Question 4 - A-Level Maths

Given that $y = \operatorname { sech } t$, show that:
1. $\frac { \mathrm { d } y } { \mathrm {~d} t } = - \operatorname { sech } t \tanh t$;
2. $\left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \operatorname { sech } ^ { 2 } t - \operatorname { sech } ^ { 4 } t$.
The diagram shows a sketch of part of the curve given parametrically by $$x = t - \tanh t \quad y = \operatorname { sech } t$$
\includegraphics[max width=\textwidth, alt={}]{1891766e-7744-49ac-82b6-7e51cb63b381-3_424_625_863_703}
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:
1. $\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \tanh ^ { 2 } t$;
2. $s = \ln \cosh t$;
3. $y = \mathrm { e } ^ { - s }$.
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)