 marks

(i) At $t = 0$ for Plant B:

(ii) For Plant A, with the calculator in radians:

h_{B} (0) = 32 cm h_{A} (0) = sin 6 + 27 \approx 26.7 cm

(b)

3 marks

Solve $h_{A} (t) = h_{B} (t)$ on the calculator for $0 \leq t \leq 9$ .

t \approx 4.01, 4.70, 5.88 weeks

(c)

3 marks

Subtract the heights.
For $t > 6$ , $t - 5 > 1$ , and $sin (2 t + 6) \geq - 1$ . So the difference is greater than 0.
So for $t > 6$ Plant A was always taller than Plant B.

h_{A} (t) - h_{B} (t) = sin (2 t + 6) + t - 5 h_{A} (t) - h_{B} (t) > 1 - 1 = 0

(d)

6 marks

The rates are $h_{A}^{'} (t) = 2 cos (2 t + 6) + 9$ and $h_{B}^{'} (t) = 8$ . Plant B grows faster when $2 cos (2 t + 6) + 9 < 8$ , that is when $cos (2 t + 6) < - \frac{1}{2}$ .
Solve $h_{A}^{'} (t) = h_{B}^{'} (t)$ on the calculator: the crossings are at about 1.19, 2.24; 4.33, 5.38; and 7.47, 8.52. Between each pair, B is growing faster.

(2.236 - 1.189) + (5.378 - 4.330) + (8.519 - 7.472) \approx 3.14 weeks

Mentor tip

Exactly: each of the three intervals has length $\frac{π}{3}$ , so the total is $π$ weeks.

Question 11

20 marks

Two airplanes: bearing, speeds, angle between flight paths and closest approach

Geometry and trigonometry

(a)

2 marks

Airplane B moves 4 km east and 2 km north per minute, so its bearing is the angle from north, measured clockwise.

tan^{- 1} \frac{4}{2} \approx 63. 4^{\circ} \Rightarrow bearing 06 3^{\circ}

(b)

2 marks

The speed is the length of the direction vector.
Since $7.48 > 4.90$ , airplane A travels faster.

∣ b_{A} ∣ = (- 6)^{2} + 2^{2} + 4^{2} = 56 \approx 7.48 km min^{- 1}, ∣ b_{B} ∣ = 4^{2} + 2^{2} + (- 2)^{2} = 24 \approx 4.90 km min^{- 1}

(c)

4 marks

Use the scalar product of the direction vectors $- 6 24$ and $42 - 2$ .
The acute angle between the lines is the supplement.

cos θ = \frac{- 24 + 4 - 8}{56 24} = \frac{- 28}{1344} \approx - 0.764 \Rightarrow θ \approx 139. 8^{\circ} 18 0^{\circ} - 139. 8^{\circ} \approx 40. 2^{\circ}

(d)

7 marks

(i) Equate positions with separate times $t_{1}$ and $t_{2}$ .

Solve the second and third equations: $t_{1} = 2$ and $t_{2} = \frac{3}{2}$ . Check the first: $19 - 12 = 7 = 1 + 6$ . Then

(ii) Airplane A reaches P at $t = 2$ and B at $t = 1.5$ .

19 - 6 t_{1} = 1 + 4 t_{2}, - 1 + 2 t_{1} = 2 t_{2}, 1 + 4 t_{1} = 12 - 2 t_{2} P (7, 3, 9) 2 - 1.5 = 0.5 minutes (30 seconds)

(e)

5 marks

The distance between the planes at time $t$ is the length of $r_{B} - r_{A}$ . Find each component: east $= (1 + 4 t) - (19 - 6 t) = 10 t - 18$ , north $= 2 t - (- 1 + 2 t) = 1$ , height $= (12 - 2 t) - (1 + 4 t) = 11 - 6 t$ .
Minimise $D (t)$ on the calculator, or set $D^{'} (t) = 0$ .

r_{B} - r_{A} = 10 t - 18 1 11 - 6 t, D (t) = (10 t - 18)^{2} + 1^{2} + (11 - 6 t)^{2} t \approx 1.81, D_{m i n} \approx 1.01 km

Question 12

21 marks

A logistic population model: maximum growth rate and an exact solution

Calculus

(a)

1 mark

$\frac{d P}{d t}$ is the rate at which the population of marsupials is changing: its growth rate, in animals per year.

