Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 746: 20


$x=4+4t; y=3+2t; z=8+12t$

Work Step by Step

We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$ Now,$v(t)=\lt 2t,2,3t^2\gt \\ \implies v(2)=\lt 4,2,12 \gt$ Thus, we have the velocity components $v_x=4,v_y=2,v_y=12$ Now, the parametric equations are as follows: $x=(4)(t)+4=4+4t; y=3+2t; z=(12)t+(8)=8+12t$ Thus, $x=4+4t; y=3+2t; z=8+12t$
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