Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 16

$${\bf{v}}\left( {\frac{\pi }{4}} \right) = \left\langle { - \sqrt 2 ,\sqrt 2 ,\frac{\pi }{2}} \right\rangle $$$${\bf{a}}\left( \pi \right) = \left\langle { - \sqrt 2 , - \sqrt 2 ,2} \right\rangle $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {2\cos t,2\sin t,{t^2}} \right\rangle ,{\text{ }}t = \frac{\pi }{4} \cr & \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {2\cos t,2\sin t,{t^2}} \right\rangle } \right] \cr & {\bf{v}}\left( t \right) = \left\langle { - 2\sin t,2\cos t,2t} \right\rangle \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {\left\langle { - 2\sin t,2\cos t,2t} \right\rangle } \right\| \cr & {\text{speed}} = \sqrt {4{{\sin }^2}t + 4{{\cos }^2}t + 4{t^2}} \cr & {\text{speed}} = \sqrt {4 + 4{t^2}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle { - 2\sin t,2\cos t,2t} \right\rangle } \right] \cr & {\bf{a}}\left( t \right) = \left\langle { - 2\cos t, - 2\sin t,2} \right\rangle \cr & \cr & \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ at }}t = \frac{\pi }{4} \cr & {\bf{v}}\left( {\frac{\pi }{4}} \right) = \left\langle { - 2\sin \left( {\frac{\pi }{4}} \right),2\cos \left( {\frac{\pi }{4}} \right),2\left( {\frac{\pi }{4}} \right)} \right\rangle \cr & {\bf{v}}\left( {\frac{\pi }{4}} \right) = \left\langle { - \sqrt 2 ,\sqrt 2 ,\frac{\pi }{2}} \right\rangle \cr & {\bf{a}}\left( {\frac{\pi }{4}} \right) = \left\langle { - 2\cos \left( {\frac{\pi }{4}} \right), - 2\sin \left( {\frac{\pi }{4}} \right),2} \right\rangle \cr & {\bf{a}}\left( \pi \right) = \left\langle { - \sqrt 2 , - \sqrt 2 ,2} \right\rangle \cr} $$