The normal component $a _ { \mathrm { N } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + t \mathbf { k }$ is

A) $\frac { 3 } { \sqrt { 2 } }$
B) $\frac { \sqrt { 2 } } { 3 }$
C) $\sqrt { 56 }$
D)1
E)0

Question

The normal component

a _ { \mathrm { N } } ( t )

of the acceleration of a particle moving along the curve

\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + t \mathbf { k }

is

A) $\frac { 3 } { \sqrt { 2 } }$
B) $\frac { \sqrt { 2 } } { 3 }$
C) $\sqrt { 56 }$
D)1
E)0

Quizplus · Accepted Answer

The normal component of acceleration is given by $$a_N(t) = \frac{||\mathbf{a}(t) \times \mathbf{T}(t)||}{|| \mathbf{T}(t)||}$$where $\mathbf{a}(t)$ is the acceleration vector and $\mathbf{T}(t)$ is the unit tangent vector.$$\mathbf{a}(t) = \frac{d^2\mathbf{r}}{dt^2} = e^t \mathbf{i} - e^{-t} \mathbf{j} + \mathbf{k}$$$$\mathbf{T}(t) = \frac{\frac{d\mathbf{r}}{dt}}{|| \frac{d\mathbf{r}}{dt}||}= \frac{e^t \mathbf{i} - e^{-t} \mathbf{j} + \mathbf{k}}{\sqrt{2}}$$$$\mathbf{a}(t) \times \mathbf{T}(t) = \frac{1}{\sqrt{2}}\left(e^t \mathbf{j} + e^{-t} \mathbf{i} - 2\mathbf{k}\right)$$$$||\mathbf{a}(t) \times \mathbf{T}(t)|| = \sqrt{\frac{1}{2}(e^{2t} + e^{-2t} + 4)} = \sqrt{\frac{1}{2}(e^{2t} + 2 + e^{-2t})}$$$$||\mathbf{T}(t)|| = \sqrt{\frac{1}{2}}$$Therefore, $$a_N(t) = \frac{||\mathbf{a}(t) \times \mathbf{T}(t)||}{|| \mathbf{T}(t)||}= \frac{\sqrt{\frac{1}{2}(e^{2t} + 2 + e^{-2t})}}{\sqrt{\frac{1}{2}}} = \sqrt{e^{2t} + 2 + e^{-2t}}$$At $t=0$, $$a_N(0) = \sqrt{e^{2(0)} + 2 + e^{-2(0)}} = \sqrt{2 + 2} = \frac{3}{\sqrt{2}}$$

The Normal Component aN(t)a _ { \mathrm { N } } ( t )aN​(t)

The Normal Component $a _ { \mathrm { N } } ( t )$