Find the unit tangent vector $T ( t )$ . $\mathbf { r } ( t ) = \langle 2 \sin t , 4 t , 2 \cos t \rangle$
A) $\left\langle\frac { \cos t } { 2 \sqrt { 5 } } , \frac { 2 } { \sqrt { 5 } } , - \frac { \sin t } { 2 \sqrt { 5 } } \right\rangle$
B) $\langle3 \cos t , 6 , - 3 \sin t \rangle$
C) $\left\langle\frac { \cos t } { \sqrt { 5 } } , \frac { 2 } { \sqrt { 5 } } , - \frac { \sin t } { \sqrt { 5 } } \right\rangle$
D) $\langle3 \sqrt { 5 } \cos t , 6 , - 3 \sqrt { 5 } \sin t \rangle$
E) $\left\langle- \frac { \cos t } { 2 \sqrt { 5 } } , \frac { 4 } { 2 \sqrt { 5 } } , \frac { 3 \sin t } { 2 \sqrt { 5 } } \right\rangle$

Question

Quizplus · Accepted Answer

The unit tangent vector is given by 
$$T(t) = \frac{r'(t)}{\|r'(t)\|}$$
where $r'(t) = \langle 2\cos t, 4, -2\sin tangle$. Thus, 
\begin{align*}
\|r'(t)\| &= \sqrt{(2\cos t)^2 + 4^2 + (-2\sin t)^2} \
&= \sqrt{4\cos^2t + 16 + 4\sin^2t}\
&= 2\sqrt{1+\sin^2t}.
\end{align*}
Therefore, 
\begin{align*}
T(t) &= \frac{1}{2\sqrt{1+\sin^2t}}\langle 2\cos t, 4, -2\sin tangle \
&= \left\langle\frac{\cos t}{\sqrt{5}}, \frac{2}{\sqrt{5}}, -\frac{\sin t}{\sqrt{5}}ightangle.
\end{align*}

Find the Unit Tangent Vector T(t)T ( t )T(t) r(t)=⟨2sin⁡t,4t,2cos⁡t⟩\mathbf { r } ( t ) = \langle 2 \sin t , 4 t , 2 \cos t \rangler(t)=⟨2sint,4t,2cost⟩

Find the Unit Tangent Vector $T ( t )$ $\mathbf { r } ( t ) = \langle 2 \sin t , 4 t , 2 \cos t \rangle$