Find the unit tangent vector of the given curve.
-$$\mathbf { r } ( \mathrm { t } ) = \left( 5 + 2 \mathrm { t } ^ { 8 } \right) \mathbf { i } + \left( 3 + 10 \mathrm { t } ^ { 8 } \right) \mathbf { j } + \left( 5 + 11 t ^ { 8 } \right) \mathbf { k }$$
A) $\mathrm { T } = \frac { 2 } { 225 } \mathrm { i } + \frac { 2 } { 45 } \mathrm { j } + \frac { 11 } { 225 } \mathbf { k }$
B) $\mathrm { T } = \frac { 2 } { 15 } \mathrm { i } + \frac { 2 } { 3 } \mathrm { j } + \frac { 11 } { 15 } \mathbf { k }$
C) $\mathbf { T } = \frac { 16 } { 15 } \mathbf { i } + \frac { 16 } { 3 } \mathbf { j } + \frac { 88 } { 15 } \mathbf { k }$
D) $\mathbf { T } = 2 \mathbf { i } + 10 \mathbf { j } + 11 \mathbf { k }$

Question

Quizplus · Accepted Answer

The unit tangent vector is given by $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$. We have
\begin{align*}
\mathbf{r}'(t) &= 16t^7\mathbf{i} + 80t^7\mathbf{j} + 88t^7\mathbf{k}, \
|\mathbf{r}'(t)| &= \sqrt{(16t^7)^2 + (80t^7)^2 + (88t^7)^2} \
&= 16t^7\sqrt{1+25^2+11^2} \
&= 225t^7\sqrt{827}.
\end{align*}
Therefore, the unit tangent vector is
\begin{align*}
\mathbf{T}(t) &= \frac{16t^7}{225t^7\sqrt{827}}\mathbf{i} + \frac{80t^7}{225t^7\sqrt{827}}\mathbf{j} + \frac{88t^7}{225t^7\sqrt{827}}\mathbf{k} \
&= \frac{2}{15}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{11}{15}\mathbf{k}.
\end{align*}

Find the Unit Tangent Vector of the Given Curve A) T=2225i+245j+11225k\mathrm { T } = \frac { 2 } { 225 } \mathrm { i } + \frac { 2 } { 45 } \mathrm { j } + \frac { 11 } { 225 } \mathbf { k }T=2252​i+452​j+22511​k

Find the Unit Tangent Vector of the Given Curve A) $\mathrm { T } = \frac { 2 } { 225 } \mathrm { i } + \frac { 2 } { 45 } \mathrm { j } + \frac { 11 } { 225 } \mathbf { k }$