AQA FP2 2011 January — Question 6 10 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJanuary
Marks10
PaperDownload PDF ↗
TopicParametric integration
TypeParametric arc length calculation
DifficultyChallenging +1.2 This is a Further Maths FP2 parametric arc length question requiring differentiation of a composite logarithmic/trigonometric expression and then arc length integration. Part (a) involves careful chain rule application with trigonometric identities (showing dx/dt = sin t tan t), while part (b) requires computing √((dx/dt)² + (dy/dt)²) and integrating—the algebra simplifies nicely to give a clean logarithmic answer. While technically demanding with multiple steps, it follows standard FP2 techniques without requiring novel insight, making it moderately above average difficulty.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.07d Differentiate/integrate: hyperbolic functions8.06a Reduction formulae: establish, use, and evaluate recursively

6
  1. Given that $$x = \ln ( \sec t + \tan t ) - \sin t$$ show that $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \sin t \tan t$$
  2. A curve is given parametrically by the equations $$x = \ln ( \sec t + \tan t ) - \sin t , \quad y = \cos t$$ The length of the arc of the curve between the points where \(t = 0\) and \(t = \frac { \pi } { 3 }\) is denoted by \(s\). Show that \(s = \ln p\), where \(p\) is an integer.

6
\begin{enumerate}[label=(\alph*)]
\item Given that

$$x = \ln ( \sec t + \tan t ) - \sin t$$

show that

$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \sin t \tan t$$
\item A curve is given parametrically by the equations

$$x = \ln ( \sec t + \tan t ) - \sin t , \quad y = \cos t$$

The length of the arc of the curve between the points where $t = 0$ and $t = \frac { \pi } { 3 }$ is denoted by $s$.

Show that $s = \ln p$, where $p$ is an integer.
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2011 Q6 [10]}}