Parametric curves with hyperbolic functions

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

1. (a) Given that \(y = \ln \left[ t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right]\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} t } = \frac { 1 } { \sqrt { } \left( 1 + t ^ { 2 } \right) }\).

The curve \(C\) has parametric equations $$x = \frac { 1 } { \sqrt { } \left( 1 + t ^ { 2 } \right) } , \quad y = \ln \left[ t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right] , \quad t \in \mathbb { R }$$ A student was asked to prove that, for \(t > 0\), the gradient of the tangent to \(C\) is negative.
The attempted proof was as follows: $$\begin{aligned} y & = \ln \left( t + \frac { 1 } { x } \right) \\ & = \ln \left( \frac { t x + 1 } { x } \right) \\ & = \ln ( t x + 1 ) - \ln x \\ \therefore \frac { \mathrm {~d} y } { \mathrm {~d} x } & = \frac { t } { t x + 1 } - \frac { 1 } { x } \\ & = \frac { \frac { t } { x } } { t + \frac { 1 } { x } } - \frac { 1 } { x } \\ & = \frac { t \sqrt { } \left( 1 + t ^ { 2 } \right) } { t + \sqrt { } \left( 1 + t ^ { 2 } \right) } - \sqrt { } \left( 1 + t ^ { 2 } \right) \\ & = - \frac { \left( 1 + t ^ { 2 } \right) } { t + \sqrt { } \left( 1 + t ^ { 2 } \right) } \end{aligned}$$ As \(\left( 1 + t ^ { 2 } \right) > 0\), and \(t + \sqrt { } \left( 1 + t ^ { 2 } \right) > 0\) for \(t > 0 , \frac { \mathrm {~d} y } { \mathrm {~d} x } < 0\) for \(t > 0\).
(b) (i) Identify the error in this attempt.
(ii) Give a correct version of the proof.
(c) Prove that \(\ln \left[ - t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right] = - \ln \left[ t + \sqrt { } \left( 1 + t ^ { 2 } \right) \right]\).
(d) Deduce that \(C\) is symmetric about the \(x\)-axis and sketch the graph of \(C\).