Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.4 Motion in Space: Velocity and Acceleration - 13.4 Exercises - Page 918: 10

Answer

$\lt -2 \sin t,3, 2 \cos t\gt$, $\lt -2 \cos t, 0, -2 \sin t\gt$ , $\sqrt {13}$

Work Step by Step

As we are given that $r(t)=\lt 2 \cos t,3t, 2 \sin t \gt$ Need to determine the velocity vector, acceleration vector and speed. We have $v(t)=r'(t)$ and $a(t)=v'(t)$ and speed is the magnitude of the velocity vector, that is $s(t)=|v(t)|$ $v(t)=r'(t)=\lt -2 \sin t,3, 2 \cos t\gt$ [ given: $r(t)=\lt 2 \cos t,3t, 2 \sin t \gt$] $a(t)=v'(t)=\lt -2 \cos t, 0, -2 \sin t\gt$ $s(t)=|v(t)|=\sqrt {(-2 \sin t)^2+(3)^2+(2 \cos t)^2}=\sqrt {13}$
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