| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| 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 standard parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)), followed by routine applications. Part (i) involves differentiating trigonometric functions and simplifying using double angle formulas—straightforward for P2 level. Parts (ii) and (iii) are direct applications of the derivative. Slightly above average due to the algebraic manipulation needed, but follows standard textbook methods with no novel insight required. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(\frac{dx}{dt} = 2\sin 2t\) | B1 | |
| Obtain \(\frac{dy}{dt} = 6\sin^2 t\cos t - 9\cos^2 t\sin t\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \Big/ \frac{dx}{dt}\) for their first derivatives | M1 | |
| Use identity \(\sin 2t = 2\sin t\cos t\) | B1 | |
| Simplify to obtain \(\frac{3}{2}\sin t - \frac{9}{4}\cos t\) with necessary detail present | A1 | |
| Total: | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate \(\frac{dy}{dx}\) to zero and obtain \(\tan t = k\) | M1 | |
| Obtain \(\tan t = \frac{3}{2}\) or equivalent | A1 | |
| Substitute value of \(t\) to obtain coordinates \((2.38,\ 2.66)\) | A1 | |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Identify \(t = \frac{1}{4}\pi\) | B1 | |
| Substitute to obtain exact value for gradient of the normal | M1 | |
| Obtain gradient \(\frac{4}{3}\sqrt{2}\), \(\dfrac{8}{3\sqrt{2}}\) or similarly simplified exact equivalent | A1 | |
| Total: | 3 |
## Question 8(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\frac{dx}{dt} = 2\sin 2t$ | B1 | |
| Obtain $\frac{dy}{dt} = 6\sin^2 t\cos t - 9\cos^2 t\sin t$ | B1 | |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \Big/ \frac{dx}{dt}$ for their first derivatives | M1 | |
| Use identity $\sin 2t = 2\sin t\cos t$ | B1 | |
| Simplify to obtain $\frac{3}{2}\sin t - \frac{9}{4}\cos t$ with necessary detail present | A1 | |
| **Total:** | **5** | |
## Question 8(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate $\frac{dy}{dx}$ to zero and obtain $\tan t = k$ | M1 | |
| Obtain $\tan t = \frac{3}{2}$ or equivalent | A1 | |
| Substitute value of $t$ to obtain coordinates $(2.38,\ 2.66)$ | A1 | |
| **Total:** | **3** | |
## Question 8(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Identify $t = \frac{1}{4}\pi$ | B1 | |
| Substitute to obtain exact value for gradient of the normal | M1 | |
| Obtain gradient $\frac{4}{3}\sqrt{2}$, $\dfrac{8}{3\sqrt{2}}$ or similarly simplified exact equivalent | A1 | |
| **Total:** | **3** | |8\\
\includegraphics[max width=\textwidth, alt={}, center]{de2f8bf3-fd03-4199-9eb2-c9cbac4d4385-10_549_495_258_824}
The diagram shows the curve with parametric equations
$$x = 2 - \cos 2 t , \quad y = 2 \sin ^ { 3 } t + 3 \cos ^ { 3 } t + 1$$
for $0 \leqslant t \leqslant \frac { 1 } { 2 } \pi$. The end-points of the curve $( 1,4 )$ and $( 3,3 )$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 2 } \sin t - \frac { 9 } { 4 } \cos t$.\\
(ii) Find the coordinates of the minimum point, giving each coordinate correct to 3 significant figures.\\
(iii) Find the exact gradient of the normal to the curve at the point for which $x = 2$.\\
\hfill \mbox{\textit{CAIE P2 2017 Q8 [11]}}