| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find normal equation at parameter |
| Difficulty | Moderate -0.5 This is a straightforward parametric differentiation question requiring the standard formula dy/dx = (dy/dt)/(dx/dt), followed by a routine normal line calculation. The differentiation involves basic exponential and product rules with simple algebraic manipulation. Both parts are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use product rule to differentiate \(y\) | M1 | |
| Obtain correct derivative in any form | A1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}\) | M1 | |
| Obtain given answer correctly | A1 | [4] |
| (ii) Substitute \(t = 0\) in \(\frac{dy}{dx}\) and both parametric equations | B1 | |
| Obtain \(\frac{dy}{dx} = 2\) and coordinates \((1, 0)\) | B1 | |
| Form equation of the normal at their point, using negative reciprocal of their \(\frac{dy}{dx}\) | M1 | |
| State correct equation of normal \(y = -\frac{1}{2}x + \frac{1}{2}\) or equivalent | A1 | [4] |
(i) Use product rule to differentiate $y$ | M1 |
Obtain correct derivative in any form | A1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt}$ | M1 |
Obtain given answer correctly | A1 | [4]
(ii) Substitute $t = 0$ in $\frac{dy}{dx}$ and both parametric equations | B1 |
Obtain $\frac{dy}{dx} = 2$ and coordinates $(1, 0)$ | B1 |
Form equation of the normal at their point, using negative reciprocal of their $\frac{dy}{dx}$ | M1 |
State correct equation of normal $y = -\frac{1}{2}x + \frac{1}{2}$ or equivalent | A1 | [4]5 The parametric equations of a curve are
$$x = \mathrm { e } ^ { 2 t } , \quad y = 4 t \mathrm { e } ^ { t }$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( t + 1 ) } { \mathrm { e } ^ { t } }$.\\
(ii) Find the equation of the normal to the curve at the point where $t = 0$.
\hfill \mbox{\textit{CAIE P2 2013 Q5 [8]}}