| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| 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 (part a) and eliminating the parameter using exponential manipulation (part b). While it involves exponentials, the algebraic steps are routine and follow predictable patterns taught in A-level Further Maths, 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dt} = (3\ln 2)2^t\) and \(\frac{dx}{dt} = (-4\ln 2)2^{-t}\) | M1, A1 | Differentiates \(2^t\) or \(2^{-t}\) to obtain \(\pm A\ln 2 \times 2^{\pm t}\); obtains both correct derivatives |
| \(\frac{dy}{dx} = \frac{(3\ln 2)2^t}{(-4\ln 2)2^{-t}} = -\frac{3}{4} \times 2^{2t}\) | R1 | Uses chain rule with correct \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\); completes rigorous argument to obtain fully correct printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2^t = \frac{y+5}{3}\), \(2^{-t} = \frac{x-3}{4}\) | M1 | Rearranges to write \(2^{-t}\) in terms of \(x\) or \(2^t\) in terms of \(y\), or writes given expression in terms of \(t\) |
| \(1 = \left(\frac{y+5}{3}\right)\left(\frac{x-3}{4}\right)\) | M1 | Eliminates \(t\), or compares coefficients PI by \(a=5\) or \(b=-3\) |
| \(12 = xy + 5x - 3y - 15\) | ||
| \(xy + 5x - 3y = 27\) | R1 | Completes rigorous argument to obtain correct values of \(a\), \(b\) and \(c\); ISW |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dt} = (3\ln 2)2^t$ and $\frac{dx}{dt} = (-4\ln 2)2^{-t}$ | M1, A1 | Differentiates $2^t$ or $2^{-t}$ to obtain $\pm A\ln 2 \times 2^{\pm t}$; obtains both correct derivatives |
| $\frac{dy}{dx} = \frac{(3\ln 2)2^t}{(-4\ln 2)2^{-t}} = -\frac{3}{4} \times 2^{2t}$ | R1 | Uses chain rule with correct $\frac{dy}{dt}$ and $\frac{dx}{dt}$; completes rigorous argument to obtain fully correct printed answer |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2^t = \frac{y+5}{3}$, $2^{-t} = \frac{x-3}{4}$ | M1 | Rearranges to write $2^{-t}$ in terms of $x$ or $2^t$ in terms of $y$, or writes given expression in terms of $t$ |
| $1 = \left(\frac{y+5}{3}\right)\left(\frac{x-3}{4}\right)$ | M1 | Eliminates $t$, or compares coefficients PI by $a=5$ or $b=-3$ |
| $12 = xy + 5x - 3y - 15$ | | |
| $xy + 5x - 3y = 27$ | R1 | Completes rigorous argument to obtain correct values of $a$, $b$ and $c$; ISW |
---5 A curve is defined by the parametric equations
$$\begin{aligned}
& x = 4 \times 2 ^ { - t } + 3 \\
& y = 3 \times 2 ^ { t } - 5
\end{aligned}$$
5
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 } { 4 } \times 2 ^ { 2 t }$\\
5
\item Find the Cartesian equation of the curve in the form $x y + a x + b y = c$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q5 [6]}}