Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 23

$$\eqalign{ & {\bf{T}}\left( 1 \right) = - \frac{{\sqrt 5 }}{5}{\bf{i}} + \frac{{2\sqrt 5 }}{5}{\bf{j}} \cr & {\bf{N}}\left( 1 \right) = - \frac{2}{{\sqrt 5 }}{\bf{i}} - \frac{1}{{\sqrt 5 }}{\bf{j}} \cr & {a_{\bf{T}}} = \frac{{14\sqrt 5 }}{5} \cr & {a_{\bf{N}}} = \frac{{8\sqrt 5 }}{5} \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {t - {t^3}} \right){\bf{i}} + 2{t^2}{\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[ {\left( {t - {t^3}} \right){\bf{i}} + 2{t^2}{\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = \left( {1 - 3{t^2}} \right){\bf{i}} + 4t{\bf{j}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left( {1 - 3{t^2}} \right){\bf{i}} + 4t{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = - 6t{\bf{i}} + 4{\bf{j}} \cr & {\bf{a}}\left( 1 \right) = - 6{\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{{\left( {1 - 3{t^2}} \right){\bf{i}} + 4t{\bf{j}}}}{{\left\| {\left( {1 - 3{t^2}} \right){\bf{i}} + 4t{\bf{j}}} \right\|}} = \frac{{\left( {1 - 3{t^2}} \right){\bf{i}} + 4t{\bf{j}}}}{{\sqrt {{{\left( {1 - 3{t^2}} \right)}^2} + {{\left( {4t} \right)}^2}} }} \cr & {\bf{T}}\left( t \right) = \frac{{\left( {1 - 3{t^2}} \right){\bf{i}} + 4t{\bf{j}}}}{{\sqrt {1 - 6{t^2} + 9{t^4} + 16{t^2}} }} \cr & {\bf{T}}\left( t \right) = \frac{{\left( {1 - 3{t^2}} \right){\bf{i}} + 4t{\bf{j}}}}{{\sqrt {9{t^4} + 10{t^2} + 1} }} \cr & {\bf{T}}\left( 1 \right) = \frac{{\left( {1 - 3{{\left( 1 \right)}^2}} \right){\bf{i}} + 4\left( 1 \right){\bf{j}}}}{{\sqrt {9{{\left( 1 \right)}^4} + 10{{\left( 1 \right)}^2} + 1} }} \cr & {\bf{T}}\left( 1 \right) = \frac{{ - 2{\bf{i}} + 4{\bf{j}}}}{{\sqrt {20} }} \cr & {\bf{T}}\left( 1 \right) = - \frac{{\sqrt 5 }}{5}{\bf{i}} + \frac{{2\sqrt 5 }}{5}{\bf{j}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{1 - 3{t^2}}}{{\sqrt {9{t^4} + 10{t^2} + 1} }}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\frac{{4t}}{{\sqrt {9{t^4} + 10{t^2} + 1} }}} \right]{\bf{j}} \cr & {\text{Differentiate by using a CAS }}\left( {{\text{WolframWebsite}}} \right) \cr & {\bf{T}}'\left( t \right) = - \frac{{16t\left( {3{t^2} + 1} \right)}}{{{{\left( {9{t^4} + 10{t^2} + 1} \right)}^{3/2}}}}{\bf{i}} + \frac{{4 - 36{t^4}}}{{{{\left( {9{t^4} + 10{t^2} + 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{{16\left( 1 \right)\left( {3{{\left( 1 \right)}^2} + 1} \right)}}{{{{\left( {9{{\left( 1 \right)}^4} + 10{{\left( 1 \right)}^2} + 1} \right)}^{3/2}}}}{\bf{i}} + \frac{{4 - 36{{\left( 1 \right)}^4}}}{{{{\left( {9{{\left( 1 \right)}^4} + 10{{\left( 1 \right)}^2} + 1} \right)}^{3/2}}}}{\bf{j}} \cr & {\bf{T}}'\left( 1 \right) = - \frac{{64}}{{{{20}^{3/2}}}}{\bf{i}} - \frac{{32}}{{{{20}^{3/2}}}}{\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{{64}}{{{{20}^{3/2}}}}{\bf{i}} - \frac{{32}}{{{{20}^{3/2}}}}{\bf{j}}}}{{\left\| { - \frac{{64}}{{{{20}^{3/2}}}}{\bf{i}} - \frac{{32}}{{{{20}^{3/2}}}}{\bf{j}}} \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 = 1 \cr & {a_{\bf{T}}} = {\bf{a}} \cdot {\bf{T}} = \left( { - 6{\bf{i}} + 4{\bf{j}}} \right)\left( { - \frac{{64}}{{{{20}^{3/2}}}}{\bf{i}} - \frac{{32}}{{{{20}^{3/2}}}}{\bf{j}}} \right) \cr & {\text{Simplify by using a calculator}} \cr & {a_{\bf{T}}} = \frac{{14\sqrt 5 }}{5} \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( { - 6{\bf{i}} + 4{\bf{j}}} \right)\left( { - \frac{2}{{\sqrt 5 }}{\bf{i}} - \frac{1}{{\sqrt 5 }}{\bf{j}}} \right) \cr & {\text{Simplify by using a calculator}} \cr & {a_{\bf{N}}} = \frac{{8\sqrt 5 }}{5} \cr & \cr & {\text{Summary}} \cr & {\bf{T}}\left( 1 \right) = - \frac{{\sqrt 5 }}{5}{\bf{i}} + \frac{{2\sqrt 5 }}{5}{\bf{j}} \cr & {\bf{N}}\left( 1 \right) = - \frac{2}{{\sqrt 5 }}{\bf{i}} - \frac{1}{{\sqrt 5 }}{\bf{j}} \cr & {a_{\bf{T}}} = \frac{{14\sqrt 5 }}{5} \cr & {a_{\bf{N}}} = \frac{{8\sqrt 5 }}{5} \cr} $$