IB Mathematics AHL 5.10 The second derivative-AI HL Paper 2- Exam Style Questions- New Syllabus
A skip is a container designed to transport waste from a construction site. For safety, the waste must not exceed the skip’s top edge, so the maximum waste volume equals the skip’s volume.
A specific skip design is modeled as a prism with a trapezoidal cross-section. To ensure transportability, it has a rectangular base measuring \(10 \, \text{m}\) in length and \(3 \, \text{m}\) in width. The sloping edge is fixed at \(4 \, \text{m}\) and forms an angle \(\theta\) with the horizontal.
The diagram below illustrates this skip (not to scale).
[Trapezoidal skip: base = 10 m, width = 3 m, sloping edge = 4 m, angle = θ]
(a) Calculate the volume of the skip,
(i) when the top edge length is \(11 \, \text{m}\).
(ii) when the skip’s height is \(3.2 \, \text{m}\).
(iii) when \(\theta = \frac{\pi}{3}\).
[9 marks]
(b) Prove that the skip’s volume, \(V \, \text{m}^3\), is expressed as
\( V = 24 \sin(\theta)\big(5 + \cos(\theta)\big) \)
[2 marks]
(c) Explain, in the context of the skip, why \(\theta \neq 0\).
[1 mark]
(d) (i) Sketch the graph of
\( V = 24 \sin(\theta)\big(5 + \cos(\theta)\big), \quad 0 < \theta < \frac{\pi}{2}. \)
(ii) Determine the maximum volume of the skip and the corresponding value of \(\theta\).
[4 marks]
(e) Demonstrate, using differentiation, that the maximum volume occurs at a \(\theta\) satisfying
\( 2 \cos^2(\theta) + 5 \cos(\theta) – 1 = 0 \).
▶️ Answer/Explanation
Base length \(=10\text{ m}\), width \(=3\text{ m}\). Sloping edge fixed \(=4\text{ m}\) at angle \(\theta\) to the horizontal. Cross-section area (trapezium): \(A=\frac{1}{2} (b_1+b_2)h\). Volume \(V = 3 \times A\) (width \(=3\)).
(a)(i) Top edge \(=11\text{ m}\). Find cross-section height using the 4-m sloping edge and 1-m horizontal offset:
\[ h=\sqrt{4^{2}-1^{2}}=\sqrt{15}\,(=3.873\ldots). \]
\[ A=\frac{1}{2}(10+11)h=\frac{1}{2} \times 21 \times \sqrt{15}. \]
\[ V=3A=3\left[\frac{1}{2}(10+11)\sqrt{4^{2}-1^{2}}\right]=\boxed{122\text{ m}^3}\ (121.998\ldots). \]
(a)(ii) Height of skip \(=3.2\text{ m}\). Horizontal offset along the base from the 4-m side:
\[ \text{offset}=\sqrt{4^{2}-3.2^{2}}=\sqrt{16-10.24}=2.4. \]
\[ \text{top edge}=10+\text{offset}=10+2.4=12.4. \]
\[ A=\frac{1}{2}(10+12.4)(3.2),\quad V=3A=3\left[\frac{1}{2}(10+10+\sqrt{4^{2}-3.2^{2}})(3.2)\right]. \]
\[ V=\boxed{108\text{ m}^3}\ (107.52\ldots). \]
(a)(iii) \(\theta=\dfrac{\pi}{3}\). Use \(h=4\sin\theta\), horizontal offset \(=4\cos\theta\), top edge \(=10+4\cos\theta\):
\[ V=3\left[\frac{1}{2}(10+10+4\cos\frac{\pi}{3})(4\sin\frac{\pi}{3})\right] =3\left[\frac{1}{2}(20+2) \times 4 \times \frac{\sqrt{3}}{2}\right] =\boxed{114\text{ m}^3}\ (114.315\ldots). \]
(b) Show \(V=24\sin\theta(5+\cos\theta)\).
\[ V=3\left[\frac{1}{2}(10+10+4\cos\theta)(4\sin\theta)\right] =3 \times \frac{1}{2} \times (20+4\cos\theta) \times 4\sin\theta \]
\[ =6\sin\theta(20+4\cos\theta)=\boxed{24\sin\theta(5+\cos\theta)}. \]
The AG (answer given) line must be seen for the final accuracy mark.
(c) \(\theta\neq 0\) because if \(\theta=0\) the sloping side is horizontal, giving height \(h=4\sin\theta=0\); the cross-section (and hence the volume) would be zero and the contents would fall out.
(d)(i) Sketch of \(V=24\sin\theta(5+\cos\theta)\) on \(0<\theta<\frac{\pi}{2}\): starts at \(0\), rises concave down, then gently falls after a maximum before \(\frac{\pi}{2}\). Axes labelled \(\theta\) (horizontal), \(V\) (vertical).
(d)(ii) Maximum (from calculus/graph):
\[ \theta_{\max}\approx \boxed{1.38\text{ rad}}\ (1.38356\ldots)\ \;=\; \boxed{79.3^\circ}\ (79.2723\ldots). \]
\[ V_{\max}\approx \boxed{122\text{ m}^3}\ (122.292\ldots). \]
If values are reversed, award A0A1; do not award a coordinate pair here.
(e) Differentiate and show the condition for the maximum.
\[ V(\theta)=24\sin\theta(5+\cos\theta),\quad \frac{dV}{d\theta}=0 \text{ at a turning point.} \]
Product rule:
\[ \frac{dV}{d\theta}=24\left[\cos\theta(5+\cos\theta)+\sin\theta(-\sin\theta)\right] =24\left(5\cos\theta+\cos^2\theta-\sin^2\theta\right). \]
Use \(\sin^2\theta=1-\cos^2\theta\):
\[ \frac{dV}{d\theta}=24\left(5\cos\theta+\cos^2\theta-(1-\cos^2\theta)\right) =24\left(5\cos\theta+2\cos^2\theta-1\right). \]
Set to zero:
\[ 2\cos^2\theta+5\cos\theta-1=0. \]
From the graph the turning point is the global maximum on \(0<\theta<\frac{\pi}{2}\).