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: 19


$x=t; y=-1; z=t+1$

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, $v(t)=\lt \cos (t), 2t+\sin (t), e^{(t)}\gt \\ \implies v(0)=\lt 1,0,1 \gt$ Thus, we have the velocity components $v_x=1,v_y=0,v_y=1$ Now, the parametric equations are as follows: $x=(1)t+0=t; y=(0)( t)+(-1)=-1$ and $z=(1)t+1=t+1$ Thus, $x=t; y=-1; z=t+1$
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