| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Properties of specific curves |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: finding dy/dx from parametric form, eliminating the parameter to get Cartesian form, and computing a volume of revolution. The angle relationship requires some trigonometric manipulation but follows directly from the gradients. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation4.08d Volumes of revolution: about x and y axes |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4}{4t} = \frac{1}{t}\) | M1 (their \(dy/dt\) ÷ \(dx/dt\); accept \(\frac{4}{4t}\) here) |
| gradient of tangent \(= \tan\theta\) | A1 (ag – need reference to gradient is \(\tan\theta\)) |
| \(\tan\theta = \frac{1}{t}\) | A1 |
| Answer | Marks |
|---|---|
| Gradient of \(QP = \frac{4t - 2t}{2t^2 - t^2 - 1} = \frac{2t}{t^2 - 1}\) | M1 (correct method for subtracting co-ordinates) |
| \(\frac{2t}{t^2 - 1}\) | A1 (correct; does not need to be cancelled) |
| \(\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}\) | M1 (either substituting \(t = 1/\tan\theta\) in above expression or substituting \(\tan\theta = 1/t\) in double angle formula for \(\tan 2\theta\)) |
| \(= \frac{2/t}{1 - 1/t^2} = \frac{2t}{t^2 - 1}\) | A1 (showing expressions are equal) |
| \(\tan\phi = \tan 2\theta\) | A1 (ag) |
| \(\phi = 2\theta\) | A1 |
| Angle \(QPR = 180° - 2\theta\) | M1 (supplementary angles or angles on a straight line oe) |
| \(\angle TPQ + 180° - 2\theta + \theta = 180°\) | M1 |
| \(\angle TPQ = \theta\) | A1 (ag) |
| Answer | Marks |
|---|---|
| \(t = \frac{y}{4}\) | M1 (eliminating \(t\) from parametric equation) |
| \(x = 2\left(\frac{y}{4}\right)^2 = \frac{y^2}{8}\) | A1 |
| \(y^2 = 8x\) | A1 (ag) |
| When \(t = 2\): \(x = 2(2)^2 = 8\), \(y = 4(2) = 8\) | B1 (for M1 allow no limits or their limits) |
| \(V = \int_0^8 \pi y^2 \, dx = \int_0^8 8\pi x \, dx\) | M1 (need correct limits but they may appear later) |
| \(= \left[4\pi x^2\right]_0^8\) | A1 (for \(4\pi x^2\) – ignore incorrect or missing limits) |
| \(= 256\pi\) | A1 (in terms of \(\pi\) only; allow SC B1 for omission of \(\pi\) throughout integral but otherwise correct) |
## (i)
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4}{4t} = \frac{1}{t}$ | M1 (their $dy/dt$ ÷ $dx/dt$; accept $\frac{4}{4t}$ here)
gradient of tangent $= \tan\theta$ | A1 (ag – need reference to gradient is $\tan\theta$)
$\tan\theta = \frac{1}{t}$ | A1
[3]
## (ii)
Gradient of $QP = \frac{4t - 2t}{2t^2 - t^2 - 1} = \frac{2t}{t^2 - 1}$ | M1 (correct method for subtracting co-ordinates)
$\frac{2t}{t^2 - 1}$ | A1 (correct; does not need to be cancelled)
$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$ | M1 (either substituting $t = 1/\tan\theta$ in above expression or substituting $\tan\theta = 1/t$ in double angle formula for $\tan 2\theta$)
$= \frac{2/t}{1 - 1/t^2} = \frac{2t}{t^2 - 1}$ | A1 (showing expressions are equal)
$\tan\phi = \tan 2\theta$ | A1 (ag)
$\phi = 2\theta$ | A1
Angle $QPR = 180° - 2\theta$ | M1 (supplementary angles or angles on a straight line oe)
$\angle TPQ + 180° - 2\theta + \theta = 180°$ | M1
$\angle TPQ = \theta$ | A1 (ag)
[8]
## (iii)
$t = \frac{y}{4}$ | M1 (eliminating $t$ from parametric equation)
$x = 2\left(\frac{y}{4}\right)^2 = \frac{y^2}{8}$ | A1
$y^2 = 8x$ | A1 (ag)
When $t = 2$: $x = 2(2)^2 = 8$, $y = 4(2) = 8$ | B1 (for M1 allow no limits or their limits)
$V = \int_0^8 \pi y^2 \, dx = \int_0^8 8\pi x \, dx$ | M1 (need correct limits but they may appear later)
$= \left[4\pi x^2\right]_0^8$ | A1 (for $4\pi x^2$ – ignore incorrect or missing limits)
$= 256\pi$ | A1 (in terms of $\pi$ only; allow SC B1 for omission of $\pi$ throughout integral but otherwise correct)
[7]1 Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve
$$x = 2 t ^ { 2 } , y = 4 t , \quad - \sqrt { 2 } \leqslant t \leqslant \sqrt { 2 } .$$
$\mathrm { P } \left( 2 t ^ { 2 } , 4 t \right)$ is a point on the curve with parameter $t$. TS is the tangent to the curve at P , and PR is the line through P parallel to the $x$-axis. Q is the point $( 2,0 )$. The angles that PS and QP make with the positive $x$-direction are $\theta$ and $\phi$ respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-1_962_1248_673_420}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) By considering the gradient of the tangent TS, show that $\tan \theta = \frac { 1 } { t }$.\\
(ii) Find the gradient of the line QP in terms of $t$. Hence show that $\phi = 2 \theta$, and that angle TPQ is equal to $\theta$.\\[0pt]
[The above result shows that if a lamp bulb is placed at Q , then the light from the bulb is reflected to produce a parallel beam of light.]
The inside surface of the headlight has the shape produced by rotating the curve about the $x$-axis.\\
(iii) Show that the curve has cartesian equation $y ^ { 2 } = 8 x$. Hence find the volume of revolution of the curve, giving your answer as a multiple of $\pi$.
\hfill \mbox{\textit{OCR MEI C4 Q1 [18]}}