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: 19



Work Step by Step

Need to determine the speed of a particle. As we know $v(t)=r'(t)$ and speed is the magnitude of the velocity vector, that is ,$s(t)=|v(t)|$. $v(t)=r'(t)=\lt 2t,5,2t-16 \gt$ [Given: $r(t)=\lt t^2, 5t, t^2-16t \gt$] Solve for speed. $s(t)=|v(t)|=\sqrt {(2t)^2+(5)^2+( 2t-16)^2}$ $s(t)=\sqrt {4t^2+25+4t^2+256-64t}$ $s(t)=\sqrt {8t^2-64t+281}$ Our aim is to find the minimum value of speed, in order to this take the extreme value of the quadratic equation. Simplify to get thje value of $t$. $t=-\dfrac{b}{2a}=-\dfrac{(-64)}{2(8)}=4$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.