| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: Cartesian curve |
| Difficulty | Challenging +1.8 This FP3 question involves multiple advanced techniques: arc length requiring algebraic manipulation of the derivative, surface area of revolution (a less common topic), centre of curvature (requiring second derivatives and a complex formula), and finding an envelope (a sophisticated Further Maths concept). While each part follows standard procedures, the combination of four distinct advanced topics, the algebraic complexity, and especially the envelope calculation (requiring parametric elimination) place this significantly above average A-level difficulty but within reach for well-prepared FP3 students. |
| 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 |
| \(\frac{dy}{dx} = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{1/2}\) | B1 | |
| \(1+\left(\frac{dy}{dx}\right)^2 = 1+(\frac{1}{2}x^{-1/2}-\frac{1}{2}x^{1/2})^2\) | M1 | |
| \(= 1+\frac{1}{4}x^{-1}-\frac{1}{2}+\frac{1}{4}x = \frac{1}{4}x^{-1}+\frac{1}{2}+\frac{1}{4}x\) | ||
| \(= (\frac{1}{2}x^{-1/2}+\frac{1}{2}x^{1/2})^2\) | A1 | |
| Arc length is \(\int_0^a (\frac{1}{2}x^{-1/2}+\frac{1}{2}x^{1/2})\,dx\) | ||
| \(= \left[x^{1/2}+\frac{1}{3}x^{3/2}\right]_0^a\) | M1 | |
| \(= a^{1/2}+\frac{1}{3}a^{3/2}\) | A1 (ag) | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curved surface area is \(\int 2\pi y\,ds\) | M1 | For \(\int y\,ds\) |
| \(= \int_0^3 2\pi(x^{1/2}-\frac{1}{3}x^{3/2})(\frac{1}{2}x^{-1/2}+\frac{1}{2}x^{1/2})\,dx\) | A1 | Correct integral form including limits |
| \(= 2\pi\int_0^3 (\frac{1}{2}+\frac{1}{3}x-\frac{1}{6}x^2)\,dx\) | ||
| \(= 2\pi\left[\frac{1}{2}x+\frac{1}{6}x^2-\frac{1}{18}x^3\right]_0^3\) | M1A1 | For \(\frac{1}{2}x+\frac{1}{6}x^2-\frac{1}{18}x^3\) |
| \(= 3\pi\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x=4\), \(\frac{dy}{dx}=-\frac{3}{4}\) | B1 | |
| Unit normal vector is \(\begin{pmatrix}-\frac{3}{5}\\-\frac{4}{5}\end{pmatrix}\) | M1, A1 ft | Finding a normal vector; correct unit normal (either direction) |
| \(\frac{d^2y}{dx^2}=-\frac{1}{4}x^{-3/2}-\frac{1}{4}x^{-1/2}\) \(\left(=-\frac{5}{32}\right)\) | B1 | |
| \(\rho = \frac{\{1+(-\frac{3}{4})^2\}^{3/2}}{(-)\frac{5}{32}}\) \(\left(=\frac{\frac{125}{64}}{\frac{5}{32}}=\frac{25}{2}\right)\) | M1, A1 ft | Applying formula for \(\rho\) or \(\kappa\) |
| \(\mathbf{c} = \begin{pmatrix}4\\-\frac{2}{3}\end{pmatrix}+\frac{25}{2}\begin{pmatrix}-\frac{3}{5}\\-\frac{4}{5}\end{pmatrix}\) | M1 | |
| \(= \begin{pmatrix}-3\frac{1}{2}\\-10\frac{2}{3}\end{pmatrix}\) | A1, A1 | [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiating partially w.r.t. \(p\): \(0=2px^{1/2}-p^2x^{3/2}\) | M1, A1 | |
| \(p=\frac{2}{x}\) | ||
| Envelope is \(y=\frac{4}{x^2}x^{1/2}-\frac{1}{3}\frac{8}{x^3}x^{3/2}\) | M1, A1 | |
| \(y=\frac{4}{3}x^{-3/2}\) | A1 | [5] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{1/2}$ | B1 | |
