June 2024 Paper 1 Q9
9 The depth of the water, \(d\) metres, in a tidal river during a given day is modelled by the equation
\(d = 1.9 + 1.1\cos(30t - 60)^\circ\)
where \(t\) is the number of hours after midnight.
(A tidal river is one whose level is influenced by tides.)
Determine the two periods of time during the day between which this boat will not be able to enter the river. Give your answers correct to the nearest minute. [5]
In reality the depth of the river decreases as this boat travels along the river. An improved model uses the equation
\(d = \mathrm{e}^{-cp}\left(1.9 + 1.1\cos(30t - 60)^\circ\right)\)
where \(c\) is a positive constant and \(p\) is the distance, in kilometres, travelled along the river after entering it.
| Scheme | Marks | AO |
|---|---|---|
| (i) 0.8 m | B1 | 3.4 |
| [1] | ||
| (ii) \(\cos(30t - 60) = -1\) \(30t - 60 = 180\) | M1 | 3.4 |
| \(t = 8\) (hours) | A1 | 3.4 |
| [2] |
Notes
(a)(i) B1: State 0.8 m, units required
Units may be given as m or metres
Could be \(\tfrac{4}{5}\) m
Could be 80 cm
(a)(ii)
M1: Identify that minimum occurs when \(\cos(30t - 60)\) is \(-1\), so need \(30t - 60 = 180\)
M1 does not require attempt at solution for \(t\)
No FT on an incorrect \(d\) being used from the previous part eg \(d = 2.45\) from using \(t = 0\)
A1: No units needed, as value of \(t\) is requested
Condone 0800 or 8am
Ignore additional values of \(t\) that are greater than 8, but A0 for a smaller positive value of \(t\) also given
| Scheme | Marks | AO |
|---|---|---|
| \(1.9 + 1.1\cos(30t - 60) = 1\) \(\cos(30t - 60) = -0.8181\ldots\) \(30t - 60 = 144.903\) | M1 | 3.3 |
| \(30t = 204.90\) \(t = 6.830\ldots\) | A1 | 1.1 |
| \(30t - 60 = 215.097,\ 504.903,\ 575.097\) \(30t = 275.097,\ 564.903,\ 635.097\) \(t = 9.169\ldots,\ 18.830\ldots,\ 21.169\ldots\) | M1 | 3.4 |
| Obtain the further 3 correct values and no others | A1 | 1.1 |
| River cannot be entered 0650 to 0910 and 1850 to 2110 | A1 | 3.2a |
| [5] |
Notes
M1: Equate model to 1, rearrange and use \(\cos^{-1}\)
As far as \(30t - 60 = k\), using correct order of operations
Allow M1 if working in radians (gives \(30t - 60 = 2.529\))
A1: Obtain correct first value of \(t\)
Implied by first time of 0650 with no errors seen
3sf or better
Ignore inequality signs if used
M1: Attempt all further values of \(t\) within \(0 \lt t \lt 24\)
Using a valid method
M0 if using radians
Values of \(t\) could also be found using the symmetry of the curve eg \(8 + (8 - 6.83) = 9.17\)
A1: Allow answers to 3sf
Ignore inequality signs if used
Correct time periods would imply \(t\) values
A1: Correct two periods, given as time intervals
Could also be given as 6:50am to 9:10am, and 6:50pm to 9:10pm
Must be given as intervals and not just times eg A0 for ‘0650 and 0910’ etc
BOD if correct intervals given following any incorrect inequality signs
Accept 0649 to 0911 and 1849 to 2111 www (from checking times and realising that rounded answers give depths of less than 1 metre)
A0 if giving answers in minutes, or hours and minutes, after midnight and not times (eg ‘410 minutes to 550 minutes’ or ‘6 hours 50 minutes to 9 hours 10 minutes’)
Condone attempt at interval notation / inequalities as long as intention is clear, and allow BOD if written as a strict inequality such as \(0650 \lt t \lt 0910\)
Special Case
If M1A1M0 awarded, then allow SC B1 for giving a correct time period eg 0650 to 0910
| Scheme | Marks | AO |
|---|---|---|
| As \(p\) increases, \(\mathrm{e}^{-cp}\) decreases so difference between max / min depths will decrease | B1 | 3.5c |
| [1] |
Notes
B1: Any sensible suggestion that suggests that amplitudes of the tides will be reduced due to the exponential term
Must refer to the effect of the exponential term in context in some way
Condone reference to the exponential term having a ‘damping’ effect on the tides
Cannot just restate the question eg ‘river gets shallower’