| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring standard techniques: substituting a given parameter value, applying the chain rule (dy/dx = (dy/dθ)/(dx/dθ)), and using a double angle identity to eliminate the parameter. All steps are routine for C4 level with no problem-solving 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 |
| When \(\theta = \pi/3\): \(x = 2\sin(\pi/3) = \sqrt{3}\) | B1 | exact only (isw all decimal answers following exact answer) |
| \(y = \cos(2\pi/3) = -\frac{1}{2}\) | B1 | |
| \(\dfrac{dx}{d\theta} = 2\cos\theta\), \(\dfrac{dy}{d\theta} = -2\sin 2\theta\) | M1 | \(dy/dx = (dy/d\theta)/(dx/d\theta)\) used; ft their derivatives if right way up (condone one further minor slip if intention clear); condone poor notation |
| \(\Rightarrow \dfrac{dy}{dx} = \dfrac{-\sin 2\theta}{\cos\theta}\) | A1 | any correct equivalent form; can isw if incorrect simplification |
| EITHER: \(= \dfrac{-\sin(2\pi/3)}{\cos(\pi/3)} = \dfrac{-\sqrt{3}/2}{1/2} = -\sqrt{3}\) | A1 | exact www |
| OR: expressing \(y\) in terms of \(x\): \(y = 1 - x^2/2\), \(\dfrac{dy}{dx} = -x\) or \(-2\sin\theta = -\sqrt{3}\) | M1, A1, A1 | exact www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 1 - 2\sin^2\theta = 1 - 2(x/2)^2 = 1 - \frac{1}{2}x^2\) | M1A1 | for M1, need correct trig identity and attempt to substitute for \(x\); or reference to (i) if used there; allow SC B1 for \(y = \cos(2\arcsin(x/2))\) or equivalent |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $\theta = \pi/3$: $x = 2\sin(\pi/3) = \sqrt{3}$ | B1 | exact only (isw all decimal answers following exact answer) |
| $y = \cos(2\pi/3) = -\frac{1}{2}$ | B1 | |
| $\dfrac{dx}{d\theta} = 2\cos\theta$, $\dfrac{dy}{d\theta} = -2\sin 2\theta$ | M1 | $dy/dx = (dy/d\theta)/(dx/d\theta)$ used; ft their derivatives if right way up (condone one further minor slip if intention clear); condone poor notation |
| $\Rightarrow \dfrac{dy}{dx} = \dfrac{-\sin 2\theta}{\cos\theta}$ | A1 | any correct equivalent form; can isw if incorrect simplification |
| **EITHER:** $= \dfrac{-\sin(2\pi/3)}{\cos(\pi/3)} = \dfrac{-\sqrt{3}/2}{1/2} = -\sqrt{3}$ | A1 | exact www |
| **OR:** expressing $y$ in terms of $x$: $y = 1 - x^2/2$, $\dfrac{dy}{dx} = -x$ or $-2\sin\theta = -\sqrt{3}$ | M1, A1, A1 | exact www |
**[5]**
---
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 1 - 2\sin^2\theta = 1 - 2(x/2)^2 = 1 - \frac{1}{2}x^2$ | M1A1 | for M1, need correct trig identity and attempt to substitute for $x$; or reference to (i) if used there; allow SC B1 for $y = \cos(2\arcsin(x/2))$ or equivalent |
**[2]**
---3 A curve has parametric equations
$$x = 2 \sin \theta , \quad y = \cos 2 \theta$$
(i) Find the exact coordinates and the gradient of the curve at the point with parameter $\theta = \frac { 1 } { 3 } \pi$.\\
(ii) Find $y$ in terms of $x$.
\hfill \mbox{\textit{OCR MEI C4 Q3 [7]}}