Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 24

$$\eqalign{ & {\bf{T}}\left( 0 \right) = - {\bf{i}} \cr & {\bf{N}}\left( 0 \right) = {\bf{j}} \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = 2 \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {{t^3} - 4t} \right){\bf{i}} + \left( {{t^2} - 1} \right){\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[ {\left( {{t^3} - 4t} \right){\bf{i}} + \left( {{t^2} - 1} \right){\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = \left( {3{t^2} - 4} \right){\bf{i}} + 2t{\bf{j}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left( {3{t^2} - 4} \right){\bf{i}} + 2t{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = 6t{\bf{i}} + 2{\bf{j}} \cr & {\bf{a}}\left( 0 \right) = 2{\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( {3{t^2} - 4} \right){\bf{i}} + 2t{\bf{j}}}}{{\left\| {\left( {3{t^2} - 4} \right){\bf{i}} + 2t{\bf{j}}} \right\|}} = \frac{{\left( {3{t^2} - 4} \right){\bf{i}} + 2t{\bf{j}}}}{{\sqrt {{{\left( {3{t^2} - 4} \right)}^2} + {{\left( {2t} \right)}^2}} }} \cr & {\bf{T}}\left( t \right) = \frac{{\left( {3{t^2} - 4} \right){\bf{i}} + 2t{\bf{j}}}}{{\sqrt {9{t^4} - 24{t^2} + 16 + 4{t^2}} }} \cr & {\bf{T}}\left( t \right) = \frac{{\left( {3{t^2} - 4} \right){\bf{i}} + 2t{\bf{j}}}}{{\sqrt {9{t^4} - 20{t^2} + 16} }} \cr & {\bf{T}}\left( 0 \right) = \frac{{\left( {3{{\left( 0 \right)}^2} - 4} \right){\bf{i}} + 2\left( 0 \right){\bf{j}}}}{{\sqrt {9{{\left( 0 \right)}^4} - 20{{\left( 0 \right)}^2} + 16} }} \cr & {\bf{T}}\left( 0 \right) = - {\bf{i}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{3{t^2} - 4}}{{\sqrt {9{t^4} - 20{t^2} + 16} }}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\frac{{2t}}{{\sqrt {9{t^4} - 20{t^2} + 16} }}} \right]{\bf{j}} \cr & {\text{Differentiate by using a CAS }}\left( {{\text{WolframWebsite}}} \right) \cr & {\bf{T}}'\left( t \right) = \frac{{4t\left( {3{t^2} + 4} \right)}}{{{{\left( {9{t^4} - 20{t^2} + 16} \right)}^{3/2}}}}{\bf{i}} + \frac{{32 - 18{t^4}}}{{{{\left( {9{t^4} - 20{t^2} + 16} \right)}^{3/2}}}}{\bf{j}} \cr & {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 0 \cr & {\bf{T}}'\left( 0 \right) = 0{\bf{i}} + \frac{1}{2}{\bf{j}} \cr & {\bf{T}}'\left( 0 \right) = \frac{1}{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 = 0 \cr & {\bf{N}}\left( 0 \right) = \frac{{{\bf{T}}'\left( 0 \right)}}{{\left\| {{\bf{T}}'\left( 0 \right)} \right\|}} = \frac{{\frac{1}{2}{\bf{j}}}}{{\left\| {\frac{1}{2}{\bf{j}}} \right\|}} = {\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{j}}} \right)\left( { - {\bf{i}}} \right) \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( {2{\bf{j}}} \right)\left( {\bf{j}} \right) \cr & {\text{Simplify by using a calculator}} \cr & {a_{\bf{N}}} = 2 \cr & \cr & {\text{Summary}} \cr & {\bf{T}}\left( 0 \right) = - {\bf{i}} \cr & {\bf{N}}\left( 0 \right) = {\bf{j}} \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = 2 \cr} $$