Find the tangential and normal components for the position function $\vec { r } ( t ) = \left\langle e ^ { t } , e ^ { - t } \right\rangle$
A) $\begin{array} { l }
a _ { T } = e ^ { 2 t } - e ^ { - 2 t } \
a _ { N } = 2
\end{array}$
B) $\begin{array} { l }
a _ { T } = \frac { 2 } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } } \
a _ { N } = \frac { e ^ { 2 t } - e ^ { - 2 t } } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } }
\end{array}$
C) $\begin{array} { l }
a _ { T } = \frac { 2 } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } } \
a _ { N } = \frac { e ^ { 2 t } + e ^ { - 2 t } } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } }
\end{array}$
D) $\begin{array} { l }
a _ { T } = \frac { e ^ { 2 t } - e ^ { - 2 t } } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } } \
a _ { N } = \frac { 2 } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } }
\end{array}$

Question

Find the tangential and normal components for the position function $\vec { r } ( t ) = \left\langle e ^ { t } , e ^ { - t } \right\rangle$
A) $\begin{array} { l } 
a _ { T } = e ^ { 2 t } - e ^ { - 2 t } \
a _ { N } = 2
\end{array}$
B) $\begin{array} { l } 
a _ { T } = \frac { 2 } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } } \
a _ { N } = \frac { e ^ { 2 t } - e ^ { - 2 t } } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } }
\end{array}$
C) $\begin{array} { l } 
a _ { T } = \frac { 2 } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } } \
a _ { N } = \frac { e ^ { 2 t } + e ^ { - 2 t } } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } }
\end{array}$
D) $\begin{array} { l } 
a _ { T } = \frac { e ^ { 2 t } - e ^ { - 2 t } } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } } \
a _ { N } = \frac { 2 } { \sqrt { e ^ { 2 t } + e ^ { - 2 t } } }
\end{array}$

Quizplus · Accepted Answer

The tangential component $a_T$ is the component of acceleration in the direction of the velocity vector, and the normal component $a_N$ is perpendicular to the velocity vector. To find these, we first calculate the velocity $\vec{v}(t)$ and acceleration $\vec{a}(t)$ vectors by differentiating $\vec{r}(t)$, then use these to find $a_T$ and $a_N$.Velocity $\vec{v}(t) = \frac{d\vec{r}}{dt} = \left\langle e^t, -e^{-t} \right\rangle$.Acceleration $\vec{a}(t) = \frac{d\vec{v}}{dt} = \left\langle e^t, e^{-t} \right\rangle$.The magnitude of velocity $|\vec{v}(t)| = \sqrt{(e^t)^2 + (-e^{-t})^2} = \sqrt{e^{2t} + e^{-2t}}$.The tangential acceleration $a_T = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|}$, and the normal acceleration $a_N = \frac{|\vec{v} \times \vec{a}|}{|\vec{v}|}$.After calculations, $a_T$ aligns with the formula given in choice D for the tangential component, and $a_N$ aligns with the formula given in choice D for the normal component, based on the proper application of these formulas and the properties of the dot and cross product in two dimensions.

Find the Tangential and Normal Components for the Position Function r⃗(t)=⟨et,e−t⟩\vec { r } ( t ) = \left\langle e ^ { t } , e ^ { - t } \right\rangler(t)=⟨et,e−t⟩

Find the Tangential and Normal Components for the Position Function $\vec { r } ( t ) = \left\langle e ^ { t } , e ^ { - t } \right\rangle$