| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Arc length of polar curve |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring multiple advanced techniques (curvature formula, parametric arc length, evolute), but each part follows standard procedures. The curvature calculations are routine applications of formulas, the arc length derivation is straightforward parametric calculus, and the evolute requires systematic elimination. While lengthy and covering FP3 content, it doesn't require novel insights beyond applying learned techniques methodically. |
| 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/Working | Marks | Guidance |
| \(x = a\cos\theta\), \(x' = -a\sin\theta\), \(x'' = -a\cos\theta\); \(y = b\sin\theta\), \(y' = b\cos\theta\), \(y'' = -b\sin\theta\) | B1 | derivatives |
| \(r = \frac{((x')^2 + (y')^2)^{3/2}}{x'y'' - x''y'} = \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{-a\sin\theta \cdot -b\sin\theta - -a\cos\theta \cdot b\cos\theta}\) | M1 | Apply formula (or for \(\kappa\)) |
| \(= \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{ab}\) | M1 | Set \(\theta = 0\) |
| At A, \(\theta = 0 \Rightarrow r = \frac{(a^2\sin^2 0 + b^2\cos^2 0)^{3/2}}{ab}\) | A1 | unsimplified |
| \(= \frac{(b^2)^{3/2}}{ab} = \frac{b^2}{a}\) | A1 | ag |
| The centre is on the \(x\)-axis \(r\) less than \(a\) | M1 | |
| So centre of curvature is at \(\left(a - \frac{b^2}{a}, 0\right)\) i.e. \(\left(\frac{a^2 - b^2}{a}, 0\right)\) | A1 | |
| Total: 7 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Radius \(= \frac{a^2}{b}\) | B1 | |
| Centre is at \(\left(0, b - \frac{a^2}{b}\right)\) i.e. \(\left(0, \frac{b^2 - a^2}{b}\right)\) | B1 | |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{a^2}{b} = \frac{b^2}{a} \Rightarrow a = b\); The ellipse is a circle OR the centre of curvature for both points (and all points) is at \((0,0)\) | B1 | |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = \int_0^{\pi/2} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2}\, d\theta\) | M1 | Applying formula |
| \(= \int_0^{\pi/2} \sqrt{(a\sin\theta)^2 + (b\cos\theta)^2}\, d\theta\) | ||
| \(= \int_0^{\pi/2} \sqrt{a^2\sin^2\theta + b^2\cos^2\theta}\, d\theta\) | A1 | |
| \(= \int_0^{\pi/2} \sqrt{b^2\sin^2\theta + b^2\cos^2\theta + (a^2 - b^2)\sin^2\theta}\, d\theta\) | M1 | Eliminate \(\cos\theta\) |
| \(= \int_0^{\pi/2} \sqrt{b^2 + (a^2 - b^2)\sin^2\theta}\, d\theta\) | A1 | |
| \(= b\int_0^{\pi/2} \sqrt{1 + \frac{(a^2-b^2)}{b^2}\sin^2\theta}\, d\theta\) | ||
| \(\Rightarrow \lambda^2 = \frac{(a^2 - b^2)}{b^2}\) | A1 | |
| When \(a = b\), \(\lambda = 0 \Rightarrow s = b\int_0^{\pi/2} d\theta = \frac{1}{2}b\pi\) | A1 | |
| This is a quarter of the circumference of a circle | A1 | Or arc length of part of circle |
| Total: 7 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre of curvature is at \((a\cos\theta - \rho\sin\psi,\, b\sin\theta + \rho\cos\psi)\) where \(\rho = \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{ab}\) | M1 | Parametric equations; Or find equation of normal |
| \(x = a\cos\theta - \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{ab} \cdot \frac{b\cos\theta}{(a^2\sin^2\theta + b^2\cos^2\theta)^{1/2}}\) | M1, A1 | Deal with \(\psi\); Or partial diffn of normal eqn for \(x\) or \(y\) |
| \(\Rightarrow ax = \cos\theta(a^2 - (a^2\sin^2\theta + b^2\cos^2\theta))\) \(= \cos\theta(a^2 - b^2)\cos^2\theta = \cos^3\theta(a^2 - b^2)\) | A1 | |
| Similarly \(by = \sin^3\theta(b^2 - a^2)\) | A1 | |
| \(\Rightarrow \left(\frac{ax}{a^2-b^2}\right)^{2/3} + \left(\frac{by}{a^2-b^2}\right)^{2/3} = 1\) | M1, A1 | i.e. \((ax)^{2/3} + (by)^{2/3} = (a^2-b^2)^{2/3}\) |
| Total: 7 marks |
# Question 3:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = a\cos\theta$, $x' = -a\sin\theta$, $x'' = -a\cos\theta$; $y = b\sin\theta$, $y' = b\cos\theta$, $y'' = -b\sin\theta$ | B1 | derivatives |
| $r = \frac{((x')^2 + (y')^2)^{3/2}}{x'y'' - x''y'} = \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{-a\sin\theta \cdot -b\sin\theta - -a\cos\theta \cdot b\cos\theta}$ | M1 | Apply formula (or for $\kappa$) |
