| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.2 This is a standard Further Pure 2 polar coordinates question requiring the tangent condition (dr/dθ formula), followed by routine area integration. While it involves multiple steps and careful algebra, the techniques are well-practiced FP2 material with no novel insights required. The 'show that' format provides guidance, making it moderately above average difficulty but not exceptional. |
| Spec | 1.07s Parametric and implicit differentiation4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((y =)\, r\sin\theta = a\sin 2\theta\sin\theta\) | M1 | |
| \(\frac{dy}{d\theta} = a(2\cos 2\theta\sin\theta + \sin 2\theta\cos\theta)\) | M1depA1 | |
| \(\frac{dy}{d\theta} = 2a\sin\theta(\cos 2\theta + \cos^2\theta)\) | M1 | |
| At \(P\): \(\frac{dy}{d\theta} = 0 \Rightarrow \sin\theta = 0\) (n/a) or \(2\cos^2\theta - 1 + \cos^2\theta = 0\), \(3\cos^2\theta = 1\) | M1 | \(\sin\theta = 0\) not needed |
| \(\cos\theta = \frac{1}{\sqrt{3}}\) ※ | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(r = a\sin 2\theta = 2a\sin\theta\cos\theta = 2a\sqrt{\left(1-\frac{1}{3}\right)}\sqrt{\frac{1}{3}} = 2a\frac{\sqrt{2}}{3}\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Area \(= \int_0^\phi \frac{1}{2}r^2\,d\theta = \frac{1}{2}a^2\int_0^\phi \sin^2 2\theta\,d\theta\) | M1 | |
| \(= \frac{1}{2}a^2\int_0^\phi \frac{1}{2}(1-\cos 4\theta)\,d\theta\) | M1 | |
| \(= \frac{1}{4}a^2\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^\phi\) | M1A1 | |
| \(= \frac{1}{4}a^2\left[\phi - \frac{1}{4}\left(4\sin\phi\cos\phi(2\cos^2\phi - 1)\right)\right]\) | M1dep | Dep on 2nd M mark |
| \(= \frac{1}{4}a^2\left[\arccos\!\left(\frac{1}{\sqrt{3}}\right) - \left(\sqrt{\frac{2}{3}}\times\frac{1}{\sqrt{3}}\times\left(\frac{2}{3}-1\right)\right)\right]\) | M1dep | Dep on all Ms |
| \(\frac{1}{36}a^2\!\left[9\arccos\!\left(\frac{1}{\sqrt{3}}\right) + \sqrt{2}\right]\) ※ | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtain \(y\) coordinate: \(y = r\sin\theta = a\sin 2\theta \sin\theta\) | M1 | For obtaining the \(y\) coordinate |
| Attempt differentiation to obtain \(\frac{dy}{d\theta}\) | M1dep | Product rule and/or chain rule must be used; \(\sin\) to become \(\pm\cos\), \(\cos\) to become \(\pm\sin\). The 2 may be omitted. Dependent on first M mark |
| \(\frac{dy}{d\theta} = a(2\cos 2\theta \sin\theta + \sin 2\theta \cos\theta)\) | A1 | For correct differentiation (oe) |
| Use \(\sin 2\theta = 2\sin\theta\cos\theta\) anywhere in solution | M1 | |
| Set \(\frac{dy}{d\theta} = 0\) and get a quadratic factor with no \(\sin^2\theta\) included | M1 | *Alternative*: Obtain a quadratic in \(\sin\theta\) or \(\tan\theta\) and complete to \(\cos\theta\) later |
| \(\cos\theta = \frac{1}{\sqrt{3}}\) or \(\cos\phi = \frac{1}{\sqrt{3}}\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Form | \(\frac{dy}{d\theta}\) | Quadratic |
| \(y = a\sin 2\theta\sin\theta\) | \(2\cos^2\theta - \sin^2\theta = 0\) | \(\tan^2\theta = 2\) |
| \(y = 2a\sin^2\theta\cos\theta\) | \(3\cos^2\theta - 1 = 0\) | \(\cos^2\theta = \frac{1}{3}\) |
| \(y = 2a(\cos\theta - \cos^3\theta)\) | \(2 - 3\sin^2\theta = 0\) | \(\sin^2\theta = \frac{2}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\sin 2\theta = 2\sin\theta\cos\theta\), \(\cos^2\theta + \sin^2\theta = 1\) and \(\cos\phi = \frac{1}{\sqrt{3}}\) in \(r = a\sin 2\theta\) to obtain numerical multiple of \(a\) for \(R\) | M1 | Need not be simplified |
