June 2023 Paper 3 Q9

AQACurrent spec12 marksModelling Parametric Equations

9 A water slide is the shape of a curve \(PQ\) as shown in Figure 1 below.

Side view of a water slide: a platform at P directly above O on the ground, with the slide curving down from P to a lowest point and then rising slightly to Q at the edge of a pool — Figure 1

The curve can be modelled by the parametric equations

\[x = t - \frac{1}{t} + 4.8\]\[y = t + \frac{2}{t}\]

where \(0.2 \leqslant t \leqslant 3\)

The horizontal distance from \(O\) is \(x\) metres.

The vertical distance above the point \(O\) at ground level is \(y\) metres.

\(P\) is the point where \(t = 0.2\) and \(Q\) is the point where \(t = 3\)

(a) To make sure speeds are safe at \(Q\), the difference in height between \(P\) and \(Q\) must be less than 7 metres.

Show that the slide meets this safety requirement. [3 marks]

(b)

(i) Find an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) [3 marks]

(ii) A vertical support, \(RS\), is to be added between the ground and the lowest point on the slide as shown in Figure 2 below.

The water slide from Figure 1 with a vertical support RS from the lowest point R on the slide down to S on the ground, between O and Q — Figure 2

Find the length of \(RS\) [4 marks]

(iii) Find the acute angle the slide makes with the horizontal at \(Q\)

Give your answer to the nearest degree. [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains \(y = 10.2\) or \(y = \dfrac{11}{3}\) OE AWFW [3.6, 3.7] for \(\dfrac{11}{3}\)	B1	3.4
Finds the difference between their two values of \(y\)	M1	1.1b
Makes a comparison between 6.53 and 7 and states that the safety requirement is met. For 6.53, accept AWFW [6.5, \(6.5\dot{3}\)] OE	R1	3.2a
	(3)

Typical solution

When \(t = 0.2\), \(y = 10.2\)

\(t = 3\), \(y = \dfrac{11}{3}\)

\[10.2 - \frac{11}{3} = 6.53 \lt 7\]

The slide is safe.

Mark scheme (b)

Scheme	Marks	AO
(i) Obtains \(1 + t^{-2}\) or \(1 - 2t^{-2}\) OE Ignore labels	B1	1.1b
Uses chain rule to obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) using their \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) Condone missing brackets	M1	1.1a
Obtains a correct expression ISW	A1	1.1b
	(3)
(ii) Forms equation for appropriate derivative equal to zero. Their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) or their \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0\)	M1	3.1a
Obtains \(t = \sqrt{2}\) Allow 1.4 or better for \(\sqrt{2}\) \(t = \sqrt{2}\) must come from correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) PI by substituting \(\sqrt{2}\) into \(y\)	A1	1.1b
Substitutes their value for \(t\) into \(y\) and obtains a value for \(y\) provided \(0.2 \lt t \lt 3\)	M1	3.4
Obtains correct length with unit e.g \(2\sqrt{2}\) metres or 2.8 metres or or AWFW [2.82, 2.83] metres Allow equivalent correct length in different units Do not ignore subsequent incorrect rounding	A1	3.2a
	(4)
(iii) States \(\tan\theta\) = value of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at \(t = 3\) OE PI by correct answer or 0.61 or better or 55°	M1	3.1a
Obtains 35° CAO	A1	3.2a
	(2)
	(12 marks)

Typical solution

(i)

\[\frac{\mathrm{d}x}{\mathrm{d}t} = 1 + t^{-2}\]\[\frac{\mathrm{d}y}{\mathrm{d}t} = 1 - 2t^{-2}\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \times \frac{\mathrm{d}t}{\mathrm{d}x}\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 - 2t^{-2}}{1 + t^{-2}}\]

(ii)

\[1 - 2t^{-2} = 0\]\[t^2 = 2\]\[t = \sqrt{2}\]\[y = \sqrt{2} + \frac{2}{\sqrt{2}}\]\[= 2\sqrt{2}\]

Length of \(RS\) = 2.83 metres

(iii)

\[\text{When } t = 3,\ \frac{\mathrm{d}y}{\mathrm{d}x} = 0.7\]\[\tan\theta = 0.7\]\[\theta = 35^\circ\]