| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Applied context modeling |
| Difficulty | Standard +0.3 This is a standard harmonic modeling question requiring verification of amplitude/vertical shift, solving simultaneous equations for phase parameters, and interpreting the model. All techniques are routine A-level applications of cosine graphs with no novel insight required, though the multi-part structure and real-world context make it slightly above average difficulty. |
| Spec | 1.02z Models in context: use functions in modelling1.05f Trigonometric function graphs: symmetries and periodicities1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | AOs |
| \(h_{\max} = 5.15 + 3.4\times1 = 8.55\) | B1 | 3.4 |
| \(h_{\min} = 5.15 - 3.4\times1 = 1.75\) | [1] | |
| These are the correct \(h\) values for high and low tide |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | AOs |
| When \(t = 1\): \(8.55 = 5.15 + 3.4\cos(a+b)\) | B1 | 3.3 |
| So \(\cos(a+b) = 1\) giving \(a + b = 0\) | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | AOs |
| Minimum when \((at+b) = 180°\) and \(t = 7\frac{1}{3}\) | B1 | 3.3 |
| So \(\frac{22}{3}a + b = 180\) | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | AOs |
| Solve simultaneously to give | M1 | 3.3 |
| \(a = 28.42\) to 2 dp | A1 [2] | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | AOs |
| Substitute \(h = 3\): \(3 = 5.15 + 3.4\cos(28.4t - 28.4)\) | M1 | 3.4 |
| \(\cos(28.4t - 28.4) = -\frac{43}{68}\) | ||
| \(28.4t - 28.4 = 129.2\), \(230.8\) | A1 | 3.4 |
| \(t = 5.55\), \(9.13\) | ||
| He does not sail between 5.33 am and 9.08 am | A1 [3] | 3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | AOs |
| EITHER: The model predicts every high tide 8.55 m. The next high tide 8.91 is higher than that so not perfect model. | B1, E1 [2] | 3.4, 3.5b |
| OR: Time difference between high tide and low tide is 6 hr 20 min, and between low tide and next high tide is 5 hours and 40 minutes. The model gives these times as equal, so not perfect model | B1, E1 [2] | |
| OR: tide reaches 8.91 m when \(\cos(at+b) = 1.105\) which is impossible | B1, E1 [2] | |
| OR: When \(t = 12.983\), \(h = 8.35\) which is less than the given value of 8.91 m so the model is not suitable | B1, E1 [2] |
## Question 14:
### Part (a):
| Answer | Marks | AOs | Guidance |
|--------|-------|-----|----------|
| $h_{\max} = 5.15 + 3.4\times1 = 8.55$ | B1 | 3.4 | Choosing $\cos t = \pm1$ to give both values must be seen; Allow without further comment |
| $h_{\min} = 5.15 - 3.4\times1 = 1.75$ | [1] | | Allow for using given $h$ values to find $\cos t = \pm1$ only if there is a comment that these are max and min values for $\cos t$ |
| These are the correct $h$ values for high and low tide | | | |
### Part (b)(i):
| Answer | Marks | AOs | Guidance |
|--------|-------|-----|----------|
| When $t = 1$: $8.55 = 5.15 + 3.4\cos(a+b)$ | B1 | 3.3 | Correctly relating high tide, $t=1$ and $\cos 0$ |
| So $\cos(a+b) = 1$ giving $a + b = 0$ | [1] | | Accept 8.55 or $\cos t = 1$ as evidence of high tide |
### Part (b)(ii):
| Answer | Marks | AOs | Guidance |
|--------|-------|-----|----------|
| Minimum when $(at+b) = 180°$ and $t = 7\frac{1}{3}$ | B1 | 3.3 | Condone the use 7.2 hours here |
| So $\frac{22}{3}a + b = 180$ | [1] | | Allow for $1.75 = 5.15 + 3.4\cos\!\left(\frac{22}{3}a + b\right)$ |
### Part (b)(iii):
| Answer | Marks | AOs | Guidance |
|--------|-------|-----|----------|
| Solve simultaneously to give | M1 | 3.3 | Attempt to solve simultaneous equations: may be BC |
| $a = 28.42$ to 2 dp | A1 [2] | 3.3 | AG (value of $b$ not needed here); $[b = -28.42]$ |
### Part (c):
| Answer | Marks | AOs | Guidance |
|--------|-------|-----|----------|
| Substitute $h = 3$: $3 = 5.15 + 3.4\cos(28.4t - 28.4)$ | M1 | 3.4 | Attempting to solve trig equation or inequality |
| $\cos(28.4t - 28.4) = -\frac{43}{68}$ | | | |
| $28.4t - 28.4 = 129.2$, $230.8$ | A1 | 3.4 | At least one correct [decimal] value for $t$ |
| $t = 5.55$, $9.13$ | | | |
| He does not sail between 5.33 am and 9.08 am | A1 [3] | 3.2a | Both times correct. Need not convert to hours and minutes. Must indicate between these times |
### Part (d):
| Answer | Marks | AOs | Guidance |
|--------|-------|-----|----------|
| **EITHER:** The model predicts every high tide 8.55 m. The next high tide 8.91 is higher than that so not perfect model. | B1, E1 [2] | 3.4, 3.5b | Allow for a comment about the maximum height being wrong. FT their values |
| **OR:** Time difference between high tide and low tide is 6 hr 20 min, and between low tide and next high tide is 5 hours and 40 minutes. The model gives these times as equal, so not perfect model | B1, E1 [2] | | Allow for a comment that the time of the next high tide is wrong. FT their values |
| **OR:** tide reaches 8.91 m when $\cos(at+b) = 1.105$ which is impossible | B1, E1 [2] | | Allow for a comment that the height predicted cannot reach 8.91 m. FT their values |
| **OR:** When $t = 12.983$, $h = 8.35$ which is less than the given value of 8.91 m so the model is not suitable | B1, E1 [2] | | Allow for $t = 13$ but not $t = 12.59$; Allow for a comment that the height predicted is not 8.91 m. FT their values |
---14 Douglas wants to construct a model for the height of the tide in Liverpool during the day, using a cosine graph to represent the way the height changes.
He knows that the first high tide of the day measures 8.55 m and the first low tide of the day measures 1.75 m .
Douglas uses $t$ for time and $h$ for the height of the tide in metres. With his graph-drawing software set to degrees, he begins by drawing the graph of $\mathrm { h } = 5.15 + 3.4$ cost.
\begin{enumerate}[label=(\alph*)]
\item Verify that this equation gives the correct values of $h$ for the high and low tide.
Douglas also knows that the first high tide of the day occurs at 1 am and the first low tide occurs at 7.20 am. He wants $t$ to represent the time in hours after midnight, so he modifies his equation to $h = 5.15 + 3.4 \cos ( a t + b )$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that Douglas's modified equation gives the first high tide of the day occurring at the correct time if $\mathrm { a } + \mathrm { b } = 0$.
\item Use the time of the first low tide of the day to form a second equation relating $a$ and $b$.
\item Hence show that $a = 28.42$ correct to 2 decimal places.
\end{enumerate}\item Douglas can only sail his boat when the height of the tide is at least 3 m .
Use the model to predict the range of times that morning when he cannot sail.
\item The next high tide occurs at 12.59 pm when the height of the tide is 8.91 m .
Comment on the suitability of Douglas's model.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q14 [9]}}