A curve is given parametrically by
$$x = \cosh ^ { 3 } \theta , \quad y = \sinh ^ { 3 } \theta$$
Show that
$$\left( \frac { \mathrm { d } x } { \mathrm {~d} \theta } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} \theta } \right) ^ { 2 } = \frac { 9 } { 4 } \sinh ^ { 2 } 2 \theta \cosh 2 \theta$$
Show that the length of the arc of the curve from the point where \(\theta = 0\) to the point where \(\theta = 1\) is
$$\frac { 1 } { 2 } \left[ ( \cosh 2 ) ^ { \frac { 3 } { 2 } } - 1 \right]$$