Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 10

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

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {t,t,\sqrt {4 - {t^2}} } \right\rangle ,{\text{ }}P\left( {1,1,\sqrt 3 } \right) \cr & {\text{Let }}t = 1 \cr & {\bf{r}}\left( 1 \right) = \left\langle {1,1,\sqrt {4 - {1^2}} } \right\rangle \cr & {\bf{r}}\left( 1 \right) = \left\langle {1,1,\sqrt 3 } \right\rangle ,{\text{ then at }}P\left( {1,1,\sqrt 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[ {\left\langle {t,t,\sqrt {4 - {t^2}} } \right\rangle } \right] \cr & {\bf{r}}'\left( t \right) = \left\langle {1,1, - \frac{t}{{\sqrt {4 - {t^2}} }}} \right\rangle \cr & {\text{At }}t = 1 \cr & {\bf{r}}'\left( 1 \right) = \left\langle {1,1, - \frac{1}{{\sqrt {4 - {1^2}} }}} \right\rangle \cr & {\bf{r}}'\left( 1 \right) = \left\langle {1,1, - \frac{1}{{\sqrt 3 }}} \right\rangle \cr & {\text{at }}t = 1 \cr & {\bf{T}}\left( 1 \right) = \frac{{{\bf{r}}'\left( 1 \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( 1 \right)} \right\|}} = \frac{{\left\langle {1,1, - \frac{1}{{\sqrt 3 }}} \right\rangle }}{{\left\| {\left\langle {1,1, - \frac{1}{{\sqrt 3 }}} \right\rangle } \right\|}} = \frac{{\left\langle {1,1, - \frac{1}{{\sqrt 3 }}} \right\rangle }}{{7/3}} \cr & {\bf{T}}\left( 1 \right) = \frac{3}{7}\left\langle {1,1, - \frac{1}{{\sqrt 3 }}} \right\rangle \cr & {\text{The direction numbers are}} \cr & a = 1,{\text{ }}b = 1,{\text{ }}c = - \frac{1}{{\sqrt 3 }} \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,\sqrt 3 } \right) = \left( {{x_1},{y_1},{z_1}} \right) \cr & x = 1 + t \cr & y = 1 + t \cr & z = 1 - \frac{1}{{\sqrt 3 }}t \cr & x = 1 + t,{\text{ }}y = 1 + t,{\text{ }}z = 1 - \frac{1}{{\sqrt 3 }}t \cr & \cr & {\text{Parametric equations: }}x = 1 + t,{\text{ }}y = 1 + t,{\text{ }}z = 1 - \frac{1}{{\sqrt 3 }}t \cr & {\text{Direction numbers: }}a = 1,{\text{ }}b = 1,{\text{ }}c = - \frac{1}{{\sqrt 3 }} \cr} $$