| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric normal then bounded area |
| Difficulty | Standard +0.8 This is a substantial multi-part parametric question requiring: (a) coordinate substitution, (b) finding dy/dx via chain rule and normal equation, (c) setting up area integral with parametric formula requiring algebraic manipulation, and (d) integration using trigonometric identities. The setup in part (c) is non-trivial as it involves decomposing the region and working with the parametric area formula. While the techniques are all Core 3/4 standard, the extended multi-step nature and the need to carefully handle the composite region makes this moderately challenging. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1, 4.5)\) | B1B1 | Either of \((1, 4.5)\); both \((1, 4.5)\); accept exact equivalents e.g. 18/4, 9/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{12\sin\theta\cos\theta}{-24\cos^2\theta\sin\theta} = \left(-\frac{1}{2\cos\theta}\right)\) | M1A1 | Attempts \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\); allow poor differentiation provided both functions changed; correct answer |
| Subs \(\theta = \frac{\pi}{3}\) into \(\frac{dy}{dx} = (-1)\) | M1 | Substitutes \(\theta = \frac{\pi}{3}\) into their \(\frac{dy}{dx}\) |
| Uses gradient of normal with \((1, 4.5) \Rightarrow (y-4.5) = 1(x-1)\) | ddM1 | Negative reciprocal of their \(\frac{dy}{dx}\) with \((1,4.5)\); dependent on both previous M marks |
| \(y = x + 3.5\) | A1* | Given answer; allow \(y = x + \frac{7}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int y\frac{dx}{d\theta}\, d\theta = \int 6\sin^2\theta \times -24\cos^2\theta\sin\theta\, d\theta\) | M1A1 | Attempts parametric area formula; allow omission of \(d\theta\) and un-simplified |
| Uses \(\sin^2\theta = 1-\cos^2\theta \Rightarrow \int y\frac{dx}{d\theta}\, d\theta = \int A(\cos^2\theta\sin\theta - \cos^4\theta\sin\theta)\, d\theta\) | dM1 | Uses identity to produce integrable form; dependent on previous M |
| Area of trapezium \(= \frac{1}{2}(3.5+4.5) = 4\) | B1 | Or \(\frac{1}{2}\times 1\times 8\); must see a calculation - cannot just state area = 4 |
| Attempts trapezium + area under curve \(= \frac{1}{2}(3.5+4.5) - 144\int_{\pi/3}^{0}\sin^3\theta\cos^2\theta\, d\theta\) | ddM1 | Dependent on both previous M marks; correct limits must be seen |
| \(\text{Area} = 4 + 144\int_0^{\pi/3}(\sin\theta\cos^2\theta - \sin\theta\cos^4\theta)\, d\theta\) | A1* | Given answer; all previous marks must have been awarded, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of \(S = 4 + 144\left[-\frac{\cos^3\theta}{3} + \frac{\cos^5\theta}{5}\right]_0^{\pi/3} = 4+144\left(\left(-\frac{1}{24}+\frac{1}{160}\right)-\left(-\frac{1}{3}+\frac{1}{5}\right)\right)\) | M1A1 | \(\int\sin\theta\cos^n\theta\, d\theta = \pm\frac{\cos^{n+1}\theta}{n+1}\); any correct un-simplified answer with appropriate limits |
| \(= \frac{181}{10}\) | A1 | cso |
# Question 14:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1, 4.5)$ | B1B1 | Either of $(1, 4.5)$; both $(1, 4.5)$; accept exact equivalents e.g. 18/4, 9/2 |
**(2 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{12\sin\theta\cos\theta}{-24\cos^2\theta\sin\theta} = \left(-\frac{1}{2\cos\theta}\right)$ | M1A1 | Attempts $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$; allow poor differentiation provided both functions changed; correct answer |
| Subs $\theta = \frac{\pi}{3}$ into $\frac{dy}{dx} = (-1)$ | M1 | Substitutes $\theta = \frac{\pi}{3}$ into their $\frac{dy}{dx}$ |
| Uses gradient of normal with $(1, 4.5) \Rightarrow (y-4.5) = 1(x-1)$ | ddM1 | Negative reciprocal of their $\frac{dy}{dx}$ with $(1,4.5)$; dependent on both previous M marks |
| $y = x + 3.5$ | A1* | Given answer; allow $y = x + \frac{7}{2}$ |
**(5 marks)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int y\frac{dx}{d\theta}\, d\theta = \int 6\sin^2\theta \times -24\cos^2\theta\sin\theta\, d\theta$ | M1A1 | Attempts parametric area formula; allow omission of $d\theta$ and un-simplified |
| Uses $\sin^2\theta = 1-\cos^2\theta \Rightarrow \int y\frac{dx}{d\theta}\, d\theta = \int A(\cos^2\theta\sin\theta - \cos^4\theta\sin\theta)\, d\theta$ | dM1 | Uses identity to produce integrable form; dependent on previous M |
| Area of trapezium $= \frac{1}{2}(3.5+4.5) = 4$ | B1 | Or $\frac{1}{2}\times 1\times 8$; must see a calculation - cannot just state area = 4 |
| Attempts trapezium + area under curve $= \frac{1}{2}(3.5+4.5) - 144\int_{\pi/3}^{0}\sin^3\theta\cos^2\theta\, d\theta$ | ddM1 | Dependent on both previous M marks; correct limits must be seen |
| $\text{Area} = 4 + 144\int_0^{\pi/3}(\sin\theta\cos^2\theta - \sin\theta\cos^4\theta)\, d\theta$ | A1* | Given answer; all previous marks must have been awarded, no errors |
**(6 marks)**
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of $S = 4 + 144\left[-\frac{\cos^3\theta}{3} + \frac{\cos^5\theta}{5}\right]_0^{\pi/3} = 4+144\left(\left(-\frac{1}{24}+\frac{1}{160}\right)-\left(-\frac{1}{3}+\frac{1}{5}\right)\right)$ | M1A1 | $\int\sin\theta\cos^n\theta\, d\theta = \pm\frac{\cos^{n+1}\theta}{n+1}$; any correct un-simplified answer with appropriate limits |
| $= \frac{181}{10}$ | A1 | cso |
**(3 marks)**
The image appears to be a blank page (page 32) from a Pearson Education mark scheme document, containing only the company registration information at the bottom and a page number. There is no mark scheme content to extract from this page.14.
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\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-48_506_812_219_571}
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\caption{Figure 6}
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\end{figure}
Figure 6 shows a sketch of the curve $C$ with parametric equations
$$x = 8 \cos ^ { 3 } \theta , \quad y = 6 \sin ^ { 2 } \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$
Given that the point $P$ lies on $C$ and has parameter $\theta = \frac { \pi } { 3 }$
\begin{enumerate}[label=(\alph*)]
\item find the coordinates of $P$.
The line $l$ is the normal to $C$ at $P$.
\item Show that an equation of $l$ is $y = x + 3.5$
The finite region $S$, shown shaded in Figure 6, is bounded by the curve $C$, the line $l$, the $y$-axis and the $x$-axis.
\item Show that the area of $S$ is given by
$$4 + 144 \int _ { 0 } ^ { \frac { \pi } { 3 } } \left( \sin \theta \cos ^ { 2 } \theta - \sin \theta \cos ^ { 4 } \theta \right) d \theta$$
\item Hence, by integration, find the exact area of $S$.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
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\hfill \mbox{\textit{Edexcel C34 2017 Q14 [16]}}