| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2010 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find stationary points of parametric curve |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt รท dx/dt), basic differentiation of ln and rational functions, and algebraic simplification. Part (ii) adds a simple application by setting the gradient to zero. Slightly easier than average as it's a standard textbook exercise with clear steps and no conceptual surprises. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{dx}{dt} = t-2\) or \(\frac{dy}{dt} = 1 - 9t^{-2}\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain given answer correctly | A1 | [3] |
| (ii) Equate derivative to zero and solve for \(t\) | M1 | |
| State or imply that \(t = 3\) is admissible c.w.o., and note \(t = -3, 2\) cases | A1 | |
| Obtain coordinates \((1, 6)\) and no others | A1 | [3] |
**(i)** State $\frac{dx}{dt} = t-2$ or $\frac{dy}{dt} = 1 - 9t^{-2}$ | B1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain given answer correctly | A1 | [3]
**(ii)** Equate derivative to zero and solve for $t$ | M1 |
State or imply that $t = 3$ is admissible c.w.o., and note $t = -3, 2$ cases | A1 |
Obtain coordinates $(1, 6)$ and no others | A1 | [3]4 The parametric equations of a curve are
$$x = 1 + \ln ( t - 2 ) , \quad y = t + \frac { 9 } { t } , \quad \text { for } t > 2$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \left( t ^ { 2 } - 9 \right) ( t - 2 ) } { t ^ { 2 } }$.\\
(ii) Find the coordinates of the only point on the curve at which the gradient is equal to 0 .
\hfill \mbox{\textit{CAIE P2 2010 Q4 [6]}}