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Date	November 2021	Marks available	2	Reference code	21N.2.AHL.TZ0.9
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	9	Adapted from	N/A

Question

The height of water, in metres, in Dungeness harbour is modelled by the function $H (t) = a \sin (b (t - c)) + d$ , where $t$ is the number of hours after midnight, and $a, b, c$ and $d$ are constants, where $a > 0, b > 0$ and $c > 0$ .

The following graph shows the height of the water for $13$ hours, starting at midnight.

The first high tide occurs at $04 : 30$ and the next high tide occurs $12$ hours later. Throughout the day, the height of the water fluctuates between $2.2 m$ and $6.8 m$ .

All heights are given correct to one decimal place.

Show that $b = \frac{π}{6}$ .

[1]

Find the value of $a$ .

[2]

Find the value of $d$ .

[2]

Find the smallest possible value of $c$ .

[3]

Find the height of the water at $12 : 00$ .

[2]

Determine the number of hours, over a 24-hour period, for which the tide is higher than $5$ metres.

[3]

A fisherman notes that the water height at nearby Folkestone harbour follows the same sinusoidal pattern as that of Dungeness harbour, with the exception that high tides (and low tides) occur $50$ minutes earlier than at Dungeness.

Find a suitable equation that may be used to model the tidal height of water at Folkestone harbour.

[2]

Markscheme

$12 = \frac{2 π}{b}$ OR $b = \frac{2 π}{12}$ A1

$b = \frac{π}{6}$ AG

[1 mark]

$a = \frac{6.8 - 2.2}{2}$ OR $a = \frac{max - min}{2}$ (M1)

$= 2.3 (m)$ A1

[2 marks]

$d = \frac{6.8 + 2.2}{2}$ OR $d = \frac{max + min}{2}$ (M1)

$= 4.5 (m)$ A1

[2 marks]

METHOD 1

substituting $t = 4.5$ and $H = 6.8$ for example into their equation for $H$ (A1)

$6.8 = 2.3 \sin (\frac{π}{6} (4.5 - c)) + 4.5$

attempt to solve their equation (M1)

$c = 1.5$ A1

METHOD 2

using horizontal translation of $\frac{12}{4}$ (M1)

$4.5 - c = 3$ (A1)

$c = 1.5$ A1

METHOD 3

$H' (t) = (2.3) (\frac{π}{6}) \cos (\frac{π}{6} (t - c))$ (A1)

attempts to solve their $H' (4.5) = 0$ for $c$ (M1)

$(2.3) (\frac{π}{6}) \cos (\frac{π}{6} (4.5 - c)) = 0$

$c = 1.5$ A1

[3 marks]

attempt to find $H$ when $t = 12$ or $t = 0$ , graphically or algebraically (M1)

$H = 2.87365 \dots$

$H = 2.87 (m)$ A1

[2 marks]

attempt to solve $5 = 2.3 \sin (\frac{π}{6} (t - 1.5)) + 4.5$ (M1)

times are $t = 1.91852 \dots$ and $t = 7.08147 \dots, (t = 13.9185 \dots, t = 19.0814 \dots)$ (A1)

total time is $2 \times (7.081 \dots - 1.919 \dots)$

$10.3258 \dots$

$= 10.3$ (hours) A1

Note: Accept $10$ .

[3 marks]

METHOD 1

substitutes $t = \frac{11}{3}$ and $H = 6.8$ into their equation for $H$ and attempts to solve for $c$ (M1)

$6.8 = 2.3 \sin (\frac{π}{6} (\frac{11}{3} - c)) + 4.5 \Rightarrow c = \frac{2}{3}$

$H (t) = 2.3 \sin (\frac{π}{6} (t - \frac{2}{3})) + 4.5$ A1

METHOD 2
uses their horizontal translation $(\frac{12}{4} = 3)$ (M1)

$\frac{11}{3} - c = 3 \Rightarrow c = \frac{2}{3}$

$H (t) = 2.3 \sin (\frac{π}{6} (t - \frac{2}{3})) + 4.5$ A1

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology

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