Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 17

Answer

$${\bf{v}}\left( 0 \right) = \left\langle {1,1,1} \right\rangle $$$${\bf{a}}\left( 0 \right) = \left\langle {0,2,1} \right\rangle $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {{e^t}\cos t,{e^t}\sin t,{e^t}} \right\rangle ,{\text{ }}t = 0 \cr & \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {{e^t}\cos t,{e^t}\sin t,{e^t}} \right\rangle } \right] \cr & {\bf{v}}\left( t \right) = \left\langle {{e^t}\cos t - {e^t}\sin t,{e^t}\sin t + {e^t}\cos t,{e^t}} \right\rangle \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {\left\langle {{e^t}\cos t - {e^t}\sin t,{e^t}\sin t + {e^t}\cos t,{e^t}} \right\rangle } \right\| \cr & {\text{speed}} = \sqrt {2{e^{2t}} - 2{e^t}\sin t\cos t + 2{e^t}\sin t\cos t + {e^{2t}}} \cr & {\text{speed}} = \sqrt {3{e^{2t}}} \cr & {\text{speed}} = \sqrt 3 {e^t} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {{e^t}\cos t - {e^t}\sin t,{e^t}\sin t + {e^t}\cos t,{e^t}} \right\rangle } \right] \cr & {\text{Differentiating by hand and replacing}} \cr & {\bf{a}}\left( t \right) = \left\langle { - 2{e^t}\sin t,2{e^t}\cos t,{e^t}} \right\rangle \cr & \cr & \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ at }}t = 0 \cr & {\bf{v}}\left( 0 \right) = \left\langle {{e^0}\cos 0 - {e^0}\sin 0,{e^0}\sin 0 + {e^0}\cos 0,{e^0}} \right\rangle \cr & {\bf{v}}\left( 0 \right) = \left\langle {1,1,1} \right\rangle \cr & {\bf{a}}\left( 0 \right) = \left\langle { - 2{e^0}\sin 0,2{e^0}\cos 0,{e^0}} \right\rangle \cr & {\bf{a}}\left( 0 \right) = \left\langle {0,2,1} \right\rangle \cr} $$
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