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 - Review Exercises - Page 863: 22

Answer

$$\eqalign{ & {\text{summary}} \cr & \left( {\bf{a}} \right)4{\bf{i}} + 2t{\bf{j}} + 4t{\bf{k}} \cr & \left( {\bf{b}} \right)2{\bf{j}} + 4{\bf{k}} \cr & \left( {\bf{c}} \right)20t \cr & \left( {\bf{d}} \right) - 16{\bf{j}} + 8{\bf{k}} \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {4t + 3} \right){\bf{i}} + {t^2}{\bf{j}} + \left( {2{t^2} + 4} \right){\bf{k}} \cr & \left( {\bf{a}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {4t + 3} \right){\bf{i}} + {t^2}{\bf{j}} + \left( {2{t^2} + 4} \right){\bf{k}}} \right] \cr & {\bf{r}}'\left( t \right) = 4{\bf{i}} + 2t{\bf{j}} + 4t{\bf{k}} \cr & \cr & \left( {\bf{b}} \right){\text{Find }}{\bf{r}}''\left( t \right) \cr & {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {4{\bf{i}} + 2t{\bf{j}} + 4t{\bf{k}}} \right] \cr & {\bf{r}}''\left( t \right) = 0{\bf{i}} + 2{\bf{j}} + 4{\bf{k}} \cr & \cr & \left( {\bf{c}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = \left( {4{\bf{i}} + 2t{\bf{j}} + 4t{\bf{k}}} \right) \cdot \left( {0{\bf{i}} + 2{\bf{j}} + 4{\bf{k}}} \right) \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = 0 + 4t + 16t \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = 20t \cr & \cr & \left( {\bf{d}} \right){\text{Find }}{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) \cr} $$ \[\begin{gathered} {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ 4&{2t}&{4t} \\ 0&2&4 \end{array}} \right| \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} {2t}&{4t} \\ 2&4 \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} 4&{4t} \\ 0&4 \end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}} 4&{2t} \\ 0&2 \end{array}} \right|{\mathbf{k}} \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 0{\mathbf{i}} - 16{\mathbf{j}} + 8{\mathbf{k}} \hfill \\ \end{gathered} \]
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