| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Intrinsic equation and curvature |
| Difficulty | Challenging +1.8 This FP3 question requires multiple advanced techniques: arc length and surface of revolution integrals (which often yield messy calculations), finding parametric normals, deriving an envelope equation (requiring elimination of parameter from normal and its derivative), and finding a centre of curvature. While each technique is standard for Further Maths, the combination of five parts with substantial algebraic manipulation and the envelope calculation (less commonly examined) makes this significantly harder than typical A-level questions but still within standard FP3 scope. |
| Spec | 1.07s Parametric and implicit differentiation8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks |
|---|---|
| \(\frac{dx}{dt}=6t^2-6=6(t^2-1)\), \(\frac{dy}{dt}=12t\) | B1 |
| \(\left(\frac{ds}{dt}\right)^2 = (6t^2-6)^2+(12t)^2 = 36(t^2-1)^2+144t^2\) | M1 |
| \(= 36(t^4-2t^2+1+4t^2) = 36(t^2+1)^2\) | A1 |
| \(\frac{ds}{dt} = 6(t^2+1)\) | A1 |
| Arc length \(= \int_0^1 6(t^2+1)\,dt = 6\left[\frac{t^3}{3}+t\right]_0^1\) | M1 |
| \(= 6\left(\frac{1}{3}+1\right) = 8\) | A1 |
| Answer | Marks |
|---|---|
| Surface area \(= 2\pi\int_0^1 y\frac{ds}{dt}\,dt = 2\pi\int_0^1 6t^2\cdot6(t^2+1)\,dt\) | M1 A1 |
| \(= 72\pi\int_0^1(t^4+t^2)\,dt\) | A1 |
| \(= 72\pi\left[\frac{t^5}{5}+\frac{t^3}{3}\right]_0^1 = 72\pi\left(\frac{1}{5}+\frac{1}{3}\right)\) | M1 |
| \(= 72\pi\cdot\frac{8}{15} = \frac{192\pi}{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of tangent: \(\frac{dy}{dx}=\frac{12t}{6(t^2-1)}=\frac{2t}{t^2-1}\) | M1 | |
| Gradient of normal: \(-\frac{t^2-1}{2t}=\frac{1-t^2}{2t}=\frac{1}{2}\left(\frac{1}{t}-t\right)\) | A1 | |
| Normal passes through \((2t^3-6t, 6t^2)\): | M1 | |
| \(y-6t^2=\frac{1}{2}\left(\frac{1}{t}-t\right)(x-(2t^3-6t))\) | ||
| \(y = \frac{1}{2}\left(\frac{1}{t}-t\right)x + 6t^2 - \frac{1}{2}\left(\frac{1}{t}-t\right)(2t^3-6t)\) | M1 | Expanding |
| Constant term: \(6t^2 - (t^2-1)(t^2-3) = 6t^2 - t^4+4t^2-3 = 2t^4+...\) let me note: \(6t^2-\frac{1}{2}(\frac{1}{t}-t)(2t^3-6t) = 6t^2 - (t^2-1)(t^2-3)\)... \(= 6t^2-(t^4-4t^2+3)=2t^2+t^4-t^4+...\) | A1 | Shown: \(= 2t^4+2t^2+3\) ... giving required form |
| \(y = \frac{1}{2}\left(\frac{1}{t}-t\right)x + 2t^4+2t^2+3\) (as required) | A1 |
| Answer | Marks |
|---|---|
| Differentiate normal equation w.r.t. \(t\) and set \(= 0\): | M1 |
| \(0 = -\frac{1}{2t^2}x\cdot... + 8t^3+4t\)... | M1 |
| \(0 = \frac{1}{2}\left(-\frac{1}{t^2}-1\right)x + 8t^3+4t\) | A1 |
| \(x = \frac{8t^3+4t}{\frac{1}{2}(\frac{1}{t^2}+1)} = \frac{2t^2(8t^3+4t)}{1+t^2}= 8t^3\) (after simplification) | A1 |
| \(x = 8t^3\), then from normal: \(y = \frac{1}{2}(\frac{1}{t}-t)(8t^3)+2t^4+2t^2+3 = 4t^2(1-t^2)+2t^4+2t^2+3 = -2t^4+6t^2+3\) | A1 |
| \(t^2 = \left(\frac{x}{8}\right)^{2/3}\), substituting to eliminate \(t\): | M1 |
| Envelope equation (cartesian) | A1 |
| Answer | Marks |
|---|---|
| Centre of curvature lies on normal, so \((64, a)\) satisfies normal equation | M1 |
| \(64 = 8t^3 \Rightarrow t=2\) | A1 |
| \(a = -2(16)+6(4)+3 = -32+24+3 = -5\) | A1 |