(b)

4 marks

Differentiate $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ with the product rule, using the chain rule for $\frac{d P}{d t}$ .
Replace $\frac{d P}{d t}$ by the original expression.

\frac{d ^{2} P}{d t ^{2}} = k \frac{d P}{d t} (1 - \frac{P}{N}) - \frac{k}{N} P \frac{d P}{d t} = k \frac{d P}{d t} (1 - \frac{2 P}{N}) \frac{d ^{2} P}{d t ^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})

(c)

5 marks

For $0 < P < N$ the factors $P$ and $1 - \frac{P}{N}$ are positive, so the sign of $\frac{d ^{2} P}{d t ^{2}}$ is the sign of $1 - \frac{2 P}{N}$ .
For $P < \frac{N}{2}$ that is positive (so the growth rate is increasing), for $P > \frac{N}{2}$ it is negative (so the growth rate is decreasing). The growth rate changes from increasing to decreasing at $P = \frac{N}{2}$ .
So the population grows at its maximum rate when $P = \frac{N}{2}$ .

(d)

2 marks

Substitute $P = \frac{N}{2}$ into $\frac{d P}{d t}$ .

\frac{d P}{d t}_{m a x} = k \cdot \frac{N}{2} (1 - \frac{1}{2}) = \frac{k N}{4}

(e)

7 marks

Separate the variables: $\frac{N}{P ( N - P )} d P = k d t$ . Use partial fractions.
At $t = 0$ , $P = P_{0}$ , so $C = ln \frac{P _{0}}{N - P _{0}}$ . Combine the logarithms.

\frac{N}{P ( N - P )} = \frac{1}{P} + \frac{1}{N - P} \Rightarrow \int (\frac{1}{P} + \frac{1}{N - P}) d P = \int k d t ln P - ln (N - P) = k t + C ln \frac{P}{N - P} - ln \frac{P _{0}}{N - P _{0}} = k t \Rightarrow k t = ln (\frac{P ( N - P _{0} )}{P _{0} ( N - P )})

(f)

2 marks

Put $N = 4 P_{0}$ , $P = 3 P_{0}$ and $t = 10$ into the formula.

10 k = ln \frac{3 P _{0} \cdot 3 P _{0}}{P _{0} \cdot P _{0}} = ln 9 \Rightarrow k = \frac{ln 9}{10} \approx 0.220

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus36 marks · 33%
Q5, Q7, Q12
Geometry and trigonometry26 marks · 24%
Q1, Q11
Number and algebra23 marks · 21%
Q2, Q3, Q6, Q8
Functions15 marks · 14%
Q10
Statistics and probability10 marks · 9%
Q4, Q9

Mentor insights

Traps to watch for

01
Check the end of the interval · Q5
The greatest speed on $0 \leq t \leq 3$ might be at an end point. Compare $∣ v ∣$ at the turning points of $v$ and at $t = 0$ and $t = 3$ .
02
Zero over zero first · Q7
A finite limit with denominator $x^{2} \to 0$ needs the numerator to tend to 0 too. That forces $k$ . Then l'Hôpital's rule applies, and may need to be used twice.
03
Rates are derivatives · Q10
In (d) compare the derivatives $h_{A}^{'} (t)$ and $h_{B}^{'} (t)$ . The condition is a cosine inequality, and each period contributes the same length of time.
04
Bearing is clockwise from north · Q11
Bearing uses north and east components only: $tan^{- 1} \frac{east}{north}$ . Speed uses the whole direction vector, including height.
05
Rate of growth maximum is the peak of $\frac{d P}{d t}$ · Q12
The maximum of $\frac{d P}{d t}$ is where its derivative, $\frac{d ^{2} P}{d t ^{2}}$ , is zero and changes sign from positive to negative.

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