| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (exponential/logarithmic) |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: finding dy/dx using the chain rule (dy/dt รท dx/dt), evaluating at a point, finding a tangent equation, and converting to Cartesian form by eliminating the parameter. The algebraic manipulation with exponentials is routine (recognizing e^(2t) = (e^t)^2), and the question guides students to the final form. Slightly above average due to the multiple parts and exponential functions, but all techniques are standard C4 material with no novel insight required. |
| 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 |
|---|---|
| \(\frac{dx}{dt} = 3e^t\), \(\frac{dy}{dt} = 2e^{2t} + 2e^{-2t}\) | M1 |
| \(\frac{dy}{dx} = \frac{2e^{2t}+2e^{-2t}}{3e^t}\) | M1 |
| At \(t=0\): \(\frac{dy}{dx} = \frac{4}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| At \(t=0\): point is \((3, 0)\); tangent: \(y = \frac{4}{3}(x-3)\) | B1 | ft their gradient |
| Answer | Marks |
|---|---|
| \(e^t = \frac{x}{3}\), so \(e^{2t} = \frac{x^2}{9}\) and \(e^{-2t} = \frac{9}{x^2}\) | M1 |
| \(y = \frac{x^2}{9} - \frac{9}{x^2}\), so \(k = 9\) | A1 |
# Question 4:
## Part (a)(i)
| $\frac{dx}{dt} = 3e^t$, $\frac{dy}{dt} = 2e^{2t} + 2e^{-2t}$ | M1 | |
|---|---|---|
| $\frac{dy}{dx} = \frac{2e^{2t}+2e^{-2t}}{3e^t}$ | M1 | |
| At $t=0$: $\frac{dy}{dx} = \frac{4}{3}$ | A1 | |
## Part (a)(ii)
| At $t=0$: point is $(3, 0)$; tangent: $y = \frac{4}{3}(x-3)$ | B1 | ft their gradient |
|---|---|---|
## Part (b)
| $e^t = \frac{x}{3}$, so $e^{2t} = \frac{x^2}{9}$ and $e^{-2t} = \frac{9}{x^2}$ | M1 | |
|---|---|---|
| $y = \frac{x^2}{9} - \frac{9}{x^2}$, so $k = 9$ | A1 | |
---4 A curve is defined by the parametric equations
$$x = 3 \mathrm { e } ^ { t } , \quad y = \mathrm { e } ^ { 2 t } - \mathrm { e } ^ { - 2 t }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of the curve at the point where $t = 0$.
\item Find an equation of the tangent to the curve at the point where $t = 0$.
\end{enumerate}\item Show that the cartesian equation of the curve can be written in the form
$$y = \frac { x ^ { 2 } } { k } - \frac { k } { x ^ { 2 } }$$
where $k$ is an integer.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2011 Q4 [6]}}