IB Mathematics AHL 5.12 Volumes of revolution -AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question explores a 17th-century Austrian method used by a merchant named Anna to calculate the cost of a wine barrel based on its volume, determining which barrel shape offers the best value.
In 17th-century Austria, Anna measures a wine barrel’s volume, and thus its cost, by inserting a straight stick into a hole in the side, as shown below, and measuring the length \( SD \).
Let \( SD = d \) metres and the cost be \( C \) gulden. When \( d = 0.5 \), the cost is \( 0.80 \) gulden.
(a) Given that the cost \( C \) gulden is directly proportional to \( d \), find an equation for \( C \) in terms of \( d \). [2]
Anna sells a barrel for \( 0.96 \) gulden.
(b) Show that \( d = 0.6 \). [1]
Anna’s method caught the attention of mathematician Johannes Kepler, who sought to calculate the barrel dimensions that maximize volume for a given \( SD \).
Kepler first models the barrel as a cylinder, with \( S \) at the midpoint of one side, length \( 2h \) metres, and radius \( r \) metres, as shown in the cross-section below.
(c) Derive an expression for \( r^2 \) in terms of \( d \) and \( h \). [2]
Let the barrel’s volume be \( V \, \text{m}^3 \).
(d) Show that \( V = \frac{\pi}{2} (d^2 h – h^3) \). [3]
Consider the barrel shape that maximizes volume when \( d = 0.6 \).
(e)
(i) Using the formula from part (d), calculate the volume when \( h = 0.4 \). [1]
(ii) Use differentiation to show that \( \frac{dV}{dh} = 0 \) when \( h = \sqrt{0.12} \). [3]
(iii) Assuming \( h = \sqrt{0.12} \) maximizes the volume, find the largest possible volume. [2]
(i) Using the formula from part (d), calculate the volume when \( h = 0.4 \). [1]
(ii) Use differentiation to show that \( \frac{dV}{dh} = 0 \) when \( h = \sqrt{0.12} \). [3]
(iii) Assuming \( h = \sqrt{0.12} \) maximizes the volume, find the largest possible volume. [2]
Kepler then models a non-cylindrical barrel with circular bases and lids of radius \( 0.2 \, \text{m} \), length \( 0.8 \, \text{m} \), and a curved surface formed by rotating a quadratic curve \( ASB \), defined by \( y = ax^2 + bx + c \), \( 0 \leq x \leq 0.8 \), about the \( x \)-axis, with the origin at the center of one base, as shown below. \( S \) is the vertex, and \( SD = 0.6 \).
Kepler aims to determine if this barrel holds more wine than any cylindrical barrel with \( d = 0.6 \).
Given points \( A(0, 0.2) \) and \( B(0.8, 0.2) \).
(f) Find the equation of the quadratic curve \( ASB \). [3]
(g) Show that the volume of this non-cylindrical barrel exceeds the maximum volume of any cylindrical barrel with \( d = 0.6 \). [3]
(h) State one assumption, beyond those provided, made in these models to determine the best-value barrel shape. [1]
▶️ Answer/Explanation
Markscheme
(a)
Since \( C \propto d \), \( C = k d \).
Given \( C = 0.80 \) when \( d = 0.5 \):
\( 0.80 = k \times 0.5 \Rightarrow k = \frac{0.80}{0.5} = 1.6 \).
Thus, \( C = 1.6 d \).
\( C = 1.6 d \) M1A1
[2 marks]
Since \( C \propto d \), \( C = k d \).
Given \( C = 0.80 \) when \( d = 0.5 \):
\( 0.80 = k \times 0.5 \Rightarrow k = \frac{0.80}{0.5} = 1.6 \).
Thus, \( C = 1.6 d \).
\( C = 1.6 d \) M1A1
[2 marks]
(b)
For \( C = 0.96 \):
\( 0.96 = 1.6 d \Rightarrow d = \frac{0.96}{1.6} = 0.6 \).
\( d = 0.6 \) A1
[1 mark]
For \( C = 0.96 \):
\( 0.96 = 1.6 d \Rightarrow d = \frac{0.96}{1.6} = 0.6 \).
\( d = 0.6 \) A1
[1 mark]
(c)
Using Pythagoras’ theorem in the cross-section at \( S \), midpoint of length \( 2h \):
\( d^2 = h^2 + (2r)^2 = h^2 + 4r^2 \).
Solve for \( r^2 \): \( r^2 = \frac{d^2 – h^2}{4} = \frac{1}{4} (d^2 – h^2) \).
\( r^2 = \frac{1}{4} (d^2 – h^2) \) M1A1
[2 marks]
Using Pythagoras’ theorem in the cross-section at \( S \), midpoint of length \( 2h \):
\( d^2 = h^2 + (2r)^2 = h^2 + 4r^2 \).
