Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } \left( \frac { \pi } { 6 } \right)$ is

A) $\sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$
B) $\sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$
C) $- \sqrt { 3 } \mathbf { i } + \mathbf { j } + 6 \mathbf { k }$
D) $- \sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$
E) $- \sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$

Question

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } \left( \frac { \pi } { 6 } \right)$ is

A) $\sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$
B) $\sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$
C) $- \sqrt { 3 } \mathbf { i } + \mathbf { j } + 6 \mathbf { k }$
D) $- \sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$
E) $- \sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$

Quizplus · Accepted Answer

The derivative of $\mathbf{r}(t)$ is $\mathbf{r}'(t) = -2\sin(2t)\mathbf{i} - 2\cos(2t)\mathbf{j} + 6\mathbf{k}$. Evaluating at $t = \frac{\pi}{6}$, we get $\mathbf{r}'\left(\frac{\pi}{6}\right) = -\sqrt{3}\mathbf{i} - \mathbf{j} + 6\mathbf{k}$.

Let r(t)=cos⁡(2t)i−sin⁡(2t)j+6tk\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }r(t)=cos(2t)i−sin(2t)j+6tk

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }$