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: 54

Answer

$${\bf{r}}\left( t \right) = \left\langle {4,2t - 1,{t^2} - 6} \right\rangle $$

Work Step by Step

$$\eqalign{ & {\bf{r}}'\left( t \right) = \left\langle {0,2,2t} \right\rangle ;\,\,\,\,\,\,{\bf{r}}\left( 1 \right) = \left\langle {4,3, - 5} \right\rangle \cr & \int {{\bf{r}}'\left( t \right)dt} = \int {\left\langle {0,2,2t} \right\rangle dt} \cr & {\bf{r}}\left( t \right) = \left\langle {k,2t,{t^2}} \right\rangle + {\bf{C}}\,\,\,\left( 1 \right) \cr & \cr & {\text{Use the initial condition}} \cr & \left\langle {4,3, - 5} \right\rangle = \left\langle {k,2\left( 1 \right),{{\left( 1 \right)}^2}} \right\rangle + {\bf{C}} \cr & \left\langle {4,3, - 5} \right\rangle = \left\langle {k,2,1} \right\rangle + {\bf{C}} \cr & {\bf{C}} = \left\langle {4,3, - 5} \right\rangle - \left\langle {k,2,1} \right\rangle \cr & {\bf{C}} = \left\langle {4 - k,3 - 2, - 5 - 1} \right\rangle \cr & {\bf{C}} = \left\langle {4 - k,1, - 6} \right\rangle \cr & \cr & {\text{Substitute the vector constant into }}\left( 1 \right) \cr & {\bf{r}}\left( t \right) = \left\langle {k,2t,{t^2}} \right\rangle + \left\langle {4 - k,1, - 6} \right\rangle \cr & {\bf{r}}\left( t \right) = \left\langle {k + 4 - k,2t + 1,{t^2} - 6} \right\rangle \cr & {\bf{r}}\left( t \right) = \left\langle {4,2t - 1,{t^2} - 6} \right\rangle \cr} $$
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