| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Moderate -0.3 Part (i) is a standard parametric differentiation exercise requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)) with straightforward derivatives. Part (ii) involves solving a quadratic equation after substituting the gradient. This is routine A-level work with no conceptual challenges, slightly easier than average due to its mechanical nature. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{dx}{dt} = \frac{-2}{1-2t}\) or \(\frac{dy}{dt} = -2t^{-2}\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain given answer correctly | A1 | [3] |
| (ii) Equate derivative to 3 and solve for \(t\) | M1 | |
| State or imply that \(t = -1\) c.w.o. | A1 | |
| Obtain coordinates (ln 3, –2) | A1 | [3] |
(i) State $\frac{dx}{dt} = \frac{-2}{1-2t}$ or $\frac{dy}{dt} = -2t^{-2}$ | B1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain given answer correctly | A1 | [3]
(ii) Equate derivative to 3 and solve for $t$ | M1 |
State or imply that $t = -1$ c.w.o. | A1 |
Obtain coordinates (ln 3, –2) | A1 | [3]4 The parametric equations of a curve are
$$x = \ln ( 1 - 2 t ) , \quad y = \frac { 2 } { t } , \quad \text { for } t < 0$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 t } { t ^ { 2 } }$.\\
(ii) Find the exact coordinates of the only point on the curve at which the gradient is 3 .
\hfill \mbox{\textit{CAIE P2 2012 Q4 [6]}}