| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Intrinsic equation and curvature |
| Difficulty | Challenging +1.2 This is a standard Further Pure 3 parametric curves question covering routine techniques: finding intersections (algebraic manipulation), radius of curvature (formula application), arc length and surface of revolution (both standard integral setups). While it requires multiple techniques and careful calculation, each part follows well-established procedures without requiring novel insight or particularly challenging integration. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x = 0\), \(t = 0, \pm\frac{1}{\sqrt{3}}\) | E1 | |
| \(y = 1\), \(y = 2\) | B1 | For both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dot{x} = 1-9t^2\), \(\dot{y} = 6t\), \(\ddot{x} = -18t\), \(\ddot{y} = 6\) | B1 | All 4 soi; \(\dot{x}=-2\), \(\dot{y}=\frac{6}{\sqrt{3}}\), \(\ddot{x}=-\frac{18}{\sqrt{3}}\), \(\ddot{y}=6\) |
| \(\rho = \frac{\left((1-9t^2)^2 + 36t^2\right)^{3/2}}{6(1-9t^2)+108t^2} = \frac{(1+9t^2)^3}{6(1+9t^2)} = \frac{(1+9t^2)^2}{6}\) | M1, A1 | Use of formula for \(\rho\) or \(\kappa\); Unsimplified; \(\rho = \frac{(4+12)^{3/2}}{-12+36}\) |
| When \(t = \frac{1}{\sqrt{3}}\), \(\rho = \frac{16}{6} = \frac{8}{3}\) | A1 | |
| \(\tan\psi = \frac{dy}{dx} = \frac{6t}{1-9t^2}\) | M1, A1 | \(\tan\psi = \frac{dy}{dx} = -\sqrt{3}\) |
| \(\sin\psi = \frac{6t}{1+9t^2}\), \(\cos\psi = \frac{1-9t^2}{1+9t^2}\) | M1, A1 | or unit normal is \(\begin{pmatrix}\frac{\sqrt{3}}{2}\\ \frac{1}{2}\end{pmatrix}\); \(\sin\psi = \frac{\sqrt{3}}{2}\), \(\cos\psi = -\frac{1}{2}\) |
| Centre of curvature is at \(\left(0 - \frac{8}{3}\times\frac{\sqrt{3}}{2},\ 2 - \frac{8}{3}\times\frac{1}{2}\right)\) | M1 | |
| i.e. \(\left(-\frac{4\sqrt{3}}{3}, \frac{2}{3}\right)\) | A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dot{x}^2 + \dot{y}^2 = (1-9t^2)^2 + (6t)^2 = (1+9t^2)^2\) | M1A1 | Soi |
| \(s = \int_0^{1/\sqrt{3}} (1+9t^2)\,dt\) | M1 | Limits not required |
| \(= \left[t + 3t^3\right]_0^{1/\sqrt{3}}\) | A1 | Including limits |
| \(= \frac{2}{\sqrt{3}} = \frac{2}{3}\sqrt{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S = 2\pi\int x\,ds = 2\pi\int_0^{1/\sqrt{3}}(t-3t^3)(1+9t^2)\,dt\) | M1, M1, A1 | Correct formula; Integral in terms of \(t\); Including limits |
| \(= 2\pi\int_0^{1/\sqrt{3}}(t + 6t^3 - 27t^5)\,dt = 2\pi\left[\frac{t^2}{2} + \frac{3}{2}t^4 - \frac{9}{2}t^6\right]_0^{1/\sqrt{3}}\) | M1, A1 | Expand and integrate; Including limits |
| \(= 2\pi\left(\frac{1}{6} + \frac{1}{6} - \frac{1}{6}\right) = \frac{\pi}{3}\) | E1 | Intermediate step required |
# Question 3 (part i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x = 0$, $t = 0, \pm\frac{1}{\sqrt{3}}$ | E1 | |
| $y = 1$, $y = 2$ | B1 | For both |
---
