Answer
(a)
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{\sqrt 5 }},0,\frac{2}{{\sqrt 5 }}} \right),}\\
{{\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right),}\\
{{\bf{B}}\left( {\frac{\pi }{2}} \right) = \left( {\frac{2}{{\sqrt 5 }},0,\frac{1}{{\sqrt 5 }}} \right)}
\end{array}$
(b) the equation of the osculating plane at the point corresponding to $t = \frac{\pi }{2}$ is
$2x + z = \pi $
Work Step by Step
(a) We have ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,2t} \right)$. So, the tangent vector is ${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t,2} \right)$.
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( { - \sin t,\cos t,2} \right)}}{{||\left( { - \sin t,\cos t,2} \right)||}}$
${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {\left( { - \sin t,\cos t,2} \right)\cdot\left( { - \sin t,\cos t,2} \right)} }}\left( { - \sin t,\cos t,2} \right)$
${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {{{\sin }^2}t + {{\cos }^2}t + 4} }}\left( { - \sin t,\cos t,2} \right) = \left( { - \frac{{\sin t}}{{\sqrt 5 }},\frac{{\cos t}}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)$
So, ${\bf{T}}'\left( t \right) = \left( { - \frac{{\cos t}}{{\sqrt 5 }}, - \frac{{\sin t}}{{\sqrt 5 }},0} \right)$
At $t = \frac{\pi }{2}$, we get ${\bf{T}}'\left( {\frac{\pi }{2}} \right) = \left( {0, - \frac{1}{{\sqrt 5 }},0} \right)$. Thus,
${\bf{N}}\left( {\frac{\pi }{2}} \right) = \frac{{{\bf{T}}'\left( {\frac{\pi }{2}} \right)}}{{||{\bf{T}}'\left( {\frac{\pi }{2}} \right)||}} = \sqrt 5 \left( {0, - \frac{1}{{\sqrt 5 }},0} \right)$
${\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right)$
The binormal vector is defined by ${\bf{B}} = {\bf{T}} \times {\bf{N}}$. So,
${\bf{B}}\left( {\frac{\pi }{2}} \right) = {\bf{T}}\left( {\frac{\pi }{2}} \right) \times {\bf{N}}\left( {\frac{\pi }{2}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - \frac{1}{{\sqrt 5 }}}&0&{\frac{2}{{\sqrt 5 }}}\\
0&{ - 1}&0
\end{array}} \right|$
${\bf{B}}\left( {\frac{\pi }{2}} \right) = \frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{k}}$
In summary, we obtain
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{\sqrt 5 }},0,\frac{2}{{\sqrt 5 }}} \right),}\\
{{\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right),}\\
{{\bf{B}}\left( {\frac{\pi }{2}} \right) = \left( {\frac{2}{{\sqrt 5 }},0,\frac{1}{{\sqrt 5 }}} \right)}
\end{array}$
(b) In part (a) we obtain ${\bf{B}}\left( {\frac{\pi }{2}} \right) = \left( {\frac{2}{{\sqrt 5 }},0,\frac{1}{{\sqrt 5 }}} \right)$.
The point corresponding to $t = \frac{\pi }{2}$ is ${\bf{r}}\left( {\frac{\pi }{2}} \right) = \left( {0,1,\pi } \right)$.
Thus, by Theorem 1 of Section 13.5, the equation of the osculating plane at the point corresponding to $t = \frac{\pi }{2}$ is
$\frac{2}{{\sqrt 5 }}\left( {x - 0} \right) + 0\left( {y - 1} \right) + \frac{1}{{\sqrt 5 }}\left( {z - \pi } \right) = 0$
$2\left( {x - 0} \right) + \left( {z - \pi } \right) = 0$
$2x + z = \pi $
