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

A) $\frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 4 } { \sqrt { 5 } } \mathbf { k }$
B) $- \frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 4 } { \sqrt { 5 } } \mathbf { k }$
C) $- \frac { 1 } { \sqrt { 5 } } \mathbf { i } - \frac { 4 } { \sqrt { 5 } } \mathbf { k }$
D) $- \frac { 4 } { \sqrt { 5 } } \mathbf { i } + \frac { 1 } { \sqrt { 5 } } \mathbf { k }$
E) $\frac { 4 } { \sqrt { 5 } } \mathbf { i } + \frac { 1 } { \sqrt { 5 } } \mathbf { k }$

Question

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

A) $\frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 4 } { \sqrt { 5 } } \mathbf { k }$
B) $- \frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 4 } { \sqrt { 5 } } \mathbf { k }$
C) $- \frac { 1 } { \sqrt { 5 } } \mathbf { i } - \frac { 4 } { \sqrt { 5 } } \mathbf { k }$
D) $- \frac { 4 } { \sqrt { 5 } } \mathbf { i } + \frac { 1 } { \sqrt { 5 } } \mathbf { k }$
E) $\frac { 4 } { \sqrt { 5 } } \mathbf { i } + \frac { 1 } { \sqrt { 5 } } \mathbf { k }$

Quizplus · Accepted Answer

The derivative of $\mathbf{r}(t)$ is $\mathbf{r}'(t) = -2\sin(2t)\mathbf{i} + 2\cos(2t)\mathbf{j} + 4\mathbf{k}$. At $t = \frac{\pi}{4}$, $\mathbf{r}'\left(\frac{\pi}{4}\right) = -2\sin(\frac{\pi}{2})\mathbf{i} + 2\cos(\frac{\pi}{2})\mathbf{j} + 4\mathbf{k} = -2\mathbf{i} + 4\mathbf{k}$. The magnitude is $\sqrt{(-2)^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$. The unit tangent vector is $\mathbf{T}\left(\frac{\pi}{4}\right) = \frac{-2\mathbf{i} + 4\mathbf{k}}{2\sqrt{5}} = -\frac{1}{\sqrt{5}}\mathbf{i} + \frac{4}{\sqrt{5}}\mathbf{k}$.

Let r(t)=cos⁡(2t)i+sin⁡(2t)j+4tk\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 4 t \mathbf { k }r(t)=cos(2t)i+sin(2t)j+4tk

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 4 t \mathbf { k }$