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: 10

Answer

$$\eqalign{ & {\bf{v}}\left( 3 \right) = 4{\bf{i}} + 4{\bf{j}} + 2{\bf{k}} \cr & {\text{speed}} = 6 \cr & {\bf{a}}\left( 3 \right) = 0 \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = 4t{\bf{i}} + 4t{\bf{j}} + 2t{\bf{k}},{\text{ }}t = 3 \cr & \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {4t{\bf{i}} + 4t{\bf{j}} + 2t{\bf{k}}} \right] \cr & {\bf{v}}\left( t \right) = 4{\bf{i}} + 4{\bf{j}} + 2{\bf{k}} \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {4{\bf{i}} + 4{\bf{j}} + 2{\bf{k}}} \right\| \cr & {\text{speed}} = \sqrt {16 + 16 + 4} = 6 \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {4{\bf{i}} + 4{\bf{j}} + 2{\bf{k}}} \right] \cr & {\bf{a}}\left( t \right) = 0 \cr & \cr & \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed at }}t = 3 \cr & {\bf{v}}\left( 3 \right) = 4{\bf{i}} + 4{\bf{j}} + 2{\bf{k}} \cr & {\text{speed}} = 6 \cr & {\bf{a}}\left( 3 \right) = 0 \cr} $$
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