| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2004 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show dy/dx simplifies to given form |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dx/dt and dy/dt, then computing dy/dx = (dy/dt)/(dx/dt), followed by routine applications (tangent equation and stationary point). The algebra is manageable and all steps follow predictable patterns, making it slightly easier than average for A-level. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State that \(\frac{dx}{dt} = 2 + \frac{1}{t}\) or \(\frac{dy}{dt} = 1 - \frac{4}{t^2}\), or equivalent | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) | M1 | |
| Obtain the given answer | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(t = 1\) in \(\frac{dy}{dx}\) and both parametric equations | M1 | |
| Obtain \(\frac{dy}{dx} = -1\) and coordinates \((2, 5)\) | A1 | |
| State equation of tangent in any correct horizontal form e.g. \(x + y = 7\) | \(\text{A1}\sqrt{}\) | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equate \(\frac{dy}{dx}\) to zero and solve for \(t\) | M1 | |
| Obtain answer \(t = 2\) | A1 | |
| Obtain answer \(y = 4\) | A1 | |
| Show by any method (but not via \(\frac{d}{dt}(y')\)) that this is a minimum point | A1 | Total: 4 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State that $\frac{dx}{dt} = 2 + \frac{1}{t}$ or $\frac{dy}{dt} = 1 - \frac{4}{t^2}$, or equivalent | B1 | |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 | |
| Obtain the given answer | A1 | **Total: 3** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $t = 1$ in $\frac{dy}{dx}$ and both parametric equations | M1 | |
| Obtain $\frac{dy}{dx} = -1$ and coordinates $(2, 5)$ | A1 | |
| State equation of tangent in any correct horizontal form e.g. $x + y = 7$ | $\text{A1}\sqrt{}$ | **Total: 3** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate $\frac{dy}{dx}$ to zero and solve for $t$ | M1 | |
| Obtain answer $t = 2$ | A1 | |
| Obtain answer $y = 4$ | A1 | |
| Show by any method (but not via $\frac{d}{dt}(y')$) that this is a minimum point | A1 | **Total: 4** |
---6 The parametric equations of a curve are
$$x = 2 t + \ln t , \quad y = t + \frac { 4 } { t }$$
where $t$ takes all positive values.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } - 4 } { t ( 2 t + 1 ) }$.\\
(ii) Find the equation of the tangent to the curve at the point where $t = 1$.\\
(iii) The curve has one stationary point. Find the $y$-coordinate of this point, and determine whether this point is a maximum or a minimum.
\hfill \mbox{\textit{CAIE P2 2004 Q6 [10]}}