Chapter 11 - Vectors and Vector-Valued Functions - 11.7 Motion in Space - 11.7 Exercises - Page 827: 15

$${\bf{v}}\left( t \right) = \left\langle {1, - 4,6} \right\rangle ,{\text{ speed}}:{\text{ }}\sqrt {53} {\text{ and }}{\bf{a}}\left( t \right) = \left\langle {0,0,0} \right\rangle $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {3 + t,2 - 4t,1 + 6t} \right\rangle ,\,\,\,\,\,{\text{for }}t \geqslant 0 \cr & \left( a \right){\text{find the velocity and speed }} \cr & {\text{the velocity vector is given by }}\,\,\left( {{\text{see page 817}}} \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {3 + t,2 - 4t,1 + 6t} \right\rangle } \right] \cr & {\text{differentiating each component}} \cr & {\bf{v}}\left( t \right) = \left\langle {\frac{d}{{dt}}\left[ {3 + t} \right],\frac{d}{{dt}}\left[ {2 - 4t} \right],\frac{d}{{dt}}\left[ {1 + 6t} \right]} \right\rangle \cr & {\bf{v}}\left( t \right) = \left\langle {1, - 4,6} \right\rangle \cr & {\text{The speed of the object is the scalar function }}\left| {{\bf{v}}\left( t \right)} \right|{\text{ then}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \left| {\left\langle {1, - 4,6} \right\rangle } \right| \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} \cr & {\text{simplifying}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {1 + 16 + 36} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {53} \cr & \cr & \left( b \right){\text{find the acceleration of the object}} \cr & {\text{the acceleration vector is given by }}\,\,{\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right)\left( {{\text{see page 817}}} \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {1, - 4,6} \right\rangle } \right] \cr & {\text{differentiating each component}} \cr & {\bf{a}}\left( t \right) = \left\langle {0,0,0} \right\rangle \cr} $$