Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 22

$$\eqalign{ & {\bf{T}}\left( 1 \right) = \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\bf{N}}\left( 1 \right) = \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {a_{\bf{T}}} = \sqrt 2 \cr & {a_{\bf{N}}} = \sqrt 2 \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + 2t{\bf{j}},{\text{ }}t = 1 \cr & {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}{\bf{a}}\left( t \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{\bf{i}} + 2t{\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = 2t{\bf{i}} + 2{\bf{j}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {2t{\bf{i}} + 2{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = 2{\bf{i}} \cr & {\bf{a}}\left( 1 \right) = 2{\bf{i}} \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\|}},{\text{ }}{\bf{v}}\left( t \right) \ne 0 \cr & {\bf{T}}\left( t \right) = \frac{{2t{\bf{i}} + 2{\bf{j}}}}{{\left\| {2t{\bf{i}} + 2{\bf{j}}} \right\|}} = \frac{{2t{\bf{i}} + 2{\bf{j}}}}{{\sqrt {4{t^2} + 4} }} = \frac{{2t{\bf{i}} + 2{\bf{j}}}}{{2\sqrt {{t^2} + 1} }} = \frac{{t{\bf{i}} + {\bf{j}}}}{{\sqrt {{t^2} + 1} }} \cr & {\bf{T}}\left( t \right) = \frac{1}{{\sqrt {{t^2} + 1} }}\left( {t{\bf{i}} + {\bf{j}}} \right) \cr & {\bf{T}}\left( 1 \right) = \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{t}{{\sqrt {{t^2} + 1} }}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\frac{1}{{\sqrt {{t^2} + 1} }}} \right]{\bf{j}} \cr & {\text{Differentiate by using a CAS }}\left( {{\text{WolframWebsite}}} \right) \cr & {\bf{T}}'\left( t \right) = \frac{{2t}}{{{{\left( {{t^4} + 1} \right)}^{3/2}}}}{\bf{i}} - \frac{t}{{{{\left( {{t^4} + 1} \right)}^{3/2}}}}{\bf{j}} \cr & {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 1 \cr & {\bf{T}}'\left( 1 \right) = \frac{{2\left( 1 \right)}}{{{{\left( {{1^4} + 1} \right)}^{3/2}}}}{\bf{i}} - \frac{1}{{{{\left( {{1^4} + 1} \right)}^{3/2}}}}{\bf{j}} \cr & {\bf{T}}'\left( 1 \right) = \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\text{Finding the Principal Unit Normal Vector}} \cr & {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr & {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = 1 \cr & {\bf{N}}\left( 1 \right) = \frac{{{\bf{T}}'\left( 1 \right)}}{{\left\| {{\bf{T}}'\left( 1 \right)} \right\|}} = \frac{{\frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}}}}{{\left\| {\frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{2\sqrt 2 }}{\bf{j}}} \right\|}} = \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\text{Find }}{a_{\bf{T}}}{\text{ and }}{a_{\bf{N}}}{\text{ at }}t = 1 \cr & {a_{\bf{T}}} = {\bf{a}} \cdot {\bf{T}} = \left( {2{\bf{i}}} \right)\left( {\frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}}} \right) \cr & {a_{\bf{T}}} = \sqrt 2 \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( {2{\bf{i}}} \right)\left( {\frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}}} \right) \cr & {a_{\bf{N}}} = \sqrt 2 \cr & \cr & {\text{Summary}} \cr & {\bf{T}}\left( 1 \right) = \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\bf{N}}\left( 1 \right) = \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {a_{\bf{T}}} = \sqrt 2 \cr & {a_{\bf{N}}} = \sqrt 2 \cr} $$