| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Sigma Notation Manipulation |
| Difficulty | Hard +2.3 This is a demanding Further Maths question combining intrinsic equations (a specialized topic), differential geometry (curvature), surface of revolution, and envelope theory. Part (a) requires understanding the relationship between arc length, tangent angle, and curvature—topics rarely seen even in Further Maths. Part (b)(ii) on envelopes requires knowledge of a technique beyond standard A-level. The multi-step nature, specialized knowledge requirements, and need for careful geometric reasoning place this well above average difficulty. |
| Spec | 4.08d Volumes of revolution: about x and y axes8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\psi = \frac{1}{3}\pi(1 - e^{-\frac{s}{2}})\) | B1 | |
| Graph: positive increasing gradient through O | B1 | Positive increasing gradient through O |
| Graph: zero gradient at P | B1 | Zero gradient at P |
| Graph: Q marked in first quadrant | B1 | Q marked in first quadrant |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At O \((\psi = \frac{1}{6}\pi)\), \(s = 2\ln 2\) | M1 | |
| At Q \((\psi = \frac{3}{10}\pi)\), \(s = 2\ln 10\) | M1 | |
| Arc length OQ is \(2\ln 10 - 2\ln 2 = 2\ln 5\) | A1 | Or \(\ln 25\) or \(3.22\) (only) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\rho = \frac{ds}{d\psi}\) | M1 | Or \((\kappa =)\frac{d\psi}{ds}\) |
| \(= \frac{6}{\pi - 3\psi}\) | A1 | Or \(\kappa = \frac{\pi}{6}e^{-\frac{s}{2}}\) and \(s = 2\ln 2\) |
| At O \((\psi = \frac{1}{6}\pi)\), radius of curvature is \(\rho = \frac{12}{\pi}\) | A1 | Accept 3.82 |
| [3] | All 3 marks can be awarded in (iv) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Centre of curvature is \((-\rho\sin\psi,\ \rho\cos\psi)\) | M1 | |
| \(\left(-\frac{6}{\pi},\ \frac{6\sqrt{3}}{\pi}\right)\) | A1A1 | FT is \(\left(-\frac{1}{2} |
| [3] | Accept \((-1.91,\ 3.31)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}\right)^2\) | B1 | |
| \(= 1 + \frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1} = \frac{1}{4}x + \frac{1}{2} + \frac{1}{4}x^{-1}\) | M1 | |
| \(= \left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)^2\) | A1 | or \(\frac{(x+1)^2}{4x}\) — Condone correct answer from inaccurate working |
| Area is \(\int 2\pi x\, ds\) | M1 | |
| \(= \int_1^4 2\pi x\left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)dx\) | A1 | Any correct form |
| \(= \pi\left[\frac{2}{5}x^{\frac{5}{2}} + \frac{2}{3}x^{\frac{3}{2}}\right]_1^4\) | A1 | |
| \(= \frac{256}{15}\pi\) | A1 | Exact answer only |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiating partially with respect to \(\lambda\) | M1 | |
| \(0 = \frac{1}{9}x^{\frac{3}{2}} - \frac{2}{9}\lambda x^{\frac{1}{2}}\) | A1 | For RHS |
| \(\lambda = \frac{1}{2}x\), so \(y = \frac{1}{9}(\frac{1}{2}x)x^{\frac{3}{2}} - \frac{1}{9}(\frac{1}{4}x^2)x^{\frac{1}{2}}\) | M1 | Eliminating \(\lambda\) |
| \(y = \frac{1}{36}x^{\frac{5}{2}}\) | A1 | Must be simplified |
| [4] |
## Question 3(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\psi = \frac{1}{3}\pi(1 - e^{-\frac{s}{2}})$ | B1 | |
| Graph: positive increasing gradient through O | B1 | Positive increasing gradient through O |
| Graph: zero gradient at P | B1 | Zero gradient at P |
| Graph: Q marked in first quadrant | B1 | Q marked in first quadrant |
| **[4]** | | |