| \(R = 2a\dfrac{\sqrt{2}}{3}\) | A1cao | Can be done on calculator. Completely correct answer with no working scores 2/2; incorrect answer with no working scores 0/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use area formula: \(\int_0^\phi \frac{1}{2}r^2\,d\theta = \frac{1}{2}a^2\int_0^\phi \sin^2 2\theta\,d\theta\) | M1 | Limits not needed |
| Prepare \(\int \sin^2 2\theta\,d\theta\) using \(\cos 2x = 1 - 2\sin^2 x\) | M1 | |
| Attempt integration: \(\cos 4\theta\) to become \(\pm\sin 4\theta\) — the \(\frac{1}{4}\) may be missing but inclusion of 4 implies differentiation — and constant to become \(k\theta\) | M1 | Limits not needed |
| \(= \frac{1}{4}a^2\left[\theta - \frac{1}{4}\sin 4\theta\right]\) | A1 | Limits not needed |
| Change integrated function to expression in \(\sin\theta\) and \(\cos\theta\) and substitute limits \(0\) and \(\phi\) | M1dep | Dependent on second M mark of (c) |
| Numerical multiple of \(a^2\) for the area | M1dep | Dependent on all previous M marks of (c) |
| \(\dfrac{1}{36}a^2\left[9\arccos\!\left(\dfrac{1}{\sqrt{3}}\right) + \sqrt{2}\right]\) | A1cso | Given answer — check carefully it can be obtained from previous step. Final 3 marks only awarded if working is shown (i.e. \(\sin 4\theta\) cannot be obtained by calculator) |
## Question 8:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(y =)\, r\sin\theta = a\sin 2\theta\sin\theta$ | M1 | |
| $\frac{dy}{d\theta} = a(2\cos 2\theta\sin\theta + \sin 2\theta\cos\theta)$ | M1depA1 | |
| $\frac{dy}{d\theta} = 2a\sin\theta(\cos 2\theta + \cos^2\theta)$ | M1 | |
| At $P$: $\frac{dy}{d\theta} = 0 \Rightarrow \sin\theta = 0$ (n/a) or $2\cos^2\theta - 1 + \cos^2\theta = 0$, $3\cos^2\theta = 1$ | M1 | $\sin\theta = 0$ not needed |
| $\cos\theta = \frac{1}{\sqrt{3}}$ ※ | A1cso | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $r = a\sin 2\theta = 2a\sin\theta\cos\theta = 2a\sqrt{\left(1-\frac{1}{3}\right)}\sqrt{\frac{1}{3}} = 2a\frac{\sqrt{2}}{3}$ | M1A1 | |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Area $= \int_0^\phi \frac{1}{2}r^2\,d\theta = \frac{1}{2}a^2\int_0^\phi \sin^2 2\theta\,d\theta$ | M1 | |
| $= \frac{1}{2}a^2\int_0^\phi \frac{1}{2}(1-\cos 4\theta)\,d\theta$ | M1 | |
| $= \frac{1}{4}a^2\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^\phi$ | M1A1 | |
| $= \frac{1}{4}a^2\left[\phi - \frac{1}{4}\left(4\sin\phi\cos\phi(2\cos^2\phi - 1)\right)\right]$ | M1dep | Dep on 2nd M mark |
| $= \frac{1}{4}a^2\left[\arccos\!\left(\frac{1}{\sqrt{3}}\right) - \left(\sqrt{\frac{2}{3}}\times\frac{1}{\sqrt{3}}\times\left(\frac{2}{3}-1\right)\right)\right]$ | M1dep | Dep on all Ms |
| $\frac{1}{36}a^2\!\left[9\arccos\!\left(\frac{1}{\sqrt{3}}\right) + \sqrt{2}\right]$ ※ | A1 | |
# Question 8:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain $y$ coordinate: $y = r\sin\theta = a\sin 2\theta \sin\theta$ | M1 | For obtaining the $y$ coordinate |
| Attempt differentiation to obtain $\frac{dy}{d\theta}$ | M1dep | Product rule and/or chain rule must be used; $\sin$ to become $\pm\cos$, $\cos$ to become $\pm\sin$. The 2 may be omitted. Dependent on first M mark |
