Answer
(a)
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( 0 \right) = \left( {0,1,0} \right)}\\
{{\bf{N}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 2 }},0, - \frac{1}{{\sqrt 2 }}} \right)}\\
{{\bf{B}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)}
\end{array}$
(b) the equation of the osculating plane at the point $\left( {1,0,0} \right)$:
$ - x + z = - 1$
Work Step by Step
(a) We have ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,\ln \left( {\cos t} \right)} \right)$. The tangent vector is ${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t, - \tan t} \right)$. So, the unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( { - \sin t,\cos t, - \tan t} \right)}}{{\sqrt {\left( { - \sin t,\cos t, - \tan t} \right)\cdot\left( { - \sin t,\cos t, - \tan t} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( { - \sin t,\cos t, - \tan t} \right)}}{{\sqrt {{{\sin }^2}t + {{\cos }^2}t + {{\tan }^2}t} }} = \frac{{\left( { - \sin t,\cos t, - \tan t} \right)}}{{\sqrt {1 + {{\tan }^2}t} }}$
Since ${\sec ^2}t - {\tan ^2}t = 1$, so
${\bf{T}}\left( t \right) = \frac{1}{{\sec t}}\left( { - \sin t,\cos t, - \tan t} \right) = \left( { - \frac{1}{2}\sin 2t,{{\cos }^2}t, - \sin t} \right)$
The point $\left( {1,0,0} \right)$ corresponds to $t=0$. So, at $t=0$, we get ${\bf{T}}\left( 0 \right) = \left( {0,1,0} \right)$.
The derivative of ${\bf{T}}\left( t \right)$ is
${\bf{T}}'\left( t \right) = \left( { - \cos 2t, - \sin 2t, - \cos t} \right)$
At $t=0$, we get ${\bf{T}}'\left( 0 \right) = \left( { - 1,0, - 1} \right)$.
The normal vector at $t=0$:
${\bf{N}}\left( 0 \right) = \frac{{{\bf{T}}'\left( 0 \right)}}{{||{\bf{T}}'\left( 0 \right)||}}$
${\bf{N}}\left( 0 \right) = \frac{{\left( { - 1,0, - 1} \right)}}{{\sqrt {\left( { - 1,0, - 1} \right)\cdot\left( { - 1,0, - 1} \right)} }} = \frac{1}{{\sqrt 2 }}\left( { - 1,0, - 1} \right)$
${\bf{N}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 2 }},0, - \frac{1}{{\sqrt 2 }}} \right)$
By Eq. (8), the binormal vector at $t=0$ is given by
${\bf{B}}\left( 0 \right) = {\bf{T}}\left( 0 \right) \times {\bf{N}}\left( 0 \right)$
${\bf{B}}\left( 0 \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
0&1&0\\
{ - \frac{1}{{\sqrt 2 }}}&0&{ - \frac{1}{{\sqrt 2 }}}
\end{array}} \right|$
${\bf{B}}\left( 0 \right) = - \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{k}}$
In summary, we obtain
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( 0 \right) = \left( {0,1,0} \right)}\\
{{\bf{N}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 2 }},0, - \frac{1}{{\sqrt 2 }}} \right)}\\
{{\bf{B}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)}
\end{array}$
(b) the point $\left( {1,0,0} \right)$ corresponds to $t=0$. In part (a) we obtain ${\bf{B}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}} \right)$.
Thus, by Theorem 1 of Section 13.5, the equation of the osculating plane at the point $\left( {1,0,0} \right)$ is
$ - \frac{1}{{\sqrt 2 }}\left( {x - 1} \right) + 0\left( {y - 0} \right) + \frac{1}{{\sqrt 2 }}\left( {z - 0} \right) = 0$
$ - \left( {x - 1} \right) + z = 0$
$ - x + z = - 1$
