## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.5 - Tangential and Normal Components of Acceleration - Exercises 13.5 - Page 771: 1

#### Answer

$a=0 \ T +|a| \ N$

#### Work Step by Step

$v(t)=\dfrac{dr}{dt}=(-a \sin t) i +(a \cos t) j +b \ k \implies |v(t)|=\sqrt {a^2+b^2}$ and $a(t)=\dfrac{d \ v(t)}{dt}=(-a \cos t) i +(-a \sin t) j + 0 \ k$ $|a(t)|=\sqrt {a^2(\cos^2 t+\sin ^2 t)} \implies |a(t)|=a$ Now, $a_{N}=\sqrt {|a|^2 -a^2_{T}}=\sqrt {a^2-0}=|a|$ So, $a=a_T T+a_{N}=0 \ T +|a| \ N$

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