Pre-U Pre-U 9794/2 2011 June — Question 8 15 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2011
SessionJune
Marks15
TopicParametric differentiation
TypeTangent parallel to axis condition
DifficultyChallenging +1.3 This is a substantial parametric calculus question requiring multiple techniques: chain rule for dy/dx, trigonometric identities (including half-angle formulas for the cot result), algebraic manipulation to reach the specified form with constraints on constants, numerical methods/sign changes for root location, and second derivatives. While each individual step uses standard A-level techniques, the combination of parts, the need to manipulate into a specific form with constrained parameters, and the extended reasoning across multiple sub-parts makes this moderately harder than average, though still within reach of well-prepared students.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.05o Trigonometric equations: solve in given intervals1.07e Second derivative: as rate of change of gradient1.07s Parametric and implicit differentiation1.09a Sign change methods: locate roots

  1. A curve \(C_1\) is defined by the parametric equations $$x = \theta - \sin \theta, \quad y = 1 - \cos \theta,$$ where the parameter \(\theta\) is measured in radians.
    1. Show that \(\frac{dy}{dx} = \cot \frac{1}{2}\theta\), except for certain values of \(\theta\), which should be identified. [5]
    2. Show that the points of intersection of the curve \(C_1\) and the line \(y = x\) are determined by an equation of the form \(\theta = 1 + A \sin(\theta - \alpha)\), where \(A\) and \(\alpha\) are constants to be found, such that \(A > 0\) and \(0 < \alpha < \frac{1}{2}\pi\). [4]
    3. Show that the equation found in part (b) has a root between \(\frac{1}{4}\pi\) and \(\pi\). [2]
  2. A curve \(C_2\) is defined by the parametric equations $$x = \theta - \frac{1}{2} \sin \theta, \quad y = 1 - \frac{1}{2} \cos \theta,$$ where the parameter \(\theta\) is measured in radians. Find the y-coordinates of all points on \(C_2\) for which \(\frac{d^2y}{dx^2} = 0\). [4]

Part (i) (a)
AnswerMarks Guidance
\(\frac{dx}{d\theta} = 1 - \cos \theta\)B1
\(\frac{dy}{d\theta} = \sin \theta\)B1
\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\sin \theta}{1 - \cos \theta}\)M1
\(= \frac{2\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta}{2\sin^2 \frac{1}{2}\theta} = \cot \frac{1}{2}\theta\)A1 AG
At least two of \(\theta = ..., -2\pi, 0, 2\pi...\) without any incorrect valuesB1 [5]
Part (i) (b)
AnswerMarks Guidance
Rearranging \(y = x\) to give \(\theta = 1 + \sin \theta - \cos \theta\)M1
\(= 1 + A\sin(\theta - \alpha)\)M1
where \(A = \sqrt{2}\)A1
and \(\alpha = \frac{\pi}{4}\)A1 [4]
Part (i) (c)
AnswerMarks Guidance
Consider sign of \(\theta - 1 - \sqrt{2}\sin\left(\theta - \frac{\pi}{4}\right)\) at \(\theta = \frac{\pi}{2}, \pi\)M1
Change of sign implies root:A1 [2]
\(\left(\frac{\pi}{2} - 2 \text{(negative)} \text{ and } \pi - 2 \text{(positive)}\right)\)
Part (ii)
AnswerMarks Guidance
\(\frac{dy}{dx} = \frac{\sin \theta}{2 - \cos \theta}\)B1
\(\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \times \frac{d\theta}{dx}\)or equivalent M1
\(= \frac{2(2\cos \theta - 1)}{(2 - \cos \theta)^3}\)AEF, unsimplified A1
\(\frac{d^2y}{dx^2} = 0 \Rightarrow y = \frac{3}{4}\)A1 [4]
**Part (i) (a)**
$\frac{dx}{d\theta} = 1 - \cos \theta$ | B1
$\frac{dy}{d\theta} = \sin \theta$ | B1
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\sin \theta}{1 - \cos \theta}$ | M1
$= \frac{2\sin \frac{1}{2}\theta \cos \frac{1}{2}\theta}{2\sin^2 \frac{1}{2}\theta} = \cot \frac{1}{2}\theta$ | A1 AG
At least two of $\theta = ..., -2\pi, 0, 2\pi...$ without any incorrect values | B1 | [5]

**Part (i) (b)**
Rearranging $y = x$ to give $\theta = 1 + \sin \theta - \cos \theta$ | M1
$= 1 + A\sin(\theta - \alpha)$ | M1
where $A = \sqrt{2}$ | A1
and $\alpha = \frac{\pi}{4}$ | A1 | [4]

**Part (i) (c)**
Consider sign of $\theta - 1 - \sqrt{2}\sin\left(\theta - \frac{\pi}{4}\right)$ at $\theta = \frac{\pi}{2}, \pi$ | M1
Change of sign implies root: | A1 | [2]
$\left(\frac{\pi}{2} - 2 \text{(negative)} \text{ and } \pi - 2 \text{(positive)}\right)$

**Part (ii)**
$\frac{dy}{dx} = \frac{\sin \theta}{2 - \cos \theta}$ | B1
$\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \times \frac{d\theta}{dx}$ | or equivalent | M1
$= \frac{2(2\cos \theta - 1)}{(2 - \cos \theta)^3}$ | AEF, unsimplified | A1
$\frac{d^2y}{dx^2} = 0 \Rightarrow y = \frac{3}{4}$ | A1 | [4]
\begin{enumerate}[label=(\roman*)]
\item A curve $C_1$ is defined by the parametric equations
$$x = \theta - \sin \theta, \quad y = 1 - \cos \theta,$$
where the parameter $\theta$ is measured in radians.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dy}{dx} = \cot \frac{1}{2}\theta$, except for certain values of $\theta$, which should be identified. [5]
\item Show that the points of intersection of the curve $C_1$ and the line $y = x$ are determined by an equation of the form $\theta = 1 + A \sin(\theta - \alpha)$, where $A$ and $\alpha$ are constants to be found, such that $A > 0$ and $0 < \alpha < \frac{1}{2}\pi$. [4]
\item Show that the equation found in part (b) has a root between $\frac{1}{4}\pi$ and $\pi$. [2]
\end{enumerate}
\item A curve $C_2$ is defined by the parametric equations
$$x = \theta - \frac{1}{2} \sin \theta, \quad y = 1 - \frac{1}{2} \cos \theta,$$
where the parameter $\theta$ is measured in radians. Find the y-coordinates of all points on $C_2$ for which $\frac{d^2y}{dx^2} = 0$. [4]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q8 [15]}}