Answer
The equation of the osculating plane at the point corresponding to $t=1$ is
$2x - 4y + 2z = - 3$
Work Step by Step
Step 1. Find ${\bf{T}}$, ${\bf{N}}$, and ${\bf{B}}$ at the point corresponding to $t=1$.
We have ${\bf{r}}\left( t \right) = \left( {\ln t,t,\frac{{{t^2}}}{2}} \right)$. So, the tangent vector is ${\bf{r}}'\left( t \right) = \left( {\frac{1}{t},1,t} \right)$.
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {\frac{1}{t},1,t} \right)}}{{||\left( {\frac{1}{t},1,t} \right)||}}$
${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {\left( {\frac{1}{t},1,t} \right)\cdot\left( {\frac{1}{t},1,t} \right)} }}\left( {\frac{1}{t},1,t} \right)$
${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {\frac{1}{{{t^2}}} + 1 + {t^2}} }}\left( {\frac{1}{t},1,t} \right) = \frac{t}{{\sqrt {1 + {t^2} + {t^4}} }}\left( {\frac{1}{t},1,t} \right)$
${\bf{T}}\left( t \right) = \left( {\frac{1}{{\sqrt {1 + {t^2} + {t^4}} }},\frac{t}{{\sqrt {1 + {t^2} + {t^4}} }},\frac{{{t^2}}}{{\sqrt {1 + {t^2} + {t^4}} }}} \right)$
To find ${\bf{T}}'\left( t \right)$:
1. Evaluate $\frac{d}{{dt}}\frac{1}{{\sqrt {1 + {t^2} + {t^4}} }}$
$\frac{d}{{dt}}\frac{1}{{\sqrt {1 + {t^2} + {t^4}} }} = - \frac{{2t + 4{t^3}}}{{2{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}} = - \frac{{t + 2{t^3}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}}$
2. Evaluate $\frac{d}{{dt}}\frac{t}{{\sqrt {1 + {t^2} + {t^4}} }}$
$\frac{d}{{dt}}\frac{t}{{\sqrt {1 + {t^2} + {t^4}} }} = \frac{{\sqrt {1 + {t^2} + {t^4}} - t\left( {\frac{{t + 2{t^3}}}{{\sqrt {1 + {t^2} + {t^4}} }}} \right)}}{{1 + {t^2} + {t^4}}}$
$\frac{d}{{dt}}\frac{t}{{\sqrt {1 + {t^2} + {t^4}} }} = \frac{{1 + {t^2} + {t^4} - {t^2} - 2{t^4}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}} = \frac{{1 - {t^4}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}}$
3. Evaluate $\frac{d}{{dt}}\frac{{{t^2}}}{{\sqrt {1 + {t^2} + {t^4}} }}$
$\frac{d}{{dt}}\frac{{{t^2}}}{{\sqrt {1 + {t^2} + {t^4}} }} = \frac{{2t\sqrt {1 + {t^2} + {t^4}} - {t^2}\left( {\frac{{t + 2{t^3}}}{{\sqrt {1 + {t^2} + {t^4}} }}} \right)}}{{1 + {t^2} + {t^4}}}$
$\frac{d}{{dt}}\frac{{{t^2}}}{{\sqrt {1 + {t^2} + {t^4}} }} = \frac{{2t\left( {1 + {t^2} + {t^4}} \right) - {t^3} - 2{t^5}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}} = \frac{{2t + {t^3}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}}$
It follows that
${\bf{T}}'\left( t \right) = \left( { - \frac{{t + 2{t^3}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}},\frac{{1 - {t^4}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}},\frac{{2t + {t^3}}}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}}} \right)$
At $t=1$, we get ${\bf{T}}'\left( 1 \right) = \left( { - \frac{1}{{\sqrt 3 }},0,\frac{1}{{\sqrt 3 }}} \right)$. Thus,
${\bf{N}}\left( 1 \right) = \frac{{{\bf{T}}'\left( 1 \right)}}{{||{\bf{T}}'\left( 1 \right)||}} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\left( { - \frac{1}{{\sqrt 3 }},0,\frac{1}{{\sqrt 3 }}} \right)$
${\bf{N}}\left( 1 \right) = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)$
The binormal vector is defined by ${\bf{B}} = {\bf{T}} \times {\bf{N}}$. So,
${\bf{B}}\left( 1 \right) = {\bf{T}}\left( 1 \right) \times {\bf{N}}\left( 1 \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\
{ - \frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}}
\end{array}} \right|$
${\bf{B}}\left( 1 \right) = \frac{1}{{\sqrt 6 }}{\bf{i}} - \frac{2}{{\sqrt 6 }}{\bf{j}} + \frac{1}{{\sqrt 6 }}{\bf{k}}$
In summary, we obtain
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( 1 \right) = \left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right),}\\
{{\bf{N}}\left( 1 \right) = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right),}\\
{{\bf{B}}\left( 1 \right) = \left( {\frac{1}{{\sqrt 6 }}, - \frac{2}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }}} \right)}
\end{array}$
Step 2. Find the equation of the osculating plane corresponding $t=1$.
From previous results we obtain ${\bf{B}}\left( 1 \right) = \left( {\frac{1}{{\sqrt 6 }}, - \frac{2}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }}} \right)$.
The point corresponding to $t=1$ is ${\bf{r}}\left( 1 \right) = \left( {0,1,\frac{1}{2}} \right)$.
Thus, by Theorem 1 of Section 13.5, the equation of the osculating plane at the point corresponding to $t=1$ is
$\frac{1}{{\sqrt 6 }}\left( {x - 0} \right) - \frac{2}{{\sqrt 6 }}\left( {y - 1} \right) + \frac{1}{{\sqrt 6 }}\left( {z - \frac{1}{2}} \right) = 0$
$\left( {x - 0} \right) - 2\left( {y - 1} \right) + \left( {z - \frac{1}{2}} \right) = 0$
$x - 2y + 2 + z - \frac{1}{2} = 0$
$2x - 4y + 2z = - 3$
