| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2009 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring standard application of dy/dx = (dy/dt)/(dx/dt), followed by solving a simple exponential equation. The derivatives are elementary (exponentials), the algebra is clean, and part (ii) is routine substitution and solving. Slightly easier than average due to the mechanical nature and lack of conceptual challenges. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| State \(\frac{dx}{dt} = e^t\) or \(\frac{dy}{dt} = e^t - e^{-t}\) | B1 |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 |
| Obtain given answer correctly | A1 |
| Answer | Marks |
|---|---|
| Substitute \(\frac{dy}{dx} = 2\) and use correct method for solving an equation of the form \(e^{2t} = a\), where \(a > 0\) | M1 |
| Obtain answer \(t = \frac{1}{2} \ln 3\), or equivalent | A1 |
**(i)**
| State $\frac{dx}{dt} = e^t$ or $\frac{dy}{dt} = e^t - e^{-t}$ | B1 |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
| Obtain given answer correctly | A1 |
**Total: [3]**
**(ii)**
| Substitute $\frac{dy}{dx} = 2$ and use correct method for solving an equation of the form $e^{2t} = a$, where $a > 0$ | M1 |
| Obtain answer $t = \frac{1}{2} \ln 3$, or equivalent | A1 |
**Total: [2]**
---4 The parametric equations of a curve are
$$x = 1 - \mathrm { e } ^ { - t } , \quad y = \mathrm { e } ^ { t } + \mathrm { e } ^ { - t }$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 t } - 1$.\\
(ii) Hence find the exact value of $t$ at the point on the curve at which the gradient is 2 .
\hfill \mbox{\textit{CAIE P2 2009 Q4 [5]}}