Find the unit tangent vector for the curve given by $\mathbf { r } ( t ) = \left\langle \frac { 1 } { 7 } t ^ { 7 } , \frac { 1 } { 3 } t ^ { 3 } , t \right\rangle$ .
A) $\frac { \left\langle t ^ { 6 } , t ^ { 2 } , 1 \right\rangle } { \sqrt { t ^ { 10 } + t ^ { 4 } } }$
B) $\frac { \left\langle t ^ { 6 } , t ^ { 2 } , 1 \right\rangle } { \sqrt { 6 t ^ { 12 } + 4 t ^ { 4 } + 1 } }$
C) $\frac { \left\langle t ^ { 6 } , t ^ { 2 } , 1 \right\rangle } { \sqrt { t ^ { 12 } + t ^ { 4 } } }$
D) $\frac { \left\langle t ^ { 6 } , t ^ { 2 } , 1 \right\rangle } { \sqrt { t ^ { 12 } + t ^ { 4 } + 1 } }$
E) None of these

Question

Quizplus · Accepted Answer

To find the unit tangent vector, first compute the derivative $\mathbf{r}'(t) = \left\langle t^6, t^2, 1 \right\rangle$. Then, find the magnitude $\|\mathbf{r}'(t)\| = \sqrt{t^{12} + t^4 + 1}$. The unit tangent vector is $\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{\left\langle t^6, t^2, 1 \right\rangle}{\sqrt{t^{12} + t^4 + 1}}$.

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