| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Standard +0.8 This is a substantial parametric equations question requiring multiple techniques: finding dy/dx using the chain rule with exponential differentiation (part a), algebraic manipulation of exponentials (part b), and elimination of the parameter to find a Cartesian equation in a specific form (part c). While the individual techniques are C4 standard, the combination—especially manipulating the exponential relationship and substituting to achieve the required form—requires careful algebraic work and insight beyond routine exercises. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{dx}{dt}\right) = -(-6)\frac{e^{2-6t}}{4}\) | B1 | ACF: \(\frac{3}{2}e^{2-6t}\) |
| \(\left(\frac{dy}{dt}\right) = \frac{pe^{3t} \cdot t + qe^{3t}}{(3t)^2}\) | M1 | From quotient rule |
| \(= \frac{3e^{3t} \cdot 3t - e^{3t} \cdot 3}{(3t)^2}\) | A1 | ACF: \(\frac{9te^{3t} - 3e^{3t}}{9t^2}\), \(\frac{e^{3t}}{t}\) – \(\frac{e^{3t}}{3t^2}\) etc. |
| \(\frac{dy}{dx} = \frac{(9\frac{2}{3}e^2 - 3e^2)/4}{(3/2)e^{-2}}\) | m1 | Using \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\) and clear evidence of an attempt to substitute \(t = \frac{2}{3}\) (must be this) done in either order. |
| \(= \frac{2}{e}e^4\) | A1 | Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \frac{4 - e^{2-6t}}{4} \Rightarrow 4x = 4 - e^{2-6t}\) leading to | M1 | Any correct expression for \(e^{-6t}\) or \(e^{6t}\). |
| \(e^{-6t} = \frac{4 - 4x}{e^2}\) | ||
| \(e^{6t} = \frac{e^2}{4(1 - x)}\) | ||
| \(e^{3t} = \frac{e}{2\sqrt{1 - x}}\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| From (b) \(e^{3t} = \frac{e}{2\sqrt{1 - x}}\) | ||
| \(\ln(e^{3t}) = \ln\left(\frac{e}{2\sqrt{1 - x}}\right)\) | ||
| \(3t = \ln e - \ln(2\sqrt{1 - x})\) | M1 | Find \(3t\) (or \(t\)) in terms of \(x\); must have used laws of logs correctly on both sides (possibly lne = 1 at this stage) |
| \(y = \frac{e}{2\sqrt{1 - x}\left[1 - \ln 2\sqrt{1 - x}\right]}\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 9 |
## Part (a)
$\left(\frac{dx}{dt}\right) = -(-6)\frac{e^{2-6t}}{4}$ | B1 | ACF: $\frac{3}{2}e^{2-6t}$
$\left(\frac{dy}{dt}\right) = \frac{pe^{3t} \cdot t + qe^{3t}}{(3t)^2}$ | M1 | From quotient rule
$= \frac{3e^{3t} \cdot 3t - e^{3t} \cdot 3}{(3t)^2}$ | A1 | ACF: $\frac{9te^{3t} - 3e^{3t}}{9t^2}$, $\frac{e^{3t}}{t}$ – $\frac{e^{3t}}{3t^2}$ etc.
$\frac{dy}{dx} = \frac{(9\frac{2}{3}e^2 - 3e^2)/4}{(3/2)e^{-2}}$ | m1 | Using $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ and clear evidence of an attempt to substitute $t = \frac{2}{3}$ (must be this) done in either order.
$= \frac{2}{e}e^4$ | A1 | Total: 5 | CAO : Accept this or $\frac{e^4}{2}$ or $0.5e^4$ only
$\frac{dy}{dx}$ found using product rule : $\frac{dy}{dt} = pe^{3t} \cdot t^{-1} + qe^{3t} \cdot t^{-2}$ M1
$= e^{3t}t^{-1} - \frac{1}{3}e^{3t}t^{-2}$ A1 $\left(= \frac{e^{3t}}{t} - \frac{e^{3t}}{3t^2}\right)$
For guidance, when $t = \frac{2}{3}$: $\frac{dx}{dt} = \frac{3}{2}e^{-2}$ and $\frac{dy}{dt} = \frac{3}{4}e^2$
## Part (b)
$x = \frac{4 - e^{2-6t}}{4} \Rightarrow 4x = 4 - e^{2-6t}$ leading to | M1 | Any correct expression for $e^{-6t}$ or $e^{6t}$.
$e^{-6t} = \frac{4 - 4x}{e^2}$ | |
$e^{6t} = \frac{e^2}{4(1 - x)}$ | |
$e^{3t} = \frac{e}{2\sqrt{1 - x}}$ | A1 | Total: 2 | AG Must see inversion step and be convinced they haven't worked backwards
Alternative for the M1 mark is a correct expression for $3t$ or $1^{-3t}$ or $\frac{1}{e^{-3t}}$ OE.
$4x = 4 - e^{2-6t} \Rightarrow 2 - 6t = \ln(4 - 4x) \Rightarrow 3t = 1 - \frac{1}{2}\ln(4 - 4x)$ M1 or
$4x = 4 - e^{2-6t} \Rightarrow e^{2-6t} = 4 - 4x \Rightarrow e^{1-3t} = \sqrt{4 - 4x}$ OE e.g. $\frac{x}{2e^{3t}} = \sqrt{1 - x}$ (OE) for M1, then further correct working needed before printed answer for A1.
## Part (c)
From (b) $e^{3t} = \frac{e}{2\sqrt{1 - x}}$ | |
$\ln(e^{3t}) = \ln\left(\frac{e}{2\sqrt{1 - x}}\right)$ | |
$3t = \ln e - \ln(2\sqrt{1 - x})$ | M1 | Find $3t$ (or $t$) in terms of $x$; must have used laws of logs correctly on both sides (possibly lne = 1 at this stage)
$y = \frac{e}{2\sqrt{1 - x}\left[1 - \ln 2\sqrt{1 - x}\right]}$ | A1 | Total: 2 | From $y = \frac{e^{3t}}{3t}$ ; must be in this form
Alt. From $x$, $t = \frac{1}{6}(2 - \ln(4 - 4x))$ M1 or an expression for $3t$ then A1 for printed answer.
Total | | 9
---A curve $C$ is defined by the parametric equations
$$x = \frac{4 - e^{-6t}}{4}, \quad y = \frac{e^{3t}}{3t}, \quad t \neq 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $\frac{dy}{dx}$ at the point on $C$ where $t = \frac{2}{3}$.
[5 marks]
\item Show that $x = \frac{4 - e^{-6t}}{4}$ can be rearranged into the form $e^{3t} = \frac{e}{2\sqrt{(1-x)}}$.
[2 marks]
\item Hence find the Cartesian equation of $C$, giving your answer in the form
$$y = \frac{e}{f(x)[1 - \ln(f(x))]}$$
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2016 Q7 [9]}}