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: 25



Work Step by Step

Our aim is to find the value of $t$. $(2,4 \sin (t/2), 3-\dfrac{t}{\pi} ) \cdot (1,-1,0) =0$ This implies that $2-4 \sin (t/2)=0$ $\implies 2=-4 \sin (t/2)$ $\implies t=\dfrac{\pi}{3}$
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