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: 28

Answer

$$\eqalign{ & \left( {\bf{a}} \right)\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr & \left( {\bf{b}} \right)\left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr & \left( {\bf{c}} \right) - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr & \left( {\bf{d}} \right)4t{\sec ^2}t\tan t - 2{\sec ^2}t + 2{e^{ - t}}{\sec ^2}t\tan t + {e^{ - t}}{\sec ^2}t \cr & {\text{ }} - 2{e^{ - t}} - 2t{e^{ - t}} \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {{e^{ - t}},{t^2},\tan t} \right\rangle \cr & \left( {\bf{a}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {{e^{ - t}},{t^2},\tan t} \right\rangle } \right] \cr & {\bf{r}}'\left( t \right) = \left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr & \cr & \left( {\bf{b}} \right){\text{Find }}{\bf{r}}''\left( t \right) \cr & {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle } \right] \cr & {\bf{r}}''\left( t \right) = \left\langle {{e^{ - t}},2,2\sec t\sec t\tan t} \right\rangle \cr & {\bf{r}}''\left( t \right) = \left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr & \cr & \left( {\bf{c}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = \left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cdot \left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr & {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr} $$ \[\begin{gathered} \left( {\mathbf{d}} \right){\text{Find }}{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ { - {e^{ - t}}}&{2t}&{{{\sec }^2}t} \\ {{e^{ - t}}}&2&{2{{\sec }^2}t\tan t} \end{array}} \right| \hfill \\ {\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}} {2t}&{{{\sec }^2}t} \\ 2&{2{{\sec }^2}t\tan t} \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} { - {e^{ - t}}}&{{{\sec }^2}t} \\ {{e^{ - t}}}&{2{{\sec }^2}t\tan t} \end{array}} \right|{\mathbf{j}} \hfill \\ {\text{ }} + \left| {\begin{array}{*{20}{c}} { - {e^{ - t}}}&{2t} \\ {{e^{ - t}}}&2 \end{array}} \right|{\mathbf{k}} \hfill \\ \end{gathered} \] $$\eqalign{ & {\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) = 4t{\sec ^2}t\tan t - 2{\sec ^2}t \cr & {\text{ }} - \left( { - 2{e^{ - t}}{{\sec }^2}t\tan t - {e^{ - t}}{{\sec }^2}t} \right) + \left( { - 2{e^{ - t}} - 2t{e^{ - t}}} \right) \cr & {\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) = 4t{\sec ^2}t\tan t - 2{\sec ^2}t + 2{e^{ - t}}{\sec ^2}t\tan t + {e^{ - t}}{\sec ^2}t \cr & {\text{ }} - 2{e^{ - t}} - 2t{e^{ - t}} \cr & \cr & {\text{Summary}} \cr & \left( {\bf{a}} \right)\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr & \left( {\bf{b}} \right)\left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr & \left( {\bf{c}} \right) - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr & \left( {\bf{d}} \right)\left\langle { - \left( {t\cos t + \sin t} \right),\cos t - t\sin t,{t^2}} \right\rangle \cr} $$ $$\eqalign{ & {\text{Summary}} \cr & \left( {\bf{a}} \right)\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr & \left( {\bf{b}} \right)\left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr & \left( {\bf{c}} \right) - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr & \left( {\bf{d}} \right)4t{\sec ^2}t\tan t - 2{\sec ^2}t + 2{e^{ - t}}{\sec ^2}t\tan t + {e^{ - t}}{\sec ^2}t \cr & {\text{ }} - 2{e^{ - t}} - 2t{e^{ - t}} \cr} $$
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