Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 25

$$\eqalign{ & {\bf{T}}\left( 0 \right) = \frac{1}{{\sqrt 5 }}{\bf{i}} - \frac{2}{{\sqrt 5 }}{\bf{j}} \cr & {\bf{N}}\left( 0 \right) = \frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{j}} \cr & {a_{\bf{T}}} = - \frac{7}{{\sqrt 5 }} \cr & {a_{\bf{N}}} = \frac{{6\sqrt 5 }}{5} \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = {e^t}{\bf{i}} + {e^{ - 2t}}{\bf{j}},{\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^{ - 2t}}{\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = {e^t}{\bf{i}} - 2{e^{ - 2t}}{\bf{j}} \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}} - 2{e^{ - 2t}}{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = {e^t}{\bf{i}} + 4{e^{ - 2t}}{\bf{j}} \cr & {\bf{a}}\left( 0 \right) = {e^0}{\bf{i}} + 4{e^0}{\bf{j}} \cr & {\bf{a}}\left( 0 \right) = {\bf{i}} + 4{\bf{j}} \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}} - 2{e^{ - 2t}}{\bf{j}}}}{{\left\| {{e^t}{\bf{i}} - 2{e^{ - 2t}}{\bf{j}}} \right\|}} = \frac{{{e^t}{\bf{i}} - 2{e^{ - 2t}}{\bf{j}}}}{{\sqrt {{e^{2t}} + 4{e^{ - 4t}}} }} \cr & {\bf{T}}\left( 0 \right) = \frac{{{e^0}{\bf{i}} - 2{e^0}{\bf{j}}}}{{\sqrt {{e^0} + 4{e^0}} }} \cr & {\bf{T}}\left( 0 \right) = \frac{1}{{\sqrt 5 }}{\bf{i}} - \frac{2}{{\sqrt 5 }}{\bf{j}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{{e^t}}}{{\sqrt {{e^{2t}} + 4{e^{ - 4t}}} }}} \right]{\bf{i}} - \frac{d}{{dt}}\left[ {\frac{{2{e^{ - 2t}}}}{{\sqrt {{e^{2t}} + 4{e^{ - 4t}}} }}} \right]{\bf{j}} \cr & {\text{Differentiate by using a CAS }}\left( {{\text{WolframWebsite}}} \right) \cr & {\bf{T}}'\left( t \right) = \frac{{12{e^{ - 3t}}}}{{{{\left[ {{e^{ - 4t}}\left( {{e^{6t}} + 4} \right)} \right]}^{3/2}}}}{\bf{i}} + \frac{6}{{{{\left[ {{e^{ - 4t}}\left( {{e^{6t}} + 4} \right)} \right]}^{3/2}}}}{\bf{j}} \cr & {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 0 \cr & {\bf{T}}'\left( 0 \right) = \frac{{12{e^0}}}{{{{\left[ {{e^0}\left( {{e^0} + 4} \right)} \right]}^{3/2}}}}{\bf{i}} + \frac{6}{{{{\left[ {{e^0}\left( {{e^0} + 4} \right)} \right]}^{3/2}}}}{\bf{j}} \cr & {\bf{T}}'\left( 0 \right) = \frac{{12}}{{5\sqrt 5 }}{\bf{i}} + \frac{6}{{5\sqrt 5 }}{\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{{12}}{{5\sqrt 5 }}{\bf{i}} + \frac{6}{{5\sqrt 5 }}{\bf{j}}}}{{\left\| {\frac{{12}}{{5\sqrt 5 }}{\bf{i}} + \frac{6}{{5\sqrt 5 }}{\bf{j}}} \right\|}} = \frac{{\frac{{12}}{{5\sqrt 5 }}{\bf{i}} + \frac{6}{{5\sqrt 5 }}{\bf{j}}}}{{6/5}} \cr & {\bf{N}}\left( 0 \right) = \frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\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}} + 4{\bf{j}}} \right)\left( {\frac{1}{{\sqrt 5 }}{\bf{i}} - \frac{2}{{\sqrt 5 }}{\bf{j}}} \right) \cr & {a_{\bf{T}}} = \frac{1}{{\sqrt 5 }} - \frac{8}{{\sqrt 5 }} = - \frac{7}{{\sqrt 5 }} \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( {{\bf{i}} + 4{\bf{j}}} \right)\left( {\frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{j}}} \right) \cr & {a_{\bf{N}}} = \frac{2}{{\sqrt 5 }} + \frac{4}{{\sqrt 5 }} = \frac{{6\sqrt 5 }}{5} \cr & \cr & {\text{Summary}} \cr & {\bf{T}}\left( 0 \right) = \frac{1}{{\sqrt 5 }}{\bf{i}} - \frac{2}{{\sqrt 5 }}{\bf{j}} \cr & {\bf{N}}\left( 0 \right) = \frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{j}} \cr & {a_{\bf{T}}} = - \frac{7}{{\sqrt 5 }} \cr & {a_{\bf{N}}} = \frac{{6\sqrt 5 }}{5} \cr} $$