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

A) $\left( 9 t ^ { 2 } - 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$
B) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } - 36 t \mathbf { j } - ( 4 + 24 t ) \mathbf { k }$
C) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$
D) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } - ( 4 + 24 t ) \mathbf { k }$
E) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } - 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$

Question

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

A) $\left( 9 t ^ { 2 } - 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$
B) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } - 36 t \mathbf { j } - ( 4 + 24 t ) \mathbf { k }$
C) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$
D) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } - ( 4 + 24 t ) \mathbf { k }$
E) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } - 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$

Quizplus · Accepted Answer

To find $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$, we use the product rule for differentiation, which in the context of vector cross products states that $\frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \frac{d\mathbf{u}}{dt} \times \mathbf{v} + \mathbf{u} \times \frac{d\mathbf{v}}{dt}$. Applying this rule to $\mathbf{v}(t)$ and $\mathbf{w}(t)$, and after computing the cross products and their derivatives, we obtain the expression corresponding to choice C.

Let v(t)=2i−(3t+1)j+4tk\mathbf { v } ( t ) = 2 \mathbf { i } - ( 3 t + 1 ) \mathbf { j } + 4 t \mathbf { k }v(t)=2i−(3t+1)j+4tk

Let $\mathbf { v } ( t ) = 2 \mathbf { i } - ( 3 t + 1 ) \mathbf { j } + 4 t \mathbf { k }$