Find the unit tangent vector $\mathbf { T } ( t )$ for $\mathbf { r } ( t ) = 6 t \mathbf { i } + 6 t \mathbf { j } + 7 t \mathbf { k }$ at $t = 2$.
Select the correct answer.
A) $\frac { 3 } { 121 } \mathbf { i } + \frac { 3 } { 121 } \mathbf { j } + \frac { 7 } { 242 } \mathbf { k }$
B) $\frac { 6 } { 11 } \mathbf { i } + \frac { 6 } { 11 } \mathbf { j } + \frac { 7 } { 11 } \mathbf { k }$
C) $12 \mathbf { i } + 12 \mathbf { j } + 14 \mathbf { k }$
D) $6 \mathbf { i } + 6 \mathbf { j } + 7 \mathbf { k }$

Question

Quizplus · Accepted Answer

The unit tangent vector is defined as

$$\mathbf { T } ( t ) = \frac {\mathbf { r' } ( t )}{|\mathbf { r' } ( t )|}$$

where $\mathbf { r } ( t )$ is the position vector and $\mathbf { r' } ( t )$ is its derivative with respect to $t$.

For the given position vector $\mathbf { r } ( t ) = 6 t \mathbf { i } + 6 t \mathbf { j } + 7 t \mathbf { k }$, its derivative is

$$\mathbf { r' } ( t ) = 6 \mathbf { i } + 6 \mathbf { j } + 7 \mathbf { k }$$

At $t = 2$, we have

$$\mathbf { r' } ( 2 ) = 6 \mathbf { i } + 6 \mathbf { j } + 7 \mathbf { k }$$

So, the unit tangent vector at $t = 2$ is

$$\mathbf { T } ( 2 ) = \frac {6 \mathbf { i } + 6 \mathbf { j } + 7 \mathbf { k }}{\sqrt {6^2 + 6^2 + 7^2}} = \frac {6}{11} \mathbf { i } + \frac {6}{11} \mathbf { j } + \frac {7}{11} \mathbf { k }$$

Therefore, the correct answer is B.

Find the Unit Tangent Vector T(t)\mathbf { T } ( t )T(t)

Find the Unit Tangent Vector $\mathbf { T } ( t )$