Chapter 14 - Calculus of Vector-Valued Functions - 14.5 Motion in 3-Space - Exercises - Page 745: 42

The decomposition of ${\bf{a}}$ at $\theta=0$: ${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$ ${\bf{a}} = \left( { - 1,0,0} \right) = 0\cdot{\bf{T}} + 1\cdot{\bf{N}}$ ${\bf{a}} = \left( { - 1,0,0} \right) = {\bf{N}}$, where ${\bf{T}} = \left( {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ and ${\bf{N}} = \left( { - 1,0,0} \right)$.

We have ${\bf{r}}\left( \theta \right) = \left( {\cos \theta ,\sin \theta ,\theta } \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( \theta \right) = {\bf{r}}'\left( \theta \right) = \left( { - \sin \theta ,\cos \theta ,1} \right)$ and ${\bf{a}}\left( \theta \right) = {\bf{r}}{\rm{''}}\left( \theta \right) = \left( { - \cos \theta , - \sin \theta ,0} \right)$, respectively. At $\theta=0$, we get ${\bf{v}}\left( 0 \right) = \left( {0,1,1} \right)$ and ${\bf{a}}\left( 0 \right) = \left( { - 1,0,0} \right)$. Thus, the unit tangent vector is ${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {0,1,1} \right)}}{{\sqrt {\left( {0,1,1} \right)\cdot\left( {0,1,1} \right)} }} = \left( {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ By Eq. (2) of Theorem 1 we have ${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( { - 1,0,0} \right)\cdot\left( {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ ${a_{\bf{T}}} = 0$ Next, we use Eq. (3) of Theorem 1 to find ${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$ ${a_{\bf{N}}}{\bf{N}} = \left( { - 1,0,0} \right)$ Since ${\bf{N}}$ is an unit vector, so ${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = 1$ To find ${\bf{N}}$ we use the equation ${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \left( { - 1,0,0} \right)$ Thus, we obtain the decomposition of ${\bf{a}}$ at $\theta=0$: ${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$ ${\bf{a}} = \left( { - 1,0,0} \right) = 0\cdot{\bf{T}} + 1\cdot{\bf{N}}$ ${\bf{a}} = \left( { - 1,0,0} \right) = {\bf{N}}$, where ${\bf{T}} = \left( {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ and ${\bf{N}} = \left( { - 1,0,0} \right)$.