Find the unit tangent vector $\mathbf { T } ( t )$ for $\mathbf { r } ( t ) = 2 t \mathbf { i } + 6 t \mathbf { j } + 3 t \mathbf { k }$ at $t = - 1$
A) $- 2 \mathbf { i } - 6 \mathbf { j } - 3 \mathbf { k }$
B) $\frac { 2 } { 7 }$ i + $\frac { 6 } { 7 }$ j + $\frac { 3 } { 7 }$ k
C) $- \frac { 2 } { 49 }$ i $- \frac { 6 } { 49 }$ j $- \frac { 3 } { 49 }$ k
D) $2 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k }$

Question

Quizplus · Accepted Answer

The unit tangent vector is given by:
$$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\left\|\mathbf{r}'(t)\right\|}$$
Differentiating $\mathbf{r}(t)$ we get:
$$\mathbf{r}'(t)=2\mathbf{i}+6\mathbf{j}+3\mathbf{k}$$
Now, putting $t=-1$, we get:
$$\mathbf{r}'(-1)=2\mathbf{i}+6\mathbf{j}+3\mathbf{k}$$
Therefore, the unit tangent vector at $t=-1$ is:
$$\mathbf{T}(-1)=\frac{2\mathbf{i}+6\mathbf{j}+3\mathbf{k}}{\left\|2\mathbf{i}+6\mathbf{j}+3\mathbf{k}\right\|}=\frac{2}{7}\mathbf{i}+\frac{6}{7}\mathbf{j}+\frac{3}{7}\mathbf{k}$$
Therefore, the answer is (B).

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

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