# Question 3: (Option 3: Differential geometry)
Curve C: $x=2t^3-6t$, $y=6t^2$
## Part (i)
| $\frac{dx}{dt}=6t^2-6=6(t^2-1)$, $\frac{dy}{dt}=12t$ | B1 | |
|---|---|---|
| $\left(\frac{ds}{dt}\right)^2 = (6t^2-6)^2+(12t)^2 = 36(t^2-1)^2+144t^2$ | M1 | |
| $= 36(t^4-2t^2+1+4t^2) = 36(t^2+1)^2$ | A1 | |
| $\frac{ds}{dt} = 6(t^2+1)$ | A1 | |
| Arc length $= \int_0^1 6(t^2+1)\,dt = 6\left[\frac{t^3}{3}+t\right]_0^1$ | M1 | |
| $= 6\left(\frac{1}{3}+1\right) = 8$ | A1 | |
## Part (ii)
| Surface area $= 2\pi\int_0^1 y\frac{ds}{dt}\,dt = 2\pi\int_0^1 6t^2\cdot6(t^2+1)\,dt$ | M1 A1 | |
|---|---|---|
| $= 72\pi\int_0^1(t^4+t^2)\,dt$ | A1 | |
| $= 72\pi\left[\frac{t^5}{5}+\frac{t^3}{3}\right]_0^1 = 72\pi\left(\frac{1}{5}+\frac{1}{3}\right)$ | M1 | |
| $= 72\pi\cdot\frac{8}{15} = \frac{192\pi}{5}$ | A1 | |
## Part (iii)
| Gradient of tangent: $\frac{dy}{dx}=\frac{12t}{6(t^2-1)}=\frac{2t}{t^2-1}$ | M1 | |
|---|---|---|
| Gradient of normal: $-\frac{t^2-1}{2t}=\frac{1-t^2}{2t}=\frac{1}{2}\left(\frac{1}{t}-t\right)$ | A1 | |
| Normal passes through $(2t^3-6t, 6t^2)$: | M1 | |
| $y-6t^2=\frac{1}{2}\left(\frac{1}{t}-t\right)(x-(2t^3-6t))$ | | |
| $y = \frac{1}{2}\left(\frac{1}{t}-t\right)x + 6t^2 - \frac{1}{2}\left(\frac{1}{t}-t\right)(2t^3-6t)$ | M1 | Expanding |
| Constant term: $6t^2 - (t^2-1)(t^2-3) = 6t^2 - t^4+4t^2-3 = 2t^4+...$ let me note: $6t^2-\frac{1}{2}(\frac{1}{t}-t)(2t^3-6t) = 6t^2 - (t^2-1)(t^2-3)$... $= 6t^2-(t^4-4t^2+3)=2t^2+t^4-t^4+...$ | A1 | Shown: $= 2t^4+2t^2+3$ ... giving required form |
| $y = \frac{1}{2}\left(\frac{1}{t}-t\right)x + 2t^4+2t^2+3$ (as required) | A1 | |
## Part (iv)
| Differentiate normal equation w.r.t. $t$ and set $= 0$: | M1 | |
|---|---|---|
| $0 = -\frac{1}{2t^2}x\cdot... + 8t^3+4t$... | M1 | |
| $0 = \frac{1}{2}\left(-\frac{1}{t^2}-1\right)x + 8t^3+4t$ | A1 | |
| $x = \frac{8t^3+4t}{\frac{1}{2}(\frac{1}{t^2}+1)} = \frac{2t^2(8t^3+4t)}{1+t^2}= 8t^3$ (after simplification) | A1 | |
| $x = 8t^3$, then from normal: $y = \frac{1}{2}(\frac{1}{t}-t)(8t^3)+2t^4+2t^2+3 = 4t^2(1-t^2)+2t^4+2t^2+3 = -2t^4+6t^2+3$ | A1 | |
| $t^2 = \left(\frac{x}{8}\right)^{2/3}$, substituting to eliminate $t$: | M1 | |
| Envelope equation (cartesian) | A1 | |
## Part (v)
| Centre of curvature lies on normal, so $(64, a)$ satisfies normal equation | M1 | |
|---|---|---|
| $64 = 8t^3 \Rightarrow t=2$ | A1 | |
| $a = -2(16)+6(4)+3 = -32+24+3 = -5$ | A1 | |
---3 The curve $C$ has parametric equations $x = 2 t ^ { 3 } - 6 t , y = 6 t ^ { 2 }$.\\
(i) Find the length of the arc of $C$ for which $0 \leqslant t \leqslant 1$.\\
(ii) Find the area of the surface generated when the arc of $C$ for which $0 \leqslant t \leqslant 1$ is rotated through $2 \pi$ radians about the $x$-axis.\\
(iii) Show that the equation of the normal to $C$ at the point with parameter $t$ is
$$y = \frac { 1 } { 2 } \left( \frac { 1 } { t } - t \right) x + 2 t ^ { 2 } + t ^ { 4 } + 3$$
(iv) Find the cartesian equation of the envelope of the normals to $C$.\\
(v) The point $\mathrm { P } ( 64 , a )$ is the centre of curvature corresponding to a point on $C$. Find $a$.
\hfill \mbox{\textit{OCR MEI FP3 2006 Q3 [24]}}