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: 22


$x=-t; y=1; z=-2t$

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