## Question 3(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At O $(\psi = \frac{1}{6}\pi)$, $s = 2\ln 2$ | M1 | |
| At Q $(\psi = \frac{3}{10}\pi)$, $s = 2\ln 10$ | M1 | |
| Arc length OQ is $2\ln 10 - 2\ln 2 = 2\ln 5$ | A1 | Or $\ln 25$ or $3.22$ (only) |
| **[3]** | | |
## Question 3(a)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\rho = \frac{ds}{d\psi}$ | M1 | Or $(\kappa =)\frac{d\psi}{ds}$ |
| $= \frac{6}{\pi - 3\psi}$ | A1 | Or $\kappa = \frac{\pi}{6}e^{-\frac{s}{2}}$ and $s = 2\ln 2$ |
| At O $(\psi = \frac{1}{6}\pi)$, radius of curvature is $\rho = \frac{12}{\pi}$ | A1 | Accept 3.82 |
| **[3]** | | All 3 marks can be awarded in (iv) |
## Question 3(a)(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre of curvature is $(-\rho\sin\psi,\ \rho\cos\psi)$ | M1 | |
| $\left(-\frac{6}{\pi},\ \frac{6\sqrt{3}}{\pi}\right)$ | A1A1 | FT is $\left(-\frac{1}{2}|\rho|,\ \frac{\sqrt{3}}{2}|\rho|\right)$ |
| **[3]** | | Accept $(-1.91,\ 3.31)$ |
## Question 3(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}\right)^2$ | B1 | |
| $= 1 + \frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1} = \frac{1}{4}x + \frac{1}{2} + \frac{1}{4}x^{-1}$ | M1 | |
| $= \left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)^2$ | A1 | or $\frac{(x+1)^2}{4x}$ — Condone correct answer from inaccurate working |
| Area is $\int 2\pi x\, ds$ | M1 | |
| $= \int_1^4 2\pi x\left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)dx$ | A1 | Any correct form |
| $= \pi\left[\frac{2}{5}x^{\frac{5}{2}} + \frac{2}{3}x^{\frac{3}{2}}\right]_1^4$ | A1 | |
| $= \frac{256}{15}\pi$ | A1 | Exact answer only |
| **[7]** | | |
## Question 3(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiating partially with respect to $\lambda$ | M1 | |
| $0 = \frac{1}{9}x^{\frac{3}{2}} - \frac{2}{9}\lambda x^{\frac{1}{2}}$ | A1 | For RHS |
| $\lambda = \frac{1}{2}x$, so $y = \frac{1}{9}(\frac{1}{2}x)x^{\frac{3}{2}} - \frac{1}{9}(\frac{1}{4}x^2)x^{\frac{1}{2}}$ | M1 | Eliminating $\lambda$ |
| $y = \frac{1}{36}x^{\frac{5}{2}}$ | A1 | Must be simplified |
| **[4]** | | |3
\begin{enumerate}[label=(\alph*)]
\item A curve has intrinsic equation $s = 2 \ln \left( \frac { \pi } { \pi - 3 \psi } \right)$ for $0 \leqslant \psi < \frac { 1 } { 3 } \pi$, where $s$ is the arc length measured from a fixed point P and $\tan \psi = \frac { \mathrm { d } y } { \mathrm {~d} x } . \mathrm { P }$ is in the third quadrant. The curve passes through the origin O , at which point $\psi = \frac { 1 } { 6 } \pi . \mathrm { Q }$ is the point on the curve at which $\psi = \frac { 3 } { 10 } \pi$.
\begin{enumerate}[label=(\roman*)]
\item Express $\psi$ in terms of $s$, and sketch the curve, indicating the points $\mathrm { O } , \mathrm { P }$ and Q .
\item Find the arc length OQ .
\item Find the radius of curvature at the point O .
\item Find the coordinates of the centre of curvature corresponding to the point O .
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the surface area of revolution formed when the curve $y = \frac { 1 } { 3 } \sqrt { x } ( x - 3 )$ for $1 \leqslant x \leqslant 4$ is rotated through $2 \pi$ radians about the $y$-axis.
\item The curve in part (b)(i) is one member of the family $y = \frac { 1 } { 9 } \lambda \sqrt { x } ( x - \lambda )$, where $\lambda$ is a positive parameter. Find the equation of the envelope of this family of curves.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP3 2014 Q3 [24]}}