Find the unit tangent vector of the given curve.
-$$r ( t ) = 3 t ^ { 5 } i - 12 t ^ { 5 } j + 4 t ^ { 5 } \mathbf { k }$$
A) $\mathrm { T } = \frac { 3 } { 13 } \mathrm { i } + \frac { 12 } { 13 } \mathrm { j } + \frac { 4 } { 13 } \mathbf { k }$
B) $T = \frac { 3 } { 13 } \mathbf { i } - \frac { 12 } { 13 } j \mathbf { j } - \frac { 4 } { 13 } \mathbf { k }$
C) $\mathbf { T } = \frac { 3 } { 169 } \mathbf { i } - \frac { 12 } { 169 } \mathbf { j } + \frac { 4 } { 169 } \mathbf { k }$
D) $\mathbf { T } = \frac { 3 } { 13 } \mathbf { i } - \frac { 12 } { 13 } \mathbf { j } + \frac { 4 } { 13 } \mathbf { k }$

Question

Quizplus · Accepted Answer

The unit tangent vector $\mathbf{T}$ is given by $\mathbf{T} = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}$, where $\mathbf{r}'(t)$ is the derivative of $\mathbf{r}(t)$ with respect to $t$.

We have:
$$\mathbf{r}'(t) = 15t^4 \mathbf{i} - 60t^4 \mathbf{j} + 20t^4 \mathbf{k}$$
and
$$\|\mathbf{r}'(t)\| = \sqrt{(15t^4)^2 + (-60t^4)^2 + (20t^4)^2} = 65t^4$$

Therefore, the unit tangent vector is
$$\mathbf{T} = \frac{15t^4 \mathbf{i} - 60t^4 \mathbf{j} + 20t^4 \mathbf{k}}{65t^4} = \frac{3}{13} \mathbf{i} - \frac{12}{13} \mathbf{j} + \frac{4}{13} \mathbf{k}.$$ So, the best choice is $\boxed{\text{D}}$.

Find the Unit Tangent Vector of the Given Curve r(t)=3t5i−12t5j+4t5kr ( t ) = 3 t ^ { 5 } i - 12 t ^ { 5 } j + 4 t ^ { 5 } \mathbf { k }r(t)=3t5i−12t5j+4t5k

Find the Unit Tangent Vector of the Given Curve $r ( t ) = 3 t ^ { 5 } i - 12 t ^ { 5 } j + 4 t ^ { 5 } \mathbf { k }$