| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show dy/dx simplifies to given form |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: differentiate both parametric equations, apply the chain rule dy/dx = (dy/dt)/(dx/dt), and simplify using trigonometric identities. Part (ii) is routine tangent line application. Slightly above average due to the algebraic manipulation needed, but follows a well-practiced method with no novel insight required. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use chain rule correctly at least once | M1 | |
| Obtain either \(\frac{dx}{dt} = \frac{3\sin t}{\cos^2 t}\) or \(\frac{dy}{dt} = 3\tan^2 \sec^2 t\), or equivalent | A1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain the given answer | A1 | [4] |
| (ii) State a correct equation for the tangent in any form | B1 | |
| Use Pythagoras | M1 | |
| Obtain the given answer | A1 | [3] |
**(i)** Use chain rule correctly at least once | M1 |
Obtain either $\frac{dx}{dt} = \frac{3\sin t}{\cos^2 t}$ or $\frac{dy}{dt} = 3\tan^2 \sec^2 t$, or equivalent | A1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain the given answer | A1 | [4]
**(ii)** State a correct equation for the tangent in any form | B1 |
Use Pythagoras | M1 |
Obtain the given answer | A1 | [3]4 The parametric equations of a curve are
$$x = \frac { 1 } { \cos ^ { 3 } t } , \quad y = \tan ^ { 3 } t$$
where $0 \leqslant t < \frac { 1 } { 2 } \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin t$.\\
(ii) Hence show that the equation of the tangent to the curve at the point with parameter $t$ is $y = x \sin t - \tan t$.
\hfill \mbox{\textit{CAIE P3 2014 Q4 [7]}}