| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 16 |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Standard +0.8 This is a multi-part parametric question requiring: (i) finding parameter limits from a diagram, (ii) chain rule differentiation, (iii) using small angle approximations to solve cos t - t sin t = 0 (non-trivial), and (iv) setting up and evaluating a parametric area integral with double angle identity and integration by parts. The small angle approximation for finding the maximum and the detailed integration push this above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x1.07s Parametric and implicit differentiation1.08f Area between two curves: using integration1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 0 \Rightarrow (t = 0\) or \(\cos t = 0)\) | M1 | Setting \(y = 0\) |
| \(k = \frac{1}{2}\pi\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dt} = \cos t - t \sin t\) | M1 | Attempt at product rule – allow sign errors |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos t - t \sin t = 0 \Rightarrow \left(1 - \frac{1}{2}t^2\right) - t(t) = 0\) | M1* | Setting \(\frac{dy}{dt} = 0\) and substituting small angle approximations for both sine and cosine |
| \(\frac{3}{2}t^2 = 1 \Rightarrow t = \ldots\) | dep*M1 | Simplify and attempt to solve for \(t\) (with correct order of operations) |
| \(t = \sqrt{\frac{2}{3}}\) | A1 | Allow \(\pm\) |
| \(x \approx 0.2\) | A1 | Condone 0.18 |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = \int t\cos t\left(\frac{1}{4}\cos t\right) dt\) | M1 | Attempted use of \(\int y\frac{dx}{dt} dt\) |
| \(= \frac{1}{4}\int t\left(\frac{1}{2}(1 + \cos 2t)\right) dt\) | M1 | Use of \(\cos^2 t = \frac{1}{2}(1 + \cos 2t)\) |
| \(= \frac{1}{8}\int_0^{\frac{1}{4}\pi} t(1 + \cos 2t) dt\) | A1FT | a = \(\frac{1}{2}\pi\) FT their \(k\) from part (i); \(b = \frac{1}{8}\) |
| Answer | Marks | Guidance |
|---|---|---|
| (DR) \(\int t\cos 2t \, dt = \alpha t \sin 2t + \beta \int \sin 2t \, dt\) | M1* | For any non-zero \(\alpha, \beta\) Must be seen |
| \(\int t\cos 2t \, dt = \frac{1}{2}t \sin 2t + \frac{1}{4}\cos 2t\) | A1 | |
| \(\int t \, dt = \frac{1}{2}t^2\) | B1 | |
| \(I = \frac{1}{16}\left[t^2\right]_0^{\frac{1}{4}\pi} + \frac{1}{8}\left[\frac{1}{2}t \sin 2t + \frac{1}{4}\cos 2t\right]_0^{\frac{1}{4}\pi}\) | dep*M1 | Use of 0 and their \(k\) in their integrated expression |
| \(= \frac{1}{64}(\pi^2 - 4)\) | A1 | Or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int t(1 + \cos 2t) dt = t\left(t + \frac{1}{2}\sin 2t\right) - \int\left(t + \frac{1}{2}\sin 2t\right) dt\) | M1* | For attempt at integration by parts |
| A1 | Correct first application | |
| A1 | Complete integration correct | |
| \(I = \left[t\left(t + \frac{1}{2}\sin 2t\right)\right]_0^{\frac{1}{4}\pi} - \left[\frac{1}{2}t^2 - \frac{1}{4}\cos 2t\right]_0^{\frac{1}{4}\pi}\) | dep*M1 | Use of 0 and their \(k\) in their integrated expression |
| \(= \frac{1}{64}(\pi^2 - 4)\) | A1 | Or exact equivalent |
