Chapter 12 - Vector-Valued Functions - Review Exercises - Page 864: 51

$$\eqalign{ & {\bf{T}}\left( 3 \right) = - \frac{1}{{\sqrt {325} }}{\bf{i}} - \frac{{18}}{{\sqrt {325} }}{\bf{j}} \cr & {\bf{N}}\left( 3 \right) = \frac{{18}}{{\sqrt {325} }}{\bf{i}} - \frac{1}{{\sqrt {325} }}{\bf{j}} \cr & {a_{\bf{T}}} = - \frac{2}{{9\sqrt {325} }} \cr & {a_{\bf{N}}} = \frac{4}{{\sqrt {325} }} \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \frac{3}{t}{\bf{i}} - 6t{\bf{j}},{\text{ }}t = 3 \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[ {\frac{3}{t}{\bf{i}} - 6t{\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = - \frac{3}{{{t^2}}}{\bf{i}} - 6{\bf{j}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ { - \frac{3}{{{t^2}}}{\bf{i}} - 6{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = \frac{6}{{{t^3}}}{\bf{i}} \cr & {\bf{a}}\left( 3 \right) = \frac{6}{{{t^3}}}{\bf{i}} = \frac{2}{9}{\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{{ - \frac{3}{{{t^2}}}{\bf{i}} - 6{\bf{j}}}}{{\left\| { - \frac{3}{{{t^2}}}{\bf{i}} - 6{\bf{j}}} \right\|}} = \frac{{ - \frac{3}{{{t^2}}}{\bf{i}} - 6{\bf{j}}}}{{\sqrt {\frac{9}{{{t^4}}} + 36} }} \cr & {\bf{T}}\left( t \right) = \frac{{{t^2}\left( { - \frac{3}{{{t^2}}}{\bf{i}} - 6{\bf{j}}} \right)}}{{\sqrt {36{t^4} + 9} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - 3{\bf{i}} - 6{t^2}{\bf{j}}}}{{\sqrt {36{t^4} + 9} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - 3{\bf{i}} - 6{t^2}{\bf{j}}}}{{3\sqrt {4{t^4} + 1} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - {\bf{i}} - 2{t^2}{\bf{j}}}}{{\sqrt {4{t^4} + 1} }} \cr & {\bf{T}}\left( 3 \right) = \frac{{ - {\bf{i}} - 2{t^2}{\bf{j}}}}{{\sqrt {4{t^4} + 1} }} = - \frac{1}{{\sqrt {325} }}{\bf{i}} - \frac{{18}}{{\sqrt {325} }}{\bf{j}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{ - {\bf{i}} - 2{t^2}{\bf{j}}}}{{\sqrt {4{t^4} + 1} }}} \right] \cr & {\bf{T}}'\left( t \right) = - \frac{d}{{dt}}\left[ {\frac{1}{{\sqrt {4{t^4} + 1} }}} \right]{\bf{i}} - \frac{d}{{dt}}\left[ {\frac{{2{t^2}}}{{\sqrt {4{t^4} + 1} }}} \right]{\bf{j}} \cr & {\text{Differentiating by hand and replacing}} \cr & {\bf{T}}'\left( t \right) = \frac{{8{t^3}}}{{{{\left( {4{t^4} + 1} \right)}^{3/2}}}}{\bf{i}} - \frac{{4t}}{{{{\left( {4{t^4} + 1} \right)}^{3/2}}}}{\bf{j}} \cr & {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 3 \cr & {\bf{T}}'\left( 3 \right) = \frac{{8{{\left( 3 \right)}^3}}}{{{{\left( {4{{\left( 3 \right)}^4} + 1} \right)}^{3/2}}}}{\bf{i}} - \frac{{4\left( 3 \right)}}{{{{\left( {4{{\left( 3 \right)}^4} + 1} \right)}^{3/2}}}}{\bf{j}} \cr & {\bf{T}}'\left( 3 \right) = \frac{{216}}{{325\sqrt {325} }}{\bf{i}} - \frac{{12}}{{325\sqrt {325} }}{\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 = 3 \cr & {\bf{N}}\left( 3 \right) = \frac{{\frac{{216}}{{325\sqrt {325} }}{\bf{i}} - \frac{{12}}{{325\sqrt {325} }}{\bf{j}}}}{{\left\| {\frac{{216}}{{325\sqrt {325} }}{\bf{i}} - \frac{{12}}{{325\sqrt {325} }}{\bf{j}}} \right\|}} \cr & {\bf{N}}\left( 3 \right) = \frac{{\frac{{216}}{{325\sqrt {325} }}{\bf{i}} - \frac{{12}}{{325\sqrt {325} }}{\bf{j}}}}{{\frac{{12}}{{325}}}} \cr & {\bf{N}}\left( 3 \right) = \frac{{18}}{{\sqrt {325} }}{\bf{i}} - \frac{1}{{\sqrt {325} }}{\bf{j}} \cr & {\text{Find }}{a_{\bf{T}}}{\text{ and }}{a_{\bf{N}}} \cr & {a_{\bf{T}}} = {\bf{a}} \cdot {\bf{T}} = \left( {\frac{2}{9}{\bf{i}}} \right)\left( { - \frac{1}{{\sqrt {325} }}{\bf{i}} - \frac{{18}}{{\sqrt {325} }}{\bf{j}}} \right) \cr & {a_{\bf{T}}} = - \frac{2}{{9\sqrt {325} }} \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( {\frac{2}{9}{\bf{i}}} \right)\left( {\frac{{18}}{{\sqrt {325} }}{\bf{i}} - \frac{1}{{\sqrt {325} }}{\bf{j}}} \right) \cr & {a_{\bf{N}}} = \frac{4}{{\sqrt {325} }} \cr & \cr & {\text{Summary}} \cr & {\bf{T}}\left( 3 \right) = - \frac{1}{{\sqrt {325} }}{\bf{i}} - \frac{{18}}{{\sqrt {325} }}{\bf{j}} \cr & {\bf{N}}\left( 3 \right) = \frac{{18}}{{\sqrt {325} }}{\bf{i}} - \frac{1}{{\sqrt {325} }}{\bf{j}} \cr & {a_{\bf{T}}} = - \frac{2}{{9\sqrt {325} }} \cr & {a_{\bf{N}}} = \frac{4}{{\sqrt {325} }} \cr} $$