| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find normal equation at parameter |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question testing standard C4 techniques. Part (a) uses the chain rule formula dy/dx = (dy/dt)/(dx/dt), part (b) requires finding a point and normal gradient (routine calculation), and part (c) involves eliminating the parameter by solving for t from the simpler equation. All three parts are textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = -3\), \(\frac{dy}{dt} = 6t^2\) | B1 | Both derivatives correct; PI |
| \(\frac{dy}{dx} = \frac{6t^2}{3} = -2t^2\) | M1 A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 1\), \(m_T = -2\), \(m_N = -\frac{1}{2}\) | M1 A1F | Substitute \(t=1\); \(m_N = -\frac{1}{m_T}\); F on gradient; \(m_T \neq \pm 1\) |
| Attempt at equation of normal using \((x, y) = (-2, 3)\) | M1 | Condone one error |
| Normal has equation \(y - 3 = -\frac{1}{2}(x + 2)\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \frac{1-x}{3}\) or \(t = \sqrt{\frac{y-1}{2}}\) | M1 | Correct expression for \(t\) in terms of \(x\) or \(y\) |
| \(y = 1 + 2\left(\frac{1-x}{3}\right)^3\) | A1 | 2 marks |
**2(a)**
$\frac{dx}{dt} = -3$, $\frac{dy}{dt} = 6t^2$ | B1 | Both derivatives correct; PI
$\frac{dy}{dx} = \frac{6t^2}{3} = -2t^2$ | M1 A1 | 3 marks | Correct use of chain rule; CSO
**2(b)**
$t = 1$, $m_T = -2$, $m_N = -\frac{1}{2}$ | M1 A1F | Substitute $t=1$; $m_N = -\frac{1}{m_T}$; F on gradient; $m_T \neq \pm 1$
Attempt at equation of normal using $(x, y) = (-2, 3)$ | M1 | Condone one error
Normal has equation $y - 3 = -\frac{1}{2}(x + 2)$ | A1 | 4 marks | CSO; ACF
**2(c)**
$t = \frac{1-x}{3}$ or $t = \sqrt{\frac{y-1}{2}}$ | M1 | Correct expression for $t$ in terms of $x$ or $y$
$y = 1 + 2\left(\frac{1-x}{3}\right)^3$ | A1 | 2 marks | ACF
---A curve is defined by the parametric equations
$$x = 1 - 3t, \quad y = 1 + 2t^3$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$ in terms of $t$. [3 marks]
\item Find an equation of the normal to the curve at the point where $t = 1$. [4 marks]
\item Find a cartesian equation of the curve. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2010 Q2 [9]}}