| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Arc length of polar curve |
| Difficulty | Challenging +1.2 This is a standard Further Maths question testing routine application of formulas for arc length in polar coordinates, surface of revolution, and curvature. Part (a) requires the polar arc length formula with a cardioid (standard example), parts (b)(i)-(iii) apply memorized formulas for surface area and curvature with straightforward calculus. While technically demanding with multiple parts, it requires no novel insight—just careful execution of learned techniques, making it moderately above average difficulty. |
| Spec | 4.08d Volumes of revolution: about x and y axes4.09c Area enclosed: by polar curve8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r^2+\left(\dfrac{dr}{d\theta}\right)^2 = a^2(1+\cos\theta)^2+(-a\sin\theta)^2\) | B1 | Condone \(\ldots+(a\sin\theta)^2\); or \(4a^2\cos^4\frac{1}{2}\theta + 4a^2\sin^2\frac{1}{2}\theta\cos^2\frac{1}{2}\theta\) |
| \(= a^2(1+2\cos\theta+\cos^2\theta+\sin^2\theta) = 2a^2(1+\cos\theta)\) | ||
| \(= 4a^2\cos^2\tfrac{1}{2}\theta\) | M1, A1 | Using \(1+\cos\theta = 2\cos^2\tfrac{1}{2}\theta\) |
| \(\text{Arc} = \displaystyle\int\sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta = \int_0^{\frac{1}{2}\pi} 2a\cos\tfrac{1}{2}\theta\,d\theta\) | M1 | For \(\displaystyle\int\sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta\) in terms of \(\theta\); Limits not required |
| \(= \left[4a\sin\tfrac{1}{2}\theta\right]_0^{\frac{1}{2}\pi}\) | A1 | For \(4a\sin\tfrac{1}{2}\theta\) |
| \(= 2\sqrt{2}\,a\) | A1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1+\left(\dfrac{dy}{dx}\right)^2 = 1+\left(\dfrac{x^2}{2}-\dfrac{1}{2x^2}\right)^2\) | B1 | |
| \(= \dfrac{x^4}{4}+\dfrac{1}{2}+\dfrac{1}{4x^4}\) | M1 | |
| \(= \left(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\right)^2\) | A1 | |
| Area \(= \displaystyle\int 2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\) | M1* | |
| \(= \displaystyle\int_1^2 2\pi\left(\dfrac{x^3}{6}+\dfrac{1}{2x}\right)\left(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\right)dx\) | A1 ft | Integral expression including limits |
| \(= 2\pi\displaystyle\int_1^2\left(\dfrac{x^5}{12}+\dfrac{x}{3}+\dfrac{1}{4x^3}\right)dx\) | M1 | Obtaining integrable form; Dep* |
| \(= 2\pi\left[\dfrac{x^6}{72}+\dfrac{x^2}{6}-\dfrac{1}{8x^2}\right]_1^2\) | A1 | Allow one error |
| \(= \dfrac{47\pi}{16}\) | A1 [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = x + \frac{1}{x^3} \left(= \frac{17}{8}\right)\) | B1 | |
| \(\rho = \dfrac{\left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)^3}{x + \dfrac{1}{x^3}}\) | M1, A1ft | Using formula for \(\rho\) or \(\kappa\); Correct expression for \(\rho\) or \(\kappa\) |
| \(= \dfrac{\left(1+\left(\dfrac{15}{8}\right)^2\right)^{\frac{3}{2}}}{2+\dfrac{1}{8}} = \dfrac{\left(\dfrac{17}{8}\right)^3}{\dfrac{17}{8}}\) | A1ft | Correct numerical expression for \(\rho\) |
| \(= \dfrac{289}{64}\) | E1 | Correctly shown |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{15}{8}\), so unit normal is \(\frac{1}{17}\begin{pmatrix}-15\\8\end{pmatrix}\) | M1, A1 | Obtaining a normal vector; Correct unit normal |
| \(\mathbf{c} = \begin{pmatrix}2\\19/12\end{pmatrix} + \frac{289}{64}\begin{pmatrix}-15/17\\8/17\end{pmatrix}\) | M1 | Allow sign errors |
| Centre of curvature is \(\left(-\dfrac{127}{64}, \dfrac{89}{24}\right)\) | A1A1 | |
| [5] |
