| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2002 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find dy/dx expression in terms of parameter |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)), finding a tangent line at a given parameter value, and locating/classifying a stationary point. All techniques are routine for P2 level with straightforward algebra, making it slightly easier than average but not trivial due to the logarithmic terms and multi-part nature. |
| 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) State \(\frac{dx}{dt} = 1 + \frac{2}{t}\), \(\frac{dy}{dt} = 2 - \frac{1}{t}\) | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) | M1 | |
| Obtain \(\frac{dy}{dx}\) in any correct form e.g. \(\frac{2t - 1}{t + 2}\) | A1 | 3 |
| (ii) Substitute \(t = 1\) in \(\frac{dy}{dx}\) and both parametric equations | M1 | |
| Obtain \(\frac{dy}{dx} = \frac{1}{3}\) and coordinates \((1, 2)\) | A1✓ | |
| Obtain equation \(3y = x + 5\), or any 3-term equivalent | A1✓ | 3 |
| (iii) Equate \(\frac{dy}{dx}\) to zero and solve for \(t\) | M1 | |
| Obtain answer \(t = \frac{1}{2}\) | A1 | |
| Obtain the given value of \(v\) correctly | A1 | |
| Show by any method that this is a minimum | A1 | 4 |
**(i)** State $\frac{dx}{dt} = 1 + \frac{2}{t}$, $\frac{dy}{dt} = 2 - \frac{1}{t}$ | B1 |
Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 |
Obtain $\frac{dy}{dx}$ in any correct form e.g. $\frac{2t - 1}{t + 2}$ | A1 | 3
**(ii)** Substitute $t = 1$ in $\frac{dy}{dx}$ and both parametric equations | M1 |
Obtain $\frac{dy}{dx} = \frac{1}{3}$ and coordinates $(1, 2)$ | A1✓ |
Obtain equation $3y = x + 5$, or any 3-term equivalent | A1✓ | 3
**(iii)** Equate $\frac{dy}{dx}$ to zero and solve for $t$ | M1 |
Obtain answer $t = \frac{1}{2}$ | A1 |
Obtain the given value of $v$ correctly | A1 |
Show by any method that this is a minimum | A1 | 47 The parametric equations of a curve are
$$x = t + 2 \ln t , \quad y = 2 t - \ln t$$
where $t$ takes all positive values.\\
(i) Express $\frac { d y } { d x }$ in terms of $t$.\\
(ii) Find the equation of the tangent to the curve at the point where $t = 1$.\\
(iii) The curve has one stationary point. Show that the $y$-coordinate of this point is $1 + \ln 2$ and determine whether this point is a maximum or a minimum.
\hfill \mbox{\textit{CAIE P2 2002 Q7 [10]}}