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Date	May 2017	Marks available	3	Reference code	17M.2.AHL.TZ1.H_8
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	H_8	Adapted from	N/A

Question

A water trough which is 10 metres long has a uniform cross-section in the shape of a semicircle with radius 0.5 metres. It is partly filled with water as shown in the following diagram of the cross-section. The centre of the circle is O and the angle KOL is $\theta$ radians.

M17/5/MATHL/HP2/ENG/TZ1/08

The volume of water is increasing at a constant rate of $0.0008{\text{ }}{{\text{m}}^3}{{\text{s}}^{ - 1}}$ .

Find an expression for the volume of water $V{\text{ }}({{\text{m}}^3})$ in the trough in terms of $\theta$ .

[3]

Calculate $\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ when $\theta = \frac{\pi }{3}$ .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

area of segment $= \frac{1}{2} \times {0.5^2} \times (\theta - \sin \theta )$ M1A1

$V = {\text{area of segment}} \times 10$

$V = \frac{5}{4}(\theta - \sin \theta )$ A1

[3 marks]

METHOD 1

$\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{5}{4}(1 - \cos \theta )\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ M1A1

$0.0008 = \frac{5}{4}\left( {1 - \cos \frac{\pi }{3}} \right)\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ (M1)

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128{\text{ }}({\text{rad}}\,{s^{ - 1}})$ A1

METHOD 2

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}V}} \times \frac{{{\text{d}}V}}{{{\text{d}}t}}$ (M1)

$\frac{{{\text{d}}V}}{{{\text{d}}\theta }} = \frac{5}{4}(1 - \cos \theta )$ A1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{4 \times 0.0008}}{{5\left( {1 - \cos \frac{\pi }{3}} \right)}}$ (M1)

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128\left( {\frac{4}{{3125}}} \right)({\text{rad }}{s^{ - 1}})$ A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.3—Differentiating polynomials, n E Z

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19M.2.SL.TZ2.T_5b:
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SPM.1.SL.TZ0.8b:
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18M.2.SL.TZ1.S_1c:
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16N.1.SL.TZ0.S_10b:
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18N.1.SL.TZ0.S_10c:
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17M.2.AHL.TZ1.H_2a:
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22M.1.SL.TZ2.7d:
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SPM.1.SL.TZ0.8c.ii:
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19M.1.AHL.TZ1.H_7:
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20N.1.SL.TZ0.T_13b:
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17N.1.SL.TZ0.S_8d:
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17M.2.AHL.TZ1.H_8b:
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19M.1.SL.TZ2.S_10d:
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17M.2.AHL.TZ1.H_12d:
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17M.2.AHL.TZ1.H_12g.i:
Hence, show that there are no solutions to $g'(x) = 0$ ;
17N.1.AHL.TZ0.H_7:
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Find the inverse function ${g^{ - 1}}$ and state its domain.
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SPM.1.SL.TZ0.8e:
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Find the value of $p$ and the value of $q$ .
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Find $f' (p)$ in terms of $k$ and $p$ .
19M.1.SL.TZ1.T_15a:
Write down an equation for the area, $A$ , of the curved surface in terms of $r$ .
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Find the value of $P$ if no vases are sold.
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Write down the $x$ -coordinates of these two points;
17N.2.SL.TZ0.T_5b.i:
Expand the expression for $f(x)$ .
17M.1.SL.TZ1.S_9c:
The line $y = kx - 5$ is a tangent to the curve of $f$ . Find the values of $k$ .
16N.2.SL.TZ0.T_6b:
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Write down, in terms of $r$ and $h$ , an equation for the volume of this water container.
16N.2.SL.TZ0.T_6e:
Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .
16N.2.SL.TZ0.T_6f:
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18N.1.SL.TZ0.S_10b.ii:
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18M.1.SL.TZ2.S_9c:
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Give your answers in standard form.
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17M.2.AHL.TZ1.H_2b:
Determine the equation of the tangent to $C$ at the point $\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$
19M.1.SL.TZ2.S_10b:
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18M.1.AHL.TZ1.H_7b:
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17M.2.AHL.TZ1.H_12g.ii:
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16N.1.SL.TZ0.T_14a:
Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .
18M.1.AHL.TZ2.H_6b:
Hence, or otherwise, find $\int\limits_0^\pi {{{\text{e}}^{ - x}}\,{\text{sin}}\,x\,{\text{d}}x}$ .
18N.2.SL.TZ0.T_6g:
Use your answer to part (f) to show that the width of Nanako’s bag is 12 cm.
19M.1.SL.TZ1.T_15c:
Find the value of $r$ when the area of the curved surface is maximized.
18N.2.SL.TZ0.T_6d:
Write down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.
19M.1.SL.TZ2.T_15b:
Differentiate $P\left( x \right)$ .
20N.1.SL.TZ0.T_13c:
Find the value of $k$ .
17M.2.SL.TZ2.T_6a:
Write down the $y$ -intercept of the graph.
17M.2.SL.TZ2.T_6c.i:
Show that $a = 8$ .
17M.2.SL.TZ2.T_6b:
Find $f'(x)$ .
20N.1.SL.TZ0.T_13a:
Write down $f' (x)$ .
19M.2.SL.TZ2.T_5h:
Determine the range of $f\left( x \right)$ for $p$ ≤ $x$ ≤ $q$ .
17N.2.SL.TZ0.T_5e:
Write down the coordinates of the point of intersection.
17M.2.SL.TZ2.T_6g:
The equation $f(x) = m$ , where $m \in \mathbb{R}$ , has four solutions. Find the possible values of $m$ .
16N.2.SL.TZ0.T_6d:
Show that $A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ .
16N.2.SL.TZ0.T_6g:
Find the value of this minimum area.
19M.1.SL.TZ1.T_15b:
Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .
18N.2.SL.TZ0.T_6e:
Use your answers to parts (c) and (d) to show that

$A = 3{x^2} + \frac{{10368}}{x}$ .
16N.2.SL.TZ0.T_6a:
Write down a formula for $A$ , the surface area to be coated.
16N.2.SL.TZ0.T_6h:
Find the least number of cans of water-resistant material that will coat the area in part (g).
17N.2.SL.TZ0.T_5d:
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17M.2.SL.TZ2.T_6c.ii:
Find $f(2)$ .
17M.2.SL.TZ2.T_6d.ii:
Write down the intervals where the gradient of the graph of $y = f(x)$ is positive.
17M.2.SL.TZ2.T_6e:
Write down the range of $f(x)$ .
18N.2.SL.TZ0.T_6b:
Calculate the volume, in cm³, of the bag.
18N.2.SL.TZ0.T_6c:
Use this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.
18N.2.SL.TZ0.T_6f:
Find $\frac{{{\text{d}}A}}{{{\text{d}}x}}$ .

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