Find the unit tangent vector for the curve given by $\mathbf { r } ( t ) = \left\langle \frac { 1 } { 5 } t ^ { 5 } , \frac { 1 } { 3 } t ^ { 3 } , t \right)$.
A) $\frac { \left\langle t ^ { 4 } , t ^ { 2 } , 1 \right) } { \sqrt { t ^ { 10 } + t ^ { 4 } } }$
B) $\frac { \left( t ^ { 4 } , t ^ { 2 } , 1 \right) } { \sqrt { 4 t ^ { 8 } + 4 t ^ { 4 } + 1 } }$
C)
$\frac { \left\langle t ^ { 4 } , t ^ { 2 } , 1 \right) } { \sqrt { t ^ { 8 } + t ^ { 4 } + 1 } }$
D)
$\frac { \left\langle t ^ { 4 } , t ^ { 2 } , 1 \right\rangle } { \sqrt { t ^ { 8 } + t ^ { 4 } } }$
E) None of these

Question

Quizplus · Accepted Answer

First, we find the velocity vector: $\mathbf{r}'(t) = \left\langle t^4, t^2, 1 ightangle.$

Next, we find the magnitude of the velocity vector: $\left\| \mathbf{r}'(t) ight\| = \sqrt{t^8 + t^4 + 1}.$

Therefore, the unit tangent vector is: $$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\left\| \mathbf{r}'(t) ight\|}=\boxed{\frac{\left\langle t^4,t^2,1ightangle}{\sqrt{t^8+t^4+1}}}	ext{.}$$

Find the Unit Tangent Vector for the Curve Given By \mathbf