Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 26

$$\eqalign{ & {\bf{T}}\left( 0 \right) = \frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}} \cr & {\bf{N}}\left( 0 \right) = \frac{{\sqrt 2 }}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}} \cr & {a_{\bf{T}}} = \frac{2}{{\sqrt 3 }} \cr & {a_{\bf{N}}} = \sqrt 2 \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = {e^t}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}},{\text{ }}t = 0 \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[ {{e^t}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}}} \right] \cr & {\bf{v}}\left( t \right) = {e^t}{\bf{i}} - {e^{ - t}}{\bf{j}} + {\bf{k}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {{e^t}{\bf{i}} - {e^{ - t}}{\bf{j}} + {\bf{k}}} \right] \cr & {\bf{a}}\left( t \right) = {e^t}{\bf{i}} + {e^{ - t}}{\bf{j}} \cr & \cr & {\bf{v}}\left( 0 \right) = {e^0}{\bf{i}} - {e^{ - 0}}{\bf{j}} + {\bf{k}} \cr & {\bf{v}}\left( 0 \right) = {\bf{i}} - {\bf{j}} + {\bf{k}} \cr & {\bf{a}}\left( 0 \right) = {e^0}{\bf{i}} + {e^0}{\bf{j}} \cr & {\bf{a}}\left( 0 \right) = {\bf{i}} + {\bf{j}} \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\|}},{\text{ }}{\bf{v}}\left( t \right) \ne 0 \cr & {\bf{T}}\left( t \right) = \frac{{{e^t}{\bf{i}} - {e^{ - t}}{\bf{j}} + {\bf{k}}}}{{\left\| {{e^t}{\bf{i}} - {e^{ - t}}{\bf{j}} + {\bf{k}}} \right\|}} = \frac{{{e^t}{\bf{i}} - {e^{ - t}}{\bf{j}} + {\bf{k}}}}{{\sqrt {{e^{2t}} + {e^{ - 2t}} + 1} }} \cr & {\bf{T}}\left( 0 \right) = \frac{{{\bf{v}}\left( 0 \right)}}{{\left\| {{\bf{v}}\left( 0 \right)} \right\|}} = \frac{{{\bf{i}} - {\bf{j}} + {\bf{k}}}}{{\left\| {{\bf{i}} - {\bf{j}} + {\bf{k}}} \right\|}} = \frac{{{\bf{i}} - {\bf{j}} + {\bf{k}}}}{{\sqrt 3 }} \cr & {\bf{T}}\left( 0 \right) = \frac{1}{{\sqrt 3 }}{\bf{i}} - \frac{1}{{\sqrt 3 }}{\bf{j}} + \frac{1}{{\sqrt 3 }}{\bf{k}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{{e^t}{\bf{i}} - {e^{ - t}}{\bf{j}} + {\bf{k}}}}{{\sqrt {{e^{2t}} + {e^{ - 2t}} + 1} }}} \right] \cr & {\text{Differentiate by using a CAS }}\left( {{\text{WolframWebsite}}} \right) \cr & {\bf{T}}'\left( t \right) = \frac{{{e^{3t}} + 2{e^t}}}{{{{\left( {{e^{2t}} + {e^{ - 2t}} + 1} \right)}^{3/2}}}}{\bf{i}} + \frac{{{e^t} + 2{e^{3t}}}}{{{{\left( {{e^{2t}} + {e^{ - 2t}} + 1} \right)}^{3/2}}}}{\bf{j}} \cr & {\text{ }} - \frac{{{e^{2t}} - {e^{ - 2t}}}}{{{{\left( {{e^{2t}} + {e^{ - 2t}} + 1} \right)}^{3/2}}}} \cr & {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 0 \cr & {\bf{T}}'\left( 0 \right) = \frac{3}{{{{\left( 3 \right)}^{3/2}}}}{\bf{i}} + \frac{3}{{{{\left( 3 \right)}^{3/2}}}}{\bf{j}} - \frac{0}{{{{\left( 3 \right)}^{3/2}}}} \cr & {\bf{T}}'\left( 0 \right) = \frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\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 = 0 \cr & {\bf{N}}\left( 0 \right) = \frac{{{\bf{T}}'\left( 0 \right)}}{{\left\| {{\bf{T}}'\left( 0 \right)} \right\|}} = \frac{{\frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}}}}{{\left\| {\frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}}} \right\|}} = \frac{{\frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}}}}{{\frac{{\sqrt 6 }}{3}}} \cr & {\bf{N}}\left( 0 \right) = \frac{{\sqrt 2 }}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}} \cr & {\text{Find }}{a_{\bf{T}}}{\text{ and }}{a_{\bf{N}}}{\text{ at }}t = 0 \cr & {a_{\bf{T}}} = {\bf{a}} \cdot {\bf{T}} = \left( {{\bf{i}} + {\bf{j}}} \right)\left( {\frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}}} \right) \cr & {a_{\bf{T}}} = \frac{2}{{\sqrt 3 }} \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( {{\bf{i}} + {\bf{j}}} \right)\left( {\frac{{\sqrt 2 }}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}}} \right) \cr & {a_{\bf{N}}} = \sqrt 2 \cr & \cr & {\text{Summary}} \cr & {\bf{T}}\left( 0 \right) = \frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}} \cr & {\bf{N}}\left( 0 \right) = \frac{{\sqrt 2 }}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}} \cr & {a_{\bf{T}}} = \frac{2}{{\sqrt 3 }} \cr & {a_{\bf{N}}} = \sqrt 2 \cr} $$