Chapter 12 - Vector-Valued Functions - 12.2 Calculus Of Vector-Valued Functions - Exercises Set 12.2 - Page 857: 48

$${\bf{y}}\left( t \right) = \left( {{t^4} + 2} \right){\bf{i}} - \left( {\frac{1}{3}{t^3} + 4} \right){\bf{j}}$$

$$\eqalign{ & {\bf{y}}''\left( t \right) = 12{t^2}{\bf{i}} - 2t{\bf{j}},{\text{ }}{\bf{y}}\left( 0 \right) = 2{\bf{i}} - 4{\bf{j}},{\text{ }}{\bf{y}}'\left( 0 \right) = 0 \cr & {\text{Calculate }}{\bf{y}}'\left( t \right) \cr & {\bf{y}}'\left( t \right) = \int {{\bf{y}}''\left( t \right)} dt \cr & {\bf{y}}'\left( t \right) = \int {\left( {12{t^2}{\bf{i}} - 2t{\bf{j}}} \right)} dt \cr & {\bf{y}}'\left( t \right) = 4{t^3}{\bf{i}} - {t^2}{\bf{j}} + {\bf{C}} \cr & {\text{Use the initial condition }}{\bf{y}}'\left( 0 \right) = 0 \cr & {\bf{y}}'\left( 0 \right) = 4{\left( 0 \right)^3}{\bf{i}} - {\left( 0 \right)^2}{\bf{j}} + {\bf{C}} \cr & 0 = {\bf{C}} \cr & {\text{Therefore,}} \cr & {\bf{y}}'\left( t \right) = 4{t^3}{\bf{i}} - {t^2}{\bf{j}} \cr & \cr & {\text{Calculate }}{\bf{y}}\left( t \right) \cr & {\bf{y}}\left( t \right) = \int {{\bf{y}}'\left( t \right)} dt \cr & {\bf{y}}\left( t \right) = \int {\left( {4{t^3}{\bf{i}} - {t^2}{\bf{j}}} \right)} dt \cr & {\bf{y}}\left( t \right) = {t^4}{\bf{i}} - \frac{1}{3}{t^3}{\bf{j}} + {\bf{C}} \cr & {\text{Use the initial condition }}{\bf{y}}\left( 0 \right) = 2{\bf{i}} - 4{\bf{j}} \cr & {\bf{y}}\left( 0 \right) = {\left( 0 \right)^4}{\bf{i}} - \frac{1}{3}{\left( 0 \right)^3}{\bf{j}} + {\bf{C}} \cr & 2{\bf{i}} - 4{\bf{j}} = {\bf{C}} \cr & {\bf{C}} = 2{\bf{i}} - 4{\bf{j}} \cr & {\text{Therefore,}} \cr & {\bf{y}}\left( t \right) = {t^4}{\bf{i}} - \frac{1}{3}{t^3}{\bf{j}} + 2{\bf{i}} - 4{\bf{j}} \cr & {\bf{y}}\left( t \right) = \left( {{t^4} + 2} \right){\bf{i}} - \left( {\frac{1}{3}{t^3} + 4} \right){\bf{j}} \cr} $$