- A particle moves in a plane so that its position vector \(\mathbf { r }\) metres at time \(t\) seconds satisfies the differential equation
$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + ( \tan t ) \mathbf { r } = \left( \cos ^ { 2 } t \right) \mathbf { i } - ( 3 \cos t ) \mathbf { j } , \quad 0 \leqslant t < \frac { \pi } { 2 }$$
When \(t = 0\), the particle is at the point with position vector \(4 \mathbf { j } \mathrm {~m}\).
Find \(\mathbf { r }\) in terms of \(t\).