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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

26 marks

Kepler's wine barrel: the cost, a cylinder and a barrel with a curved side

Number and algebra
Geometry and trigonometry
Calculus

(a)

3 marks

Direct proportion means $C = k d$ . Use $C = 0.80$ when $d = 0.5$ .

0.80 = 0.5 k \Rightarrow k = 1.6 \Rightarrow C = 1.6 d

(b)

1 mark

Put $C = 0.96$ .

0.96 = 1.6 d \Rightarrow d = 0.6

(c)

3 marks

The stick runs from the midpoint S of the top side to the far corner D at the bottom. Its horizontal part is $h$ and its vertical part is $2 r$ .

d^{2} = h^{2} + (2 r)^{2} \Rightarrow r^{2} = \frac{d ^{2} - h ^{2}}{4}

(d)

2 marks

The volume of a cylinder of radius $r$ and length $2 h$ is $π r^{2} (2 h)$ .

V = π \cdot \frac{d ^{2} - h ^{2}}{4} \cdot 2 h = \frac{π ( d ^{2} h - h ^{3} )}{2}

(e)(i)

2 marks

Put $d = 0.6$ and $h = 0.4$ .

V = \frac{π ( 0.36 ( 0.4 ) - 0. 4 ^{3} )}{2} = \frac{π ( 0.144 - 0.064 )}{2} \approx 0.126 m^{3}

(e)(ii)

3 marks

Differentiate with respect to $h$ .
Set it to zero with $d = 0.6$ .

\frac{d V}{d h} = \frac{π}{2} (d^{2} - 3 h^{2}) 0.36 - 3 h^{2} = 0 \Rightarrow h^{2} = 0.12 \Rightarrow h = 0.12 = 0.346

(e)(iii)

2 marks

Substitute $h = 0.12$ .

V = \frac{π ( 0.36 0.12 - 0.12 0.12 )}{2} = \frac{π ( 0.24 ) 0.12}{2} \approx 0.131 m^{3}

(f)

6 marks

The curve $y = a x^{2} + b x + c$ passes through $A (0, 0.2)$ , so $c = 0.2$ , and it is symmetric about $x = 0.4$ since $B$ is $(0.8, 0.2)$ . The vertex is $S = (0.4, y_{S})$ .
The stick goes from S to $D (0, - 0.2)$ and has length 0.6.
Write the curve in vertex form and use $A$ .

0. 4^{2} + (y_{S} + 0.2)^{2} = 0. 6^{2} \Rightarrow (y_{S} + 0.2)^{2} = 0.2 \Rightarrow y_{S} = 0.2 - 0.2 = 0.2472 y = a (x - 0.4)^{2} + 0.2472, 0.2 = 0.16 a + 0.2472 \Rightarrow a = - 0.2951 y = - 0.295 x^{2} + 0.236 x + 0.2

(g)

3 marks

The volume of the barrel is the volume of revolution of the curve about the $x$ -axis.
The largest cylinder with $d = 0.6$ holds about $0.1306 m^{3}$ , and $0.1352 > 0.1306$ , so the curved barrel holds more.

V = π \int_{0}^{0.8} (- 0.295 x^{2} + 0.236 x + 0.2)^{2} d x \approx 0.1352 m^{3}

(h)

1 mark

For example: the wood of the barrel is assumed to have negligible thickness, so the volume inside is the volume calculated, or the barrel is a perfect solid of revolution about its axis.

Question 2

29 marks

Air traffic control: vector methods for aircraft paths

Geometry and trigonometry

(a)

2 marks

Subtract the position vectors.

P Q = (215 - 100)^{2} + (- 197 + 82)^{2} + 0^{2} = 11 5^{2} + 11 5^{2} \approx 163 km

(b)

2 marks

The speed is the magnitude of the velocity vector.

speed = (- 640)^{2} + 64 0^{2} + 0^{2} = 6402 \approx 905 km h^{- 1}

(c)

3 marks

Time = distance divided by speed.
This is more than 10 minutes, so the aircraft are not in conflict.

\frac{162.6}{905.1} = 0.1797 h = 10.8 minutes

(d)

1 mark

A starts at P with the given velocity.

r_{A} = 100 - 82 10.7 + t - 640 6400

(e)(i)

5 marks

The vector from A to C is $r_{C} - r_{A}$ .
Set its length equal to 10.
Solve.

r_{C} - r_{A} = - 500 + 500 t 41 - 36 t - 1.6 + 2 t (- 500 + 500 t)^{2} + (41 - 36 t)^{2} + (- 1.6 + 2 t)^{2} = 100 t = 0.983 and t = 1.018

(e)(ii)

5 marks

The aircraft are less than 10 km apart only for $0.983 < t < 1.018$ . The vertical separation is $∣ - 1.6 + 2 t ∣$ km.
At $t = 0.983$ it is $0.367$ km and at $t = 1.018$ it is $0.436$ km, and it increases between them, so it is always more than 0.3 km (300 m) while they are less than 10 km apart.
The regulations only require 10 km separation when the vertical distance is less than 300 m. This never happens at the same time, so the aircraft do not break the regulations.

