Standard linear first order - vector form

3. The position vector \(\mathbf { r }\) of a particle \(P\) at time \(t\) satisfies the vector differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 4 \mathbf { i }$$ Given that the position vector of \(P\) at time \(t = 0\) is \(2 \mathbf { j }\), find the position vector of \(P\) at time \(t\).
(Total 6 marks)

2. With respect to a fixed origin \(O\), the position vector, \(\mathbf { r }\) metres, of a particle \(P\) at time \(t\) seconds satisfies $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \mathbf { r } = ( \mathbf { i } - \mathbf { j } ) \mathrm { e } ^ { - 2 t } .$$ Given that \(P\) is at \(O\) when \(t = 0\), find

1. \(\mathbf { r }\) in terms of \(t\),
2. a cartesian equation of the path of \(P\).

2. At time \(t\) seconds the position vector of a particle \(P\), relative to a fixed origin \(O\), is \(\mathbf { r }\) metres, where \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 3 \mathrm { e } ^ { - t } \mathbf { j }$$ Given that \(\mathbf { r } = 2 \mathbf { i } - \mathbf { j }\) when \(t = 0\), find \(\mathbf { r }\) in terms of \(t\).
(Total 7 marks)

2. The velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) of a particle \(P\) at time \(t\) seconds satisfies the vector differential equation $$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 4 \mathbf { v } = \mathbf { 0 }$$ The position vector of \(P\) at time \(t\) seconds is \(\mathbf { r }\) metres.
Given that at \(t = 0 , \mathbf { r } = ( \mathbf { i } - \mathbf { j } )\) and \(\mathbf { v } = ( - 8 \mathbf { i } + 4 \mathbf { j } )\), find \(\mathbf { r }\) at time \(t\) seconds.

1. At time \(t = 0\), the position vector of a particle \(P\) is \(- 3 \mathbf { j } \mathrm {~m}\). At time \(t\) seconds, the position vector of \(P\) is \(\mathbf { r }\) metres and the velocity of \(P\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). Given that

$$\mathbf { v } - 2 \mathbf { r } = 4 \mathrm { e } ^ { t } \mathbf { j }$$ find the time when \(P\) passes through the origin.

1. Solve the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - 2 \mathbf { r } = \mathbf { 0 }$$ given that when \(t = 0 , \mathbf { r } . \mathbf { j } = 0\) and \(\mathbf { r } \times \mathbf { j } = \mathbf { i } + \mathbf { k }\).

1. 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 \(t\).

2. A particle \(P\) moves so that its position vector, \(\mathbf { r }\) metres, at time \(t\) seconds, where \(0 \leqslant t < \frac { \pi } { 2 }\), satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - ( \tan t ) \mathbf { r } = ( \sin t ) \mathbf { i }$$ When \(t = 0 , \mathbf { r } = - \frac { 1 } { 2 } \mathbf { i }\).
Find \(\mathbf { r }\) in terms of \(t\).

3. A particle \(P\) moves in the \(x y\)-plane in such a way that its position vector \(\mathbf { r }\) metres at time \(t\) seconds, where \(0 \leqslant t < \pi\), satisfies the differential equation $$\sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \sec ^ { 3 } \left( \frac { 1 } { 2 } t \right) \sin \left( \frac { 1 } { 2 } t \right) \mathbf { r } = \sin \left( \frac { 1 } { 2 } t \right) \mathbf { i } + \sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \mathbf { j }$$ When \(t = 0\), the particle is at the point with position vector \(( - \mathbf { i } + \mathbf { j } ) \mathrm { m }\).
Find \(\mathbf { r }\) in terms of \(t\).