| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Stationary points of curves |
| Difficulty | Standard +0.3 This is a standard C3 harmonic form question requiring the routine R sin(y+α) transformation followed by implicit differentiation. Part (a) is textbook application of harmonic identities, and part (b) involves straightforward implicit differentiation and solving a trigonometric equation within a given range. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R\cos\alpha=3\), \(R\sin\alpha=3\) | ||
| \(R=3\sqrt{2}\) or \(\sqrt{18}\) or awrt 4.24 | B1 | Accept \(\pm 3\sqrt{2}\) but not just \(-3\sqrt{2}\); no working needed |
| \(\tan\alpha=1 \Rightarrow \alpha=\frac{\pi}{4}\) or 0.785 | M1, A1 (3) | M1: \(\tan\alpha=\pm\frac{3}{3}\); A1: accept awrt 0.785; 45 degrees is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{dx}{dy}\right)=3\cos y - 3\sin y\) or \(3\sqrt{2}\cos\left(y+\frac{\pi}{4}\right)\) | M1 A1 | M1: attempts differentiation (may be sign errors); A1: correct in either form; A0 if clearly in degrees |
| Puts \(3\sqrt{2}\cos(y+\alpha)=2\) | B1 | Obtains equation; \(3\cos y - 3\sin y = 2\) without further work is B0 |
| \(\cos(y+\alpha)=\frac{\sqrt{2}}{3}\) and \(y="1.0799"-\alpha\) | M1 | Allow in degrees or radians for \(\arccos\left(\frac{\pm 2}{R}\right)\pm\alpha\) or \(\arcsin\left(\frac{\pm 2}{R}\right)\pm\alpha\) |
| \(y=0.295\) and \(x=3.742\) (or 3.743) | A1 A1 (6) | A1: one correct answer; A1: two correct answers (accept awrt in both cases) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dy} = 3\cos y - 3\sin y\) (as before) | M1A1 | |
| \(3\cos y - 3\sin y = 2\), so \(3\frac{1-t^2}{1+t^2} - 3\frac{2t}{1+t^2} = 2\) | B1 | |
| Attempt to solve \(5t^2 + 6t - 1 = 0\) and use \(y = 2 \times \arctan(0.148)\) | M1 | |
| \(y = 0.295\) and \(x = 3.742\) (or \(3.743\)) | A1 A1 | (6) total [9] |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R\cos\alpha=3$, $R\sin\alpha=3$ | | |
| $R=3\sqrt{2}$ or $\sqrt{18}$ or awrt 4.24 | B1 | Accept $\pm 3\sqrt{2}$ but not just $-3\sqrt{2}$; no working needed |
| $\tan\alpha=1 \Rightarrow \alpha=\frac{\pi}{4}$ or 0.785 | M1, A1 (3) | M1: $\tan\alpha=\pm\frac{3}{3}$; A1: accept awrt 0.785; 45 degrees is A0 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{dx}{dy}\right)=3\cos y - 3\sin y$ or $3\sqrt{2}\cos\left(y+\frac{\pi}{4}\right)$ | M1 A1 | M1: attempts differentiation (may be sign errors); A1: correct in either form; A0 if clearly in degrees |
| Puts $3\sqrt{2}\cos(y+\alpha)=2$ | B1 | Obtains equation; $3\cos y - 3\sin y = 2$ without further work is B0 |
| $\cos(y+\alpha)=\frac{\sqrt{2}}{3}$ and $y="1.0799"-\alpha$ | M1 | Allow in degrees or radians for $\arccos\left(\frac{\pm 2}{R}\right)\pm\alpha$ or $\arcsin\left(\frac{\pm 2}{R}\right)\pm\alpha$ |
| $y=0.295$ and $x=3.742$ (or 3.743) | A1 A1 (6) | A1: one correct answer; A1: two correct answers (accept awrt in both cases) |
# Question 5 (Alternative method, last 4 marks of part b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dy} = 3\cos y - 3\sin y$ (as before) | M1A1 | |
| $3\cos y - 3\sin y = 2$, so $3\frac{1-t^2}{1+t^2} - 3\frac{2t}{1+t^2} = 2$ | B1 | |
| Attempt to solve $5t^2 + 6t - 1 = 0$ and use $y = 2 \times \arctan(0.148)$ | M1 | |
| $y = 0.295$ and $x = 3.742$ (or $3.743$) | A1 A1 | (6) total [9] |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f22eb1b-21de-45f1-9a8a-deac7ac8d0b0-14_646_1013_207_532}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve shown in Figure 1 has equation
$$x = 3 \sin y + 3 \cos y , \quad - \frac { \pi } { 4 } < y < \frac { \pi } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Express the equation of the curve in the form\\
$x = R \sin ( y + \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$
\item Find the coordinates of the point on the curve where the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ is $\frac { 1 } { 2 }$. Give your answers to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q5 [9]}}