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 - 12.2 Exercises - Page 830: 26

Answer

$$\eqalign{ & \left( {\bf{a}} \right)3{t^2}{\bf{i}} + 4t{\bf{j}} + 3{\bf{k}} \cr & \left( {\bf{b}} \right)6t{\bf{i}} + 4{\bf{j}} \cr & \left( {\bf{c}} \right)18{t^3} + 16t \cr & \left( {\bf{d}} \right) - 12{\bf{i}} + 18t{\bf{j}} - 12{t^2}{\bf{k}} \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = {t^3}{\bf{i}} + \left( {2{t^2} + 3} \right){\bf{j}} + \left( {3t - 5} \right){\bf{k}} \cr & \left( {\bf{a}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3}{\bf{i}} + \left( {2{t^2} + 3} \right){\bf{j}} + \left( {3t - 5} \right){\bf{k}}} \right] \cr & {\bf{r}}'\left( t \right) = 3{t^2}{\bf{i}} + 4t{\bf{j}} + 3{\bf{k}} \cr & \cr & \left( {\bf{b}} \right){\text{Find }}{\bf{r}}''\left( t \right) \cr & {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {3{t^2}{\bf{i}} + 4t{\bf{j}} + 3{\bf{k}}} \right] \cr & {\bf{r}}''\left( t \right) = 6t{\bf{i}} + 4{\bf{j}} \cr & \cr & \left( {\bf{c}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = \left( {3{t^2}{\bf{i}} + 4t{\bf{j}} + 3{\bf{k}}} \right) \cdot \left( {6t{\bf{i}} + 4{\bf{j}}} \right) \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = 18{t^3} + 16t + 0 \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = 18{t^3} + 16t \cr} $$ \[\begin{gathered} \left( {\mathbf{d}} \right){\text{Find }}{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ {3{t^2}}&{4t}&3 \\ {6t}&4&0 \end{array}} \right| \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} {4t}&3 \\ 4&0 \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} {3{t^2}}&3 \\ {6t}&0 \end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}} {3{t^2}}&{4t} \\ {6t}&4 \end{array}} \right|{\mathbf{k}} \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = - 12{\mathbf{i}} + 18t{\mathbf{j}} - 12{t^2}{\mathbf{k}} \hfill \\ \end{gathered} \] $$\eqalign{ & {\text{summary}} \cr & \left( {\bf{a}} \right)3{t^2}{\bf{i}} + 4t{\bf{j}} + 3{\bf{k}} \cr & \left( {\bf{b}} \right)6t{\bf{i}} + 4{\bf{j}} \cr & \left( {\bf{c}} \right)18{t^3} + 16t \cr & \left( {\bf{d}} \right) - 12{\bf{i}} + 18t{\bf{j}} - 12{t^2}{\bf{k}} \cr} $$
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