| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Surface area with arc length identity |
| Difficulty | Challenging +1.2 This is a Further Maths FP3 question covering standard parametric curve techniques (arc length, surface of revolution, curvature). Part (i) involves routine differentiation and integration with a helpful simplification shown. Parts (ii-iv) apply standard formulas requiring careful algebraic manipulation but no novel insights. The multi-part structure and FP3 content place it moderately above average A-level difficulty. |
| Spec | 1.07s Parametric and implicit differentiation8.06b Arc length and surface area: of revolution, cartesian or parametric |
## 3(i)
$$x^2 + y^2 = (24t^2)^2 + (18t - 8t^3)^2$$
$$= 576t^4 + 324t^2 - 288t^4 + 64t^6$$
$$= 324t^2 + 288t^4 + 64t^6$$
$$= (18t + 8t^3)^2$$
**B1**
**M1**
**A1 ag**
Arc length is $\int_0^2 (18t + 8t^3) dt$
**M1**
$$= \left[9t^2 + 2t^4\right]_0^2$$
**A1**
$$= 68$$
**A1**
**Note:** $\int_0^2 (18t + 8t^3) dt = \left[18t + 2t^4\right]_0^2 = 68$ earns M1A0A0
**Mark: 6**
## 3(ii)
Curved surface area is $\int 2\pi y \, ds$
**M1**
$$= \int_0^2 2\pi(9t^2 - 2t^4)(18t + 8t^3) dt$$
**M1** - Using $ds = (18t + 8t^3) dt$
**A1** - Correct integral expression including limits (may be implied by later work)
$$= \int_0 \pi(324t^3 + 72t^5 - 32t^7) dt$$
**M1**
$$= \pi \left[8t^4 + 12t^6 - 4t^8\right]_0^2$$
**M1**
$$= 1040\pi \quad (= 3267)$$
**A1**
**Mark: 6**
## 3(iii)
$$\kappa = \frac{\ddot{x}\dot{y} - \dot{x}\ddot{y}}{(x^2 + y^2)^{3/2}} = \frac{(24t^2)(18 - 24t^2) - (48t)(18 - 8t^3)}{(18t + 8t^3)^3}$$
**M1** - Using formula for $\kappa$ (or $\rho$)
**A1 A1** - For numerator and denominator
$$= \frac{48t^2(9 + 4t^2) - 48t^2(9 + 4t^2)}{8t^3(9 + 4t^2)^3} = \frac{-48t^2(9 + 4t^2)}{8t^3(9 + 4t^2)^3}$$
**M1** - Simplifying the numerator
$$= \frac{-6}{t(4t^2 + 9)^2}$$
**A1 ag**
**Mark: 5**
## 3(iv)
When $t = 1$, $x = 8$, $y = 7$, $\kappa = -\frac{6}{169}$
$\rho = (-)\frac{169}{6}$
**B1**
$\frac{dy}{dx} = \frac{y}{x} = \frac{18t - 8t^3}{24t^2} = \frac{10}{24}$
**M1** - Finding gradient (or tangent vector)
**M1**
**A1**
$$\mathbf{n} = \begin{pmatrix} \frac{12}{13} \\ -\frac{5}{13} \end{pmatrix}$$
**M1** - Finding direction of the normal. Correct unit normal (either direction)
$$\mathbf{c} = \begin{pmatrix} 8 \\ 7 \end{pmatrix} + \frac{169}{6} \begin{pmatrix} \frac{12}{13} \\ -\frac{5}{13} \end{pmatrix}$$
**M1**
Centre of curvature is $(18\frac{5}{6}, -19)$
**A1 A1**
**Mark: 7**
---3 The curve $C$ has parametric equations $x = 8 t ^ { 3 } , y = 9 t ^ { 2 } - 2 t ^ { 4 }$, for $t \geqslant 0$.\\
(i) Show that $\dot { x } ^ { 2 } + \dot { y } ^ { 2 } = \left( 18 t + 8 t ^ { 3 } \right) ^ { 2 }$. Find the length of the arc of $C$ for which $0 \leqslant t \leqslant 2$.\\
(ii) Find the area of the surface generated when the arc of $C$ for which $0 \leqslant t \leqslant 2$ is rotated through $2 \pi$ radians about the $x$-axis.\\
(iii) Show that the curvature at a general point on $C$ is $\frac { - 6 } { t \left( 4 t ^ { 2 } + 9 \right) ^ { 2 } }$.\\
(iv) Find the coordinates of the centre of curvature corresponding to the point on $C$ where $t = 1$.
\hfill \mbox{\textit{OCR MEI FP3 2008 Q3 [24]}}