| 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: finding dy/dx using the chain rule (dy/dθ ÷ dx/dθ) and substituting a given value, then eliminating the parameter using the double angle formula sin 2θ = 2sin θ cos θ and the identity cos²θ = 1 - sin²θ. Both parts are routine applications of Core 4 material 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 |
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\cos 2\theta}{\cos\theta}\) | M1 A1 | their \(dy/d\theta\) / their \(dx/d\theta\); correct (can isw) |
| When \(\theta = \pi/6\): \(\frac{dy}{dx} = \frac{2\cos(\pi/3)}{\cos(\pi/6)}\) | DM1 | subst \(\theta = \pi/6\) in theirs |
| \(= \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}\) | A1 | oe exact only (but not \(1/\sqrt{3}/2\)) |
| OR: \(y = 2x\sqrt{1-x^2}\) | M1 | full method for differentiation including product rule and function of a function |
| \(\frac{dy}{dx} = -2x^2(1-x^2)^{-1/2} + 2(1-x^2)^{1/2}\) | A1 | oe cao (condone lack of consideration of sign) |
| at \(\theta = \pi/6\), \(\sin\pi/6 = 1/2\): \(\frac{dy}{dx} = \frac{-2}{4}(1-\frac{1}{4})^{-1/2} + 2(\frac{3}{4})^{1/2} = \frac{2}{\sqrt{3}}\) | DM1, A1 [4] | subst \(\sin\pi/6 = 1/2\); oe exact only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \sin 2\theta = 2\sin\theta\cos\theta\) | M1 | using \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(\Rightarrow y^2 = 4\sin^2\theta\cos^2\theta = 4x^2(1-x^2)\) | M1 | using \(\cos^2\theta = 1 - \sin^2\theta\) to eliminate \(\cos\theta\) |
| \(= 4x^2 - 4x^4\) | A1 [3] | AG — need to see sufficient working or A0 |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\cos 2\theta}{\cos\theta}$ | M1 A1 | their $dy/d\theta$ / their $dx/d\theta$; correct (can isw) |
| When $\theta = \pi/6$: $\frac{dy}{dx} = \frac{2\cos(\pi/3)}{\cos(\pi/6)}$ | DM1 | subst $\theta = \pi/6$ in theirs |
| $= \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$ | A1 | oe exact only (but not $1/\sqrt{3}/2$) |
| **OR:** $y = 2x\sqrt{1-x^2}$ | M1 | full method for differentiation including product rule and function of a function |
| $\frac{dy}{dx} = -2x^2(1-x^2)^{-1/2} + 2(1-x^2)^{1/2}$ | A1 | oe cao (condone lack of consideration of sign) |
| at $\theta = \pi/6$, $\sin\pi/6 = 1/2$: $\frac{dy}{dx} = \frac{-2}{4}(1-\frac{1}{4})^{-1/2} + 2(\frac{3}{4})^{1/2} = \frac{2}{\sqrt{3}}$ | DM1, A1 [4] | subst $\sin\pi/6 = 1/2$; oe exact only |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \sin 2\theta = 2\sin\theta\cos\theta$ | M1 | using $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $\Rightarrow y^2 = 4\sin^2\theta\cos^2\theta = 4x^2(1-x^2)$ | M1 | using $\cos^2\theta = 1 - \sin^2\theta$ to eliminate $\cos\theta$ |
| $= 4x^2 - 4x^4$ | A1 [3] | AG — need to see sufficient working or A0 |
---4 The parametric equations of a curve are
$$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$
(i) Find the exact value of the gradient of the curve at the point where $\theta = \frac { 1 } { 6 } \pi$.\\
(ii) Show that the cartesian equation of the curve is $y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }$.
\hfill \mbox{\textit{OCR MEI C4 Q4 [7]}}