| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Moderate -0.3 This is a standard C4 parametric differentiation question requiring routine application of the chain rule (dy/dx = (dy/dθ)/(dx/dθ)) and product rule. Finding coordinates from parametric equations and writing a tangent equation are basic skills. The algebra is straightforward with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\theta = 0\) at the origin | B1 | For the correct value |
| \(A\) is \((0, a\pi)\) | B1 | For the correct y-coordinate at \(A\) |
| \(B\) is \((a, 0)\) | B1 | For the correct x-coordinate at \(B\) |
| 3 | ||
| (ii) \(\frac{dx}{d\theta} = a\cos\theta\) | B1 | For correct differentiation of \(x\) |
| \(\frac{dy}{d\theta} = a(\cos\theta - \theta\sin\theta)\) | M1 | For differentiating \(y\) using product rule |
| \(\frac{dy}{dx} = \frac{\cos\theta - \theta\sin\theta}{\cos\theta} = 1 - \theta\tan\theta\) | M1 | For use of \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) |
| Gradient of tangent at the origin is \(1\) | A1 | For given result correctly obtained |
| Hence equation is \(y = x\) | M1 | For using \(\theta = 0\) |
| A1 | For correct equation | |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(L_1: \mathbf{r} = 3\mathbf{i} + 6\mathbf{j} + k + s(2\mathbf{i} + 3\mathbf{j} - \mathbf{k})\) | M1 | For correct RHS structure for either line |
| \(L_2: \mathbf{r} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k} + t(i - 2\mathbf{j} + \mathbf{k})\) | A1 | For both lines correct |
| 2 | ||
| (ii) \(3 + 2s = 3 + t, 6 + 3s = -1 - 2t, 1 - s = 4 + t\) | M1 | For at least 2 equations with two parameters |
| First pair of equations give \(s = -1, t = -2\) | M1 | For solving any relevant pair of equations |
| Third equation checks: \(1 + 1 = 4 - 2\) | A1 | For explicit check in unused equation |
| Point of intersection is \((1, 3, 2)\) | A1 | For correct coordinates |
| 5 | ||
| (iii) \(2 \times 1 + 3(-2) + (-1) \times 1 = (\sqrt{14})(\sqrt{6})\cos\theta\) | B1 | For scalar product of correct direction vectors |
| B1 | For correct magnitudes \(\sqrt{14}\) and \(\sqrt{6}\) | |
| M1 | For correct process for \(\cos\theta\) with any pair of vectors relevant to these lines | |
| Hence acute angle is \(56.9°\) | A1 | For correct acute angle |
| 4 |
**(i)** $\theta = 0$ at the origin | B1 | For the correct value
$A$ is $(0, a\pi)$ | B1 | For the correct y-coordinate at $A$
$B$ is $(a, 0)$ | B1 | For the correct x-coordinate at $B$
| **3** |
**(ii)** $\frac{dx}{d\theta} = a\cos\theta$ | B1 | For correct differentiation of $x$
$\frac{dy}{d\theta} = a(\cos\theta - \theta\sin\theta)$ | M1 | For differentiating $y$ using product rule
$\frac{dy}{dx} = \frac{\cos\theta - \theta\sin\theta}{\cos\theta} = 1 - \theta\tan\theta$ | M1 | For use of $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$
Gradient of tangent at the origin is $1$ | A1 | For given result correctly obtained
Hence equation is $y = x$ | M1 | For using $\theta = 0$
| A1 | For correct equation
| **6** |
## Question 6 (Second question):
**(i)** $L_1: \mathbf{r} = 3\mathbf{i} + 6\mathbf{j} + k + s(2\mathbf{i} + 3\mathbf{j} - \mathbf{k})$ | M1 | For correct RHS structure for either line
$L_2: \mathbf{r} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k} + t(i - 2\mathbf{j} + \mathbf{k})$ | A1 | For both lines correct
| **2** |
**(ii)** $3 + 2s = 3 + t, 6 + 3s = -1 - 2t, 1 - s = 4 + t$ | M1 | For at least 2 equations with two parameters
First pair of equations give $s = -1, t = -2$ | M1 | For solving any relevant pair of equations
Third equation checks: $1 + 1 = 4 - 2$ | A1 | For explicit check in unused equation
Point of intersection is $(1, 3, 2)$ | A1 | For correct coordinates
| **5** |
**(iii)** $2 \times 1 + 3(-2) + (-1) \times 1 = (\sqrt{14})(\sqrt{6})\cos\theta$ | B1 | For scalar product of correct direction vectors
| B1 | For correct magnitudes $\sqrt{14}$ and $\sqrt{6}$
| M1 | For correct process for $\cos\theta$ with any pair of vectors relevant to these lines
Hence acute angle is $56.9°$ | A1 | For correct acute angle
| **4** |6\\
\includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-3_766_611_251_703}
The diagram shows the curve with parametric equations
$$x = a \sin \theta , \quad y = a \theta \cos \theta$$
where $a$ is a positive constant and $- \pi \leqslant \theta \leqslant \pi$. The curve meets the positive $y$-axis at $A$ and the positive $x$-axis at $B$.\\
(i) Write down the value of $\theta$ corresponding to the origin, and state the coordinates of $A$ and $B$.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - \theta \tan \theta$, and hence find the equation of the tangent to the curve at the origin.
\hfill \mbox{\textit{OCR C4 Q6 [9]}}