| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| 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 standard C4 parametric equations question requiring routine differentiation using the chain rule, solving a trigonometric equation, and converting to Cartesian form using the identity sin²θ = x. All techniques are textbook exercises 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 differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{\text{attempt at } \frac{dy}{d\theta}}{\text{attempt at } \frac{dx}{d\theta}}\) but not \(\frac{4 - 3\sin^2\theta}{2\sin\theta}\) | M1 | Alternative: Change to Cartesian form, differentiate and resubstitute; correct differentiation of correct equation |
| \(4\cos\theta - 3\sin^2\theta\cos\theta\) seen | B1 | indep |
| \(\left(\frac{dy}{dx}=\right)\frac{4\cos\theta - 3\sin^2\theta\cos\theta}{2\sin\theta\cos\theta} = \frac{4-3\sin^2\theta}{2\sin\theta}\) AG | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equating given \(\frac{dy}{dx}\) to 2 and producing quadratic equation | M1 | |
| \(\sin\theta = \frac{2}{3}\) | A1 | Ignore any other given value |
| \(P\) is \(\left(\frac{4}{9}, \frac{64}{27}\right)\) | A1 | Accept \(0.444\ldots\) and \(2.37\ldots\) or better |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Identify problem as solving \(4 - 3\sin^2\theta = 0\) | M1 | |
| Show convincingly that \(4 - 3\sin^2\theta = 0\) has no solutions | A1 | Consider magnitude of \(\sin\theta\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to eliminate \(\sin\theta\) from the 2 given equations | M1 | e.g. \(y = 4\sqrt{x} - (\sqrt{x})^3\) |
| Produce \(y^2 = x(4-x)^2\) or \(16x - 8x^2 + x^3\) | A1 ISW | |
| [2] |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{\text{attempt at } \frac{dy}{d\theta}}{\text{attempt at } \frac{dx}{d\theta}}$ but not $\frac{4 - 3\sin^2\theta}{2\sin\theta}$ | M1 | Alternative: Change to Cartesian form, differentiate and resubstitute; correct differentiation of correct equation |
| $4\cos\theta - 3\sin^2\theta\cos\theta$ seen | B1 | indep |
| $\left(\frac{dy}{dx}=\right)\frac{4\cos\theta - 3\sin^2\theta\cos\theta}{2\sin\theta\cos\theta} = \frac{4-3\sin^2\theta}{2\sin\theta}$ **AG** | A1 | |
| **[3]** | | |
---
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equating given $\frac{dy}{dx}$ to 2 and producing quadratic equation | M1 | |
| $\sin\theta = \frac{2}{3}$ | A1 | Ignore any other given value |
| $P$ is $\left(\frac{4}{9}, \frac{64}{27}\right)$ | A1 | Accept $0.444\ldots$ and $2.37\ldots$ or better |
| **[3]** | | |
---
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Identify problem as solving $4 - 3\sin^2\theta = 0$ | M1 | |
| Show convincingly that $4 - 3\sin^2\theta = 0$ has no solutions | A1 | Consider magnitude of $\sin\theta$ |
| **[2]** | | |
---
## Question 8(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to eliminate $\sin\theta$ from the 2 given equations | M1 | e.g. $y = 4\sqrt{x} - (\sqrt{x})^3$ |
| Produce $y^2 = x(4-x)^2$ or $16x - 8x^2 + x^3$ | A1 ISW | |
| **[2]** | | |8 A curve is defined by the parametric equations
$$x = \sin ^ { 2 } \theta , \quad y = 4 \sin \theta - \sin ^ { 3 } \theta ,$$
where $- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 - 3 \sin ^ { 2 } \theta } { 2 \sin \theta }$.\\
(ii) Find the coordinates of the point on the curve at which the gradient is 2 .\\
(iii) Show that the curve has no stationary points.\\
(iv) Find a cartesian equation of the curve, giving your answer in the form $y ^ { 2 } = \mathrm { f } ( x )$.
\hfill \mbox{\textit{OCR C4 2012 Q8 [10]}}