| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find dy/dx at a point |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: differentiation using dy/dx = (dy/dt)/(dx/dt) for part (a), and algebraic manipulation to eliminate the parameter for part (b). Both parts are routine C4 exercises with no conceptual challenges—slightly easier than average due to the simple algebraic forms involved. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = t\) | B1 | |
| \(\frac{dy}{dt} = -\frac{4}{t^2}\) | B1 | |
| \(\frac{dy}{dx} = \frac{-4/t^2}{t} = \frac{-4}{t^3}\), at \(t=2\): \(\frac{-4}{8} = -\frac{1}{2}\) | A1 | Must substitute \(t=2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| From \(x = \frac{t^2}{2}+1\): \(t^2 = 2(x-1)\), so \(t = \sqrt{2(x-1)}\) | M1 | Eliminating \(t\) |
| \(y = \frac{4}{t}-1 = \frac{4}{\sqrt{2(x-1)}}-1\) or equivalent e.g. \(y+1 = \frac{4}{\sqrt{2(x-1)}}\) | A1 | Accept any correct form |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = t$ | B1 | |
| $\frac{dy}{dt} = -\frac{4}{t^2}$ | B1 | |
| $\frac{dy}{dx} = \frac{-4/t^2}{t} = \frac{-4}{t^3}$, at $t=2$: $\frac{-4}{8} = -\frac{1}{2}$ | A1 | Must substitute $t=2$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| From $x = \frac{t^2}{2}+1$: $t^2 = 2(x-1)$, so $t = \sqrt{2(x-1)}$ | M1 | Eliminating $t$ |
| $y = \frac{4}{t}-1 = \frac{4}{\sqrt{2(x-1)}}-1$ or equivalent e.g. $y+1 = \frac{4}{\sqrt{2(x-1)}}$ | A1 | Accept any correct form |
---1 A curve is defined by the parametric equations $x = \frac { t ^ { 2 } } { 2 } + 1 , y = \frac { 4 } { t } - 1$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient at the point on the curve where $t = 2$.
\item Find a Cartesian equation of the curve.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-02_1730_1709_977_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2014 Q1 [5]}}