### (i)
$y = 0 \Rightarrow (t = 0$ or $\cos t = 0)$ | M1 | Setting $y = 0$
$k = \frac{1}{2}\pi$ | A1 |
**[2 marks total]**
### (ii)
$\frac{dy}{dt} = \cos t - t \sin t$ | M1 | Attempt at product rule – allow sign errors
| A1 |
**[2 marks total]**
### (iii)
$\cos t - t \sin t = 0 \Rightarrow \left(1 - \frac{1}{2}t^2\right) - t(t) = 0$ | M1* | Setting $\frac{dy}{dt} = 0$ and substituting small angle approximations for both sine and cosine
$\frac{3}{2}t^2 = 1 \Rightarrow t = \ldots$ | dep*M1 | Simplify and attempt to solve for $t$ (with correct order of operations)
$t = \sqrt{\frac{2}{3}}$ | A1 | Allow $\pm$
$x \approx 0.2$ | A1 | Condone 0.18
**[4 marks total]**
### (iv) (a)
$I = \int t\cos t\left(\frac{1}{4}\cos t\right) dt$ | M1 | Attempted use of $\int y\frac{dx}{dt} dt$
$= \frac{1}{4}\int t\left(\frac{1}{2}(1 + \cos 2t)\right) dt$ | M1 | Use of $\cos^2 t = \frac{1}{2}(1 + \cos 2t)$
$= \frac{1}{8}\int_0^{\frac{1}{4}\pi} t(1 + \cos 2t) dt$ | A1FT | a = $\frac{1}{2}\pi$ FT their $k$ from part (i); $b = \frac{1}{8}$
**[3 marks total]**
### (iv) (b)
**(DR)** $\int t\cos 2t \, dt = \alpha t \sin 2t + \beta \int \sin 2t \, dt$ | M1* | For any non-zero $\alpha, \beta$ Must be seen
$\int t\cos 2t \, dt = \frac{1}{2}t \sin 2t + \frac{1}{4}\cos 2t$ | A1 |
$\int t \, dt = \frac{1}{2}t^2$ | B1 |
$I = \frac{1}{16}\left[t^2\right]_0^{\frac{1}{4}\pi} + \frac{1}{8}\left[\frac{1}{2}t \sin 2t + \frac{1}{4}\cos 2t\right]_0^{\frac{1}{4}\pi}$ | dep*M1 | Use of 0 and their $k$ in their integrated expression
$= \frac{1}{64}(\pi^2 - 4)$ | A1 | Or exact equivalent
**Alternative method:**
$\int t(1 + \cos 2t) dt = t\left(t + \frac{1}{2}\sin 2t\right) - \int\left(t + \frac{1}{2}\sin 2t\right) dt$ | M1* | For attempt at integration by parts
| A1 | Correct first application
| A1 | Complete integration correct
$I = \left[t\left(t + \frac{1}{2}\sin 2t\right)\right]_0^{\frac{1}{4}\pi} - \left[\frac{1}{2}t^2 - \frac{1}{4}\cos 2t\right]_0^{\frac{1}{4}\pi}$ | dep*M1 | Use of 0 and their $k$ in their integrated expression
$= \frac{1}{64}(\pi^2 - 4)$ | A1 | Or exact equivalent
**[5 marks total]**
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{28beb431-45d5-4300-88fe-00d05d78790b-06_463_702_264_685}
The diagram shows the curve $C$ with parametric equations
$$x = \frac { 1 } { 4 } \sin t , \quad y = t \cos t$$
where $0 \leqslant t \leqslant k$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $k$.
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} t }$ in terms of $t$.
The maximum point on $C$ is denoted by $P$.
\item Using your answer to part (ii) and the standard small angle approximations, find an approximation for the $x$-coordinate of $P$.
\item (a) Show that the area of the finite region bounded by $C$ and the $x$-axis is given by
$$b \int _ { 0 } ^ { a } t ( 1 + \cos 2 t ) \mathrm { d } t$$
where $a$ and $b$ are constants to be determined.\\
(b) In this question you must show detailed reasoning.
Hence find the exact area of the finite region bounded by $C$ and the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q6 [16]}}