Using the identities
$$\cosh ^ { 2 } t - \sinh ^ { 2 } t = 1 , \quad \tanh t = \frac { \sinh t } { \cosh t } \quad \text { and } \quad \operatorname { sech } t = \frac { 1 } { \cosh t }$$
show that:
\(\tanh ^ { 2 } t + \operatorname { sech } ^ { 2 } t = 1\);
\(\frac { \mathrm { d } } { \mathrm { d } t } ( \tanh t ) = \operatorname { sech } ^ { 2 } t\);
\(\frac { \mathrm { d } } { \mathrm { d } t } ( \operatorname { sech } t ) = - \operatorname { sech } t \tanh t\).
A curve \(C\) is given parametrically by
$$x = \operatorname { sech } t , y = 4 - \tanh t$$
Show that the arc length, \(s\), of \(C\) between the points where \(t = 0\) and \(t = \frac { 1 } { 2 } \ln 3\) is given by
$$s = \int _ { 0 } ^ { \frac { 1 } { 2 } \ln 3 } \operatorname { sech } t \mathrm {~d} t$$
Using the substitution \(u = \mathrm { e } ^ { t }\), find the exact value of \(s\).