| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: finding axis intercepts, computing dy/dx using the chain rule, and converting to Cartesian form by making t the subject. All steps are routine C3/C4 procedures with no novel problem-solving 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t(t-4)=0 \Rightarrow t=4\), hence \(x=\frac{20\times4}{2\times4+1}=\frac{80}{9}\) | M1A1 | M1: attempts to find \(x\) when \(t=4\). A1: \(\frac{80}{9}\) (not 8.88… unless \(\frac{80}{9}\) seen). Ignore attempts at \(t=0\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\frac{20t}{2t+1} \Rightarrow \frac{dx}{dt}=\frac{20(2t+1)-20t\times2}{(2t+1)^2}=\frac{20}{(2t+1)^2}\) | M1A1 | M1: quotient rule on \(\frac{20t}{2t+1}\) with \(u=20t\), \(v=2t+1\); or product/chain rule. A1: correct \(\frac{dx}{dt}\) |
| \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{(2t-4)}{\frac{20}{(2t+1)^2}}=\frac{(t-2)(2t+1)^2}{10}\) | M1A1, A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\frac{20t}{2t+1}\Rightarrow 2tx+x=20t \Rightarrow t(20-2x)=x \Rightarrow t=\frac{x}{20-2x}\) or \(\frac{-x}{2x-20}\) | M1A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sub \(t=\frac{x}{20-2x}\) into \(y=t(t-4)\): \(y=\frac{x}{20-2x}\!\left(\frac{x}{20-2x}-4\right)\) | M1 | |
| \(\Rightarrow y=\frac{x}{20-2x}\times\left(\frac{x-4(20-2x)}{20-2x}\right)\) | dM1 | Correct processing |
| \(\Rightarrow y=\frac{x}{20-2x}\times\left(\frac{9x-80}{20-2x}\right)\) | ||
\(\Rightarrow y=\frac{x(9x-80)}{(20-2x)^2}\), oe, \(\quad 0| A1, B1 |
(4) |
|
| (13 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{(2t-4)}{20/(2t+1)^2}\) | A1 | Correct and un-simplified |
| \(\frac{dy}{dx} = \frac{(t-2)(2t+1)^2}{10}\) | A1 | Allow "invisible" brackets recovered if correct answer appears |
| Answer | Marks | Guidance |
|---|---|---|
| Reach \(t = \frac{\pm x}{\pm 20 \pm 2x}\) | M1 | Full attempt to make \(t\) subject of \(x = \frac{20t}{2t+1}\); sign slips only |
| \(t = \frac{x}{20-2x}\) or e.g. \(t = \frac{-x}{2x-20}\) | A1 | Or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Substitutes \(t = \frac{x}{20-2x}\) into \(y = t(t-4)\) | M1 | To find \(y\) in terms of \(x\) |
| Uses correct common denominator, adapting numerator | dM1 | Condone sign errors only when combining fractions; dependent on first M1 |
| \(y = \frac{x(9x-80)}{(20-2x)^2}\), accept e.g. \(y = \frac{x(9x-80)}{4(10-x)^2}\), \(y = \frac{9x^2-80x}{4(10-x)^2}\), \(y = \frac{9x^2-80x}{400-80x+4x^2}\) | A1 | Exact alternatives accepted |
| Domain: \(0 < x < 10\) or \(k = 10\) | B1 | Accept either form |
# Question 10:
**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t(t-4)=0 \Rightarrow t=4$, hence $x=\frac{20\times4}{2\times4+1}=\frac{80}{9}$ | M1A1 | M1: attempts to find $x$ when $t=4$. A1: $\frac{80}{9}$ (not 8.88… unless $\frac{80}{9}$ seen). Ignore attempts at $t=0$ |
| | **(2)** | |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\frac{20t}{2t+1} \Rightarrow \frac{dx}{dt}=\frac{20(2t+1)-20t\times2}{(2t+1)^2}=\frac{20}{(2t+1)^2}$ | M1A1 | M1: quotient rule on $\frac{20t}{2t+1}$ with $u=20t$, $v=2t+1$; or product/chain rule. A1: correct $\frac{dx}{dt}$ |
| $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{(2t-4)}{\frac{20}{(2t+1)^2}}=\frac{(t-2)(2t+1)^2}{10}$ | M1A1, A1 | **(5)** |
**(c)(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\frac{20t}{2t+1}\Rightarrow 2tx+x=20t \Rightarrow t(20-2x)=x \Rightarrow t=\frac{x}{20-2x}$ or $\frac{-x}{2x-20}$ | M1A1 | **(2)** |
**(c)(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $t=\frac{x}{20-2x}$ into $y=t(t-4)$: $y=\frac{x}{20-2x}\!\left(\frac{x}{20-2x}-4\right)$ | M1 | |
| $\Rightarrow y=\frac{x}{20-2x}\times\left(\frac{x-4(20-2x)}{20-2x}\right)$ | dM1 | Correct processing |
| $\Rightarrow y=\frac{x}{20-2x}\times\left(\frac{9x-80}{20-2x}\right)$ | | |
| $\Rightarrow y=\frac{x(9x-80)}{(20-2x)^2}$, oe, $\quad 0<x<10$ or $k=10$ | A1, B1 | **(4)** |
| | **(13 marks)** | |
# Question (parametric differentiation - preceding Q11):
## dy/dx part:
| $\frac{dy}{dx} = \frac{(2t-4)}{20/(2t+1)^2}$ | A1 | Correct and un-simplified |
| $\frac{dy}{dx} = \frac{(t-2)(2t+1)^2}{10}$ | A1 | Allow "invisible" brackets recovered if correct answer appears |
## Part (c)(i):
| Reach $t = \frac{\pm x}{\pm 20 \pm 2x}$ | M1 | Full attempt to make $t$ subject of $x = \frac{20t}{2t+1}$; sign slips only |
| $t = \frac{x}{20-2x}$ or e.g. $t = \frac{-x}{2x-20}$ | A1 | Or equivalent |
## Part (c)(ii):
| Substitutes $t = \frac{x}{20-2x}$ into $y = t(t-4)$ | M1 | To find $y$ in terms of $x$ |
| Uses correct common denominator, adapting numerator | dM1 | Condone sign errors only when combining fractions; dependent on first M1 |
| $y = \frac{x(9x-80)}{(20-2x)^2}$, accept e.g. $y = \frac{x(9x-80)}{4(10-x)^2}$, $y = \frac{9x^2-80x}{4(10-x)^2}$, $y = \frac{9x^2-80x}{400-80x+4x^2}$ | A1 | Exact alternatives accepted |
| Domain: $0 < x < 10$ or $k = 10$ | B1 | Accept either form |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2a6d0dba-d948-4124-9740-a88c17b0be65-32_556_716_237_607}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with parametric equations
$$x = \frac { 20 t } { 2 t + 1 } \quad y = t ( t - 4 ) , \quad t > 0$$
The curve cuts the $x$-axis at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item Find the $x$ coordinate of $P$.
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( t - A ) ( 2 t + 1 ) ^ { 2 } } { B }$ where $A$ and $B$ are constants to be found.
\item \begin{enumerate}[label=(\roman*)]
\item Make $t$ the subject of the formula
$$x = \frac { 20 t } { 2 t + 1 }$$
\item Hence find a cartesian equation of the curve $C$. Write your answer in the form
$$y = \mathrm { f } ( x ) , \quad 0 < x < k$$
where $\mathrm { f } ( x )$ is a single fraction and $k$ is a constant to be found.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q10 [13]}}