(f)

3 marks

The position of D relative to R stays at a distance $6. 4^{2} cos^{2} (38 t) + 6. 4^{2} sin^{2} (38 t) = 6.4$ km.
D flies in a horizontal circle of radius 6.4 km, centred on R (2 km above the tower), turning anticlockwise at 38 radians per hour. One lap takes $\frac{2 π}{38}$ hours, about 10 minutes.

(g)(i)

2 marks

Take the scalar product of $R E = (20 - λ 10 + λ)$ with $b = (- 1 1)$ .

R E \cdot b = - (20 - λ) + (10 + λ) = 2 λ - 10

(g)(ii)

2 marks

The distance from R to the line is smallest when $R E$ is perpendicular to the direction $b$ .

2 λ - 10 = 0 \Rightarrow λ = 5

(g)(iii)

2 marks

With $λ = 5$ , $R E = (15, 15)$ .

minimum distance = 1 5^{2} + 1 5^{2} = 152 \approx 21.2 km

(g)(iv)

2 marks

D and E are at the same height. D is always within 6.4 km of R and E is always at least 21.2 km from R.
So D and E are always more than 10 km apart and never break the regulations.

distance D E \geq 21.2 - 6.4 = 14.8 km > 10 km

Our analysis

Where the marks are

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

Geometry and trigonometry36 marks · 65%
Q1, Q2
Calculus15 marks · 27%
Q1
Number and algebra4 marks · 7%
Q1

Mentor insights

Traps to watch for

01
Pythagoras in the cross-section · Q1
The stick SD is the diagonal of the rectangle formed by half the length $h$ and the full diameter $2 r$ .
02
Maximise with calculus · Q1
Differentiate $V (h)$ with $d$ fixed, set it to zero and take the positive root.
03
Same flight path and time gap · Q2
Divide the distance between the aircraft by their speed to find the time gap.
04
Two regulations · Q2
Aircraft not on the same path need either at least 300 m vertical separation, or more than 10 km distance.
05
Distance from a point to a line · Q2
The foot of the perpendicular has $R E \cdot b = 0$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

26 marks

Kepler's wine barrel: the cost, a cylinder and a barrel with a curved side

Number and algebra
Geometry and trigonometry
Calculus

(a)

3 marks

Direct proportion means $C = k d$ . Use $C = 0.80$ when $d = 0.5$ .

0.80 = 0.5 k \Rightarrow k = 1.6 \Rightarrow C = 1.6 d

(b)

1 mark

Put $C = 0.96$ .

0.96 = 1.6 d \Rightarrow d = 0.6

(c)

3 marks

The stick runs from the midpoint S of the top side to the far corner D at the bottom. Its horizontal part is $h$ and its vertical part is $2 r$ .

d^{2} = h^{2} + (2 r)^{2} \Rightarrow r^{2} = \frac{d ^{2} - h ^{2}}{4}

(d)

2 marks

The volume of a cylinder of radius $r$ and length $2 h$ is $π r^{2} (2 h)$ .

V = π \cdot \frac{d ^{2} - h ^{2}}{4} \cdot 2 h = \frac{π ( d ^{2} h - h ^{3} )}{2}

(e)(i)

2 marks

Put $d = 0.6$ and $h = 0.4$ .

V = \frac{π ( 0.36 ( 0.4 ) - 0. 4 ^{3} )}{2} = \frac{π ( 0.144 - 0.064 )}{2} \approx 0.126 m^{3}

(e)(ii)

3 marks

Differentiate with respect to $h$ .
Set it to zero with $d = 0.6$ .

\frac{d V}{d h} = \frac{π}{2} (d^{2} - 3 h^{2}) 0.36 - 3 h^{2} = 0 \Rightarrow h^{2} = 0.12 \Rightarrow h = 0.12 = 0.346

(e)(iii)

2 marks

Substitute $h = 0.12$ .

V = \frac{π ( 0.36 0.12 - 0.12 0.12 )}{2} = \frac{π ( 0.24 ) 0.12}{2} \approx 0.131 m^{3}

(f)

6 marks

The curve $y = a x^{2} + b x + c$ passes through $A (0, 0.2)$ , so $c = 0.2$ , and it is symmetric about $x = 0.4$ since $B$ is $(0.8, 0.2)$ . The vertex is $S = (0.4, y_{S})$ .
The stick goes from S to $D (0, - 0.2)$ and has length 0.6.
Write the curve in vertex form and use $A$ .

