Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 746: 11

$$\eqalign{ & {\bf{v}}\left( t \right) = - 2\sin t{\bf{i}} + 3\cos t{\bf{j}} + 4{\bf{k}} \cr & {\bf{a}}\left( t \right) = - 2\cos t{\bf{i}} - 3\sin t{\bf{j}} \cr & {\text{speed}}:2\sqrt 5 \cr & {\text{direction}}: - \frac{1}{{\sqrt 5 }}{\bf{i}} + \frac{2}{{\sqrt 5 }}{\bf{k}} \cr & {\bf{v}}\left( {\frac{\pi }{2}} \right) = 2\sqrt 5 \left( { - \frac{1}{{\sqrt 5 }}{\bf{i}} + \frac{2}{{\sqrt 5 }}{\bf{k}}} \right) \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {2\cos t} \right){\bf{i}} + \left( {3\sin t} \right){\bf{j}} + 4t{\bf{k}},\,\,\,\,\,t = \pi /2 \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( {2\cos t} \right){\bf{i}} + \left( {3\sin t} \right){\bf{j}} + 4t{\bf{k}}} \right] \cr & {\bf{v}}\left( t \right) = - 2\sin t{\bf{i}} + 3\cos t{\bf{j}} + 4{\bf{k}} \cr & \cr & {\text{calculate }}{\bf{v}}\left( {\pi /2} \right) \cr & {\bf{v}}\left( {\frac{\pi }{2}} \right) = - 2\sin \left( {\frac{\pi }{2}} \right){\bf{i}} + 3\cos \left( {\frac{\pi }{2}} \right){\bf{j}} + 4{\bf{k}} \cr & {\bf{v}}\left( {\frac{\pi }{2}} \right) = - 2{\bf{i}} + 4{\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[ { - 2\sin t{\bf{i}} + 3\cos t{\bf{j}} + 4{\bf{k}}} \right] \cr & {\bf{a}}\left( t \right) = - 2\cos t{\bf{i}} - 3\sin t{\bf{j}} \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| { - 2\sin t{\bf{i}} + 3\cos t{\bf{j}} + 4{\bf{k}}} \right| \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( { - 2\sin t} \right)}^2} + {{\left( {3\cos t} \right)}^2} + {{\left( 4 \right)}^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {4{{\sin }^2}t + 9{{\cos }^2}t + 16} \cr & {\text{find the particle's speed at }}t = \pi /2 \cr & {\text{speed}}:\left| {{\bf{v}}\left( {\frac{\pi }{2}} \right)} \right| = \sqrt {4{{\sin }^2}\left( {\frac{\pi }{2}} \right) + 9{{\cos }^2}\left( {\frac{\pi }{2}} \right) + 16} \cr & {\text{speed}}:\left| {{\bf{v}}\left( {\frac{\pi }{2}} \right)} \right| = \sqrt {4 + 16} \cr & {\text{speed}}:\sqrt {20} \cr & {\text{speed}}:2\sqrt 5 \cr & \cr & {\text{Calculate the direction of motion at the given value of }}t \cr & {\text{direction}}:\frac{{{\bf{v}}\left( {\pi /2} \right)}}{{\left| {{\bf{v}}\left( {\pi /2} \right)} \right|}} \cr & {\text{direction}}:\frac{{ - 2{\bf{i}} + 4{\bf{k}}}}{{2\sqrt 5 }} \cr & {\text{direction}}: - \frac{1}{{\sqrt 5 }}{\bf{i}} + \frac{2}{{\sqrt 5 }}{\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( {\frac{\pi }{2}} \right) = 2\sqrt 5 \left( { - \frac{1}{{\sqrt 5 }}{\bf{i}} + \frac{2}{{\sqrt 5 }}{\bf{k}}} \right) \cr} $$