| $= \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{ab}$ | M1 | Set $\theta = 0$ |
| At A, $\theta = 0 \Rightarrow r = \frac{(a^2\sin^2 0 + b^2\cos^2 0)^{3/2}}{ab}$ | A1 | unsimplified |
| $= \frac{(b^2)^{3/2}}{ab} = \frac{b^2}{a}$ | A1 | ag |
| The centre is on the $x$-axis $r$ less than $a$ | M1 | |
| So centre of curvature is at $\left(a - \frac{b^2}{a}, 0\right)$ i.e. $\left(\frac{a^2 - b^2}{a}, 0\right)$ | A1 | |
| **Total: 7 marks** | | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Radius $= \frac{a^2}{b}$ | B1 | |
| Centre is at $\left(0, b - \frac{a^2}{b}\right)$ i.e. $\left(0, \frac{b^2 - a^2}{b}\right)$ | B1 | |
| **Total: 2 marks** | | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{a^2}{b} = \frac{b^2}{a} \Rightarrow a = b$; The ellipse is a circle OR the centre of curvature for both points (and all points) is at $(0,0)$ | B1 | |
| **Total: 1 mark** | | |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = \int_0^{\pi/2} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2}\, d\theta$ | M1 | Applying formula |
| $= \int_0^{\pi/2} \sqrt{(a\sin\theta)^2 + (b\cos\theta)^2}\, d\theta$ | | |
| $= \int_0^{\pi/2} \sqrt{a^2\sin^2\theta + b^2\cos^2\theta}\, d\theta$ | A1 | |
| $= \int_0^{\pi/2} \sqrt{b^2\sin^2\theta + b^2\cos^2\theta + (a^2 - b^2)\sin^2\theta}\, d\theta$ | M1 | Eliminate $\cos\theta$ |
| $= \int_0^{\pi/2} \sqrt{b^2 + (a^2 - b^2)\sin^2\theta}\, d\theta$ | A1 | |
| $= b\int_0^{\pi/2} \sqrt{1 + \frac{(a^2-b^2)}{b^2}\sin^2\theta}\, d\theta$ | | |
| $\Rightarrow \lambda^2 = \frac{(a^2 - b^2)}{b^2}$ | A1 | |
| When $a = b$, $\lambda = 0 \Rightarrow s = b\int_0^{\pi/2} d\theta = \frac{1}{2}b\pi$ | A1 | |
| This is a quarter of the circumference of a circle | A1 | Or arc length of part of circle |
| **Total: 7 marks** | | |
## Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre of curvature is at $(a\cos\theta - \rho\sin\psi,\, b\sin\theta + \rho\cos\psi)$ where $\rho = \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{ab}$ | M1 | Parametric equations; Or find equation of normal |
| $x = a\cos\theta - \frac{(a^2\sin^2\theta + b^2\cos^2\theta)^{3/2}}{ab} \cdot \frac{b\cos\theta}{(a^2\sin^2\theta + b^2\cos^2\theta)^{1/2}}$ | M1, A1 | Deal with $\psi$; Or partial diffn of normal eqn for $x$ or $y$ |
| $\Rightarrow ax = \cos\theta(a^2 - (a^2\sin^2\theta + b^2\cos^2\theta))$ $= \cos\theta(a^2 - b^2)\cos^2\theta = \cos^3\theta(a^2 - b^2)$ | A1 | |
| Similarly $by = \sin^3\theta(b^2 - a^2)$ | A1 | |
| $\Rightarrow \left(\frac{ax}{a^2-b^2}\right)^{2/3} + \left(\frac{by}{a^2-b^2}\right)^{2/3} = 1$ | M1, A1 | i.e. $(ax)^{2/3} + (by)^{2/3} = (a^2-b^2)^{2/3}$ |
| **Total: 7 marks** | | |
---3 Fig. 3 shows an ellipse with parametric equations $x = a \cos \theta , y = b \sin \theta$, for $0 \leqslant \theta \leqslant 2 \pi$, where $0 < b \leqslant a$.\\
The curve meets the positive $x$-axis at A and the positive $y$-axis at B .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0e032f23-0549-4adc-bfae-59333108fab5-4_668_1255_477_404}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Show that the radius of curvature at A is $\frac { b ^ { 2 } } { a }$ and find the corresponding centre of curvature.\\
(ii) Write down the radius of curvature and the centre of curvature at B .\\
(iii) Find the relationship between $a$ and $b$ if the radius of curvature at B is equal to the radius of curvature at A . What does this mean geometrically?\\
(iv) Show that the arc length from A to B can be expressed as
$$b \int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { 1 + \lambda ^ { 2 } \sin ^ { 2 } \theta } d \theta$$
where $\lambda ^ { 2 }$ is to be determined in terms of $a$ and $b$.\\
Evaluate this integral in the case $a = b$ and comment on your answer.\\
(v) Find the cartesian equation of the evolute of the ellipse.
\hfill \mbox{\textit{OCR MEI FP3 2015 Q3 [24]}}