Answer
At $t = \frac{\pi }{4}$:
${\bf{a}}\left( {\frac{\pi }{4}} \right) = \left( { - \frac{1}{2}\sqrt 2 , - 4} \right) = \frac{1}{2}\sqrt 2 {\bf{T}}\left( {\frac{\pi }{4}} \right) + 4{\bf{N}}\left( {\frac{\pi }{4}} \right)$,
where ${\bf{T}}\left( {\frac{\pi }{4}} \right) = \left( { - 1,0} \right)$ and ${\bf{N}}\left( {\frac{\pi }{4}} \right) = \left( {0, - 1} \right)$.
Work Step by Step
We have ${\bf{r}}\left( \theta \right) = \left( {\cos \theta ,\sin 2\theta } \right)$. So, the tangential and the acceleration vectors are
${\bf{r}}'\left( \theta \right) = {\bf{v}}\left( \theta \right) = \left( { - \sin \theta ,2\cos 2\theta } \right)$
and ${\bf{r}}{\rm{''}}\left( \theta \right) = {\bf{a}}\left( \theta \right) = \left( { - \cos \theta , - 4\sin 2\theta } \right)$,
respectively.
At $\theta = \frac{\pi }{4}$, we get
${\bf{v}}\left( {\frac{\pi }{4}} \right) = \left( { - \frac{1}{2}\sqrt 2 ,0} \right)$
${\bf{a}}\left( {\frac{\pi }{4}} \right) = \left( { - \frac{1}{2}\sqrt 2 , - 4} \right)$
The unit tangent vector at $\theta = \frac{\pi }{4}$ is
${\bf{T}}\left( {\frac{\pi }{4}} \right) = \frac{{{\bf{r}}'\left( {\frac{\pi }{4}} \right)}}{{||{\bf{r}}'\left( {\frac{\pi }{4}} \right)||}} = \frac{{\left( { - \frac{1}{2}\sqrt 2 ,0} \right)}}{{\sqrt {\left( { - \frac{1}{2}\sqrt 2 ,0} \right)\cdot\left( { - \frac{1}{2}\sqrt 2 ,0} \right)} }}$
${\bf{T}}\left( {\frac{\pi }{4}} \right) = \frac{1}{{\sqrt {\frac{1}{2}} }}\left( { - \frac{1}{2}\sqrt 2 ,0} \right) = \left( { - 1,0} \right)$
By Eq. (2) of Theorem 1 (Section 14.5), the tangential component of acceleration is
${a_{\bf{T}}} = {\bf{a}}\left( {\frac{\pi }{4}} \right)\cdot{\bf{T}}\left( {\frac{\pi }{4}} \right) = \left( { - \frac{1}{2}\sqrt 2 , - 4} \right)\cdot\left( { - 1,0} \right)$
${a_{\bf{T}}} = \frac{1}{2}\sqrt 2 $
By Eq. (3) of Theorem 1 (Section 14.5),
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$
$ = \left( { - \frac{1}{2}\sqrt 2 , - 4} \right) - \frac{1}{2}\sqrt 2 \left( { - 1,0} \right)$
$ = \left( { - \frac{1}{2}\sqrt 2 , - 4} \right) - \left( { - \frac{1}{2}\sqrt 2 ,0} \right)$
$ = \left( {0, - 4} \right)$
Since ${\bf{N}}$ is unit vector, so
$||{a_{\bf{N}}}{\bf{N}}|| = {a_{\bf{N}}} = \sqrt {\left( {0, - 4} \right)\cdot\left( {0, - 4} \right)} = 4$
So,
${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \frac{1}{4}\left( {0, - 4} \right) = \left( {0, - 1} \right)$
Finally, we obtain the decomposition of ${\bf{a}}$:
${\bf{a}}\left( t \right) = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$
at $t = \frac{\pi }{4}$:
${\bf{a}}\left( {\frac{\pi }{4}} \right) = \left( { - \frac{1}{2}\sqrt 2 , - 4} \right) = \frac{1}{2}\sqrt 2 {\bf{T}}\left( {\frac{\pi }{4}} \right) + 4{\bf{N}}\left( {\frac{\pi }{4}} \right)$,
where ${\bf{T}}\left( {\frac{\pi }{4}} \right) = \left( { - 1,0} \right)$ and ${\bf{N}}\left( {\frac{\pi }{4}} \right) = \left( {0, - 1} \right)$.
