| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Gradient condition leads to trig equation |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question with routine techniques: finding stationary points by setting dx/dθ=0, computing dy/dx using the chain rule, and using R-cos(θ+α) form to solve a trigonometric equation. Each part follows textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| At A, \(y = 0\): \(x\)-coord of \(A = 2\times\frac{\pi}{2} - \sin\frac{\pi}{2} = \pi - 1\) | B1 | for either A or B/C |
| \(x\)-coord of \(B = 2\pi - \sin\pi = 2\pi\) | B1 | for both A and B/C |
| \(OA = \pi - 1\), \(AC = 2\pi - \pi + 1 = \pi + 1\) | M1 | |
| \(\theta = 0\), ratio is \((\pi-1):(\pi+1)\) | A1 | |
| \(= \frac{\pi}{2}\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{d\theta} = -4\sin\theta\) | B1 | either \(dx/d\theta\) or \(dy/d\theta\) |
| \(\frac{dx}{d\theta} = 2 - \cos\theta\) | ||
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{4\sin\theta}{2-\cos\theta}\) | M1 A1 | |
| At A, gradient \(= -\frac{4\sin(\pi/2)}{2-\cos(\pi/2)} = -2\) | A1 | www |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 1 \Rightarrow -\frac{4\sin\theta}{2-\cos\theta} = 1\) | M1 | their \(dy/dx = 1\) |
| \(-4\sin\theta = 2 - \cos\theta\) | ||
| \(\cos\theta - 4\sin\theta = 2\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\theta - 4\sin\theta = R\cos(\theta + \alpha) = R(\cos\theta\cos\alpha - \sin\theta\sin\alpha)\) | M1 | correct pairs |
| \(R\cos\alpha = 1\), \(R\sin\alpha = 4\) | B1 | |
| \(R^2 = 1^2 + 4^2 = 17\), \(R = \sqrt{17}\) | M1 A1 | accept \(76.0°\), \(1.33\) radians |
| \(\tan\alpha = 4\), \(\alpha = 1.326\) | ||
| \(\sqrt{17}\cos(\theta + 1.326) = 2\) | ||
| \(\cos(\theta + 1.326) = \frac{2}{\sqrt{17}}\) | M1 | inv cos \(\frac{2}{\sqrt{17}}\) ft their \(R\) for method |
| \(\theta + 1.326 = 1.064, 5.219, 7.348\) | ||
| \(\theta = (-0.262), 3.89, 6.02\) | A1 A1 | \(-1\) extra solutions in the range |
## Question 4(i):
At A, $y = 0$: $x$-coord of $A = 2\times\frac{\pi}{2} - \sin\frac{\pi}{2} = \pi - 1$ | B1 | for either A or B/C |
$x$-coord of $B = 2\pi - \sin\pi = 2\pi$ | B1 | for both A and B/C |
$OA = \pi - 1$, $AC = 2\pi - \pi + 1 = \pi + 1$ | M1 | |
$\theta = 0$, ratio is $(\pi-1):(\pi+1)$ | A1 | |
$= \frac{\pi}{2}$ | E1 | |
## Question 4(ii):
$\frac{dy}{d\theta} = -4\sin\theta$ | B1 | either $dx/d\theta$ or $dy/d\theta$ |
$\frac{dx}{d\theta} = 2 - \cos\theta$ | | |
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{4\sin\theta}{2-\cos\theta}$ | M1 A1 | |
At A, gradient $= -\frac{4\sin(\pi/2)}{2-\cos(\pi/2)} = -2$ | A1 | www |
## Question 4(iii):
$\frac{dy}{dx} = 1 \Rightarrow -\frac{4\sin\theta}{2-\cos\theta} = 1$ | M1 | their $dy/dx = 1$ |
$-4\sin\theta = 2 - \cos\theta$ | | |
$\cos\theta - 4\sin\theta = 2$ | E1 | |
## Question 4(iv):
$\cos\theta - 4\sin\theta = R\cos(\theta + \alpha) = R(\cos\theta\cos\alpha - \sin\theta\sin\alpha)$ | M1 | correct pairs |
$R\cos\alpha = 1$, $R\sin\alpha = 4$ | B1 | |
$R^2 = 1^2 + 4^2 = 17$, $R = \sqrt{17}$ | M1 A1 | accept $76.0°$, $1.33$ radians |
$\tan\alpha = 4$, $\alpha = 1.326$ | | |
$\sqrt{17}\cos(\theta + 1.326) = 2$ | | |
$\cos(\theta + 1.326) = \frac{2}{\sqrt{17}}$ | M1 | inv cos $\frac{2}{\sqrt{17}}$ ft their $R$ for method |
$\theta + 1.326 = 1.064, 5.219, 7.348$ | | |
$\theta = (-0.262), 3.89, 6.02$ | A1 A1 | $-1$ extra solutions in the range |
---4 Part of the track of a roller-coaster is modelled by a curve with the parametric equations
$$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$
This is shown in Fig. 8. B is a minimum point, and BC is vertical.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9ac55ae6-7a7f-47d0-a363-92da179be4ca-3_591_1437_433_391}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the values of the parameter at A and B .
Hence show that the ratio of the lengths OA and AC is $( \pi - 1 ) : ( \pi + 1 )$.\\
(ii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$. Find the gradient of the track at A .\\
(iii) Show that, when the gradient of the track is $1 , \theta$ satisfies the equation
$$\cos \theta - 4 \sin \theta = 2$$
(iv) Express $\cos \theta - 4 \sin \theta$ in the form $R \cos ( \theta + \alpha )$.
Hence solve the equation $\cos \theta - 4 \sin \theta = 2$ for $0 \leqslant \theta \leqslant 2 \pi$.
\hfill \mbox{\textit{OCR MEI C4 Q4 [18]}}