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 - Practice Exercises - Page 777: 22


$a(0)=2 \sqrt 2T +2 \sqrt 3N$

Work Step by Step

$v(t)=\dfrac{dr}{dt}=1i +(1+4t) j +2 t \ k$ and $a(t)=\dfrac{dv(t)}{dt}=4 j +2k$ $|a(t)|=\sqrt {4^2+(2)^2} \implies |a(t)|=2 \sqrt 5$ and $a_{T}(0)=2 \sqrt 2$ Now, $a_{N}=\sqrt {|a(0)|^2 -a^2_{T} (0)}=\sqrt {(2 \sqrt {5})^2 -(2 \sqrt 2)^2}=2 \sqrt 3$ So, $a(0)=a_T(0) T+a_{N} N=2 \sqrt 2T +2 \sqrt 3N$
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