| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (tan/sec/cot/cosec identities) |
| Difficulty | Standard +0.8 This C4 parametric question requires multiple sophisticated steps: algebraic manipulation with reciprocals, eliminating parameters using a Pythagorean identity (sec²θ + cosec²θ), differentiation of trigonometric functions, and chain rule application. While the techniques are standard C4 content, the multi-part structure and non-obvious algebraic manipulations (particularly finding the Cartesian equation from the given expressions) make this moderately challenging, requiring more insight than routine parametric conversions. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x + \frac{1}{x} = \sec\theta + \tan\theta + \frac{1}{\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}\) | M1 | |
| \(= \frac{\sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta + 1}{\sec\theta + \tan\theta} = \frac{2\sec^2\theta + 2\sec\theta\tan\theta}{\sec\theta + \tan\theta}\) | M1 | |
| \(= \frac{2\sec\theta(\sec\theta + \tan\theta)}{\sec\theta + \tan\theta} = 2\sec\theta\) | M1 A1 | |
| (ii) \(\frac{x^2 + 1}{x} = \frac{2}{\cos\theta} \Rightarrow \cos\theta = \frac{2x}{x^2 + 1}\) | M1 | |
| \(\frac{y^2 + 1}{y} = \frac{2}{\sin\theta} \Rightarrow \sin\theta = \frac{2y}{y^2 + 1}\); \(\therefore \frac{4x^2}{(x^2 + 1)^2} + \frac{4y^2}{(y^2 + 1)^2} = 1\) | M1 A1 | |
| (iii) \(\frac{dv}{d\theta} = \sec\theta\tan\theta + \sec^2\theta = \sec\theta(\tan\theta + \sec\theta) = \frac{x^2 + 1}{2x} \times x = \frac{1}{2}(x^2 + 1)\) | M1 A1 | |
| (iv) \(\frac{dv}{d\theta} = -\cosec\theta\cot\theta - \cosec^2\theta = -\cosec\theta(\cot\theta + \cosec\theta) = -\frac{y^2 + 1}{2y} \times y = -\frac{1}{2}(y^2 + 1)\) | M1 | |
| \(\therefore \frac{dy}{dx} = \frac{y^2 + 1}{x^2 + 1}\) | M1 A1 | (13) |
**(i)** $x + \frac{1}{x} = \sec\theta + \tan\theta + \frac{1}{\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}$ | M1 |
$= \frac{\sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta + 1}{\sec\theta + \tan\theta} = \frac{2\sec^2\theta + 2\sec\theta\tan\theta}{\sec\theta + \tan\theta}$ | M1 |
$= \frac{2\sec\theta(\sec\theta + \tan\theta)}{\sec\theta + \tan\theta} = 2\sec\theta$ | M1 A1 |
**(ii)** $\frac{x^2 + 1}{x} = \frac{2}{\cos\theta} \Rightarrow \cos\theta = \frac{2x}{x^2 + 1}$ | M1 |
$\frac{y^2 + 1}{y} = \frac{2}{\sin\theta} \Rightarrow \sin\theta = \frac{2y}{y^2 + 1}$; $\therefore \frac{4x^2}{(x^2 + 1)^2} + \frac{4y^2}{(y^2 + 1)^2} = 1$ | M1 A1 |
**(iii)** $\frac{dv}{d\theta} = \sec\theta\tan\theta + \sec^2\theta = \sec\theta(\tan\theta + \sec\theta) = \frac{x^2 + 1}{2x} \times x = \frac{1}{2}(x^2 + 1)$ | M1 A1 |
**(iv)** $\frac{dv}{d\theta} = -\cosec\theta\cot\theta - \cosec^2\theta = -\cosec\theta(\cot\theta + \cosec\theta) = -\frac{y^2 + 1}{2y} \times y = -\frac{1}{2}(y^2 + 1)$ | M1 |
$\therefore \frac{dy}{dx} = \frac{y^2 + 1}{x^2 + 1}$ | M1 A1 | **(13)**
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**Total** **(72)**9. A curve has parametric equations
$$x = \sec \theta + \tan \theta , \quad y = \operatorname { cosec } \theta + \cot \theta , \quad 0 < \theta < \frac { \pi } { 2 }$$
(i) Show that $x + \frac { 1 } { x } = 2 \sec \theta$.
Given that $y + \frac { 1 } { y } = 2 \operatorname { cosec } \theta$,\\
(ii) find a cartesian equation for the curve.\\
(iii) Show that $\frac { \mathrm { d } x } { \mathrm {~d} \theta } = \frac { 1 } { 2 } \left( x ^ { 2 } + 1 \right)$.\\
(iv) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
\hfill \mbox{\textit{OCR C4 Q9 [13]}}