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: 21


$a(0)=10T +6N$

Work Step by Step

$v(t)=\dfrac{dr}{dt}=(6t+3) i +(4+8t) j +6 \sin t \ k$ and $a(t)=\dfrac{dv(t)}{dt}=6 i+8 j +6k \cos t$ $|a(t)|=2 \sqrt {9 \cos^2 t +25} \implies |a(0)|=2 \sqrt {9 \cos^2 (0) +25} =2 \sqrt {34}$ and $a_{T}(0)=10$ Now, $a_{N}=\sqrt {|a(0)|^2 -a^2_{T} (0)}=\sqrt {(2 \sqrt {34})^2 -(10)^2}=6$ So, $a(0)=a_T(0) T+a_{N} N=10T +6N$
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