Chapter 11 - Vectors and Vector-Valued Functions - 11.9 Curvature and Normal Vectors - 11.9 Exercises - Page 852: 41

$${\bf{B}} = - {\bf{k}}{\text{ and }}\tau = 0$$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {2\sin t,2\cos t} \right\rangle \cr & {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}\left| {{\bf{v}}\left( t \right)} \right| \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left\langle {2\sin t,2\cos t} \right\rangle \cr & {\bf{v}}\left( t \right) = \left\langle {2\cos t, - 2\sin t} \right\rangle \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {4{{\cos }^2}t + 4{{\sin }^2}t} \cr & \left| {{\bf{v}}\left( t \right)} \right| = 2 \cr & \cr & {\text{Find the unit tangent vector }}{\bf{T}}\left( t \right) \cr & {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} = \frac{{\left\langle {2\cos t, - 2\sin t} \right\rangle }}{2} \cr & {\bf{T}}\left( t \right) = \left\langle {\cos t, - \sin t} \right\rangle \cr & \cr & {\text{Find the principal unit normal vector }}{\bf{N}}\left( t \right) \cr & {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} \cr & {\bf{T}}'\left( t \right) = \left\langle { - \sin t, - \cos t} \right\rangle \cr & \left| {{\bf{T}}'\left( t \right)} \right| = \sqrt {{{\sin }^2}t + {{\cos }^2}t} = 1 \cr & {\text{Then}} \cr & {\bf{N}}\left( t \right) = \frac{{\left\langle { - \sin t, - \cos t} \right\rangle }}{1} \cr & {\bf{N}}\left( t \right) = \left\langle { - \sin t, - \cos t} \right\rangle \cr & \cr & {\text{Calculate the unit binormal vector:}}\,{\text{ }}{\bf{B}} = {\bf{T}} \times {\bf{N}} \cr & {\bf{B}} = \left\langle {\cos t, - \sin t} \right\rangle \times \left\langle { - \sin t, - \cos t} \right\rangle \cr} $$ \[{\mathbf{B}} = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ {\cos t}&{ - \sin t}&0 \\ { - \sin t}&{ - \cos t}&0 \end{array}} \right|\] $$\eqalign{ & {\bf{B}} = 0{\bf{i}} - {\bf{j}} + \left( { - {{\cos }^2}t - {{\sin }^2}t} \right){\bf{k}} \cr & {\bf{B}} = - {\bf{k}} \cr & \cr & {\text{The next step is to determine }}\frac{{d{\bf{B}}}}{{ds}} \cr & \frac{{d{\bf{B}}}}{{ds}} = \frac{{d{\bf{B}}/dt}}{{ds/dt}} \cr & {\text{In this case }}\frac{{ds}}{{dt}} = \left| {{\bf{r}}'\left( t \right)} \right| = \left| {{\bf{v}}\left( t \right)} \right| \cr & \left| {{\bf{r}}'\left( t \right)} \right| = 2 \cr & \frac{{d{\bf{B}}}}{{dt}} = \frac{d}{{dt}}\left\langle {0,0, - 1} \right\rangle \cr & \frac{{d{\bf{B}}}}{{dt}} = \left\langle {0,0,0} \right\rangle \cr & \frac{{d{\bf{B}}}}{{dt}} = \frac{{\left\langle {0,0,0} \right\rangle }}{2} = \left\langle {0,0,0} \right\rangle \cr & \cr & {\text{Calculate }}\tau \cr & \tau = - \frac{{d{\bf{B}}}}{{ds}} \cdot {\bf{N}} \cr & \tau = - \left\langle {0,0,0} \right\rangle \cdot \left\langle { - \sin t, - \cos t} \right\rangle \cr & \tau = 0 \cr} $$