Calculus: Early Transcendentals 8th Edition

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

Chapter 13 - Section 13.4 - Motion in Space: Velocity and Acceleration - 13.4 Exercise - Page 878: 15

Answer

$r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$

Work Step by Step

As we are given that $a(t)=2i+2tk$ and $v(0)=3i-j$ Since, $v(t)=\int a(t)$ and $r(t)=\int v(t)$ Thus, $v(t)=\int (2i+2tk) dt$ $\implies v(t)=(2t+3)i-j+t^2k$ Now, $r(t)=\int [(2t+3)i-j+t^2k]=(t^2+3t)i-t j+\frac{t^3}{3}k+c$ Here, $c$ represents a constant of integration. That is, $r(0)=j+k \implies c= j+k$ $r(t)=\int [(2t+3)i-j+t^2k]=(t^2+3t)i-t j+\frac{t^3}{3}k+j+k$ Hence, $r(t)=(t^2+3t)i+(1-t) j+(\dfrac{t^3}{3}+1)k$
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