| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Moderate -0.8 This is a straightforward parametric equations question requiring only routine techniques: substitution for part (a), simple algebraic manipulation using y=1/t to eliminate the parameter in part (b), and standard implicit differentiation or dy/dx = (dy/dt)/(dx/dt) for part (c). All steps are mechanical with no problem-solving insight needed. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(t = \frac{1}{2}\), \(x = 3\), \(y = 2\) | B1B1 | |
| (b) \(t = \frac{1}{y}\) | M1 | Attempt to eliminate \(t\) |
| \(x = 2 + \frac{1}{y} + y\) | m1 | SC verification using \(t = \frac{1}{2}\) |
| \(xy = 2 + y^2\) or \(xy - y^2 = 2\) | A1 | AG; convincingly found |
| (c) Implicit differentiation: \(x\frac{dy}{dx} + y, -2y\frac{dy}{dx} = 0\) | M1, A1A1 | Attempt at equation with \(\frac{dy}{dx}\) but not "\(\frac{dy}{du} = ...\)" |
| At \((3, 2)\): \(3\frac{dy}{dx} + 2 - 4\frac{dy}{dx} = 0\) | B1, m1 | RHS = 0; Use of \((3, 2)\) |
| \(\frac{dy}{dx} = 2\) | A1 | AG; convincingly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Parametric differentiation: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{t^2} \div \left(2 - \frac{1}{t^2}\right)\) | (M1), (A1A1) | Attempt chain rule, PI |
| \(\frac{1}{t^2} = y^2\) | (B1) | Or sub \(t = \frac{1}{2}\); -4 in numerator is sufficient for this |
| \(y = 2\), \(\frac{dy}{dx} = \frac{-4}{2-4} = 2\) | M1A1 | AG; convincingly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = y + 2y^{-1}\) | M1 | \(\frac{dx}{dy} = 1 - 2y^{-2}\) |
**(a)** $t = \frac{1}{2}$, $x = 3$, $y = 2$ | B1B1 |
**(b)** $t = \frac{1}{y}$ | M1 | Attempt to eliminate $t$
$x = 2 + \frac{1}{y} + y$ | m1 | SC verification using $t = \frac{1}{2}$
$xy = 2 + y^2$ or $xy - y^2 = 2$ | A1 | AG; convincingly found
**(c)** Implicit differentiation: $x\frac{dy}{dx} + y, -2y\frac{dy}{dx} = 0$ | M1, A1A1 | Attempt at equation with $\frac{dy}{dx}$ but not "$\frac{dy}{du} = ...$"
At $(3, 2)$: $3\frac{dy}{dx} + 2 - 4\frac{dy}{dx} = 0$ | B1, m1 | RHS = 0; Use of $(3, 2)$
$\frac{dy}{dx} = 2$ | A1 | AG; convincingly obtained
**OR**
Parametric differentiation: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{t^2} \div \left(2 - \frac{1}{t^2}\right)$ | (M1), (A1A1) | Attempt chain rule, PI
$\frac{1}{t^2} = y^2$ | (B1) | Or sub $t = \frac{1}{2}$; -4 in numerator is sufficient for this
$y = 2$, $\frac{dy}{dx} = \frac{-4}{2-4} = 2$ | M1A1 | AG; convincingly obtained
**OR**
$x = y + 2y^{-1}$ | M1 | $\frac{dx}{dy} = 1 - 2y^{-2}$ | M1A1A1 | $= 1 - \frac{1}{2} = \frac{1}{2}$ | B1 | $\rightarrow \frac{dy}{dx} = 2$ | A1 |
**Total: 10 marks**
---5 A curve is defined by the parametric equations
$$x = 2 t + \frac { 1 } { t } , \quad y = \frac { 1 } { t } , \quad t \neq 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the point on the curve where $t = \frac { 1 } { 2 }$.
\item Show that the cartesian equation of the curve can be written as
$$x y - y ^ { 2 } = 2$$
\item Show that the gradient of the curve at the point $( 3,2 )$ is 2 .
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2005 Q5 [10]}}