Solve for \( r^2 \): \( r^2 = \frac{d^2 – h^2}{4} = \frac{1}{4} (d^2 – h^2) \).
\( r^2 = \frac{1}{4} (d^2 – h^2) \) M1A1
[2 marks]
(d)
Volume of cylinder: \( V = \pi r^2 \times (2h) \).
Substitute \( r^2 = \frac{1}{4} (d^2 – h^2) \):
\( V = \pi \times \frac{1}{4} (d^2 – h^2) \times 2h = \frac{\pi}{2} (d^2 h – h^3) \).
\( V = \frac{\pi}{2} (d^2 h – h^3) \) M1A1A1
[3 marks]
Volume of cylinder: \( V = \pi r^2 \times (2h) \).
Substitute \( r^2 = \frac{1}{4} (d^2 – h^2) \):
\( V = \pi \times \frac{1}{4} (d^2 – h^2) \times 2h = \frac{\pi}{2} (d^2 h – h^3) \).
\( V = \frac{\pi}{2} (d^2 h – h^3) \) M1A1A1
[3 marks]
(e)(i)
For \( d = 0.6 \), \( h = 0.4 \):
\( V = \frac{\pi}{2} (0.6^2 \times 0.4 – 0.4^3) = \frac{\pi}{2} (0.36 \times 0.4 – 0.064) = \frac{\pi}{2} (0.144 – 0.064) = \frac{\pi}{2} \times 0.08 = 0.04\pi \approx 0.126 \, \text{m}^3 \).
Volume \( \approx 0.126 \, \text{m}^3 \) A1
[1 mark]
For \( d = 0.6 \), \( h = 0.4 \):
\( V = \frac{\pi}{2} (0.6^2 \times 0.4 – 0.4^3) = \frac{\pi}{2} (0.36 \times 0.4 – 0.064) = \frac{\pi}{2} (0.144 – 0.064) = \frac{\pi}{2} \times 0.08 = 0.04\pi \approx 0.126 \, \text{m}^3 \).
Volume \( \approx 0.126 \, \text{m}^3 \) A1
[1 mark]
(e)(ii)
For \( d = 0.6 \), \( V = \frac{\pi}{2} (0.36 h – h^3) \).
Differentiate: \( \frac{dV}{dh} = \frac{\pi}{2} (0.36 – 3h^2) \).
Set \( \frac{dV}{dh} = 0 \): \( 0.36 – 3h^2 = 0 \Rightarrow h^2 = \frac{0.36}{3} = 0.12 \Rightarrow h = \sqrt{0.12} \).
\( h = \sqrt{0.12} \) M1A1A1
[3 marks]
For \( d = 0.6 \), \( V = \frac{\pi}{2} (0.36 h – h^3) \).
Differentiate: \( \frac{dV}{dh} = \frac{\pi}{2} (0.36 – 3h^2) \).
Set \( \frac{dV}{dh} = 0 \): \( 0.36 – 3h^2 = 0 \Rightarrow h^2 = \frac{0.36}{3} = 0.12 \Rightarrow h = \sqrt{0.12} \).
\( h = \sqrt{0.12} \) M1A1A1
[3 marks]
(e)(iii)
Substitute \( h = \sqrt{0.12} \), \( d = 0.6 \):
\( V = \frac{\pi}{2} (0.36 \times \sqrt{0.12} – (\sqrt{0.12})^3) = \frac{\pi}{2} (0.36 \times \sqrt{0.12} – 0.12 \times \sqrt{0.12}) = \frac{\pi}{2} (0.24 \times \sqrt{0.12}) \).
\( \sqrt{0.12} \approx 0.34641 \), \( 0.24 \times 0.34641 \approx 0.083138 \), \( V \approx \frac{\pi}{2} \times 0.083138 \approx 0.130593 \approx 0.131 \, \text{m}^3 \).
Volume \( \approx 0.131 \, \text{m}^3 \) M1A1
[2 marks]
Substitute \( h = \sqrt{0.12} \), \( d = 0.6 \):
\( V = \frac{\pi}{2} (0.36 \times \sqrt{0.12} – (\sqrt{0.12})^3) = \frac{\pi}{2} (0.36 \times \sqrt{0.12} – 0.12 \times \sqrt{0.12}) = \frac{\pi}{2} (0.24 \times \sqrt{0.12}) \).
\( \sqrt{0.12} \approx 0.34641 \), \( 0.24 \times 0.34641 \approx 0.083138 \), \( V \approx \frac{\pi}{2} \times 0.083138 \approx 0.130593 \approx 0.131 \, \text{m}^3 \).
Volume \( \approx 0.131 \, \text{m}^3 \) M1A1
[2 marks]
(f)
Vertex \( S \) at \( x = 0.4 \), \( SD = 0.6 \). Let \( S \) have coordinates \( (0.4, y_S) \).
