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: 25

Answer

$$\eqalign{ & {\text{}} \cr & \left( {\bf{a}} \right)t{\bf{i}} - {\bf{j}} + \frac{1}{2}{t^2}{\bf{k}} \cr & \left( {\bf{b}} \right){\bf{i}} + t{\bf{k}} \cr & \left( {\bf{c}} \right)t + \frac{1}{2}{t^3} \cr & \left( {\bf{d}} \right) - t{\bf{i}} - \frac{1}{2}{t^2}{\bf{j}} + {\bf{k}} \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = \frac{1}{2}{t^2}{\bf{i}} - t{\bf{j}} + \frac{1}{6}{t^3}{\bf{k}} \cr & \left( {\bf{a}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{2}{t^2}{\bf{i}} - t{\bf{j}} + \frac{1}{6}{t^3}{\bf{k}}} \right] \cr & {\bf{r}}'\left( t \right) = t{\bf{i}} - {\bf{j}} + \frac{1}{2}{t^2}{\bf{k}} \cr & \cr & \left( {\bf{b}} \right){\text{Find }}{\bf{r}}''\left( t \right) \cr & {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} - {\bf{j}} + \frac{1}{2}{t^2}{\bf{k}}} \right] \cr & {\bf{r}}''\left( t \right) = {\bf{i}} + t{\bf{k}} \cr & \cr & \left( {\bf{c}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = \left( {t{\bf{i}} - {\bf{j}} + \frac{1}{2}{t^2}{\bf{k}}} \right) \cdot \left( {{\bf{i}} + t{\bf{k}}} \right) \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = t + 0 + \frac{1}{2}{t^3} \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = t + \frac{1}{2}{t^3} \cr} $$ \[\begin{gathered} \left( {\mathbf{d}} \right){\text{Find }}{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ t&{ - 1}&{\frac{1}{2}{t^2}} \\ 1&0&t \end{array}} \right| \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} { - 1}&{\frac{1}{2}{t^2}} \\ 0&t \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} t&{\frac{1}{2}{t^2}} \\ 1&t \end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}} t&{ - 1} \\ 1&0 \end{array}} \right|{\mathbf{k}} \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = - t{\mathbf{i}} - \frac{1}{2}{t^2}{\mathbf{j}} + {\mathbf{k}} \hfill \\ \end{gathered} \] $$\eqalign{ & {\text{summary}} \cr & \left( {\bf{a}} \right)t{\bf{i}} - {\bf{j}} + \frac{1}{2}{t^2}{\bf{k}} \cr & \left( {\bf{b}} \right){\bf{i}} + t{\bf{k}} \cr & \left( {\bf{c}} \right)t + \frac{1}{2}{t^3} \cr & \left( {\bf{d}} \right) - t{\bf{i}} - \frac{1}{2}{t^2}{\bf{j}} + {\bf{k}} \cr} $$
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