Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 746: 14

$$\eqalign{ & {\bf{v}}\left( t \right) = - {e^{ - t}}{\bf{i}} - 6\sin 3t{\bf{j}} + 6\cos 3t{\bf{k}} \cr & {\bf{a}}\left( t \right) = {e^{ - t}}{\bf{i}} - 18\cos 3t{\bf{j}} - 18\sin 3t{\bf{k}} \cr & {\text{speed}}:\sqrt {37} \cr & {\text{direction}}: - \frac{1}{{\sqrt {37} }}{\bf{i}} + \frac{6}{{\sqrt {37} }}{\bf{k}} \cr & {\bf{v}}\left( 0 \right) = \sqrt {37} \left( { - \frac{1}{{\sqrt {37} }}{\bf{i}} + \frac{6}{{\sqrt {37} }}{\bf{k}}} \right) \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {{e^{ - t}}} \right){\bf{i}} + \left( {2\cos 3t} \right){\bf{j}} + \left( {2\sin 3t} \right){\bf{k}},\,\,\,\,\,t = 0 \cr & {\text{Find }}{\bf{v}}\left( t \right){\text{ by differentiating }}{\bf{r}}\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left( {{e^{ - t}}} \right){\bf{i}} + \left( {2\cos 3t} \right){\bf{j}} + \left( {2\sin 3t} \right){\bf{k}}} \right] \cr & {\bf{v}}\left( t \right) = - {e^{ - t}}{\bf{i}} - 6\sin 3t{\bf{j}} + 6\cos 3t{\bf{k}} \cr & \cr & {\text{calculate }}{\bf{v}}\left( 0 \right) \cr & {\bf{v}}\left( 0 \right) = - {e^{ - 0}}{\bf{i}} - 6\sin 3\left( 0 \right){\bf{j}} + 6\cos 3\left( 0 \right){\bf{k}} \cr & {\bf{v}}\left( 0 \right) = - {\bf{i}} + 6{\bf{k}} \cr & \cr & {\text{Find }}{\bf{a}}\left( t \right){\text{ by differentiating }}{\bf{v}}\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ { - {e^{ - t}}{\bf{i}} - 6\sin 3t{\bf{j}} + 6\cos 3t{\bf{k}}} \right] \cr & {\bf{a}}\left( t \right) = {e^{ - t}}{\bf{i}} - 18\cos 3t{\bf{j}} - 18\sin 3t{\bf{k}} \cr & \cr & {\text{The speed of the particle is }}\left| {{\bf{v}}\left( t \right)} \right| \cr & \left| {{\bf{v}}\left( t \right)} \right| = \left| { - {e^{ - t}}{\bf{i}} - 6\sin 3t{\bf{j}} + 6\cos 3t{\bf{k}}} \right| \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( { - {e^{ - t}}} \right)}^2} + {{\left( { - 6\sin 3t} \right)}^2} + {{\left( {6\cos 3t} \right)}^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{e^{ - 2t}} + 36{{\sin }^2}3t + 36{{\cos }^2}3t} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{e^{ - 2t}} + 36} \cr & {\text{find the particle's speed at }}t = 0 \cr & {\text{speed}}:\left| {{\bf{v}}\left( 0 \right)} \right| = \sqrt {{e^{ - 2\left( 0 \right)}} + 36} \cr & {\text{speed}}:\sqrt {37} \cr & \cr & {\text{Calculate the direction of motion at the given value of }}t \cr & {\text{direction}}:\frac{{{\bf{v}}\left( 0 \right)}}{{\left| {{\bf{v}}\left( 0 \right)} \right|}} \cr & {\text{direction}}:\frac{{ - {\bf{i}} + 6{\bf{k}}}}{{\sqrt {37} }} \cr & {\text{direction}}: - \frac{1}{{\sqrt {37} }}{\bf{i}} + \frac{6}{{\sqrt {37} }}{\bf{k}} \cr & \cr & {\text{Write the particle's velocity at that time as the product of its }} \cr & {\text{speed and direction}} \cr & {\bf{v}}\left( 0 \right) = \sqrt {37} \left( { - \frac{1}{{\sqrt {37} }}{\bf{i}} + \frac{6}{{\sqrt {37} }}{\bf{k}}} \right) \cr} $$