| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| 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 form question with straightforward application. Part (a) is routine R cos(θ+α) conversion using Pythagorean identity and arctan. Parts (b)-(d) involve substituting initial conditions, reading max/min from the harmonic form, and solving a simple trigonometric equation. The modeling context adds no conceptual difficulty—it's essentially bookwork with one substitution. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = \sqrt{109}\) | B1 | Do not allow decimal equivalents |
| \(\tan\alpha = \frac{3}{10}\) | M1 | Allow \(\tan\alpha = \pm\frac{3}{10}\) |
| \(\alpha = 16.70°\) so \(\sqrt{109}\cos(\theta+16.70°)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H = 11-10\cos(80t)°+3\sin(80t)°\) or \(H=11-\sqrt{109}\cos(80t+16.70)°\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(11+\sqrt{109}\) or \(21.44\) m | B1ft | Their \(11+\) their \(\sqrt{109}\); allow decimals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(80t + \text{"16.70"} = 540\) | M1 | Follow through on their \(16.70\) |
| \(t = \frac{540-\text{"16.70"}}{80} = (6.54)\) | M1 | Solves \(80t+\text{"16.70"}=540\) correctly |
| \(t = 6\) mins \(32\) seconds | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Increase the '\(80\)' in the formula; e.g. use \(H=11-10\cos(90t)°+3\sin(90t)°\) | B1 | States that to increase speed of wheel the \(80\)'s in the equation would need to be increased |
# Question 13(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{109}$ | B1 | Do not allow decimal equivalents |
| $\tan\alpha = \frac{3}{10}$ | M1 | Allow $\tan\alpha = \pm\frac{3}{10}$ |
| $\alpha = 16.70°$ so $\sqrt{109}\cos(\theta+16.70°)$ | A1 | |
# Question 13(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = 11-10\cos(80t)°+3\sin(80t)°$ or $H=11-\sqrt{109}\cos(80t+16.70)°$ | B1 | |
# Question 13(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $11+\sqrt{109}$ or $21.44$ m | B1ft | Their $11+$ their $\sqrt{109}$; allow decimals |
# Question 13(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $80t + \text{"16.70"} = 540$ | M1 | Follow through on their $16.70$ |
| $t = \frac{540-\text{"16.70"}}{80} = (6.54)$ | M1 | Solves $80t+\text{"16.70"}=540$ correctly |
| $t = 6$ mins $32$ seconds | A1 | |
# Question 13(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Increase the '$80$' in the formula; e.g. use $H=11-10\cos(90t)°+3\sin(90t)°$ | B1 | States that to increase speed of wheel the $80$'s in the equation would need to be increased |13. (a) Express $10 \cos \theta - 3 \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < 90 ^ { \circ }$ Give the exact value of $R$ and give the value of $\alpha$, in degrees, to 2 decimal places.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a75c9ef7-b648-47be-bad1-fc8b315be3df-18_396_1329_388_367}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The height above the ground, $H$ metres, of a passenger on a Ferris wheel $t$ minutes after the wheel starts turning, is modelled by the equation
$$H = a - 10 \cos ( 80 t ) ^ { \circ } + 3 \sin ( 80 t ) ^ { \circ }$$
where $a$ is a constant.\\
Figure 3 shows the graph of $H$ against $t$ for two complete cycles of the wheel.\\
Given that the initial height of the passenger above the ground is 1 metre,\\
(b) (i) find a complete equation for the model,\\
(ii) hence find the maximum height of the passenger above the ground.\\
(c) Find the time taken, to the nearest second, for the passenger to reach the maximum height on the second cycle.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
It is decided that, to increase profits, the speed of the wheel is to be increased.\\
(d) How would you adapt the equation of the model to reflect this increase in speed?
\hfill \mbox{\textit{Edexcel Paper 2 Q13 [9]}}