| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: finding dy/dx using the chain rule (dy/dθ ÷ dx/dθ), solving a simple trigonometric equation, and eliminating the parameter using sin²θ + cos²θ = 1. All steps are routine C4 material 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 |
| their \(\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\) | M1 | |
| \(\frac{dy}{dx} = \frac{2\sin\theta}{3\cos\theta}\) | A1 | |
| their \(\frac{dy}{dx} = \frac{1}{2}\) | M1 | |
| \(\tan\theta = \frac{3}{4}\) | A1 | If \(\tan\theta = \frac{3}{4}\) not seen, award A1 only if coords are correct |
| \((3.8, -0.6)\) or \(\left(\frac{19}{5}, -\frac{3}{5}\right)\) or \(x = 3.8\), \(y = -0.6\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Manipulating equations into form \(\sin\theta = f(x)\) and \(\cos\theta = g(y)\) and using \(\sin^2\theta + \cos^2\theta = 1\) | M1 | If part (ii) attempted first then part (i): B1 for \(\frac{dy}{dx} = \frac{4(x-2)}{9(y-1)}\); M1 equating to \(\frac{1}{2}\); A1 for \(9y - 8x = -7\); M1 eliminating \(x\) or \(y\); A1 for \((3.8, -0.6)\) |
| \(\frac{(x-2)^2}{9} + \frac{(1-y)^2}{4} = 1\) oe | A1 | Accept e.g. \(\left(\frac{x-2}{3}\right)^2\); and their Cartesian equation |
| \(4x^2 + 9y^2 - 16x - 18y - 11 = 0\) | ||
| [2] |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| their $\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$ | M1 | |
| $\frac{dy}{dx} = \frac{2\sin\theta}{3\cos\theta}$ | A1 | |
| their $\frac{dy}{dx} = \frac{1}{2}$ | M1 | |
| $\tan\theta = \frac{3}{4}$ | A1 | If $\tan\theta = \frac{3}{4}$ not seen, award A1 only if coords are correct |
| $(3.8, -0.6)$ or $\left(\frac{19}{5}, -\frac{3}{5}\right)$ or $x = 3.8$, $y = -0.6$ | A1 | |
| **[5]** | | |
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# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Manipulating equations into form $\sin\theta = f(x)$ and $\cos\theta = g(y)$ and using $\sin^2\theta + \cos^2\theta = 1$ | M1 | If part (ii) attempted first then part (i): B1 for $\frac{dy}{dx} = \frac{4(x-2)}{9(y-1)}$; M1 equating to $\frac{1}{2}$; A1 for $9y - 8x = -7$; M1 eliminating $x$ or $y$; A1 for $(3.8, -0.6)$ |
| $\frac{(x-2)^2}{9} + \frac{(1-y)^2}{4} = 1$ oe | A1 | Accept e.g. $\left(\frac{x-2}{3}\right)^2$; and their Cartesian equation |
| $4x^2 + 9y^2 - 16x - 18y - 11 = 0$ | | |
| **[2]** | | |
---5 The parametric equations of a curve are
$$x = 2 + 3 \sin \theta \text { and } y = 1 - 2 \cos \theta \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$$
(i) Find the coordinates of the point on the curve where the gradient is $\frac { 1 } { 2 }$.\\
(ii) Find the cartesian equation of the curve.
\hfill \mbox{\textit{OCR C4 2013 Q5 [7]}}