Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 901: 42


$r(t)=(\frac{t^2}{2}+1)i+e^tj+(te^t-e^t +2)k$ or $r(t)=\lt \frac{t^2}{2}+1,e^t,te^t-e^t +2 \gt$

Work Step by Step

Consider $I=\int (ti+e^tj+te^t k)dt$ $I=Pi+ Qj+Rk$ ... (a) $P=2t=\frac{1}{2}t^2+C$ $Q=e^t=e^t+C'$ and $R=te^t=te^t-\int e^t(1) dt=e^t(t-1) C''$ From equation (a): $I=(\frac{1}{2}t^2+C)i+(e^t+C')j+(e^t(t-1) +C'')k$;(C,C',C'' are the constant of integration.) From question, $r(0)=i+j+k$ gives $\lt 1,1,1 \gt$. $r(0)=(\frac{1}{2}t^2+C)i+(e^t+C')j+(e^t(t-1)+ C'')k=j-k$ The desired result is: $r(t)=(\frac{t^2}{2}+1)i+e^tj+(te^t-e^t +2)k$ or $r(t)=\lt \frac{t^2}{2}+1,e^t,te^t-e^t +2 \gt$
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