Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 8

$$\eqalign{ & {\text{Parametric equations: }}x = 1 + 2t,{\text{ }}y = 1 + t,{\text{ }}z = \frac{4}{3} \cr & {\text{Direction numbers: }}a = 2,{\text{ }}b = 0,{\text{ }}c = 1 \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + t{\bf{j}} + \frac{4}{3}{\bf{k}},{\text{ }}P\left( {1,1,\frac{4}{3}} \right) \cr & {\text{Let }}t = 1 \cr & {\bf{r}}\left( 1 \right) = {\left( 1 \right)^2}{\bf{i}} + \left( 1 \right){\bf{j}} + \frac{4}{3}{\bf{k}} \cr & {\bf{r}}\left( 0 \right) = {\bf{i}} + {\bf{j}} + \frac{4}{3}{\bf{k}},{\text{ then at }}P\left( {1,1,\frac{4}{3}} \right) \to {\text{ }}t = 1 \cr & {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{\bf{i}} + t{\bf{j}} + \frac{4}{3}{\bf{k}}} \right] \cr & {\bf{r}}'\left( t \right) = 2t{\bf{i}} + {\bf{j}} + 0{\bf{k}} \cr & {\text{At }}t = 1 \cr & {\bf{r}}'\left( 1 \right) = 2{\bf{i}} + {\bf{j}} \cr & {\text{at }}t = 1 \cr & {\bf{T}}\left( 1 \right) = \frac{{{\bf{r}}'\left( 0 \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( 0 \right)} \right\|}} = \frac{{2{\bf{i}} + {\bf{j}}}}{{\left\| {2{\bf{i}} + {\bf{j}}} \right\|}} = \frac{{2{\bf{i}} + {\bf{j}}}}{{\sqrt 5 }} \cr & {\bf{T}}\left( 1 \right) = \frac{1}{{\sqrt 5 }}\left( {2{\bf{i}} + {\bf{j}} + 0{\bf{k}}} \right) \cr & {\text{The direction numbers are}} \cr & a = 2,{\text{ }}b = 0,{\text{ }}c = 1 \cr & {\text{We can obtain the parametric equations:}} \cr & x = {x_1} + at \cr & y = {y_1} + bt \cr & z = {z_1} + ct \cr & \left( {1,1,\frac{4}{3}} \right) = \left( {{x_1},{y_1},{z_1}} \right) \cr & x = 1 + 2t \cr & y = 1 + t \cr & z = \frac{4}{3} + 0t \cr & x = 1 + 2t,{\text{ }}y = 1 + t,{\text{ }}z = \frac{4}{3} \cr & \cr & {\text{Parametric equations: }}x = 1 + 2t,{\text{ }}y = 1 + t,{\text{ }}z = \frac{4}{3} \cr & {\text{Direction numbers: }}a = 2,{\text{ }}b = 0,{\text{ }}c = 1 \cr} $$