Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Exercises - Page 735: 43

$\begin{array}{*{20}{c}} {{\bf{T}}\left( 1 \right) = \left( {\frac{1}{3},\frac{2}{3},\frac{2}{3}} \right)}\\ {{\bf{N}}\left( 1 \right) = \left( { - \frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)}\\ {{\bf{B}}\left( 1 \right) = \left( {\frac{2}{3}, - \frac{2}{3},\frac{1}{3}} \right)} \end{array}$

We have ${\bf{r}}\left( t \right) = \left( {t,{t^2},\frac{2}{3}{t^3}} \right)$. The point $\left( {1,1,\frac{2}{3}} \right)$ corresponds to $t=1$. The tangent vector is ${\bf{r}}'\left( t \right) = \left( {1,2t,2{t^2}} \right)$. So, the unit tangent vector is ${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {1,2t,2{t^2}} \right)}}{{\sqrt {\left( {1,2t,2{t^2}} \right)\cdot\left( {1,2t,2{t^2}} \right)} }}$ ${\bf{T}}\left( t \right) = \left( {\frac{1}{{\sqrt {1 + 4{t^2} + 4{t^4}} }},\frac{{2t}}{{\sqrt {1 + 4{t^2} + 4{t^4}} }},\frac{{2{t^2}}}{{\sqrt {1 + 4{t^2} + 4{t^4}} }}} \right)$ ${\bf{T}}\left( t \right) = \left( {\frac{1}{{\sqrt {{{\left( {1 + 2{t^2}} \right)}^2}} }},\frac{{2t}}{{\sqrt {{{\left( {1 + 2{t^2}} \right)}^2}} }},\frac{{2{t^2}}}{{\sqrt {{{\left( {1 + 2{t^2}} \right)}^2}} }}} \right)$ ${\bf{T}}\left( t \right) = \left( {\frac{1}{{1 + 2{t^2}}},\frac{{2t}}{{1 + 2{t^2}}},\frac{{2{t^2}}}{{1 + 2{t^2}}}} \right)$ 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 $\frac{d}{{dt}}\left( {\frac{1}{{1 + 2{t^2}}}} \right)$ $\frac{d}{{dt}}\left( {\frac{1}{{1 + 2{t^2}}}} \right) = \frac{d}{{dt}}\left( {{{\left( {1 + 2{t^2}} \right)}^{ - 1}}} \right)$ $\frac{d}{{dt}}\left( {\frac{1}{{1 + 2{t^2}}}} \right) = - \left( {4t} \right){\left( {1 + 2{t^2}} \right)^{ - 2}} = - \frac{{4t}}{{{{\left( {1 + 2{t^2}} \right)}^2}}}$ 2. Evaluate $\frac{d}{{dt}}\left( {\frac{{2t}}{{1 + 2{t^2}}}} \right)$ $\frac{d}{{dt}}\left( {\frac{{2t}}{{1 + 2{t^2}}}} \right) = \frac{{2\left( {1 + 2{t^2}} \right) - \left( {2t} \right)\left( {4t} \right)}}{{{{\left( {1 + 2{t^2}} \right)}^2}}}$ $\frac{d}{{dt}}\left( {\frac{{2t}}{{1 + 2{t^2}}}} \right) = \frac{{2 - 4{t^2}}}{{{{\left( {1 + 2{t^2}} \right)}^2}}}$ 3. Evaluate $\frac{d}{{dt}}\left( {\frac{{2{t^2}}}{{1 + 2{t^2}}}} \right)$ $\frac{d}{{dt}}\left( {\frac{{2{t^2}}}{{1 + 2{t^2}}}} \right) = \frac{{4t\left( {1 + 2{t^2}} \right) - \left( {2{t^2}} \right)\left( {4t} \right)}}{{{{\left( {1 + 2{t^2}} \right)}^2}}}$ $\frac{d}{{dt}}\left( {\frac{{2{t^2}}}{{1 + 2{t^2}}}} \right) = \frac{{4t}}{{{{\left( {1 + 2{t^2}} \right)}^2}}}$ Thus, ${\bf{T}}'\left( t \right) = \left( { - \frac{{4t}}{{{{\left( {1 + 2{t^2}} \right)}^2}}},\frac{{2 - 4{t^2}}}{{{{\left( {1 + 2{t^2}} \right)}^2}}},\frac{{4t}}{{{{\left( {1 + 2{t^2}} \right)}^2}}}} \right)$ At $t=1$, we get ${\bf{T}}'\left( 1 \right) = \left( { - \frac{4}{9}, - \frac{2}{9},\frac{4}{9}} \right)$ $||{\bf{T}}'\left( 1 \right)|{|^2} = \frac{{16}}{{81}} + \frac{4}{{81}} + \frac{{16}}{{81}} = \frac{{36}}{{81}} = \frac{4}{9}$ $||{\bf{T}}'\left( 1 \right)|| = \frac{2}{3}$ The normal vector at $t=1$, ${\bf{N}}\left( 1 \right) = \frac{{{\bf{T}}'\left( 1 \right)}}{{||{\bf{T}}'\left( 1 \right)||}}$ ${\bf{N}}\left( 1 \right) = \frac{3}{2}\left( { - \frac{4}{9}, - \frac{2}{9},\frac{4}{9}} \right)$ ${\bf{N}}\left( 1 \right) = \left( { - \frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)$ 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 $\begin{array}{*{20}{c}} {{\bf{T}}\left( 1 \right) = \left( {\frac{1}{3},\frac{2}{3},\frac{2}{3}} \right)}\\ {{\bf{N}}\left( 1 \right) = \left( { - \frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)}\\ {{\bf{B}}\left( 1 \right) = \left( {\frac{2}{3}, - \frac{2}{3},\frac{1}{3}} \right)} \end{array}$