Challenging +1.8 This is a substantial Further Maths question requiring multiple techniques: hyperbolic differentiation, parametric arc length, and surface of revolution. While each individual step follows standard methods (differentiate sech, apply arc length formula, use surface area formula), the question requires sustained multi-step reasoning across hyperbolic identities and careful algebraic manipulation. The final elegant result suggests deliberate design rather than routine calculation, placing it well above average difficulty but not at the extreme end for FP2.
The curve meets the \(y\)-axis at the point \(K\), and \(P ( x , y )\) is a general point on the curve. The arc length \(K P\) is denoted by \(s\). Show that:
\(\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \tanh ^ { 2 } t\);
\(s = \ln \cosh t\);
\(y = \mathrm { e } ^ { - s }\).
The arc \(K P\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that the surface area generated is
$$2 \pi \left( 1 - \mathrm { e } ^ { - S } \right)$$
(4 marks)
4
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \operatorname { sech } t$, show that:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} t } = - \operatorname { sech } t \tanh t$;
\item $\left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \operatorname { sech } ^ { 2 } t - \operatorname { sech } ^ { 4 } t$.
\end{enumerate}\item The diagram shows a sketch of part of the curve given parametrically by
$$x = t - \tanh t \quad y = \operatorname { sech } t$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{1891766e-7744-49ac-82b6-7e51cb63b381-3_424_625_863_703}
\end{center}
The curve meets the $y$-axis at the point $K$, and $P ( x , y )$ is a general point on the curve. The arc length $K P$ is denoted by $s$. Show that:
\begin{enumerate}[label=(\roman*)]
\item $\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \tanh ^ { 2 } t$;
\item $s = \ln \cosh t$;
\item $y = \mathrm { e } ^ { - s }$.
\end{enumerate}\item The arc $K P$ is rotated through $2 \pi$ radians about the $x$-axis. Show that the surface area generated is
$$2 \pi \left( 1 - \mathrm { e } ^ { - S } \right)$$
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2007 Q4 [18]}}