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: 21


$x=t; y=\dfrac{1}{3}t; z=t$

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, $v(t)=\lt \dfrac{1}{t},(t^2+t+1)(t+2)^{-(2)},1+\ln t\gt \\ \implies v(1)=\lt 1,\dfrac{1}{3},1\gt$ Thus, we have the velocity components $v_x=1,v_y=\dfrac{1}{3},v_y=1$ Now, the parametric equations are as follows: $x=(1)(t)+0=t; y=(\dfrac{1}{3}) t+(0)=\dfrac{1}{3}t; z=(1)(t)+(0)=t$ Thus, $x=t; y=\dfrac{1}{3}t; z=t$
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