| $\frac{dy}{d\theta} = a(2\cos 2\theta \sin\theta + \sin 2\theta \cos\theta)$ | A1 | For correct differentiation (oe) |
| Use $\sin 2\theta = 2\sin\theta\cos\theta$ anywhere in solution | M1 | |
| Set $\frac{dy}{d\theta} = 0$ and get a quadratic factor with no $\sin^2\theta$ included | M1 | *Alternative*: Obtain a quadratic in $\sin\theta$ or $\tan\theta$ and complete to $\cos\theta$ later |
| $\cos\theta = \frac{1}{\sqrt{3}}$ or $\cos\phi = \frac{1}{\sqrt{3}}$ | A1cso | |
**Variations:**
| Form | $\frac{dy}{d\theta}$ | Quadratic | Result |
|---|---|---|---|
| $y = a\sin 2\theta\sin\theta$ | $2\cos^2\theta - \sin^2\theta = 0$ | $\tan^2\theta = 2$ | $\cos\theta = \pm\frac{1}{\sqrt{3}}$ |
| $y = 2a\sin^2\theta\cos\theta$ | $3\cos^2\theta - 1 = 0$ | $\cos^2\theta = \frac{1}{3}$ | $\cos\theta = \pm\frac{1}{\sqrt{3}}$ |
| $y = 2a(\cos\theta - \cos^3\theta)$ | $2 - 3\sin^2\theta = 0$ | $\sin^2\theta = \frac{2}{3}$ | $\cos\theta = \pm\frac{1}{\sqrt{3}}$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\sin 2\theta = 2\sin\theta\cos\theta$, $\cos^2\theta + \sin^2\theta = 1$ and $\cos\phi = \frac{1}{\sqrt{3}}$ in $r = a\sin 2\theta$ to obtain numerical multiple of $a$ for $R$ | M1 | Need not be simplified |
| $R = 2a\dfrac{\sqrt{2}}{3}$ | A1cao | Can be done on calculator. Completely correct answer with no working scores 2/2; incorrect answer with no working scores 0/2 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use area formula: $\int_0^\phi \frac{1}{2}r^2\,d\theta = \frac{1}{2}a^2\int_0^\phi \sin^2 2\theta\,d\theta$ | M1 | Limits not needed |
| Prepare $\int \sin^2 2\theta\,d\theta$ using $\cos 2x = 1 - 2\sin^2 x$ | M1 | |
| Attempt integration: $\cos 4\theta$ to become $\pm\sin 4\theta$ — the $\frac{1}{4}$ may be missing but inclusion of 4 implies differentiation — and constant to become $k\theta$ | M1 | Limits not needed |
| $= \frac{1}{4}a^2\left[\theta - \frac{1}{4}\sin 4\theta\right]$ | A1 | Limits not needed |
| Change integrated function to expression in $\sin\theta$ and $\cos\theta$ and substitute limits $0$ and $\phi$ | M1dep | Dependent on second M mark of (c) |
| Numerical multiple of $a^2$ for the area | M1dep | Dependent on all previous M marks of (c) |
| $\dfrac{1}{36}a^2\left[9\arccos\!\left(\dfrac{1}{\sqrt{3}}\right) + \sqrt{2}\right]$ | A1cso | Given answer — check carefully it can be obtained from previous step. Final 3 marks only awarded if working is **shown** (i.e. $\sin 4\theta$ cannot be obtained by calculator) |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6b8b399d-ba16-4fcb-be45-0ba40a7ae09d-13_542_748_205_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a curve $C$ with polar equation $r = a \sin 2 \theta , 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$, and a half-line $l$.\\
The half-line $l$ meets $C$ at the pole $O$ and at the point $P$. The tangent to $C$ at $P$ is parallel to the initial line. The polar coordinates of $P$ are $( R , \phi )$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \phi = \frac { 1 } { \sqrt { 3 } }$
\item Find the exact value of $R$.
The region $S$, shown shaded in Figure 1, is bounded by $C$ and $l$.
\item Use calculus to show that the exact area of $S$ is
$$\frac { 1 } { 36 } a ^ { 2 } \left( 9 \arccos \left( \frac { 1 } { \sqrt { 3 } } \right) + \sqrt { 2 } \right)$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q8 [15]}}