| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Applied context modeling |
| Difficulty | Standard +0.3 This is a standard harmonic form question with multiple routine parts. Part (i) is a textbook R-formula conversion, parts (ii)-(iv) involve reading max/min values and sketching, part (v) requires solving a simple trig inequality, and part (vi) asks for estimation from a graph. While multi-part with 6 sections, each individual step uses well-practiced C4 techniques without requiring novel insight or complex problem-solving. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.07b Gradient as rate of change: dy/dx notation1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R\left(\frac{2}{R}\sin 30t+\frac{1.5}{R}\cos 30t\right)=R(\sin 30t\cos\alpha+\cos 30t\sin\alpha)\) | M1 | |
| \(R=\sqrt{2^2+1.5^2}=2.5\) | A1 | \(R\) |
| \(\cos\alpha=\frac{2}{2.5} \Rightarrow \alpha=36.9°\) | A1 | \(\alpha\) |
| \(\Rightarrow 2.5\sin(30t+36.9)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum is \(2.5+3=5.5\) | B1 | |
| Occurs when \(\sin(30t+\alpha)=1\) | M1 | |
| \(\Rightarrow 30t+\alpha=90 \Rightarrow 30t=53.1 \Rightarrow t\approx 1.77\) | A1, A1 | Follow through |
| \(\Rightarrow\) time is approx. 0146 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| High water \(=3+2.5=5.5\); Low water \(=3-2.5=0.5\); Range \(=5\) metres | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sinusoidal curve with correct shape | B1 | Shape |
| Maximum and minimum correctly marked | B1 | Maximum and minimum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3+2.5\sin(30t+36.9)=2\) | M1 | |
| \(\Rightarrow\sin(30t+36.9)=-0.4\) | A1 | f.t. |
| \(\Rightarrow 30t+36.9=203.6\) or \(336.4\) | A1 | Both |
| \(\Rightarrow t=5.56\) or \(9.98\) | A1 | Both |
| i.e. Midnight to 0533 and 0959 to midday | A1 | f.t. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The time when water level is falling fastest is 3 hrs after high water, i.e. at 0446 | B1 | |
| From graph, rate is 5 metres per 4 hours or approx 1.25 metres per hour or approx 2 cm per minute. (Calculation gives 1.31 metres per hour, allow anything from 1.25–1.35 metres per hour.) | M1, A1 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R\left(\frac{2}{R}\sin 30t+\frac{1.5}{R}\cos 30t\right)=R(\sin 30t\cos\alpha+\cos 30t\sin\alpha)$ | M1 | |
| $R=\sqrt{2^2+1.5^2}=2.5$ | A1 | $R$ |
| $\cos\alpha=\frac{2}{2.5} \Rightarrow \alpha=36.9°$ | A1 | $\alpha$ |
| $\Rightarrow 2.5\sin(30t+36.9)$ | | |
**Total: 3 marks**
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum is $2.5+3=5.5$ | B1 | |
| Occurs when $\sin(30t+\alpha)=1$ | M1 | |
| $\Rightarrow 30t+\alpha=90 \Rightarrow 30t=53.1 \Rightarrow t\approx 1.77$ | A1, A1 | Follow through |
| $\Rightarrow$ time is approx. 0146 | | |
**Total: 4 marks**
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| High water $=3+2.5=5.5$; Low water $=3-2.5=0.5$; Range $=5$ metres | B1 | |
**Total: 1 mark**
## Question 8(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sinusoidal curve with correct shape | B1 | Shape |
| Maximum and minimum correctly marked | B1 | Maximum and minimum |
**Total: 2 marks**
## Question 8(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3+2.5\sin(30t+36.9)=2$ | M1 | |
| $\Rightarrow\sin(30t+36.9)=-0.4$ | A1 | f.t. |
| $\Rightarrow 30t+36.9=203.6$ or $336.4$ | A1 | Both |
| $\Rightarrow t=5.56$ or $9.98$ | A1 | Both |
| i.e. Midnight to 0533 and 0959 to midday | A1 | f.t. |
**Total: 5 marks**
## Question 8(vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The time when water level is falling fastest is 3 hrs after high water, i.e. at 0446 | B1 | |
| From graph, rate is 5 metres per 4 hours or approx 1.25 metres per hour or approx 2 cm per minute. (Calculation gives 1.31 metres per hour, allow anything from 1.25–1.35 metres per hour.) | M1, A1 | |
**Total: 3 marks**8 The height of tide at the entrance to a harbour on a particular day may be modelled by the function $h = 3 + 2 \sin 30 t + 1.5 \cos 30 t$ where $h$ is measured in metres, $t$ in hours after midnight and $30 t$ is in degrees.\\[0pt]
[The values 2 and 1.5 represent the relative effects of the moon and sun respectively.]\\
(i) Show that $2 \sin 30 t + 1.5 \cos 30 t$ can be written in the form $2.5 \sin ( 30 t + \alpha )$, where $\alpha$ is to be determined.\\
(ii) Find the height of tide at high water and the first time that this occurs after midnight.\\
(iii) Find the range of tide during the day.\\
(iv) Sketch the graph of $h$ against $t$ for $0 \leq t \leq 12$, indicating the maximum and minimum points.\\
(v) A sailing boat may enter the harbour only if there is at least 2 metres of water. Find the times during this morning when it may enter the harbour.\\
(vi) From your graph estimate the time at which the water falling fastest and the rate at which it is falling.
\hfill \mbox{\textit{OCR MEI C4 Q8 [18]}}