Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.5 Motion in 3-Space - Exercises - Page 744: 11

Answer

The velocity vector: ${\bf{v}}\left( t \right) = \left( {\frac{1}{2}{t^2} + \frac{1}{3},4t - 2} \right)$

Work Step by Step

Find the velocity vector: ${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {t,4} \right){\rm{d}}t = \left( {\frac{1}{2}{t^2},4t} \right) + {{\bf{c}}_0}$ The initial condition ${\bf{v}}\left( 0 \right) = \left( {\frac{1}{3}, - 2} \right)$ gives $\left( {\frac{1}{3}, - 2} \right) = {{\bf{c}}_0}$ Thus, ${\bf{v}}\left( t \right) = \left( {\frac{1}{2}{t^2},4t} \right) + \left( {\frac{1}{3}, - 2} \right)$ ${\bf{v}}\left( t \right) = \left( {\frac{1}{2}{t^2} + \frac{1}{3},4t - 2} \right)$
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