Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 854: 29

Answer

$y=e^{x/2}$ ; $z=e^x$ and $z=y^2$

Work Step by Step

Here, we have $x=2t; y=e^t; z=e^{2t}$ Now, $y=e^t=e^{2t/2}=e^{x/2}$ From $z=e^{2t}$, we have $z=e^x$ Also, $z=e^{2t} \implies z=(e^t)^2=y^2$ Hence, we have $y=e^{x/2}$ ; $z=e^x$ and $z=y^2$
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