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 13.5 - Tangential and Normal Components of Acceleration - Exercises 13.5 - Page 771: 3


$a=\dfrac{4}{3} \ T + \dfrac{2 \sqrt 5}{3} \ N$

Work Step by Step

We calculate the velocity and acceleration as follows: $v(t)=\dfrac{dr}{dt}=i+2j+2tk \implies |v(t)|=\sqrt {1^2+(2)^2+(2t)^2}=\sqrt {4t^2+5}$ and $a(t)=\dfrac{d \ v(t)}{dt}= \dfrac{4t}{\sqrt {4t^2+5}} $ $|a(1)|=\dfrac{4(1)}{\sqrt {4 (1)^2+5}}=\dfrac{4}{3}$ Now, $a_{N}=\sqrt {|a|^2 -a^2_{T}}=\sqrt {2^2 -(\dfrac{4}{3})^2}=\dfrac{2 \sqrt 5}{3}$ So, $a=a_T T+a_{N}=\dfrac{4}{3} \ T + \dfrac{2 \sqrt 5}{3} \ N$
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