| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find normal equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dθ ÷ dx/dθ) and compound angle formula to show the given form. Part (ii) requires finding a specific point and normal equation. Standard techniques with no novel insight needed, slightly easier than average due to clear structure and routine methods. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use correct addition formula for either \(\cos(\theta + \frac{1}{6}\pi)\) or, after differentiation, \(\sin(\theta + \frac{1}{6}\pi)\) | B1 | Condone 'missing brackets' |
| Differentiate to obtain \(\frac{dy}{d\theta}\) of form \(k_1\sin\theta + k_2\cos\theta\) or \(k\sin(\theta + \frac{1}{6}\pi)\) | M1 | |
| Divide attempt at \(\frac{dy}{d\theta}\) by attempt at \(\frac{dx}{d\theta}\) | M1 | |
| Obtain \(\dfrac{-\frac{3\sqrt{3}}{2}\sin\theta - \frac{3}{2}\cos\theta}{4\cos\theta}\) or equivalent | A1 | |
| Simplify to obtain \(-\frac{3}{8}(1 + \sqrt{3}\tan\theta)\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identify \(\theta = 0\) | B1 | soi |
| Substitute \(0\) into formula for \(\frac{dy}{dx}\) and take negative reciprocal | M1 | Be implied by \(y = 1 + \dfrac{3\sqrt{3}}{2}\) or \(3.6\); Must be from correct (i) |
| Obtain gradient of normal \(\frac{8}{3}\) | A1 | |
| Form equation of normal through point \(\left(0, 1 + \frac{3\sqrt{3}}{2}\right)\) | M1 | |
| Obtain \(y = \frac{8}{3}x + 1 + \frac{3\sqrt{3}}{2}\) or equivalent | A1 | |
| [5] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use correct addition formula for either $\cos(\theta + \frac{1}{6}\pi)$ or, after differentiation, $\sin(\theta + \frac{1}{6}\pi)$ | **B1** | Condone 'missing brackets' |
| Differentiate to obtain $\frac{dy}{d\theta}$ of form $k_1\sin\theta + k_2\cos\theta$ or $k\sin(\theta + \frac{1}{6}\pi)$ | **M1** | |
| Divide attempt at $\frac{dy}{d\theta}$ by attempt at $\frac{dx}{d\theta}$ | **M1** | |
| Obtain $\dfrac{-\frac{3\sqrt{3}}{2}\sin\theta - \frac{3}{2}\cos\theta}{4\cos\theta}$ or equivalent | **A1** | |
| Simplify to obtain $-\frac{3}{8}(1 + \sqrt{3}\tan\theta)$ | **A1** | |
| | **[5]** | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identify $\theta = 0$ | **B1** | soi |
| Substitute $0$ into formula for $\frac{dy}{dx}$ and take negative reciprocal | **M1** | Be implied by $y = 1 + \dfrac{3\sqrt{3}}{2}$ or $3.6$; Must be from correct **(i)** |
| Obtain gradient of normal $\frac{8}{3}$ | **A1** | |
| Form equation of normal through point $\left(0, 1 + \frac{3\sqrt{3}}{2}\right)$ | **M1** | |
| Obtain $y = \frac{8}{3}x + 1 + \frac{3\sqrt{3}}{2}$ or equivalent | **A1** | |
| | **[5]** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{9bbcee46-c5b8-4836-a4b4-f317bf8b1c0a-3_533_698_735_717}
The diagram shows the curve with parametric equations
$$x = 4 \sin \theta , \quad y = 1 + 3 \cos \left( \theta + \frac { 1 } { 6 } \pi \right)$$
for $0 \leqslant \theta < 2 \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x }$ can be expressed in the form $k ( 1 + ( \sqrt { } 3 ) \tan \theta )$ where the exact value of $k$ is to be determined.\\
(ii) Find the equation of the normal to the curve at the point where the curve crosses the positive $y$-axis. Give your answer in the form $y = m x + c$, where the constants $m$ and $c$ are exact.
\hfill \mbox{\textit{CAIE P2 2016 Q7 [10]}}