| 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 chain rule (dy/dx = dy/dt รท dx/dt), double angle formula for cos 2t, and algebraic simplification. The derivatives are straightforward, and showing the given result is a routine A-level exercise with clear steps, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07m Tangents and normals: gradient and equations1.07s 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 \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) 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 $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ 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]{bdc467f6-105e-4429-95c6-701eaa43deff-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 are $( 1,4 )$ 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]}}