| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find normal equation |
| Difficulty | Moderate -0.8 This is a straightforward parametric equations question testing standard techniques: basic differentiation (dx/dt, dy/dt are trivial), chain rule application, finding a normal line, and eliminating the parameter using simple algebraic substitution (t = (x-1)/2 substituted into y). All steps are routine C4 procedures with no problem-solving insight required, making it easier than average. |
| 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 |
|---|---|---|
| \(\frac{dx}{dt} = 2\), \(\frac{dy}{dt} = -8t\) | B1, B1 (2 marks) | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{-8t}{2} = -4t\) | M1, A1F (2 marks) | Chain rule in correct form; ft on sign/coefficient errors (not power of \(t\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m_T = -4\), \(m_N = \frac{1}{4}\); \(x=3\), \(y=-3\) | B1F, B1F | |
| \(\frac{y+3}{x-3} = \frac{1}{4}\) | M1 | Use candidate's \((x,y)\) and \(m_N\) |
| Any correct form | A1 (4 marks) | ISW; CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \frac{x-1}{2}\) | M1 | |
| \(y = 1 - 4\left(\frac{x-1}{2}\right)^2\) | M1A1 (3 marks) | Substitute for \(t\); simplification not required but CAO. Equivalent forms: \(y = 2x - x^2\), \(t^2 = \frac{1-y}{4}\), \(\left(\frac{x-1}{2}\right)^2 = \frac{1-y}{4}\) |
## Question 1:
**Part (a)(i)**
$\frac{dx}{dt} = 2$, $\frac{dy}{dt} = -8t$ | B1, B1 (2 marks) | CAO
**Part (a)(ii)**
$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{-8t}{2} = -4t$ | M1, A1F (2 marks) | Chain rule in correct form; ft on sign/coefficient errors (not power of $t$)
**Part (b)**
$m_T = -4$, $m_N = \frac{1}{4}$; $x=3$, $y=-3$ | B1F, B1F |
$\frac{y+3}{x-3} = \frac{1}{4}$ | M1 | Use candidate's $(x,y)$ and $m_N$
Any correct form | A1 (4 marks) | ISW; CAO
**Part (c)**
$t = \frac{x-1}{2}$ | M1 |
$y = 1 - 4\left(\frac{x-1}{2}\right)^2$ | M1A1 (3 marks) | Substitute for $t$; simplification not required but CAO. Equivalent forms: $y = 2x - x^2$, $t^2 = \frac{1-y}{4}$, $\left(\frac{x-1}{2}\right)^2 = \frac{1-y}{4}$
**Total: 11 marks**
---1 A curve is defined by the parametric equations
$$x = 1 + 2 t , \quad y = 1 - 4 t ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} t }$ and $\frac { \mathrm { d } y } { \mathrm {~d} t }$.\\
(2 marks)
\item Hence find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
\end{enumerate}\item Find an equation of the normal to the curve at the point where $t = 1$.
\item Find a cartesian equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q1 [11]}}