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 745: 1


$y= x^2-2x$; $v(1)=i+2j$ and $a(1)=2j$

Work Step by Step

We know that the velocity and acceleration is given by: velocity is : $v(t)=r'(t)$ and acceleration is: $a(t)=v'(t)$ In the given problem, $x=t+1$ and, $y=t^2-1=(t-1)(t+1)$ Re-write as: $y=(x-2)x \implies x^2-2x$ Now, $v(t)=i+2tj \implies v(1)=i+2j$ we know that acceleration $a(t)=2j \implies a(1)=2j$ Thus, we have $y= x^2-2x$; $v(1)=i+2j$ and $a(1)=2j$
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