# Question 3 (part ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dot{x} = 1-9t^2$, $\dot{y} = 6t$, $\ddot{x} = -18t$, $\ddot{y} = 6$ | B1 | All 4 soi; $\dot{x}=-2$, $\dot{y}=\frac{6}{\sqrt{3}}$, $\ddot{x}=-\frac{18}{\sqrt{3}}$, $\ddot{y}=6$ |
| $\rho = \frac{\left((1-9t^2)^2 + 36t^2\right)^{3/2}}{6(1-9t^2)+108t^2} = \frac{(1+9t^2)^3}{6(1+9t^2)} = \frac{(1+9t^2)^2}{6}$ | M1, A1 | Use of formula for $\rho$ or $\kappa$; Unsimplified; $\rho = \frac{(4+12)^{3/2}}{-12+36}$ |
| When $t = \frac{1}{\sqrt{3}}$, $\rho = \frac{16}{6} = \frac{8}{3}$ | A1 | |
| $\tan\psi = \frac{dy}{dx} = \frac{6t}{1-9t^2}$ | M1, A1 | $\tan\psi = \frac{dy}{dx} = -\sqrt{3}$ |
| $\sin\psi = \frac{6t}{1+9t^2}$, $\cos\psi = \frac{1-9t^2}{1+9t^2}$ | M1, A1 | or unit normal is $\begin{pmatrix}\frac{\sqrt{3}}{2}\\ \frac{1}{2}\end{pmatrix}$; $\sin\psi = \frac{\sqrt{3}}{2}$, $\cos\psi = -\frac{1}{2}$ |
| Centre of curvature is at $\left(0 - \frac{8}{3}\times\frac{\sqrt{3}}{2},\ 2 - \frac{8}{3}\times\frac{1}{2}\right)$ | M1 | |
| i.e. $\left(-\frac{4\sqrt{3}}{3}, \frac{2}{3}\right)$ | A1A1 | |
---
# Question 3 (part iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dot{x}^2 + \dot{y}^2 = (1-9t^2)^2 + (6t)^2 = (1+9t^2)^2$ | M1A1 | Soi |
| $s = \int_0^{1/\sqrt{3}} (1+9t^2)\,dt$ | M1 | Limits not required |
| $= \left[t + 3t^3\right]_0^{1/\sqrt{3}}$ | A1 | Including limits |
| $= \frac{2}{\sqrt{3}} = \frac{2}{3}\sqrt{3}$ | A1 | |
---
# Question 3 (part iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = 2\pi\int x\,ds = 2\pi\int_0^{1/\sqrt{3}}(t-3t^3)(1+9t^2)\,dt$ | M1, M1, A1 | Correct formula; Integral in terms of $t$; Including limits |
| $= 2\pi\int_0^{1/\sqrt{3}}(t + 6t^3 - 27t^5)\,dt = 2\pi\left[\frac{t^2}{2} + \frac{3}{2}t^4 - \frac{9}{2}t^6\right]_0^{1/\sqrt{3}}$ | M1, A1 | Expand and integrate; Including limits |
| $= 2\pi\left(\frac{1}{6} + \frac{1}{6} - \frac{1}{6}\right) = \frac{\pi}{3}$ | E1 | Intermediate step required |
---3 Fig. 3 shows the curve with parametric equations $x = t - 3 t ^ { 3 } , y = 1 + 3 t ^ { 2 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{07eaad51-dc00-44d2-8bff-8652d62902ec-4_634_1294_388_386}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Show that the values of $t$ where the curve cuts the $y$-axis are $t = 0 , \pm \frac { 1 } { \sqrt { 3 } }$. Write down the corresponding values of $y$.\\
(ii) Find the radius and centre of curvature when $t = \frac { 1 } { \sqrt { 3 } }$.
The arc of the curve given by $0 \leqslant t \leqslant \frac { 1 } { \sqrt { 3 } }$ is denoted by $C$.\\
(iii) Find the length of $C$.\\
(iv) Show that the area of the curved surface generated when $C$ is rotated about the $y$-axis through $2 \pi$ radians is $\frac { \pi } { 3 }$.
\hfill \mbox{\textit{OCR MEI FP3 2016 Q3 [24]}}