Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Exercises - Page 736: 55

(a) ${\bf{T}}\left( t \right) = \left( {\frac{1}{{\sqrt {2 + 4{t^2}} }}, - \frac{1}{{\sqrt {2 + 4{t^2}} }},\frac{{2t}}{{\sqrt {2 + 4{t^2}} }}} \right)$ ${\bf{N}}\left( t \right) = \left( { - \frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}},\frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}},\frac{1}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}}} \right)$ (b) ${\bf{B}}\left( t \right) = - \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}}$ (c) From the result in part (b), we get 1. ${\bf{B}}$ is not a function of $t$, that is, it is constant. Since ${\bf{B}}$ is perpendicular to the osculating plane, we conclude that the osculating plane is fixed for all $t$. 2. ${\bf{B}}$ is parallel to the $xy$-plane. Since ${\bf{B}}$ is perpendicular to the osculating plane, we conclude that the osculating plane is perpendicular to the $xy$-plane.

(a) We have ${\bf{r}}\left( t \right) = \left( {t,1 - t,{t^2}} \right)$. The tangent vector is ${\bf{r}}'\left( t \right) = \left( {1, - 1,2t} \right)$. So, the unit tangent vector is ${\bf{T}}\left( t \right) = \frac{{\left( {1, - 1,2t} \right)}}{{\sqrt {\left( {1, - 1,2t} \right)\cdot\left( {1, - 1,2t} \right)} }} = \frac{{\left( {1, - 1,2t} \right)}}{{\sqrt {2 + 4{t^2}} }}$ ${\bf{T}}\left( t \right) = \left( {\frac{1}{{\sqrt {2 + 4{t^2}} }}, - \frac{1}{{\sqrt {2 + 4{t^2}} }},\frac{{2t}}{{\sqrt {2 + 4{t^2}} }}} \right)$ Find derivative of ${\bf{T}}\left( t \right)$: 1. Evaluate $\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2 + 4{t^2}} }}} \right)$ $\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2 + 4{t^2}} }}} \right) = \frac{d}{{dt}}\left( {{{\left( {2 + 4{t^2}} \right)}^{ - 1/2}}} \right)$ $\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2 + 4{t^2}} }}} \right) = - \frac{{8t}}{{2{{\left( {2 + 4{t^2}} \right)}^{3/2}}}} = - \frac{{4t}}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}}$ 2. Evaluate $\frac{d}{{dt}}\left( {\frac{{2t}}{{\sqrt {2 + 4{t^2}} }}} \right)$ $\frac{d}{{dt}}\left( {\frac{{2t}}{{\sqrt {2 + 4{t^2}} }}} \right) = \frac{{2\sqrt {2 + 4{t^2}} - \left( {2t} \right)\left( {\frac{{8t}}{{2\sqrt {2 + 4{t^2}} }}} \right)}}{{2 + 4{t^2}}}$ $\frac{d}{{dt}}\left( {\frac{{2t}}{{\sqrt {2 + 4{t^2}} }}} \right) = \frac{{2\left( {2 + 4{t^2}} \right) - 8{t^2}}}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}} = \frac{4}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}}$ Thus, ${\bf{T}}'\left( t \right) = \left( { - \frac{{4t}}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}},\frac{{4t}}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}},\frac{4}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}}} \right)$ $||{\bf{T}}'\left( t \right)|{|^2} = \frac{{16{t^2}}}{{{{\left( {2 + 4{t^2}} \right)}^3}}} + \frac{{16{t^2}}}{{{{\left( {2 + 4{t^2}} \right)}^3}}} + \frac{{16}}{{{{\left( {2 + 4{t^2}} \right)}^3}}}$ $||{\bf{T}}'\left( t \right)|{|^2} = \frac{{32{t^2} + 16}}{{{{\left( {2 + 4{t^2}} \right)}^3}}} = \frac{{8\left( {2 + 4{t^2}} \right)}}{{{{\left( {2 + 4{t^2}} \right)}^3}}} = \frac{8}{{{{\left( {2 + 4{t^2}} \right)}^2}}}$ $||{\bf{T}}'\left( t \right)|| = \frac{{\sqrt 8 }}{{2 + 4{t^2}}}$ The normal vector at $t$ is ${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$ ${\bf{N}}\left( t \right) = \left( {\frac{{2 + 4{t^2}}}{{\sqrt 8 }}} \right)\left( { - \frac{{4t}}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}},\frac{{4t}}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}},\frac{4}{{{{\left( {2 + 4{t^2}} \right)}^{3/2}}}}} \right)$ ${\bf{N}}\left( t \right) = \left( { - \frac{{2t}}{{\sqrt 2 {{\left( {2 + 4{t^2}} \right)}^{1/2}}}},\frac{{2t}}{{{{\left( {2 + 4{t^2}} \right)}^{1/2}}}},\frac{2}{{{{\left( {2 + 4{t^2}} \right)}^{1/2}}}}} \right)$ ${\bf{N}}\left( t \right) = \left( { - \frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}},\frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}},\frac{1}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}}} \right)$ (b) In part (a) we obtain ${\bf{T}}\left( t \right) = \left( {\frac{1}{{\sqrt {2 + 4{t^2}} }}, - \frac{1}{{\sqrt {2 + 4{t^2}} }},\frac{{2t}}{{\sqrt {2 + 4{t^2}} }}} \right)$ ${\bf{N}}\left( t \right) = \left( { - \frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}},\frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}},\frac{1}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}}} \right)$ By Eq. (8), the binormal vector ${\bf{B}}$ as a function of t is given by ${\bf{B}}\left( t \right) = {\bf{T}}\left( t \right) \times {\bf{N}}\left( t \right)$ ${\bf{B}}\left( t \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ {\frac{1}{{\sqrt {2 + 4{t^2}} }}}&{ - \frac{1}{{\sqrt {2 + 4{t^2}} }}}&{\frac{{2t}}{{\sqrt {2 + 4{t^2}} }}}\\ { - \frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}}}&{\frac{t}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}}}&{\frac{1}{{{{\left( {1 + 2{t^2}} \right)}^{1/2}}}}} \end{array}} \right|$ ${\bf{B}}\left( t \right) = \left( { - \frac{1}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}} - \frac{{2{t^2}}}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}}} \right){\bf{i}}$ $ - \left( {\frac{1}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}} + \frac{{2{t^2}}}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}}} \right){\bf{j}} + \left( {\frac{t}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}} - \frac{t}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}}} \right){\bf{k}}$ ${\bf{B}}\left( t \right) = \left( { - \frac{{1 + 2{t^2}}}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}}} \right){\bf{i}} - \left( {\frac{{1 + 2{t^2}}}{{\sqrt 2 \left( {1 + 2{t^2}} \right)}}} \right){\bf{j}}$ ${\bf{B}}\left( t \right) = - \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}}$ (c) In part (b), we obtain ${\bf{B}}\left( t \right) = - \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}}$. We notice that: 1. ${\bf{B}}$ is not a function of $t$, that is, it is constant. Since ${\bf{B}}$ is perpendicular to the osculating plane, we conclude that the osculating plane is fixed for all $t$. 2. ${\bf{B}}$ is parallel to the $xy$-plane. Since ${\bf{B}}$ is perpendicular to the osculating plane, we conclude that the osculating plane is perpendicular to the $xy$-plane.