| $1+\left(\frac{dy}{dx}\right)^2 = 1+(\frac{1}{2}x^{-1/2}-\frac{1}{2}x^{1/2})^2$ | M1 | |
| $= 1+\frac{1}{4}x^{-1}-\frac{1}{2}+\frac{1}{4}x = \frac{1}{4}x^{-1}+\frac{1}{2}+\frac{1}{4}x$ | | |
| $= (\frac{1}{2}x^{-1/2}+\frac{1}{2}x^{1/2})^2$ | A1 | |
| Arc length is $\int_0^a (\frac{1}{2}x^{-1/2}+\frac{1}{2}x^{1/2})\,dx$ | | |
| $= \left[x^{1/2}+\frac{1}{3}x^{3/2}\right]_0^a$ | M1 | |
| $= a^{1/2}+\frac{1}{3}a^{3/2}$ | A1 (ag) | **[5]** |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curved surface area is $\int 2\pi y\,ds$ | M1 | For $\int y\,ds$ |
| $= \int_0^3 2\pi(x^{1/2}-\frac{1}{3}x^{3/2})(\frac{1}{2}x^{-1/2}+\frac{1}{2}x^{1/2})\,dx$ | A1 | Correct integral form including limits |
| $= 2\pi\int_0^3 (\frac{1}{2}+\frac{1}{3}x-\frac{1}{6}x^2)\,dx$ | | |
| $= 2\pi\left[\frac{1}{2}x+\frac{1}{6}x^2-\frac{1}{18}x^3\right]_0^3$ | M1A1 | For $\frac{1}{2}x+\frac{1}{6}x^2-\frac{1}{18}x^3$ |
| $= 3\pi$ | A1 | **[5]** |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=4$, $\frac{dy}{dx}=-\frac{3}{4}$ | B1 | |
| Unit normal vector is $\begin{pmatrix}-\frac{3}{5}\\-\frac{4}{5}\end{pmatrix}$ | M1, A1 ft | Finding a normal vector; correct unit normal (either direction) |
| $\frac{d^2y}{dx^2}=-\frac{1}{4}x^{-3/2}-\frac{1}{4}x^{-1/2}$ $\left(=-\frac{5}{32}\right)$ | B1 | |
| $\rho = \frac{\{1+(-\frac{3}{4})^2\}^{3/2}}{(-)\frac{5}{32}}$ $\left(=\frac{\frac{125}{64}}{\frac{5}{32}}=\frac{25}{2}\right)$ | M1, A1 ft | Applying formula for $\rho$ or $\kappa$ |
| $\mathbf{c} = \begin{pmatrix}4\\-\frac{2}{3}\end{pmatrix}+\frac{25}{2}\begin{pmatrix}-\frac{3}{5}\\-\frac{4}{5}\end{pmatrix}$ | M1 | |
| $= \begin{pmatrix}-3\frac{1}{2}\\-10\frac{2}{3}\end{pmatrix}$ | A1, A1 | **[9]** |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiating partially w.r.t. $p$: $0=2px^{1/2}-p^2x^{3/2}$ | M1, A1 | |
| $p=\frac{2}{x}$ | | |
| Envelope is $y=\frac{4}{x^2}x^{1/2}-\frac{1}{3}\frac{8}{x^3}x^{3/2}$ | M1, A1 | |
| $y=\frac{4}{3}x^{-3/2}$ | A1 | **[5]** |
---3 A curve $C$ has equation $y = x ^ { \frac { 1 } { 2 } } - \frac { 1 } { 3 } x ^ { \frac { 3 } { 2 } }$, for $x \geqslant 0$.\\
(i) Show that the arc of $C$ for which $0 \leqslant x \leqslant a$ has length $a ^ { \frac { 1 } { 2 } } + \frac { 1 } { 3 } a ^ { \frac { 3 } { 2 } }$.\\
(ii) Find the area of the surface generated when the arc of $C$ for which $0 \leqslant x \leqslant 3$ is rotated through $2 \pi$ radians about the $x$-axis.\\
(iii) Find the coordinates of the centre of curvature corresponding to the point $\left( 4 , - \frac { 2 } { 3 } \right)$ on $C$.
The curve $C$ is one member of the family of curves defined by
$$y = p ^ { 2 } x ^ { \frac { 1 } { 2 } } - \frac { 1 } { 3 } p ^ { 3 } x ^ { \frac { 3 } { 2 } } \quad ( \text { for } x \geqslant 0 )$$
where $p$ is a parameter (and $p > 0$ ).\\
(iv) Find the equation of the envelope of this family of curves.
\hfill \mbox{\textit{OCR MEI FP3 2010 Q3 [24]}}