Calculus: Early Transcendentals 8th Edition

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

Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 854: 17

Answer

$r(t) =\lt 2+4t, 2t,-2t \gt ; 0 \leq t \leq 1$ $x=2+4t;\\ y= 2t; \\z=-2t ; \\0 \leq t \leq 1$

Work Step by Step

The vector line equation for the given two points is as follows: $r(t)=(1-t) r_0+tr_1=(1-t) \lt 2,0,0 \gt +t \lt 6,2,-2 \gt$ or, $=\lt 2,0,0 \gt - \lt 2t, 0t,0t \gt+\lt 6t ,2t, -2t \gt$ or, $=\lt 2-2t,0,0 \gt +\lt 6t ,2t, -2t \gt$ Thus, we have $r(t) =\lt 2+4t, 2t,-2t \gt ; 0 \leq t \leq 1$ Now, the parametric equations are: $x=2+4t;\\ y= 2t; \\z=-2t ; \\0 \leq t \leq 1$
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