Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises - Page 815: 57

Answer

$${\bf{r}}\left( t \right) = \left\langle {\frac{1}{2}{e^{2t}} + \frac{1}{2},t + 2{e^{ - t}} - 1,t - 2{e^t} + 3} \right\rangle $$

Work Step by Step

$$\eqalign{ & {\bf{r}}'\left( t \right) = \left\langle {{e^{2t}},1 - 2{e^{ - t}},1 - 2{e^t}} \right\rangle ;\,\,\,\,\,\,{\bf{r}}\left( 0 \right) = \left\langle {1,1,1} \right\rangle \cr & \int {{\bf{r}}'\left( t \right)dt} = \int {\left\langle {{e^{2t}},1 - 2{e^{ - t}},1 - 2{e^t}} \right\rangle dt} \cr & {\bf{r}}\left( t \right) = \left\langle {\frac{1}{2}{e^{2t}},t + 2{e^{ - t}},t - 2{e^t}} \right\rangle + {\bf{C}}\,\,\,\left( 1 \right) \cr & \cr & {\text{Use the initial condition}} \cr & \left\langle {1,1,1} \right\rangle = \left\langle {\frac{1}{2}{e^{2\left( 0 \right)}},0 + 2{e^{ - 0}},0 - 2{e^0}} \right\rangle + {\bf{C}} \cr & \left\langle {1,1,1} \right\rangle = \left\langle {\frac{1}{2},2, - 2} \right\rangle + {\bf{C}} \cr & {\bf{C}} = \left\langle {1,1,1} \right\rangle - \left\langle {\frac{1}{2},2, - 2} \right\rangle \cr & {\bf{C}} = \left\langle {\frac{1}{2}, - 1,3} \right\rangle \cr & \cr & {\text{Substitute the vector constant into }}\left( 1 \right) \cr & {\bf{r}}\left( t \right) = \left\langle {\frac{1}{2}{e^{2t}},t + 2{e^{ - t}},t - 2{e^t}} \right\rangle + \left\langle {\frac{1}{2}, - 1,3} \right\rangle \cr & {\bf{r}}\left( t \right) = \left\langle {\frac{1}{2}{e^{2t}} + \frac{1}{2},t + 2{e^{ - t}} - 1,t - 2{e^t} + 3} \right\rangle \cr} $$
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