| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find dy/dx expression in terms of parameter |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring multiple techniques (parametric differentiation, finding stationary points, self-intersection analysis, and Cartesian conversion), but each individual part follows standard methods. The geometric setup and investigation of different cases elevates it above routine exercises, though it's well-scaffolded with clear sub-parts guiding students through the analysis. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal projection of \(QP = k\cos\theta\) | B1 | |
| Vertical projection of \(QP = k\sin\theta\) | B1 | |
| Subtract \(OQ = \tan\theta\) | B1 | Clearly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k=2\): Loop | G1 | |
| \(k=1\): Cusp | G1 | |
| \(k=\tfrac{1}{2}\): [smooth curve] | G1 | |
| \(k=-1\): [reflected curve] | G1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (A) For all \(k\), \(y\)-axis is an asymptote | B1 | Both (A) and (B) |
| (B) \(k=1\) | B1 | |
| (C) \(k>1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Crosses itself at \((1,0)\) | ||
| \(k=2 \Rightarrow \cos\theta=\tfrac{1}{2} \Rightarrow \theta=60°\) | M1 | Obtaining a value of \(\theta\) |
| \(\Rightarrow\) curve crosses itself at \(120°\) | A1 | Accept \(240°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 8\sin\theta - \tan\theta\) | ||
| \(\Rightarrow \dfrac{dy}{d\theta} = 8\cos\theta - \sec^2\theta\) | ||
| \(\Rightarrow 8\cos\theta - \dfrac{1}{\cos^2\theta}=0\) at highest point | ||
| \(\Rightarrow \cos^3\theta=\dfrac{1}{8} \Rightarrow \cos\theta=\pm\dfrac{1}{2} \Rightarrow \theta=60°\) at top | M1, A1 | Complete method giving \(\theta\) |
| \(\Rightarrow x=4\) | A1 | Both |
| \(y=3\sqrt{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{RHS} = \dfrac{(k\cos\theta-1)^2}{k^2\cos^2\theta}(k^2-k^2\cos^2\theta)\) | M1 | Expressing one side in terms of \(\theta\) |
| \(= \dfrac{(k\cos\theta-1)^2}{k^2\cos^2\theta}\times k^2\sin^2\theta\) | ||
| \(= (k\cos\theta-1)^2\tan^2\theta\) | M1 | Using trig identities |
| \(= \bigl((k\cos\theta-1)\tan\theta\bigr)^2\) | ||
| \(= (k\sin\theta-\tan\theta)^2 = \text{LHS}\) | E1 |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal projection of $QP = k\cos\theta$ | B1 | |
| Vertical projection of $QP = k\sin\theta$ | B1 | |
| Subtract $OQ = \tan\theta$ | B1 | Clearly obtained |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k=2$: Loop | G1 | |
| $k=1$: Cusp | G1 | |
| $k=\tfrac{1}{2}$: [smooth curve] | G1 | |
| $k=-1$: [reflected curve] | G1 | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (A) For all $k$, $y$-axis is an asymptote | B1 | Both (A) and (B) |
| (B) $k=1$ | B1 | |
| (C) $k>1$ | B1 | |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Crosses itself at $(1,0)$ | | |
| $k=2 \Rightarrow \cos\theta=\tfrac{1}{2} \Rightarrow \theta=60°$ | M1 | Obtaining a value of $\theta$ |
| $\Rightarrow$ curve crosses itself at $120°$ | A1 | Accept $240°$ |
## Part (v):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 8\sin\theta - \tan\theta$ | | |
| $\Rightarrow \dfrac{dy}{d\theta} = 8\cos\theta - \sec^2\theta$ | | |
| $\Rightarrow 8\cos\theta - \dfrac{1}{\cos^2\theta}=0$ at highest point | | |
| $\Rightarrow \cos^3\theta=\dfrac{1}{8} \Rightarrow \cos\theta=\pm\dfrac{1}{2} \Rightarrow \theta=60°$ at top | M1, A1 | Complete method giving $\theta$ |
| $\Rightarrow x=4$ | A1 | Both |
| $y=3\sqrt{3}$ | A1 | |
## Part (vi):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{RHS} = \dfrac{(k\cos\theta-1)^2}{k^2\cos^2\theta}(k^2-k^2\cos^2\theta)$ | M1 | Expressing one side in terms of $\theta$ |
| $= \dfrac{(k\cos\theta-1)^2}{k^2\cos^2\theta}\times k^2\sin^2\theta$ | | |
| $= (k\cos\theta-1)^2\tan^2\theta$ | M1 | Using trig identities |
| $= \bigl((k\cos\theta-1)\tan\theta\bigr)^2$ | | |
| $= (k\sin\theta-\tan\theta)^2 = \text{LHS}$ | E1 | |5 A line PQ is of length $k$ (where $k > 1$ ) and it passes through the point ( 1,0 ). PQ is inclined at angle $\theta$ to the positive $x$-axis. The end Q moves along the $y$-axis. See Fig. 5. The end P traces out a locus.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d43d1e11-3173-47c4-88c9-0397c8630a39-4_639_977_552_584}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that the locus of P may be expressed parametrically as follows.
$$x = k \cos \theta \quad y = k \sin \theta - \tan \theta$$
You are now required to investigate curves with these parametric equations, where $k$ may take any non-zero value and $- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi$.
\item Use your calculator to sketch the curve in each of the cases $k = 2 , k = 1 , k = \frac { 1 } { 2 }$ and $k = - 1$.
\item For what value(s) of $k$ does the curve have\\
(A) an asymptote (you should state what the asymptote is),\\
(B) a cusp,\\
(C) a loop?
\item For the case $k = 2$, find the angle at which the curve crosses itself.
\item For the case $k = 8$, find in an exact form the coordinates of the highest point on the loop.
\item Verify that the cartesian equation of the curve is
$$y ^ { 2 } = \frac { ( x - 1 ) ^ { 2 } } { x ^ { 2 } } \left( k ^ { 2 } - x ^ { 2 } \right) .$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2010 Q5 [18]}}