## Question 3:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r^2+\left(\dfrac{dr}{d\theta}\right)^2 = a^2(1+\cos\theta)^2+(-a\sin\theta)^2$ | B1 | Condone $\ldots+(a\sin\theta)^2$; or $4a^2\cos^4\frac{1}{2}\theta + 4a^2\sin^2\frac{1}{2}\theta\cos^2\frac{1}{2}\theta$ |
| $= a^2(1+2\cos\theta+\cos^2\theta+\sin^2\theta) = 2a^2(1+\cos\theta)$ | | |
| $= 4a^2\cos^2\tfrac{1}{2}\theta$ | M1, A1 | Using $1+\cos\theta = 2\cos^2\tfrac{1}{2}\theta$ |
| $\text{Arc} = \displaystyle\int\sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta = \int_0^{\frac{1}{2}\pi} 2a\cos\tfrac{1}{2}\theta\,d\theta$ | M1 | For $\displaystyle\int\sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta$ in terms of $\theta$; Limits not required |
| $= \left[4a\sin\tfrac{1}{2}\theta\right]_0^{\frac{1}{2}\pi}$ | A1 | For $4a\sin\tfrac{1}{2}\theta$ |
| $= 2\sqrt{2}\,a$ | A1 [6] | |
---
### Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1+\left(\dfrac{dy}{dx}\right)^2 = 1+\left(\dfrac{x^2}{2}-\dfrac{1}{2x^2}\right)^2$ | B1 | |
| $= \dfrac{x^4}{4}+\dfrac{1}{2}+\dfrac{1}{4x^4}$ | M1 | |
| $= \left(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\right)^2$ | A1 | |
| Area $= \displaystyle\int 2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx$ | M1* | |
| $= \displaystyle\int_1^2 2\pi\left(\dfrac{x^3}{6}+\dfrac{1}{2x}\right)\left(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\right)dx$ | A1 ft | Integral expression including limits |
| $= 2\pi\displaystyle\int_1^2\left(\dfrac{x^5}{12}+\dfrac{x}{3}+\dfrac{1}{4x^3}\right)dx$ | M1 | Obtaining integrable form; Dep* |
| $= 2\pi\left[\dfrac{x^6}{72}+\dfrac{x^2}{6}-\dfrac{1}{8x^2}\right]_1^2$ | A1 | Allow one error |
| $= \dfrac{47\pi}{16}$ | A1 [8] | |
## Question 3(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = x + \frac{1}{x^3} \left(= \frac{17}{8}\right)$ | B1 | |
| $\rho = \dfrac{\left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)^3}{x + \dfrac{1}{x^3}}$ | M1, A1ft | Using formula for $\rho$ or $\kappa$; Correct expression for $\rho$ or $\kappa$ |
| $= \dfrac{\left(1+\left(\dfrac{15}{8}\right)^2\right)^{\frac{3}{2}}}{2+\dfrac{1}{8}} = \dfrac{\left(\dfrac{17}{8}\right)^3}{\dfrac{17}{8}}$ | A1ft | Correct numerical expression for $\rho$ |
| $= \dfrac{289}{64}$ | E1 | Correctly shown |
| **[5]** | | |
## Question 3(b)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{15}{8}$, so unit normal is $\frac{1}{17}\begin{pmatrix}-15\\8\end{pmatrix}$ | M1, A1 | Obtaining a normal vector; Correct unit normal | Allow M1 for $\begin{pmatrix}\pm8\\\pm15\end{pmatrix}$ or $\begin{pmatrix}\pm15\\\pm8\end{pmatrix}$ |
| $\mathbf{c} = \begin{pmatrix}2\\19/12\end{pmatrix} + \frac{289}{64}\begin{pmatrix}-15/17\\8/17\end{pmatrix}$ | M1 | Allow sign errors | Must use a unit vector |
| Centre of curvature is $\left(-\dfrac{127}{64}, \dfrac{89}{24}\right)$ | A1A1 | |
| **[5]** | | |3
\begin{enumerate}[label=(\alph*)]
\item Find the length of the arc of the polar curve $r = a ( 1 + \cos \theta )$ for which $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.
\item A curve $C$ has cartesian equation $y = \frac { x ^ { 3 } } { 6 } + \frac { 1 } { 2 x }$.
\begin{enumerate}[label=(\roman*)]
\item The arc of $C$ for which $1 \leqslant x \leqslant 2$ is rotated through $2 \pi$ radians about the $x$-axis to form a surface of revolution. Find the area of this surface.
For the point on $C$ at which $x = 2$,
\item show that the radius of curvature is $\frac { 289 } { 64 }$,
\item find the coordinates of the centre of curvature.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP3 2013 Q3 [24]}}