Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.1 - Vector Fields - 16.1 Exercise - Page 1074: 33

Answer

$S = \lt{2.04},{1.03}\gt$

Work Step by Step

$V(x,y) =\lt{x^2}, {x+y^2}\gt$ The particle's initial position is $\lt2,1\gt$, therefore: $x=2$ $y=1$ $V(2,1) =\lt{4}, {3}\gt$ We can consider the velocity as being constant, taking into account how small $\triangle{t}$ is. $\triangle{S} =0,01\cdot\lt{4}, {3}\gt$ $\triangle{S} =\lt{0.04}, {0.03}\gt$ The particle's final position can then be estimated at: $\lt{2.04},{1.03}\gt$
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