Find the unit tangent vector T(t) to the curve r (t) = $\left( t ^ { 2 } - 1,3 t ^ { 2 } - t ^ { 4 } , \frac { 2 } { t } \right)$ when t = 1.
A) $\left( \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } , - \frac { 1 } { \sqrt { 3 } } \right)$

B) $\left( \frac { 1 } { \sqrt { 3 } } , - \frac { 1 } { \sqrt { 3 } } , - \frac { 1 } { \sqrt { 3 } } \right)$
C) $( 0,1,0)$
D) $( 0,0,1 )$
E) $\left( \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } \right)$
F) $\left( \frac { 1 } { \sqrt { 2 } } , 0 , \frac { 1 } { \sqrt { 2 } } \right)$

G) $\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , 0 \right)$

H) $\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right)$

Question

Find the unit tangent vector T(t) to the curve r (t) = $\left( t ^ { 2 } - 1,3 t ^ { 2 } - t ^ { 4 } , \frac { 2 } { t } \right)$ when t = 1.
A) $\left( \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } , - \frac { 1 } { \sqrt { 3 } } \right)$

B) $\left( \frac { 1 } { \sqrt { 3 } } , - \frac { 1 } { \sqrt { 3 } } , - \frac { 1 } { \sqrt { 3 } } \right)$ 
C) $( 0,1,0)$
D) $( 0,0,1 )$
E) $\left( \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } \right)$  
F) $\left( \frac { 1 } { \sqrt { 2 } } , 0 , \frac { 1 } { \sqrt { 2 } } \right)$ 
 
G) $\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , 0 \right)$

H) $\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right)$

Quizplus · Accepted Answer

To find the unit tangent vector $T(t)$ at $t = 1$, first compute the derivative of $r(t)$ with respect to $t$, which gives the tangent vector. Then, normalize this vector to get the unit tangent vector.The derivative of $r(t) = (t^2 - 1, 3t^2 - t^4, \frac{2}{t})$ is $r'(t) = (2t, 6t - 4t^3, -\frac{2}{t^2})$. At $t = 1$, $r'(1) = (2, 2, -2)$.To normalize, divide by the magnitude: $\sqrt{2^2 + 2^2 + (-2)^2} = \sqrt{12} = 2\sqrt{3}$.Thus, the unit tangent vector is $\left(\frac{2}{2\sqrt{3}}, \frac{2}{2\sqrt{3}}, \frac{-2}{2\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$.

Find the Unit Tangent Vector T(t) to the Curve R (t2−1,3t2−t4,2t)\left( t ^ { 2 } - 1,3 t ^ { 2 } - t ^ { 4 } , \frac { 2 } { t } \right)(t2−1,3t2−t4,t2​)

Find the Unit Tangent Vector T(t) to the Curve R $\left( t ^ { 2 } - 1,3 t ^ { 2 } - t ^ { 4 } , \frac { 2 } { t } \right)$