| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dy/dx using the chain rule, evaluating at a given point for the tangent, and solving dy/dx = 0 for the stationary point. While it involves multiple steps and some algebraic manipulation (including solving a transcendental equation numerically), these are routine A-level procedures without requiring novel insight or particularly challenging problem-solving. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| Use product rule to differentiate \(y\) obtaining \(k_1e^{2t} + k_2te^{2t}\) | M1 | |
| Obtain correct \(3e^{2t} + 6te^{2t}\) | A1 | |
| State derivative of \(x\) is \(1 + \frac{1}{t+1}\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) with \(t=0\) to find gradient | M1 | |
| Obtain \(y = \frac{3}{2}x\) or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate \(\frac{dy}{dx}\) or \(\frac{dy}{dt}\) to zero and solve for \(t\) | M1 | Allow full marks if correct solution is obtained but \(\frac{dx}{dt}\) is incorrect |
| Obtain \(t = -\frac{1}{2}\) | A1 | |
| Obtain \(x = -1.19\) | A1 | |
| Obtain \(y = -0.55\) | A1 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use product rule to differentiate $y$ obtaining $k_1e^{2t} + k_2te^{2t}$ | M1 | |
| Obtain correct $3e^{2t} + 6te^{2t}$ | A1 | |
| State derivative of $x$ is $1 + \frac{1}{t+1}$ | B1 | |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ with $t=0$ to find gradient | M1 | |
| Obtain $y = \frac{3}{2}x$ or equivalent | A1 | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate $\frac{dy}{dx}$ or $\frac{dy}{dt}$ to zero and solve for $t$ | M1 | Allow full marks if correct solution is obtained but $\frac{dx}{dt}$ is incorrect |
| Obtain $t = -\frac{1}{2}$ | A1 | |
| Obtain $x = -1.19$ | A1 | |
| Obtain $y = -0.55$ | A1 | |
---5 A curve has parametric equations
$$x = t + \ln ( t + 1 ) , \quad y = 3 t \mathrm { e } ^ { 2 t }$$
(i) Find the equation of the tangent to the curve at the origin.\\
(ii) Find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places. [4]\\
\hfill \mbox{\textit{CAIE P2 2018 Q5 [9]}}