0. 4^{2} + (y_{S} + 0.2)^{2} = 0. 6^{2} \Rightarrow (y_{S} + 0.2)^{2} = 0.2 \Rightarrow y_{S} = 0.2 - 0.2 = 0.2472 y = a (x - 0.4)^{2} + 0.2472, 0.2 = 0.16 a + 0.2472 \Rightarrow a = - 0.2951 y = - 0.295 x^{2} + 0.236 x + 0.2

(g)

3 marks

The volume of the barrel is the volume of revolution of the curve about the $x$ -axis.
The largest cylinder with $d = 0.6$ holds about $0.1306 m^{3}$ , and $0.1352 > 0.1306$ , so the curved barrel holds more.

V = π \int_{0}^{0.8} (- 0.295 x^{2} + 0.236 x + 0.2)^{2} d x \approx 0.1352 m^{3}

(h)

1 mark

For example: the wood of the barrel is assumed to have negligible thickness, so the volume inside is the volume calculated, or the barrel is a perfect solid of revolution about its axis.

Question 2

29 marks

Air traffic control: vector methods for aircraft paths

Geometry and trigonometry

(a)

2 marks

Subtract the position vectors.

P Q = (215 - 100)^{2} + (- 197 + 82)^{2} + 0^{2} = 11 5^{2} + 11 5^{2} \approx 163 km

(b)

2 marks

The speed is the magnitude of the velocity vector.

speed = (- 640)^{2} + 64 0^{2} + 0^{2} = 6402 \approx 905 km h^{- 1}

(c)

3 marks

Time = distance divided by speed.
This is more than 10 minutes, so the aircraft are not in conflict.

\frac{162.6}{905.1} = 0.1797 h = 10.8 minutes

(d)

1 mark

A starts at P with the given velocity.

r_{A} = 100 - 82 10.7 + t - 640 6400

(e)(i)

5 marks

The vector from A to C is $r_{C} - r_{A}$ .
Set its length equal to 10.
Solve.

r_{C} - r_{A} = - 500 + 500 t 41 - 36 t - 1.6 + 2 t (- 500 + 500 t)^{2} + (41 - 36 t)^{2} + (- 1.6 + 2 t)^{2} = 100 t = 0.983 and t = 1.018

(e)(ii)

5 marks

The aircraft are less than 10 km apart only for $0.983 < t < 1.018$ . The vertical separation is $∣ - 1.6 + 2 t ∣$ km.
At $t = 0.983$ it is $0.367$ km and at $t = 1.018$ it is $0.436$ km, and it increases between them, so it is always more than 0.3 km (300 m) while they are less than 10 km apart.
The regulations only require 10 km separation when the vertical distance is less than 300 m. This never happens at the same time, so the aircraft do not break the regulations.

(f)

3 marks

The position of D relative to R stays at a distance $6. 4^{2} cos^{2} (38 t) + 6. 4^{2} sin^{2} (38 t) = 6.4$ km.
D flies in a horizontal circle of radius 6.4 km, centred on R (2 km above the tower), turning anticlockwise at 38 radians per hour. One lap takes $\frac{2 π}{38}$ hours, about 10 minutes.

(g)(i)

2 marks

Take the scalar product of $R E = (20 - λ 10 + λ)$ with $b = (- 1 1)$ .

R E \cdot b = - (20 - λ) + (10 + λ) = 2 λ - 10

(g)(ii)

2 marks

The distance from R to the line is smallest when $R E$ is perpendicular to the direction $b$ .

2 λ - 10 = 0 \Rightarrow λ = 5

(g)(iii)

2 marks

With $λ = 5$ , $R E = (15, 15)$ .

minimum distance = 1 5^{2} + 1 5^{2} = 152 \approx 21.2 km

(g)(iv)

2 marks

D and E are at the same height. D is always within 6.4 km of R and E is always at least 21.2 km from R.
So D and E are always more than 10 km apart and never break the regulations.

distance D E \geq 21.2 - 6.4 = 14.8 km > 10 km

Our analysis

Where the marks are

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

Geometry and trigonometry36 marks · 65%
Q1, Q2
Calculus15 marks · 27%
Q1
Number and algebra4 marks · 7%
Q1

Mentor insights

Traps to watch for

01
Pythagoras in the cross-section · Q1
The stick SD is the diagonal of the rectangle formed by half the length $h$ and the full diameter $2 r$ .
02
Maximise with calculus · Q1
Differentiate $V (h)$ with $d$ fixed, set it to zero and take the positive root.
03
Same flight path and time gap · Q2
Divide the distance between the aircraft by their speed to find the time gap.
04
Two regulations · Q2
Aircraft not on the same path need either at least 300 m vertical separation, or more than 10 km distance.
05
Distance from a point to a line · Q2
The foot of the perpendicular has $R E \cdot b = 0$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers