| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (tan/sec/cot/cosec identities) |
| Difficulty | Moderate -0.3 This is a straightforward parametric-to-Cartesian conversion using the standard trigonometric identity sec²θ = 1 + tan²θ. Part (a) requires simple substitution of the boundary value, part (b) is direct application of the identity, and part (c) is routine differentiation. While it requires knowledge of multiple techniques, each step follows standard procedures with no novel insight needed, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = 3\) | B1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sec^2\theta = 1 + \tan^2\theta \Rightarrow y = 1 + \left(\frac{x}{\sqrt{3}}\right)^2\) | M1 | Attempts to use \(1 + \tan^2\theta = \sec^2\theta\) with given parametric equations to obtain an equation in terms of \(x\) and \(y\) only |
| \(\Rightarrow y = 1 + \frac{1}{3}x^2\) | A1 | \(y = 1 + \frac{1}{3}x^2\) or \(f(x) = 1 + \frac{1}{3}x^2\). Allow \(y/f(x) = \frac{3+x^2}{3}\) but not \(y = 1 + \left(\frac{x}{\sqrt{3}}\right)^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{2}{3}x\) | M1 | Differentiates their \(f(x)\) with evidence of \(x^n \rightarrow x^{n-1}\) or differentiating to a correct form for their function |
| \(\left(\frac{dy}{dx}\right)_{x=1} = \frac{2}{3}(1) = \ldots\) | M1 | Attempts to find their \(\frac{dy}{dx}\) at \(x = 1\) (or their attempt at \(x\)) |
| Gradient \(= \frac{2}{3}\) | A1 | For \(\frac{2}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{d\theta} = \sqrt{3}\sec^2\theta,\ \frac{dy}{d\theta} = 2\sec^2\theta\tan\theta\) \(\frac{dy}{dx} = \frac{2\sec^2\theta\tan\theta}{\sqrt{3}\sec^2\theta}\) | M1 | Attempts \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\ldots\sec^2\theta\tan\theta}{\ldots\sec^2\theta}\) |
| \(\frac{dy}{dx} = \frac{2\sec^2\!\left(\frac{\pi}{6}\right)\tan\!\left(\frac{\pi}{6}\right)}{\sqrt{3}\sec^2\!\left(\frac{\pi}{6}\right)}\) | M1 | Attempts to find their \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{6}\) |
| Gradient \(= \frac{2}{3}\) | A1 | For \(\frac{2}{3}\) |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = 3$ | B1 | Correct value |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sec^2\theta = 1 + \tan^2\theta \Rightarrow y = 1 + \left(\frac{x}{\sqrt{3}}\right)^2$ | M1 | Attempts to use $1 + \tan^2\theta = \sec^2\theta$ with given parametric equations to obtain an equation in terms of $x$ and $y$ only |
| $\Rightarrow y = 1 + \frac{1}{3}x^2$ | A1 | $y = 1 + \frac{1}{3}x^2$ or $f(x) = 1 + \frac{1}{3}x^2$. Allow $y/f(x) = \frac{3+x^2}{3}$ but not $y = 1 + \left(\frac{x}{\sqrt{3}}\right)^2$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{2}{3}x$ | M1 | Differentiates their $f(x)$ with evidence of $x^n \rightarrow x^{n-1}$ or differentiating to a correct form for their function |
| $\left(\frac{dy}{dx}\right)_{x=1} = \frac{2}{3}(1) = \ldots$ | M1 | Attempts to find their $\frac{dy}{dx}$ at $x = 1$ (or their attempt at $x$) |
| Gradient $= \frac{2}{3}$ | A1 | For $\frac{2}{3}$ |
## Part (c) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = \sqrt{3}\sec^2\theta,\ \frac{dy}{d\theta} = 2\sec^2\theta\tan\theta$ $\frac{dy}{dx} = \frac{2\sec^2\theta\tan\theta}{\sqrt{3}\sec^2\theta}$ | M1 | Attempts $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\ldots\sec^2\theta\tan\theta}{\ldots\sec^2\theta}$ |
| $\frac{dy}{dx} = \frac{2\sec^2\!\left(\frac{\pi}{6}\right)\tan\!\left(\frac{\pi}{6}\right)}{\sqrt{3}\sec^2\!\left(\frac{\pi}{6}\right)}$ | M1 | Attempts to find their $\frac{dy}{dx}$ at $\theta = \frac{\pi}{6}$ |
| Gradient $= \frac{2}{3}$ | A1 | For $\frac{2}{3}$ |
---3. A curve $C$ has parametric equations
$$x = \sqrt { 3 } \tan \theta , \quad y = \sec ^ { 2 } \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 3 }$$
The cartesian equation of $C$ is
$$y = \mathrm { f } ( x ) , \quad 0 \leqslant x \leqslant k , \quad \text { where } k \text { is a constant }$$
\begin{enumerate}[label=(\alph*)]
\item State the value of $k$.
\item Find $\mathrm { f } ( x )$ in its simplest form.
\item Hence, or otherwise, find the gradient of the curve at the point where $\theta = \frac { \pi } { 6 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q3 [6]}}