Answer
His acceleration vector at $t = \frac{\pi }{2}$:
${\bf{a}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{2\sqrt 2 }}, - \frac{9}{2},\frac{1}{{2\sqrt 2 }}} \right)$
Work Step by Step
We have the path traced by the runner ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,t} \right)$.
At $t = \frac{\pi }{2}$, we have his speed $v\left( {\frac{\pi }{2}} \right) = 3$. He is accelerating with $v'\left( t \right) = {a_{\bf{T}}}\left( t \right) = \frac{1}{2}$.
1. Find the unit tangent vector and normal vector of the path:
${\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{||{\bf{v}}\left( t \right)||}} = \frac{{\left( { - \sin t,\cos t,1} \right)}}{{\sqrt {\left( { - \sin t,\cos t,1} \right)\cdot\left( { - \sin t,\cos t,1} \right)} }}$
${\bf{T}}\left( t \right) = \left( { - \frac{1}{{\sqrt 2 }}\sin t,\frac{1}{{\sqrt 2 }}\cos t,\frac{1}{{\sqrt 2 }}} \right)$
${\bf{T}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)$
We find the unit normal vector using
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}} = \frac{{\left( { - \frac{1}{{\sqrt 2 }}\cos t, - \frac{1}{{\sqrt 2 }}\sin t,0} \right)}}{{\sqrt {\left( { - \frac{1}{{\sqrt 2 }}\cos t, - \frac{1}{{\sqrt 2 }}\sin t,0} \right)\cdot\left( { - \frac{1}{{\sqrt 2 }}\cos t, - \frac{1}{{\sqrt 2 }}\sin t,0} \right)} }}$
${\bf{N}}\left( t \right) = \left( { - \cos t, - \sin t,0} \right)$
${\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right)$
2. Next, we evaluate the curvature of the path by using Eq. (3) of Theorem 1 of Section 14.4:
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
$\kappa \left( {\frac{\pi }{2}} \right) = \frac{{||\left( { - 1,0,1} \right) \times \left( {0, - 1,0} \right)||}}{{{{\left( {\sqrt {\left( { - 1,0,1} \right)\cdot\left( { - 1,0,1} \right)} } \right)}^3}}} = \frac{1}{2}$
By Eq. (1), the normal component of the acceleration is
${a_{\bf{N}}} = \kappa \left( t \right)v{\left( t \right)^2}$
At $t = \frac{\pi }{2}$, we get ${a_{\bf{N}}} = \frac{1}{2}{\left( 3 \right)^2} = \frac{9}{2}$.
Thus, the acceleration vector of the runner is decomposed into
${\bf{a}}\left( t \right) = {a_{\bf{T}}}{\bf{T}}\left( t \right) + {a_{\bf{N}}}{\bf{N}}\left( t \right)$
${\bf{a}}\left( {\frac{\pi }{2}} \right) = \frac{1}{2}{\bf{T}}\left( {\frac{\pi }{2}} \right) + \frac{9}{2}{\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{2\sqrt 2 }}, - \frac{9}{2},\frac{1}{{2\sqrt 2 }}} \right)$,
where ${\bf{T}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)$ and ${\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right)$.
