| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (sin/cos identities) |
| Difficulty | Standard +0.3 This is a multi-part question requiring standard techniques: expressing trigonometric forms using R cos(θ+α), converting parametric to Cartesian form, and solving a separable differential equation. While lengthy (5 parts), each step uses routine C4 methods with clear guidance ('hence', 'verify', 'deduce'). The connection between the two approaches is elegant but the question scaffolds the solution heavily, 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) etc1.07s Parametric and implicit differentiation1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\cos\theta - \sin\theta \equiv R\cos(\theta+\alpha) \equiv R\cos\theta\cos\alpha - R\sin\theta\sin\alpha\) | M1 | |
| \(\Rightarrow \begin{cases}2 = R\cos\alpha \\ 1 = R\sin\alpha\end{cases} \Rightarrow R^2 = 5\) | A1 | |
| \(R = \sqrt{5}\) (taking \(R>0\)) | ||
| \(\alpha = \cos^{-1}\frac{2}{\sqrt{5}}\) (since both sin and cos are +ve) \(= 26.56...°\) | A1 | |
| \(R\sin(\theta+\alpha) \equiv R\sin\theta\cos\alpha + R\cos\theta\sin\alpha \equiv 2\sin\theta + \cos\theta\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = R\cos(\theta+\alpha)+2,\ y = R\sin(\theta+\alpha)-1\) where \(R\) and \(\alpha\) as above | M1 | |
| \(\Rightarrow (x-2)^2 + (y+1)^2 = R^2 = 5\) | A1 | circle |
| This is a circle of centre \((2,-1)\) and radius \(\sqrt{5}\) | A1 | radius and centre |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(\theta=0\): \(x = 2-0+2 = 4\), \(y = 1+0-1 = 0\), i.e. \((4,0)\) | E1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) | B1 | Stated or implied |
| \(= \frac{-\sin\theta + 2\cos\theta}{-2\sin\theta - \cos\theta}\) | B1, B1 | Numerator; Denominator |
| \(= -\frac{x-2}{y+1}\) | E1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Rightarrow (y+1)\frac{dy}{dx} = -(x-2)\) | M1 | Solving |
| \(\Rightarrow \frac{1}{2}(y+1)^2 = -\frac{1}{2}(x-2)^2 + c\) | A1, M1 | |
| \(x=4,\ y=0\), so \(c = \frac{5}{2}\) | A1 | Finding \(c\) |
| Thus \((y+1)^2 + (x-2)^2 = 5\) | E1 | |
| Total: 5 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\cos\theta - \sin\theta \equiv R\cos(\theta+\alpha) \equiv R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$ | M1 | |
| $\Rightarrow \begin{cases}2 = R\cos\alpha \\ 1 = R\sin\alpha\end{cases} \Rightarrow R^2 = 5$ | A1 | |
| $R = \sqrt{5}$ (taking $R>0$) | | |
| $\alpha = \cos^{-1}\frac{2}{\sqrt{5}}$ (since both sin and cos are +ve) $= 26.56...°$ | A1 | |
| $R\sin(\theta+\alpha) \equiv R\sin\theta\cos\alpha + R\cos\theta\sin\alpha \equiv 2\sin\theta + \cos\theta$ | A1 | |
| **Total: 4** | | |
---
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = R\cos(\theta+\alpha)+2,\ y = R\sin(\theta+\alpha)-1$ where $R$ and $\alpha$ as above | M1 | |
| $\Rightarrow (x-2)^2 + (y+1)^2 = R^2 = 5$ | A1 | circle |
| This is a circle of centre $(2,-1)$ and radius $\sqrt{5}$ | A1 | radius and centre |
| **Total: 3** | | |
---
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $\theta=0$: $x = 2-0+2 = 4$, $y = 1+0-1 = 0$, i.e. $(4,0)$ | E1 | |
| **Total: 1** | | |
---
## Question 9(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ | B1 | Stated or implied |
| $= \frac{-\sin\theta + 2\cos\theta}{-2\sin\theta - \cos\theta}$ | B1, B1 | Numerator; Denominator |
| $= -\frac{x-2}{y+1}$ | E1 | |
| **Total: 4** | | |
---
## Question 9(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Rightarrow (y+1)\frac{dy}{dx} = -(x-2)$ | M1 | Solving |
| $\Rightarrow \frac{1}{2}(y+1)^2 = -\frac{1}{2}(x-2)^2 + c$ | A1, M1 | |
| $x=4,\ y=0$, so $c = \frac{5}{2}$ | A1 | Finding $c$ |
| Thus $(y+1)^2 + (x-2)^2 = 5$ | E1 | |
| **Total: 5** | | |9 Two astronomers wish to model the path of motion of a particle in two dimensions.\\
Experimental results show that the position of the particle can be found using the parametric equations
$$x = 2 \cos \theta - \sin \theta + 2 \quad y = \cos \theta + 2 \sin \theta - 1 \quad \left( 0 \leq \theta \leq 360 ^ { \circ } \right)$$
One astronomer uses trigonometry.\\
(i) Express $2 \cos \theta - \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R$ and $\alpha$ are constants to be determined.
Show also that, for the same values of $R$ and $\alpha$,
$$\cos \theta + 2 \sin \theta = R \sin ( \theta + \alpha )$$
(ii) Hence, or otherwise, show that the path of particle may be written in the form
$$( x - 2 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 5$$
Describe the path of the particle.
The second astronomer sets up a first order differential equation with the condition that $x = 4$ when $y = 0$.\\
(iii) Verify that the point with parameter $\theta = 0$ has coordinates $( 4,0 )$.\\
(iv) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$. Deduce that $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x - 2 } { y + 1 }$$
(v) Solve this differential equation, using the condition that $y = 0$ when $x = 4$.
Hence show that the two solutions give the same cartesian equation for the path of particle.
\hfill \mbox{\textit{OCR MEI C4 Q9 [17]}}