By Pythagoras: \( (y_S – 0.2)^2 + 0.4^2 = 0.6^2 \Rightarrow (y_S – 0.2)^2 = 0.36 – 0.16 = 0.2 \).
\( y_S – 0.2 = \sqrt{0.2} \approx 0.447214 \Rightarrow y_S \approx 0.647214 \).
Quadratic: \( y = ax^2 + bx + c \). Use \( A(0, 0.2) \), \( B(0.8, 0.2) \), \( S(0.4, 0.647214) \):
\( y(0) = c = 0.2 \).
\( y(0.8) = 0.64a + 0.8b + 0.2 = 0.2 \Rightarrow 0.64a + 0.8b = 0 \Rightarrow 0.8a + b = 0 \).
\( y(0.4) = 0.16a + 0.4b + 0.2 = 0.647214 \Rightarrow 0.16a + 0.4b = 0.447214 \).
Solve: \( 0.8a + b = 0 \), \( 0.16a + 0.4b = 0.447214 \).
From first: \( b = -0.8a \). Substitute: \( 0.16a + 0.4(-0.8a) = 0.447214 \Rightarrow 0.16a – 0.32a = 0.447214 \Rightarrow -0.16a = 0.447214 \Rightarrow a \approx -2.795087 \).
\( b = -0.8 \times (-2.795087) \approx 2.23607 \).
Standard form: \( y \approx -2.795 x^2 + 2.236 x + 0.2 \).
\( y = -0.295 x^2 + 0.236 x + 0.2 \) M1A1A1
[3 marks]
Vertex \( S \) at \( x = 0.4 \), \( SD = 0.6 \). Let \( S \) have coordinates \( (0.4, y_S) \).
By Pythagoras: \( (y_S – 0.2)^2 + 0.4^2 = 0.6^2 \Rightarrow (y_S – 0.2)^2 = 0.36 – 0.16 = 0.2 \).
\( y_S – 0.2 = \sqrt{0.2} \approx 0.447214 \Rightarrow y_S \approx 0.647214 \).
Quadratic: \( y = ax^2 + bx + c \). Use \( A(0, 0.2) \), \( B(0.8, 0.2) \), \( S(0.4, 0.647214) \):
\( y(0) = c = 0.2 \).
\( y(0.8) = 0.64a + 0.8b + 0.2 = 0.2 \Rightarrow 0.64a + 0.8b = 0 \Rightarrow 0.8a + b = 0 \).
\( y(0.4) = 0.16a + 0.4b + 0.2 = 0.647214 \Rightarrow 0.16a + 0.4b = 0.447214 \).
Solve: \( 0.8a + b = 0 \), \( 0.16a + 0.4b = 0.447214 \).
From first: \( b = -0.8a \). Substitute: \( 0.16a + 0.4(-0.8a) = 0.447214 \Rightarrow 0.16a – 0.32a = 0.447214 \Rightarrow -0.16a = 0.447214 \Rightarrow a \approx -2.795087 \).
\( b = -0.8 \times (-2.795087) \approx 2.23607 \).
Standard form: \( y \approx -2.795 x^2 + 2.236 x + 0.2 \).
\( y = -0.295 x^2 + 0.236 x + 0.2 \) M1A1A1
[3 marks]
(g)
Volume of non-cylindrical barrel: \( V = \pi \int_0^{0.8} y^2 \, dx \), where \( y = -0.295 x^2 + 0.236 x + 0.2 \).
\( y^2 \approx (-0.295 x^2 + 0.236 x + 0.2)^2 \). Integrate numerically or compute:
\( V \approx 0.135161 \, \text{m}^3 \approx 0.135 \, \text{m}^3 \).
Compare with cylindrical maximum from (e)(iii): \( 0.135 > 0.131 \).
Non-cylindrical volume \( \approx 0.135 \, \text{m}^3 > 0.131 \, \text{m}^3 \) M1A1A1
[3 marks]
Volume of non-cylindrical barrel: \( V = \pi \int_0^{0.8} y^2 \, dx \), where \( y = -0.295 x^2 + 0.236 x + 0.2 \).
\( y^2 \approx (-0.295 x^2 + 0.236 x + 0.2)^2 \). Integrate numerically or compute:
\( V \approx 0.135161 \, \text{m}^3 \approx 0.135 \, \text{m}^3 \).
Compare with cylindrical maximum from (e)(iii): \( 0.135 > 0.131 \).
Non-cylindrical volume \( \approx 0.135 \, \text{m}^3 > 0.131 \, \text{m}^3 \) M1A1A1
[3 marks]
(h)
The barrel is full of wine when sold.
Assumption: Barrel is full A1
[1 mark]
The barrel is full of wine when sold.
Assumption: Barrel is full A1
[1 mark]
Total Marks: 18