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: 27

Answer

\[\begin{align} & \left( \mathbf{a} \right)\left\langle t\cos t,t\sin t,1 \right\rangle \\ & \left( \mathbf{b} \right)\left\langle \cos t-t\sin t,\sin t+t\cos t,0 \right\rangle \\ & \left( \mathbf{c} \right)\text{ }t \\ & \left( \mathbf{d} \right)\left\langle -\sin t-t\cos t,\cos t-t\sin t,{{t}^{2}} \right\rangle \\ \end{align}\]

Work Step by Step

\[\begin{align} & \mathbf{r}\left( t \right)=\left\langle \cos t+t\sin t,\sin t-t\cos t,t \right\rangle \\ & \left( \mathbf{a} \right)\text{Find }\mathbf{r}'\left( t \right) \\ & \mathbf{r}'\left( t \right)=\frac{d}{dt}\left[ \left\langle \cos t+t\sin t,\sin t-t\cos t,t \right\rangle \right] \\ & \mathbf{r}'\left( t \right)=\left\langle -\sin t+t\cos t+\sin t,\cos t+t\sin t-\cos t,1 \right\rangle \\ & \mathbf{r}'\left( t \right)=\left\langle t\cos t,t\sin t,1 \right\rangle \\ & \\ & \left( \mathbf{b} \right)\text{Find }\mathbf{r}''\left( t \right) \\ & \mathbf{r}''\left( t \right)=\frac{d}{dt}\left[ \left\langle t\cos t,t\sin t,1 \right\rangle \right] \\ & \mathbf{r}''\left( t \right)=\left\langle \cos t-t\sin t,t\cos t+\sin t,0 \right\rangle \\ & \\ & \left( \mathbf{c} \right)\text{Find }\mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right) \\ & \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=\left\langle t\cos t,t\sin t,1 \right\rangle \cdot \left\langle \cos t-t\sin t,t\cos t+\sin t,0 \right\rangle \\ & \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=t{{\cos }^{2}}t-{{t}^{2}}\sin t\cos t+{{t}^{2}}\sin t\cos t+t{{\sin }^{2}}t+0 \\ & \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=t{{\cos }^{2}}t+t{{\sin }^{2}}t \\ & \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=t \\ & \\ & \left( \mathbf{d} \right)\text{Find }\mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ t\cos t & t\sin t & 1 \\ \cos t-t\sin t & t\cos t+\sin t & 0 \\ \end{matrix} \right| \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix} t\sin t & 1 \\ t\cos t+\sin t & 0 \\ \end{matrix} \right|\mathbf{i}-\left| \begin{matrix} t\cos t & 1 \\ \cos t-t\sin t & 0 \\ \end{matrix} \right|\mathbf{j} \\ & \text{ }+\left| \begin{matrix} t\cos t & t\sin t \\ \cos t-t\sin t & t\cos t+\sin t \\ \end{matrix} \right|\mathbf{k} \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=-\left( t\cos t+\sin t \right)\mathbf{i}+\left( \cos t-t\sin t \right)\mathbf{j} \\ & +\left( {{t}^{2}}{{\cos }^{2}}t+t\sin t\cos t-t\sin t\cos t+{{t}^{2}}{{\sin }^{2}}t \right) \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=-\left( t\cos t+\sin t \right)\mathbf{i}+\left( \cos t-t\sin t \right)\mathbf{j}+{{t}^{2}}\mathbf{k} \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left\langle -\left( t\cos t+\sin t \right),\cos t-t\sin t,{{t}^{2}} \right\rangle \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left\langle -\sin t-t\cos t,\cos t-t\sin t,{{t}^{2}} \right\rangle \\ \end{align}\]
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