Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 896: 29

Answer

$\bf{Graph{(IV)}}$

Work Step by Step

The parametric equations of a circle having radius $r$ are; $x=r \cos t ; y =r \sin t$ Here, we have $x= \cos t , y=\sin t, z=\cos 2t$ This shows that the projection of the curve on the xy plane will be a circle of radius $1$. The term $z=e^{0.8} t$ tells us that this is greater than $0$ for all $t$. This matches with $\bf{Graph{(IV)}}$.
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