Chapter 12 - Vector-Valued Functions - 12.6 Motion Along A Curve - Exercises Set 12.6 - Page 891: 6

$$\eqalign{ & {\bf{v}}\left( t \right) = 3{\bf{i}} - 4{\bf{j}} + {\bf{k}} \cr & {\bf{v}}\left( 2 \right) = 3{\bf{i}} - 4{\bf{j}} + {\bf{k}} \cr & \left\| {{\bf{v}}\left( 2 \right)} \right\| = \sqrt {26} \cr & {\bf{a}}\left( t \right) = 0 \cr & {\bf{a}}\left( 2 \right) = 0 \cr} $$

$$\eqalign{ & x = 1 + 3t,\,\,\,y = 2 - 4t,\,\,\,z = 7 + t,\,\,\,\,t = 2 \cr & {\text{Calculate the velocity at }}t = 2 \cr & {\bf{v}}\left( t \right) = \frac{{dx}}{{dt}}{\bf{i}} + \frac{{dy}}{{dt}}{\bf{j}} + \frac{{dz}}{{dt}}{\bf{k}} \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {1 + 3t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {2 - 4t} \right]{\bf{j}} + \frac{{dz}}{{dt}}\left[ {7 + t} \right]{\bf{k}} \cr & {\bf{v}}\left( t \right) = 3{\bf{i}} - 4{\bf{j}} + {\bf{k}} \cr & {\bf{v}}\left( 2 \right) = 3{\bf{i}} - 4{\bf{j}} + {\bf{k}} \cr & \cr & {\text{Calculate the speed at }}t = 2 \cr & \left\| {{\bf{v}}\left( t \right)} \right\| = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} \cr & \left\| {{\bf{v}}\left( t \right)} \right\| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 1 \right)}^2}} \cr & \left\| {{\bf{v}}\left( t \right)} \right\| = \sqrt {26} \cr & \left\| {{\bf{v}}\left( 2 \right)} \right\| = \sqrt {26} \cr & \cr & {\text{Calculate the acceleration at }}t = 2 \cr & {\bf{a}}\left( t \right) = \frac{{d{\bf{v}}}}{{dt}} \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {3{\bf{i}} - 4{\bf{j}} + {\bf{k}}} \right] \cr & {\bf{a}}\left( t \right) = 0 \cr & {\bf{a}}\left( 2 \right) = 0 \cr} $$