Exam 12: Vector Functions

Let $\mathbf { v } ( t ) = t ^ { 2 } \mathbf { i } - \mathbf { j } + 3 t \mathbf { k }$ and $\mathbf { w } ( t ) = 2 t \mathbf { i } + 3 \mathbf { j } - t \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

Let $y = 4 x - x ^ { 2 }$ . Then the curvature K at (1, 3) is

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

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

Let $\mathbf { r } ( t ) = 6 t \mathbf { i } + \cos t \mathbf { j } - \sin t \mathbf { k }$ Then the unit tangent vector $\mathbf { T } \left( \frac { \pi } { 6 } \right)$ is

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

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

Which of the following is Newton's Law of Universal Gravitation?

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

Let $\mathbf { v } ( t ) = 2 \mathbf { i } - 3 t \mathbf { j } + t ^ { 2 } \mathbf { k }$ and $\mathbf { w } ( t ) = - \mathbf { i } + 3 t \mathbf { j } + 5 t ^ { 2 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

Let $\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } - 4 t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } \left( \frac { \pi } { 4 } \right)$ is

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

Newton's Second Law Motion states that ​

The radius of curvature of $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + t \mathbf { k }$ at the point corresponding to t = 0 is

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + \cos t \mathbf { j } + \ln ( 5 - t ) \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

In $\mathbf { F } ( t ) = - \frac { \operatorname { GmM } \mathbf { r } ( t ) } { \| \mathbf { r } ( t ) \| ^ { 3 } }$ , F represents

In $\mathbf { F } ( t ) = - \frac { \operatorname { GmM } \mathbf { r } ( t ) } { \| \mathbf { r } ( t ) \| ^ { 3 } }$ , r represents

Newton's Law of Universal Gravitation states that ​

The domain of the vector function $\mathbf { v } ( t ) = \frac { 3 } { \sqrt { t ^ { 2 } - 4 } } \mathbf { i } - \sin ( 2 t ) \mathbf { j } + \frac { 1 } { t } \mathbf { k }$ is

Let $\mathbf { v } ( t ) = 2 \mathbf { i } - ( 3 t + 1 ) \mathbf { j } + 4 t \mathbf { k }$ and $\mathbf { w } ( t ) = 4 t \mathbf { i } + 5 \mathbf { j } - t ^ { 2 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is