1 The curve \(C\) is defined parametrically by $$x = 2 \cos ^ { 3 } t \quad \text { and } \quad y = 2 \sin ^ { 3 } t , \quad \text { for } 0 < t < \frac { 1 } { 2 } \pi .$$ Show that, at the point with parameter \(t\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 1 } { 6 } \sec ^ { 4 } t \operatorname { cosec } t$$