Standard +0.3 This is a standard Further Maths harmonic form question with routine application to a real-world context. Part (a) uses the standard R sin(θ-α) conversion formula, parts (b)-(d) involve straightforward substitution and finding maxima, and part (e) requires recognizing continuity issues between models—all textbook techniques with no novel insight required. Slightly easier than average A-level due to its structured, step-by-step nature.
13. (a) Express \(2 \sin \theta - 1.5 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\)
State the value of \(R\) and give the value of \(\alpha\) to 4 decimal places.
Tom models the depth of water, \(D\) metres, at Southview harbour on 18th October 2017 by the formula
$$D = 6 + 2 \sin \left( \frac { 4 \pi t } { 25 } \right) - 1.5 \cos \left( \frac { 4 \pi t } { 25 } \right) , \quad 0 \leqslant t \leqslant 24$$
where \(t\) is the time, in hours, after 00:00 hours on 18th October 2017.
Use Tom's model to
(b) find the depth of water at 00:00 hours on 18th October 2017,
(c) find the maximum depth of water,
(d) find the time, in the afternoon, when the maximum depth of water occurs. Give your answer to the nearest minute.
Tom's model is supported by measurements of \(D\) taken at regular intervals on 18th October 2017. Jolene attempts to use a similar model in order to model the depth of water at Southview harbour on 19th October 2017.
Jolene models the depth of water, \(H\) metres, at Southview harbour on 19th October 2017 by the formula
$$H = 6 + 2 \sin \left( \frac { 4 \pi x } { 25 } \right) - 1.5 \cos \left( \frac { 4 \pi x } { 25 } \right) , \quad 0 \leqslant x \leqslant 24$$
where \(x\) is the time, in hours, after 00:00 hours on 19th October 2017.
By considering the depth of water at 00:00 hours on 19th October 2017 for both models,
(e) (i) explain why Jolene's model is not correct,
(ii) hence find a suitable model for \(H\) in terms of \(x\).
\(t = 16.9052\ldots \Rightarrow\) Time \(= 16{:}54\) or \(4{:}54\) pm
A1
AO 3.2a
Part (e)(i):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Attempts to find depth of water at 00:00 on 19th October 2017 for at least one model
M1
AO 3.4
Tom: \(D = 3.72\ldots\) m and Jolene: \(H = 4.5\) m; since \(4.5 \neq 3.72\), Jolene's model is not true / not continuous / does not continue from where Tom's model ended
A1
AO 3.5a
Part (e)(ii):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
E.g. \(H = 5.22 + 2\sin\!\left(\frac{4\pi x}{25}\right) - 1.5\cos\!\left(\frac{4\pi x}{25}\right),\ 0 \leq x < 24\) or \(H = 6 + 2\sin\!\left(\frac{4\pi(x+24)}{25}\right) - 1.5\cos\!\left(\frac{4\pi(x+24)}{25}\right),\ 0 \leq x < 24\)
B1
AO 3.3
Total: 11 marks
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 2.5$ | B1 | AO 1.1b; condone $R = \sqrt{6.25}$ |
| $\tan\alpha = \frac{1.5}{2}$ | M1 | AO 1.1b; accept $\tan\alpha = \frac{1.5}{2}$ or $-\frac{1.5}{2}$ or $\frac{2}{1.5}$ or $-\frac{2}{1.5}$ |
| $\alpha = 0.6435$, so $2.5\sin(\theta - 0.6435)$ | A1 | AO 1.1b; $\alpha$ = awrt 0.6435 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D = 6 + 2\sin\!\left(\frac{4\pi(0)}{25}\right) - 1.5\cos\!\left(\frac{4\pi(0)}{25}\right) = 4.5\text{ m}$ | B1 | AO 3.4; uses Tom's model at $t=0$ on 18th Oct 2017 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D_{\max} = 6 + 2.5 = 8.5\text{ m}$ | B1ft | AO 3.4; either 8.5 or follow through "$6 + \text{their } R$" |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{4\pi t}{25} - \text{"0.6435"} = \frac{5\pi}{2}$ or $\frac{\pi}{2}$ | M1 | AO 1.1b |
| Afternoon solution: $\frac{4\pi t}{25} - \text{"0.6435"} = \frac{5\pi}{2} \Rightarrow t = \frac{25}{4\pi}\!\left(\frac{5\pi}{2} + \text{"0.6435"}\right)$ | M1 | AO 3.1b |
| $t = 16.9052\ldots \Rightarrow$ Time $= 16{:}54$ or $4{:}54$ pm | A1 | AO 3.2a |
**Alt 1 for (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{4\pi t}{25} - \text{"0.6435"} = \frac{\pi}{2}$; period $= 2\pi \div \frac{4\pi}{25} = 12.5$ | M1 | AO 1.1b / 3.1b |
| Afternoon solution: $t = 12.5 + \frac{25}{4\pi}\!\left(\frac{\pi}{2} + \text{"0.6435"}\right)$ | M1 | AO 3.1b |
| $t = 16.9052\ldots \Rightarrow$ Time $= 16{:}54$ or $4{:}54$ pm | A1 | AO 3.2a |
### Part (e)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find depth of water at 00:00 on 19th October 2017 for at least one model | M1 | AO 3.4 |
| Tom: $D = 3.72\ldots$ m and Jolene: $H = 4.5$ m; since $4.5 \neq 3.72$, Jolene's model is not true / not continuous / does not continue from where Tom's model ended | A1 | AO 3.5a |
### Part (e)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $H = 5.22 + 2\sin\!\left(\frac{4\pi x}{25}\right) - 1.5\cos\!\left(\frac{4\pi x}{25}\right),\ 0 \leq x < 24$ or $H = 6 + 2\sin\!\left(\frac{4\pi(x+24)}{25}\right) - 1.5\cos\!\left(\frac{4\pi(x+24)}{25}\right),\ 0 \leq x < 24$ | B1 | AO 3.3 |
**Total: 11 marks**
13. (a) Express $2 \sin \theta - 1.5 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$
State the value of $R$ and give the value of $\alpha$ to 4 decimal places.
Tom models the depth of water, $D$ metres, at Southview harbour on 18th October 2017 by the formula
$$D = 6 + 2 \sin \left( \frac { 4 \pi t } { 25 } \right) - 1.5 \cos \left( \frac { 4 \pi t } { 25 } \right) , \quad 0 \leqslant t \leqslant 24$$
where $t$ is the time, in hours, after 00:00 hours on 18th October 2017.
Use Tom's model to\\
(b) find the depth of water at 00:00 hours on 18th October 2017,\\
(c) find the maximum depth of water,\\
(d) find the time, in the afternoon, when the maximum depth of water occurs. Give your answer to the nearest minute.
Tom's model is supported by measurements of $D$ taken at regular intervals on 18th October 2017. Jolene attempts to use a similar model in order to model the depth of water at Southview harbour on 19th October 2017.
Jolene models the depth of water, $H$ metres, at Southview harbour on 19th October 2017 by the formula
$$H = 6 + 2 \sin \left( \frac { 4 \pi x } { 25 } \right) - 1.5 \cos \left( \frac { 4 \pi x } { 25 } \right) , \quad 0 \leqslant x \leqslant 24$$
where $x$ is the time, in hours, after 00:00 hours on 19th October 2017.\\
By considering the depth of water at 00:00 hours on 19th October 2017 for both models,\\
(e) (i) explain why Jolene's model is not correct,\\
(ii) hence find a suitable model for $H$ in terms of $x$.
\hfill \mbox{\textit{Edexcel Paper 1 Q13 [11]}}