June 2022 Paper 3 Q6
6 A design for a surfboard is shown in Figure 1.

The curve of the top half of the surfboard can be modelled by the parametric equations
\[x = -2t^2\]\[y = 9t - 0.7t^2\]for \(0 \leqslant t \leqslant 9.5\) as shown in Figure 2, where \(x\) and \(y\) are measured in centimetres.

(a) Find the length of the surfboard. [2 marks]
(b)
(i) Find an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\). [3 marks]
(ii) Hence, show that the width of the surfboard is approximately one third of its length. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(t\) = 9.5 into \(x = -2t^2\) or \(2t^2\) | M1 | 3.4 |
| Obtains 180.5 Condone incorrect or missing units ISW | A1 | 1.1b |
| (2) |
Typical solution
\[t = 9.5 \Rightarrow x = -2 \times 9.5^2 = -180.5\]Length = 180.5 cm
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \(9 - 1.4t\) or \(-4t\) OE Ignore labels | B1 | 1.1b |
| Uses chain rule to obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) Condone sign error | M1 | 3.1a |
| Obtains a correct expression Do not ISW | A1 | 1.1b |
| (3) | ||
| (ii) Equates their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) or their numerator of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to 0 PI by correct \(t\) from correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 | 3.1a |
| Obtains correct value for \(t\) ACF eg \(t = 6.4\) or \(\dfrac{9}{1.4}\) Must come from correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | A1 | 1.1b |
| Substitutes their value for \(t\) into the model for \(y\) and obtains a value for \(y\) provided \(0 \lt t \lt 9.5\) | M1 | 3.4 |
| Compares correct width and correct length and \(\dfrac{1}{3}\) or 3 with a correct concluding statement OE CSO Allow 180 for length | R1 | 3.2a |
| (4) | ||
| (9 marks) |
Typical solution
(i)
\[\frac{\mathrm{d}y}{\mathrm{d}t} = 9 - 1.4t\]\[\frac{\mathrm{d}x}{\mathrm{d}t} = -4t\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \times \frac{\mathrm{d}t}{\mathrm{d}x}\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9 - 1.4t}{-4t}\](ii)
\[\frac{9 - 1.4t}{-4t} = 0\]\[t = \frac{45}{7} = 6.43\]\[y = 9 \times 6.43 - 0.7 \times (6.43)^2\]\[= 28.9\]Width of surfboard = 58 cm
\[180.5 \div 3 = 60.2 \approx 58\]Hence the width is approximately one third of the length