Answer
(a)
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( 1 \right) = \left( {\frac{1}{3},\frac{2}{3},\frac{2}{3}} \right)}\\
{{\bf{N}}\left( 1 \right) = \left( { - \frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)}\\
{{\bf{B}}\left( 1 \right) = \left( {\frac{2}{3}, - \frac{2}{3},\frac{1}{3}} \right)}
\end{array}$
(b) the equation of the osculating plane at the point corresponding to $t=1$:
$6x - 6y + 3z = 1$
Work Step by Step
(a) We have ${\bf{r}}\left( t \right) = \left( {t,\frac{4}{3}{t^{3/2}},{t^2}} \right)$. The tangent vector is ${\bf{r}}'\left( t \right) = \left( {1,2{t^{1/2}},2t} \right)$. So, the unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {1,2{t^{1/2}},2t} \right)}}{{\sqrt {\left( {1,2{t^{1/2}},2t} \right)\cdot\left( {1,2{t^{1/2}},2t} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( {1,2{t^{1/2}},2t} \right)}}{{\sqrt {1 + 4t + 4{t^2}} }} = \frac{{\left( {1,2{t^{1/2}},2t} \right)}}{{\sqrt {{{\left( {1 + 2t} \right)}^2}} }}$
${\bf{T}}\left( t \right) = \left( {\frac{1}{{1 + 2t}},\frac{{2{t^{1/2}}}}{{1 + 2t}},\frac{{2t}}{{1 + 2t}}} \right)$
At $t=1$, we get ${\bf{T}}\left( 1 \right) = \left( {\frac{1}{3},\frac{2}{3},\frac{2}{3}} \right)$.
To evaluate ${\bf{T}}'\left( t \right)$:
1. Evaluate $\frac{d}{{dt}}\left( {\frac{1}{{1 + 2t}}} \right)$
$\frac{d}{{dt}}\left( {\frac{1}{{1 + 2t}}} \right) = \frac{d}{{dt}}\left( {{{\left( {1 + 2t} \right)}^{ - 1}}} \right) = - \frac{2}{{{{\left( {1 + 2t} \right)}^2}}}$
2. Evaluate $\frac{d}{{dt}}\left( {\frac{{2{t^{1/2}}}}{{1 + 2t}}} \right)$
$\frac{d}{{dt}}\left( {\frac{{2{t^{1/2}}}}{{1 + 2t}}} \right) = \frac{{{t^{ - 1/2}}\left( {1 + 2t} \right) - 2{t^{1/2}}\left( 2 \right)}}{{{{\left( {1 + 2t} \right)}^2}}}$
$\frac{d}{{dt}}\left( {\frac{{2{t^{1/2}}}}{{1 + 2t}}} \right) = \frac{{\left( {1 + 2t} \right) - 4t}}{{{t^{1/2}}{{\left( {1 + 2t} \right)}^2}}} = \frac{{1 - 2t}}{{{t^{1/2}}{{\left( {1 + 2t} \right)}^2}}}$
3. Evaluate $\frac{d}{{dt}}\left( {\frac{{2t}}{{1 + 2t}}} \right)$
$\frac{d}{{dt}}\left( {\frac{{2t}}{{1 + 2t}}} \right) = \frac{{2\left( {1 + 2t} \right) - \left( {2t} \right)\left( 2 \right)}}{{{{\left( {1 + 2t} \right)}^2}}} = \frac{2}{{{{\left( {1 + 2t} \right)}^2}}}$
Thus,
${\bf{T}}'\left( t \right) = \left( { - \frac{2}{{{{\left( {1 + 2t} \right)}^2}}},\frac{{1 - 2t}}{{{t^{1/2}}{{\left( {1 + 2t} \right)}^2}}},\frac{2}{{{{\left( {1 + 2t} \right)}^2}}}} \right)$
At $t=1$, we get
${\bf{T}}'\left( 1 \right) = \left( { - \frac{2}{9}, - \frac{1}{9},\frac{2}{9}} \right)$
$||{\bf{T}}'\left( 1 \right)|{|^2} = \frac{4}{{81}} + \frac{1}{{81}} + \frac{4}{{81}} = \frac{9}{{81}} = \frac{1}{9}$
$||{\bf{T}}'\left( 1 \right)|| = \frac{1}{3}$
The normal vector at $t=1$:
${\bf{N}}\left( 1 \right) = \frac{{{\bf{T}}'\left( 1 \right)}}{{||{\bf{T}}'\left( 1 \right)||}}$
${\bf{N}}\left( 1 \right) = 3\left( { - \frac{2}{9}, - \frac{1}{9},\frac{2}{9}} \right)$
${\bf{N}}\left( 1 \right) = \left( { - \frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)$
By Eq. (8), the binormal vector at $t=1$ is given by
${\bf{B}}\left( 1 \right) = {\bf{T}}\left( 1 \right) \times {\bf{N}}\left( 1 \right)$
${\bf{B}}\left( 1 \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{1}{3}}&{\frac{2}{3}}&{\frac{2}{3}}\\
{ - \frac{2}{3}}&{ - \frac{1}{3}}&{\frac{2}{3}}
\end{array}} \right|$
${\bf{B}}\left( 1 \right) = \left( {\frac{4}{9} + \frac{2}{9}} \right){\bf{i}} - \left( {\frac{2}{9} + \frac{4}{9}} \right){\bf{j}} + \left( { - \frac{1}{9} + \frac{4}{9}} \right){\bf{k}}$
${\bf{B}}\left( 1 \right) = \frac{2}{3}{\bf{i}} - \frac{2}{3}{\bf{j}} + \frac{1}{3}{\bf{k}}$
In summary, we obtain
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( 1 \right) = \left( {\frac{1}{3},\frac{2}{3},\frac{2}{3}} \right)}\\
{{\bf{N}}\left( 1 \right) = \left( { - \frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)}\\
{{\bf{B}}\left( 1 \right) = \left( {\frac{2}{3}, - \frac{2}{3},\frac{1}{3}} \right)}
\end{array}$
(b) In part (a) we get ${\bf{B}}\left( 1 \right) = \left( {\frac{2}{3}, - \frac{2}{3},\frac{1}{3}} \right)$. The point corresponding to $t=1$ is ${\bf{r}}\left( 1 \right) = \left( {1,\frac{4}{3},1} \right)$.
Thus, by Theorem 1 of Section 13.5, the equation of the osculating plane at the point corresponding to $t=1$ is
$\frac{2}{3}\left( {x - 1} \right) - \frac{2}{3}\left( {y - \frac{4}{3}} \right) + \frac{1}{3}\left( {z - 1} \right) = 0$
$2\left( {x - 1} \right) - 2\left( {y - \frac{4}{3}} \right) + z - 1 = 0$
$6x - 6y + 3z = 1$
