| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dx/dt and dy/dt, then dividing to get dy/dx, followed by routine applications (stationary points where dy/dx=0, gradient evaluation). The algebra involves basic trigonometric identities (double angle formulas) that are standard at this level. While it's a multi-part question worth several marks, each step follows a predictable pattern with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05l Double angle formulae: and compound angle formulae1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Obtain \(12 \sin t \cos t\) or equivalent for \(\frac{dx}{dt}\) | B1 | |
| Obtain \(4 \cos 2t - 6 \sin 2t\) or equivalent for \(\frac{dy}{dt}\) | B1 | |
| Obtain expression for \(\frac{dy}{dx}\) in terms of \(t\). Use \(2 \sin t \cos t = \sin 2t\) | M1 A1 | |
| Confirm given answer \(\frac{dy}{dx} = \frac{2}{3} \cot 2t - 1\) with no errors seen | A1 | [5] |
| (ii) State or imply \(\tan 2t = \frac{2}{3}\) | B1 | |
| Obtain \(t = 0.294\). Obtain \(t = 1.865\) | B1 B1 | [3] |
| (iii) Attempt solution of \(2 \sin 2t + 3 \cos 2t = 0\) at least as far as \(\tan 2t = \ldots\) | M1 | |
| Obtain \(\tan 2t = -\frac{3}{2}\) or equivalent | A1 | |
| Substitute to obtain \(-\frac{13}{9}\) | A1 | [3] |
(i) Obtain $12 \sin t \cos t$ or equivalent for $\frac{dx}{dt}$ | B1 |
Obtain $4 \cos 2t - 6 \sin 2t$ or equivalent for $\frac{dy}{dt}$ | B1 |
Obtain expression for $\frac{dy}{dx}$ in terms of $t$. Use $2 \sin t \cos t = \sin 2t$ | M1 A1 |
Confirm given answer $\frac{dy}{dx} = \frac{2}{3} \cot 2t - 1$ with no errors seen | A1 | [5]
(ii) State or imply $\tan 2t = \frac{2}{3}$ | B1 |
Obtain $t = 0.294$. Obtain $t = 1.865$ | B1 B1 | [3]
(iii) Attempt solution of $2 \sin 2t + 3 \cos 2t = 0$ at least as far as $\tan 2t = \ldots$ | M1 |
Obtain $\tan 2t = -\frac{3}{2}$ or equivalent | A1 |
Substitute to obtain $-\frac{13}{9}$ | A1 | [3]7\\
\includegraphics[max width=\textwidth, alt={}, center]{250b4df9-2646-4246-bb6d-2be92bf29598-3_553_689_258_726}
The parametric equations of a curve are
$$x = 6 \sin ^ { 2 } t , \quad y = 2 \sin 2 t + 3 \cos 2 t$$
for $0 \leqslant t < \pi$. The curve crosses the $x$-axis at points $B$ and $D$ and the stationary points are $A$ and $C$, as shown in the diagram.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { 3 } \cot 2 t - 1$.\\
(ii) Find the values of $t$ at $A$ and $C$, giving each answer correct to 3 decimal places.\\
(iii) Find the value of the gradient of the curve at $B$.
\hfill \mbox{\textit{CAIE P2 2015 Q7 [11]}}