Exam 12: Vector Functions

Let $\mathbf { a } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = - \mathbf { i } + \mathbf { j }$ then $\mathbf { v } ( t )$ is

The fact that $\frac { d A } { d t }$ is a constant is used to explain

Let $\mathbf { r } ( t ) = e ^ { t } \sin ( t ) \mathbf { i } - e ^ { t } \cos ( t ) \mathbf { j } + ( 1 - t ) ^ { 2 } \mathbf { k }$ . Then the unit tangent vector $\mathbf { T } ( 0 )$ is

The radius of curvature of $\mathbf { r } ( t ) = 2 t \mathbf { i } + \left( t ^ { 2 } - 1 \right) \mathbf { j }$ at the point corresponding to t = 1 is

Let $\mathbf { r } ( t ) = e ^ { t } \cos ( t ) \mathbf { i } - e ^ { t } \sin ( t ) \mathbf { j } + ( 2 - t ) ^ { 2 } \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( 0 )$ is

The reason that the eccentricity of the orbit is neither equal to nor greater than 1 is because ​

Let $y = e ^ { - x }$ . Then the curvature K at (0, 1) is

In $\mathbf { F } ( t ) = - \frac { G m M \mathbf { r } ( t ) } { \| \mathbf { r } ( t ) \| ^ { 3 } }$ , G represents

The normal component $a _ { \mathrm { N } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \sin ( t ) \mathbf { j } + \cos ( t ) \mathbf { k }$ is

The magnitude of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }$ is

Let $\mathbf { v } ( t ) = \sin t \mathbf { i } + \cos t \mathbf { j } + t \mathbf { k }$ and $\mathbf { w } ( t ) = \sin t \mathbf { i } + \cos t \mathbf { j } + t ^ { 3 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \bullet \mathbf { w } ( t ) )$ is

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 4 t \mathbf { k }$ Then the unit tangent vector $\mathbf { T } \left( \frac { \pi } { 4 } \right)$ is

Let $\mathbf { r } ( t ) = 2 t \mathbf { i } + \sin ( 2 t ) \mathbf { j } - \cos ( 2 t ) \mathbf { k }$ Then the curvature K(t) is

Let $\mathbf { r } ( t ) = - 2 t \mathbf { i } - \cos ( t ) \mathbf { j } + \sin ( t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } ( 0 )$ is

Let $\mathbf { a } ( t ) = - 9.5 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = 2 \mathbf { i } - 3 \mathbf { k }$ then the speed $\| \mathbf { v } ( t ) \|$ is

Let $\mathbf { a } ( t ) = - 9.5 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = 2 \mathbf { i } - 3 \mathbf { j }$ then $\mathbf { v } ( t )$ is

Let $\mathbf { r } ( t ) = - \cos t \mathbf { i } + \sin t \mathbf { j } - e ^ { - t } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

Let $\mathbf { a } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = \mathbf { i } + \mathbf { j }$ then $\mathbf { v } ( t )$ is

The speed of a particle moving along the curve $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + \cos ( t ) \mathbf { j } - \sin ( t ) \mathbf { k }$ is

Let $\mathbf { a } ( t ) = - 32 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = \mathbf { i } + 2 \mathbf { j } + 0 \mathbf { k }$ , then $\mathbf { v } ( t )$ is