| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find stationary points of parametric curve |
| Difficulty | Standard +0.3 This is a standard parametric equations question covering routine techniques: finding coordinates by substitution, computing dy/dx using the chain rule, finding stationary points by setting dy/dx = 0, and converting to Cartesian form using a trigonometric identity. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.03h Parametric equations: in modelling contexts1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Puts \(x = 0\) and obtains \(\theta = -\frac{\pi}{6}\) | B1 | awrt \(-0.52\); may be awarded for \(\cos\theta = \frac{\sqrt{3}}{2}\) or \(\sec\theta = \frac{2}{\sqrt{3}}\) if \(\theta\) not explicitly found |
| Substitutes their \(\theta\) to obtain \(y = \frac{10\sqrt{3}}{3}\), or \(\left(0, \frac{10\sqrt{3}}{3}\right)\) | M1 A1 | cao; accept \(y = \frac{10}{\sqrt{3}}\); correct answer with no incorrect working scores all 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{5\sec\theta\tan\theta}{\sqrt{3}\sec^2\theta}\) | M1 A1 | M1: attempts to differentiate both \(x\) and \(y\) wrt \(\theta\) and establishes \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\); A1: correct derivatives and correct fraction |
| \(= \frac{5\times\sin\theta/\cos\theta}{\sqrt{3}\times 1/\cos\theta}\) | B1 | For \(\lambda = \frac{5}{\sqrt{3}}\) seen explicitly or use of \(\sec\theta = \frac{1}{\cos\theta}\) |
| \(= \frac{5}{\sqrt{3}}\sin\theta\) or \(\lambda = \frac{5}{\sqrt{3}}\) oe | A1 | Fully correct solution showing all relevant steps with correct notation, no mixed variables and no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Puts \(\frac{dy}{dx} = 0\) and obtains \(\theta\) and calculates \(x\) and \(y\) or deduces correct answer | M1 | Sets their \(\frac{dy}{dx}=0\) and proceeds to find \((x,y)\) from their \(\theta\) |
| Obtains \((1, 5)\) | A1 | for \((1,5)\) or \(x=1\), \(y=5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{x-1}{\sqrt{3}}\) and \(\sec\theta = \frac{y}{5}\) | M1 | Attempt to obtain \(\tan\theta\) and \(\sec\theta\) in terms of \(x\) and \(y\); allow \(\tan\theta = \frac{x\pm1}{\sqrt{3}}\), \(\sec\theta = \frac{y}{5}\) |
| Uses \(1+\tan^2\theta = \sec^2\theta\) to give \(1 + ''\left(\frac{x-1}{\sqrt{3}}\right)^2'' = ''\left(\frac{y}{5}\right)^2''\) | M1 | Uses \(1+\tan^2\theta = \sec^2\theta\) with their expressions |
| \(\frac{3+x^2-2x+1}{3} = \left(\frac{y}{5}\right)^2\) so \(y = \frac{5}{3}\sqrt{3}\sqrt{x^2-2x+4}\) * | A1* | Obtains printed answer with no errors and with \(k = \frac{5}{3}\sqrt{3}\) only |
# Question 13:
## Part (a)
| Puts $x = 0$ and obtains $\theta = -\frac{\pi}{6}$ | B1 | awrt $-0.52$; may be awarded for $\cos\theta = \frac{\sqrt{3}}{2}$ or $\sec\theta = \frac{2}{\sqrt{3}}$ if $\theta$ not explicitly found |
| Substitutes their $\theta$ to obtain $y = \frac{10\sqrt{3}}{3}$, or $\left(0, \frac{10\sqrt{3}}{3}\right)$ | M1 A1 | cao; accept $y = \frac{10}{\sqrt{3}}$; correct answer with no incorrect working scores all 3 marks |
## Part (b)
| $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{5\sec\theta\tan\theta}{\sqrt{3}\sec^2\theta}$ | M1 A1 | M1: attempts to differentiate both $x$ and $y$ wrt $\theta$ and establishes $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$; A1: correct derivatives and correct fraction |
| $= \frac{5\times\sin\theta/\cos\theta}{\sqrt{3}\times 1/\cos\theta}$ | B1 | For $\lambda = \frac{5}{\sqrt{3}}$ seen explicitly or use of $\sec\theta = \frac{1}{\cos\theta}$ |
| $= \frac{5}{\sqrt{3}}\sin\theta$ or $\lambda = \frac{5}{\sqrt{3}}$ oe | A1 | Fully correct solution showing all relevant steps with correct notation, no mixed variables and no errors |
## Part (c)
| Puts $\frac{dy}{dx} = 0$ and obtains $\theta$ and calculates $x$ and $y$ **or** deduces correct answer | M1 | Sets their $\frac{dy}{dx}=0$ and proceeds to find $(x,y)$ from their $\theta$ |
| Obtains $(1, 5)$ | A1 | for $(1,5)$ or $x=1$, $y=5$ |
## Part (d)
| $\tan\theta = \frac{x-1}{\sqrt{3}}$ and $\sec\theta = \frac{y}{5}$ | M1 | Attempt to obtain $\tan\theta$ and $\sec\theta$ in terms of $x$ and $y$; allow $\tan\theta = \frac{x\pm1}{\sqrt{3}}$, $\sec\theta = \frac{y}{5}$ |
| Uses $1+\tan^2\theta = \sec^2\theta$ to give $1 + ''\left(\frac{x-1}{\sqrt{3}}\right)^2'' = ''\left(\frac{y}{5}\right)^2''$ | M1 | Uses $1+\tan^2\theta = \sec^2\theta$ with their expressions |
| $\frac{3+x^2-2x+1}{3} = \left(\frac{y}{5}\right)^2$ so $y = \frac{5}{3}\sqrt{3}\sqrt{x^2-2x+4}$ * | A1* | Obtains printed answer with no errors and with $k = \frac{5}{3}\sqrt{3}$ only |13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e30f0c28-1695-40a1-8e9a-6ea7e29042bf-24_515_750_264_598}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
The curve $C$ shown in Figure 4 has parametric equations
$$x = 1 + \sqrt { 3 } \tan \theta , \quad y = 5 \sec \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$$
The curve $C$ crosses the $y$-axis at $A$ and has a minimum turning point at $B$, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $A$.
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \lambda \sin \theta$, giving the exact value of the constant $\lambda$.
\item Find the coordinates of $B$.
\item Show that the cartesian equation for the curve $C$ can be written in the form
$$y = k \sqrt { \left( x ^ { 2 } - 2 x + 4 \right) }$$
where $k$ is a simplified surd to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q13 [12]}}